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view texts/XML/echo/fr/Varignon_1687_TP04WPNS.xml @ 13:facea8c79160
DE Specs Version 2.1.1 Autumn 2011
| author | Klaus Thoden <kthoden@mpiwg-berlin.mpg.de> |
|---|---|
| date | Thu, 02 May 2013 11:29:00 +0200 |
| parents | 22d6a63640c6 |
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<?xml version="1.0" encoding="utf-8"?><echo xmlns="http://www.mpiwg-berlin.mpg.de/ns/echo/1.0/" xmlns:de="http://www.mpiwg-berlin.mpg.de/ns/de/1.0/" xmlns:dcterms="http://purl.org/dc/terms" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:echo="http://www.mpiwg-berlin.mpg.de/ns/echo/1.0/" xmlns:xhtml="http://www.w3.org/1999/xhtml" xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink" version="1.0RC"> <metadata> <dcterms:identifier>ECHO:TP04WPNS.xml</dcterms:identifier> <dcterms:creator identifier="GND:119062372">Varignon, Pierre</dcterms:creator> <dcterms:title xml:lang="fr">Projet d' une nouvelle mechanique : avec un examen de l' opinion de M. Borelli sur les propriétez des poids suspendus par des cordes</dcterms:title> <dcterms:alternative xml:lang="fr">Projet d' une nouvelle mechanique : avec un examen de l' opinion de M. Borelli sur les propriétés des poids suspendus par des cordes</dcterms:alternative> <dcterms:date xsi:type="dcterms:W3CDTF">1687</dcterms:date> <dcterms:language xsi:type="dcterms:ISO639-3">fra</dcterms:language> <dcterms:rights>CC-BY-SA</dcterms:rights> <dcterms:license xlink:href="http://creativecommons.org/licenses/by-sa/3.0/">CC-BY-SA</dcterms:license> <dcterms:rightsHolder xlink:href="http://www.mpiwg-berlin.mpg.de">Max Planck Institute for the History of Science, Library</dcterms:rightsHolder> <log>escape sequence { not only in fractions but also for two-line text</log> </metadata> <text xml:lang="fr" type="free"> <div xml:id="echoid-div1" type="section" level="1" n="1"><pb file="0001" n="1"/> <pb file="0002" n="2"/> <pb file="0003" n="3"/> <pb file="0004" n="4"/> <pb file="0005" n="5"/> <pb file="0006" n="6"/> <pb file="0007" n="7"/> <pb file="0008" n="8"/> <handwritten/> <handwritten/> <handwritten/> <pb file="0009" n="9"/> </div> <div xml:id="echoid-div2" type="section" level="1" n="2"> <head xml:id="echoid-head1" xml:space="preserve">PROJET <lb/>D’UNE NOUVELLE <lb/>MECHANIQUE</head> <head xml:id="echoid-head2" style="it" xml:space="preserve">AVEC</head> <p> <s xml:id="echoid-s1" xml:space="preserve">Un Examen de l’opinion de M. </s> <s xml:id="echoid-s2" xml:space="preserve">BORELLI, <lb/>ſur les propriétez des Poids ſuſpendus <lb/>par des Cordes.</s> <s xml:id="echoid-s3" xml:space="preserve"/> </p> <figure> <image file="0009-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0009-01"/> </figure> </div> <div xml:id="echoid-div3" type="section" level="1" n="3"> <head xml:id="echoid-head3" xml:space="preserve">A PARIS,</head> <p> <s xml:id="echoid-s4" xml:space="preserve">Chez {la Vcuve d’<emph style="sc">Edme</emph> <emph style="sc">Martin</emph>, \\ <emph style="sc">Jean</emph> <emph style="sc">Boudot</emph>, \\ & </s> <s xml:id="echoid-s5" xml:space="preserve"><emph style="sc">Estienne</emph> <emph style="sc">Martin</emph>,} ruë S. </s> <s xml:id="echoid-s6" xml:space="preserve">Jaques, \\ au Soleil d’or.</s> <s xml:id="echoid-s7" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div4" type="section" level="1" n="4"> <head xml:id="echoid-head4" xml:space="preserve">M. DC. LXXXVII. <lb/>AVEC PRIVILEGE DU ROI.</head> <pb file="0010" n="10"/> <pb file="0011" n="11"/> <figure> <image file="0011-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0011-01"/> </figure> </div> <div xml:id="echoid-div5" type="section" level="1" n="5"> <head xml:id="echoid-head5" xml:space="preserve">A MESSIEURS <lb/>DE <lb/>L’ACADEMIE <lb/>ROYALE <lb/>DES SCIENCES@</head> <p style="it"> <s xml:id="echoid-s8" xml:space="preserve">MESSIEVRS, <lb/>Fe n’ay pas crû devoir expoſer au <lb/>jugement du public ce Projet d’une <pb file="0012" n="12" rhead="EPITRE."/> Nouvelle Mècbanique, ſans m’ap-<lb/>puyer d’une auſſi grande autboritè <lb/>que la vôtre, moy qui n’ay encore au-<lb/>cun nom dans les Lettres, & </s> <s xml:id="echoid-s9" xml:space="preserve">qui dois <lb/>par conſèquent me dèfier de ces pre-<lb/>miers mouvemens que @amour des <lb/>Sciences inſpire àceux qui commencent <lb/>à s’y apliquer. </s> <s xml:id="echoid-s10" xml:space="preserve">Sans cela on pourroit <lb/>juſtement m’accuſer de quelque tèmèri-<lb/>tè, d’avoir entrepris de dècouvrir dans <lb/>cette matière ce que tant de ſc1;</s> <s xml:id="echoid-s11" xml:space="preserve">avans <lb/>Autbeurs n’ont pas dècouvert; </s> <s xml:id="echoid-s12" xml:space="preserve">& </s> <s xml:id="echoid-s13" xml:space="preserve">je <lb/>craindrois de m’être laiſſè tromper <lb/>par ces illuſions flateuſes de la nou-<lb/>veautè qui abuſent d’ordinaire les <lb/>bommes, lorſqu’ils ſe piquent d’avoir <lb/>des opinions particulières. </s> <s xml:id="echoid-s14" xml:space="preserve">Fe puis dire <lb/>neanmoins, <emph style="sc">Messieurs</emph>, que ce <lb/>n’eſt pas l’ambition de me ſignaler par <lb/>des idèes extraordinaires qui m’a <pb file="0013" n="13" rhead="EPITRE."/> pouſſè à ècrire ce petit traitè; </s> <s xml:id="echoid-s15" xml:space="preserve">c’eſt un <lb/>Eſſay que j’ay voulu faire de mes for-<lb/>ces pour être connu de vous, & </s> <s xml:id="echoid-s16" xml:space="preserve">pour <lb/>vous donner occaſion de m’encourager <lb/>dans l’ètude que j’ay embraſsèe. </s> <s xml:id="echoid-s17" xml:space="preserve">Si je <lb/>n’ay pas tout ce qui eſt nècèſſaire pour <lb/>inſtruire les autres, j’ay du moins <lb/>toute la docilitè qu’il faut pour être <lb/>inſtruit: </s> <s xml:id="echoid-s18" xml:space="preserve">je ne me flatte point auſſi <lb/>d’avoir ètabli des principes certains <lb/>dans ce Projet, n’y d’en pouvoir <lb/>tirer des conſèquences infaillibles: <lb/></s> <s xml:id="echoid-s19" xml:space="preserve">Vous enjugerez mieux que perſonne, <lb/><emph style="sc">Messieurs</emph>, Vous qui pènètrez ſi <lb/>avant dans les Sciences les plus rele-<lb/>vèes. </s> <s xml:id="echoid-s20" xml:space="preserve">On ſçait que rien n’ècbape à <lb/>vos ſoins & </s> <s xml:id="echoid-s21" xml:space="preserve">à vôtre èxactitude; </s> <s xml:id="echoid-s22" xml:space="preserve">La <lb/>Nature ſi avare aux autres de ſes <lb/>treſors & </s> <s xml:id="echoid-s23" xml:space="preserve">ſi obſtinèe à ſe cacber, n’a <lb/>pû ſe dèfendre contre la pènètration <lb/>de vôtre eſprit & </s> <s xml:id="echoid-s24" xml:space="preserve">contre la ſubtilitè <pb file="0014" n="14" rhead="EPITRE."/> de vos recbercbes; </s> <s xml:id="echoid-s25" xml:space="preserve">vous en avez plus <lb/>dècouvert en vingt ans, qu’on n’a-<lb/>voit fait en pluſieurs Siècles. </s> <s xml:id="echoid-s26" xml:space="preserve">Vos <lb/>Obſervations Aſtronomiques ont dè-<lb/>voilè (pour ainſi dire) des Planettes <lb/>qui ſe dèroboient à nos yeux; </s> <s xml:id="echoid-s27" xml:space="preserve">vos <lb/>meſures ſi prèciſes ſur la terre, par <lb/>raport à celles que vous preniez en <lb/>même tems dans le Ciel, ont rècti-<lb/>fiè mille erreurs de nos anciens Geo-<lb/>grapbes. </s> <s xml:id="echoid-s28" xml:space="preserve">La Pbyſique vous doit <lb/>ce qu’elle a de plus curieux, ſoit <lb/>dans la diſſection du Corps bumain <lb/>& </s> <s xml:id="echoid-s29" xml:space="preserve">des Animaux, ſoit dans la deſ-<lb/>cription & </s> <s xml:id="echoid-s30" xml:space="preserve">dans l’analyſe des Plan-<lb/>tes, des Eaux & </s> <s xml:id="echoid-s31" xml:space="preserve">des Mineraux. <lb/></s> <s xml:id="echoid-s32" xml:space="preserve">Lue ne vous doivent point auſſi les <lb/>Matbematiques en gènèral pour <lb/>tant d’ouvrages celebres que vous <lb/>avez mis au jour? </s> <s xml:id="echoid-s33" xml:space="preserve">Enfin, il n’y a <pb file="0015" n="15" rhead="EPITRE."/> point de Science que vous n’ayez <lb/>perfectionnèe & </s> <s xml:id="echoid-s34" xml:space="preserve">que vous n’enri-<lb/>cbiſſiez de tems en tems par vos <lb/>travaux. </s> <s xml:id="echoid-s35" xml:space="preserve">Lue n’attend on pas en-<lb/>core de vous, animez comme vous <lb/>êtes par les bienfaits d’un Grand <lb/>Roy, qui veut renàre ſon Regne <lb/>auſſi glorieux par les Sciences & </s> <s xml:id="echoid-s36" xml:space="preserve">par <lb/>les Arts, qu’il l’eſt dèja par ſes <lb/>prodigieuſes Conqueſtes, & </s> <s xml:id="echoid-s37" xml:space="preserve">par tou-<lb/>tes ſes Hèroϊques actions? </s> <s xml:id="echoid-s38" xml:space="preserve">A quoy <lb/>ne devez vous pas aſpirer vous-<lb/>meſmes aujourd’buy ſous la protè-<lb/>ction d’un Miniſtre ſi ſage & </s> <s xml:id="echoid-s39" xml:space="preserve">ſi vi-<lb/>gilant, qui excite tout le monde par <lb/>ſes ordres & </s> <s xml:id="echoid-s40" xml:space="preserve">par ſon exemple à <lb/>illuſtrer & </s> <s xml:id="echoid-s41" xml:space="preserve">à cèlèbrer un Regne ſi <lb/>plein de merveilles ? </s> <s xml:id="echoid-s42" xml:space="preserve">Souffrez donc, <lb/><emph style="sc">Messieurs</emph>, s’il vous plaît, <lb/>vous qui êtes comme à la ſource de <pb file="0016" n="16" rhead="EPITRE."/> toutes les Sciences bumaines, & </s> <s xml:id="echoid-s43" xml:space="preserve">à qui <lb/>rien ne manque pour continuer vos <lb/>recbercbes, & </s> <s xml:id="echoid-s44" xml:space="preserve">pour augmenter vos <lb/>connoiſſances, que j’oſe vous offrir <lb/>& </s> <s xml:id="echoid-s45" xml:space="preserve">mettre au jour ce que j’ay puisè <lb/>dans cette ſource, & </s> <s xml:id="echoid-s46" xml:space="preserve">qu’en eſſayant <lb/>de vous ſuivre & </s> <s xml:id="echoid-s47" xml:space="preserve">de vous imiter, je <lb/>puiſſe quelquefois profiter de vos <lb/>lumières, & </s> <s xml:id="echoid-s48" xml:space="preserve">vous aſſeurer que je ſuis <lb/>avec une parfaite vènèration,</s> </p> <p> <s xml:id="echoid-s49" xml:space="preserve">MESSIEVRS,</s> </p> <p> <s xml:id="echoid-s50" xml:space="preserve">Vôtre tres-humble & </s> <s xml:id="echoid-s51" xml:space="preserve">tres-<lb/>obéϊſſant ſerviteur</s> </p> <p> <s xml:id="echoid-s52" xml:space="preserve">V<emph style="sc">ARIGNON.</emph></s> </p> <pb file="0017" n="17"/> <figure> <image file="0017-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0017-01"/> </figure> </div> <div xml:id="echoid-div6" type="section" level="1" n="6"> <head xml:id="echoid-head6" xml:space="preserve">PREFACE.</head> <p> <s xml:id="echoid-s53" xml:space="preserve">A L’ouverture du ſecond Tome des <lb/>Lettres de Monſieur Deſcartes, je <lb/>tombai ſur un endroit de la 24. </s> <s xml:id="echoid-s54" xml:space="preserve">où <lb/>il dit que c’eſt une cboſe ridicule, <lb/>que de vouloir employer la raiſon du Le-<lb/>vier dans la Poulie. </s> <s xml:id="echoid-s55" xml:space="preserve">Cette réfléxion m’en <lb/>fit faire une autre@<unsure/>: </s> <s xml:id="echoid-s56" xml:space="preserve">Sçavoir s’il eſt plus rai-<lb/>ſonnable de s’imaginer un levier dans un <lb/>poids qui eſt ſur un plan incliné, que dans <lb/>une poulie. </s> <s xml:id="echoid-s57" xml:space="preserve">Aprés y avoir penſé, il me <lb/>ſembla que ces deux machines étant pour le <lb/>moins auſſi ſimples que le levier, elles n’en <lb/>devoient avoir aucune dépendance, & </s> <s xml:id="echoid-s58" xml:space="preserve">que <lb/>ceux qui les y rapportoient, n’y étoient forcez <lb/>que parce que leurs principes n’avoient pas aſſez <lb/>d’étenduë pour en pouvoir démontrer les pro-<lb/>priétez indépendamment les unes des autres.</s> <s xml:id="echoid-s59" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s60" xml:space="preserve">En effet en éxaminant ces principes un peu <pb file="0018" n="18" rhead="PREFACE."/> de prês, il me parut qu’ils ne pouvoient ſervir, <lb/>tout au plus, qu’à démontrer que l’equilibre <lb/>ſe trouve toujours dans un levier auquel ſont <lb/>appliquez deux poids qui ſont entr’ eux en <lb/>raiſon rèciproque des diſtances de leurs lignes <lb/>de direction à ſon point d’appui; </s> <s xml:id="echoid-s61" xml:space="preserve">encore n’é-<lb/>toit-ce qu’en ce cas: </s> <s xml:id="echoid-s62" xml:space="preserve">1°. </s> <s xml:id="echoid-s63" xml:space="preserve">Que ce levier fût <lb/>droit. </s> <s xml:id="echoid-s64" xml:space="preserve">2°. </s> <s xml:id="echoid-s65" xml:space="preserve">Que ſon point d’appui fût entre les <lb/>lignes de direction des poids qui y ſont appli-<lb/>quez. </s> <s xml:id="echoid-s66" xml:space="preserve">3°. </s> <s xml:id="echoid-s67" xml:space="preserve">Lue ces mêmes lignes fuſſent pa-<lb/>ralleles entr’elles, & </s> <s xml:id="echoid-s68" xml:space="preserve">perpendiculaιres à ce <lb/>levier. </s> <s xml:id="echoid-s69" xml:space="preserve">Auſſi Guid-Ubalde, & </s> <s xml:id="echoid-s70" xml:space="preserve">les autres qui <lb/>s’en tiennent à la démonſtration d’Archimede, <lb/>ont-ils été obligez de faire@<unsure/>revenir de gré ou <lb/>de force toutes ſortes de machines à cette <lb/>eſpece de levier, & </s> <s xml:id="echoid-s71" xml:space="preserve">de réduire de même tous <lb/>les autres cas à celui-ci.</s> <s xml:id="echoid-s72" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s73" xml:space="preserve">C’eſt peut-être ce qui a porté M. </s> <s xml:id="echoid-s74" xml:space="preserve">Deſcartes, <lb/>& </s> <s xml:id="echoid-s75" xml:space="preserve">M. </s> <s xml:id="echoid-s76" xml:space="preserve">Vvallis a prendre une autre route; <lb/></s> <s xml:id="echoid-s77" xml:space="preserve">quoi qu’il en ſoit, ce n’a pas été ſans ſuccez: </s> <s xml:id="echoid-s78" xml:space="preserve"><lb/>puiſque celle qu’ils ont ſuivie, conduit éga-<lb/>lement à la connoiſſance des uſages de chacune <lb/>de ces machines, ſans être obligé de les faire <lb/>dépendre l’une de l’autre; </s> <s xml:id="echoid-s79" xml:space="preserve">outre qu’elle à mené <lb/>M. </s> <s xml:id="echoid-s80" xml:space="preserve">Vvallis beaucoup plus loin qu’aucun <pb file="0019" n="19" rhead="PREFACE."/> Autheur, que je ſçache, n’eût encore été de <lb/>ce côté-là.</s> <s xml:id="echoid-s81" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s82" xml:space="preserve">La comparaiſon que je fis de ces deux <lb/>fortes de principes, me fit ſentir que ceux d’Ar-<lb/>chimede n’étoient ny ſi étendus, ny ſi con-<lb/>vainquants que ceux de M. </s> <s xml:id="echoid-s83" xml:space="preserve">Deſcartes, & </s> <s xml:id="echoid-s84" xml:space="preserve">de <lb/>M. </s> <s xml:id="echoid-s85" xml:space="preserve">Vvallis; </s> <s xml:id="echoid-s86" xml:space="preserve">mais je ne ſentis point que les uns <lb/>ny les autres m’éclairaſſent beaucoup: </s> <s xml:id="echoid-s87" xml:space="preserve">J’en <lb/>cherchai la raiſon, & </s> <s xml:id="echoid-s88" xml:space="preserve">ce défaut me parut venir <lb/>de ce que ces Autheurs ſe ſont tous plus atta-<lb/>chez à prouver la néceſſité de l’équilibre, qu’à <lb/>montrer la maniére dont il ſe fait.</s> <s xml:id="echoid-s89" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s90" xml:space="preserve">Ce fut ce qui me fit réſoudre à prendre le <lb/>parti d’épier moi-même la nature, & </s> <s xml:id="echoid-s91" xml:space="preserve">d’eſſayer <lb/>ſi en la ſuivant pas à pas, je ne pourrois point <lb/>apercevoir comment elle s’y prend pour faire <lb/>que deux puiſſances, ſoit égales, ſoit inégales, <lb/>demeurent en équilibre. </s> <s xml:id="echoid-s92" xml:space="preserve">Enfin je m’appliquai <lb/>à chercher l’équilibre lui-même dans ſa ſource, <lb/>ou pour mieux dire, dans ſa génération.</s> <s xml:id="echoid-s93" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s94" xml:space="preserve">Le premier objet qui me vint à l’eſprit, ce <lb/>fut un poids qu’une puiſſance ſoutient ſur <lb/>un plan incliné; </s> <s xml:id="echoid-s95" xml:space="preserve">D’abord je me le repréſentai <pb file="0020" n="20" rhead="PREFACE."/> de telle figure que le concours de ſa ligne de <lb/>direction avec celle de cette puiſſance, ſe fit <lb/>dans quelqu’un de ſes points: </s> <s xml:id="echoid-s96" xml:space="preserve">De-là je vis que <lb/>leur concours d’action ſe faiſant auſſi par ce <lb/>moyen dans ce ſeul point, il devenoit alors <lb/>ſon centre de direction: </s> <s xml:id="echoid-s97" xml:space="preserve">De ſorte que ſi ce <lb/>plan eut manqué tout d’un coup, ce corps au-<lb/>roit néceſſairement ſuivi l’impreſſion de ce <lb/>point. </s> <s xml:id="echoid-s98" xml:space="preserve">Je cherchai enſuite quelle devoit être <lb/>cette impreſſion, & </s> <s xml:id="echoid-s99" xml:space="preserve">j’aperçus que celles que <lb/>faiſoient ſur ce point, & </s> <s xml:id="echoid-s100" xml:space="preserve">la peſanteur de ce <lb/>poids, & </s> <s xml:id="echoid-s101" xml:space="preserve">la puiſſance qui le retenoit, étant <lb/>les mêmes que s’il eut été pouſſé en même-<lb/>tems par deux forces qui leur euſſent été éga-<lb/>les, & </s> <s xml:id="echoid-s102" xml:space="preserve">qui euſſent agi ſuivant leurs lignes de <lb/>direction: </s> <s xml:id="echoid-s103" xml:space="preserve">J’aperçus, dî-je, qu’il lui en réſultoit <lb/>une impreſſion compoſée ſuivant une ligne <lb/>qui étoit la diagonale d’un parallelogramme <lb/>fait ſous des parties de ces lignes de direction, <lb/>qui étoient entr’elles, comme ce poids & </s> <s xml:id="echoid-s104" xml:space="preserve"><lb/>cette puiſſance: </s> <s xml:id="echoid-s105" xml:space="preserve">D’où je vis que l’impreſſion <lb/>de ce corps ſe faiſoit alors ſuivant cette diago-<lb/>nale, qui devenoit en ce cas ſa ligne de di-<lb/>rection; </s> <s xml:id="echoid-s106" xml:space="preserve">mais que ce plan lui étant perpendi-<lb/>culairement oppoſé, il la ſoutenoit toute en-<lb/>tiére; </s> <s xml:id="echoid-s107" xml:space="preserve">ce qui faiſoit que ce poids ainſi pouſſé <pb file="0021" n="21" rhead="PREF ACE."/> par le concours d’action de ſa peſanteur & </s> <s xml:id="echoid-s108" xml:space="preserve">de <lb/>la puiſſance qui lui étoit appliquée, demeuroit <lb/>ſur ce plan incliné de même que s’il eut été <lb/>horizontal, & </s> <s xml:id="echoid-s109" xml:space="preserve">que cette impreſſion compoſée <lb/>n’eut été qu’un effet de ſa ſeule peſanteur.</s> <s xml:id="echoid-s110" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s111" xml:space="preserve">De cette penſée j’en vis naître pluſieurs <lb/>autres, & </s> <s xml:id="echoid-s112" xml:space="preserve">jem’aperçus 1°. </s> <s xml:id="echoid-s113" xml:space="preserve">Que toute l’impreſ-<lb/>ſion que ce plan recevoit alors de ce poids <lb/>ainſi ſoutenu par cette puiſſance, ſe faiſoit <lb/>ſuivant cette diagonale. </s> <s xml:id="echoid-s114" xml:space="preserve">2°. </s> <s xml:id="echoid-s115" xml:space="preserve">Que ſa charge, <lb/>c’eſt-à-dire, la force de cette même impreſ-<lb/>ſion, étoit à ce poids & </s> <s xml:id="echoid-s116" xml:space="preserve">à cette puiſſance, <lb/>comme cette même diagonale à chacun des <lb/>côtez qui les répréſentent dans ſon parallelo-<lb/>gramme. </s> <s xml:id="echoid-s117" xml:space="preserve">3°. </s> <s xml:id="echoid-s118" xml:space="preserve">Que ce poids, & </s> <s xml:id="echoid-s119" xml:space="preserve">cette puiſſance <lb/>étoient toujours entr’eux comme ces mêmes <lb/>côtez, c’eſt-à-dire, en raiſon réciproque des <lb/>ſinus des angles que font leurs lignes de direc-<lb/>tion avec cette diagonale, ou (ce qui revient <lb/>au même ) en raiſon réciproque des diſtances <lb/>de quelque point que ce ſoit de cette diago-<lb/>nale à leur lignes de direction. </s> <s xml:id="echoid-s120" xml:space="preserve">Je vis enfin <lb/>preſque tout à la fois quantité de choſes tou-<lb/>tes nouvelles qu’on verra dans les Corollaires <lb/>de la propoſition des ſurfaces.</s> <s xml:id="echoid-s121" xml:space="preserve"/> </p> <pb file="0022" n="22" rhead="PREF ACE."/> <p> <s xml:id="echoid-s122" xml:space="preserve">Aprés avoir ainſi trouvé la maniére dont <lb/>l’équilibre ſefait ſur des plans inclinez, je cher-<lb/>chai par le même chemin comment des poids <lb/>ſoutenus avec des cordes ſeulement, ou appli-<lb/>quez à des poulies, ou bien à des leviers, font <lb/>équilibre entr’eux, ou avec les puiſſances qui <lb/>les ſoutiennent; </s> <s xml:id="echoid-s123" xml:space="preserve">& </s> <s xml:id="echoid-s124" xml:space="preserve">j’aperçus de même que tout <lb/>cela ſe faiſoit encore par la voye des mouve-<lb/>mens compoſez, & </s> <s xml:id="echoid-s125" xml:space="preserve">avec tant d’uniformité <lb/>que je ne pûs m’empêcher de croire que cette <lb/>voye ne fût véritablement celle que ſuit la <lb/>nature dans le concours d’action de deux <lb/>poids, ou de deux puiſſances, en faiſant que <lb/>leurs impreſſions particulieres, quelque pro-<lb/>portion qu’elles ayent, ſe confondent en un@ <lb/>ſeule qui ſe décharge toute entiére ſur le point <lb/>ou ſe fait cét équilibre: </s> <s xml:id="echoid-s126" xml:space="preserve">De ſorte que la <lb/>raiſon Phyſique des effets qu’on admire le <lb/>plus dans les machines me parut être juſte-<lb/>ment celle des mouvemens compoſez.</s> <s xml:id="echoid-s127" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s128" xml:space="preserve">Je me démontrai d’abord par cett@ mé-<lb/>thode, & </s> <s xml:id="echoid-s129" xml:space="preserve">ſans le ſecours d’aucune machine, <lb/>les propriétez des poids ſuſpendus avec des <lb/>cordes, en quelque nombre qu’elles ſoient, &</s> <s xml:id="echoid-s130" xml:space="preserve"> <pb file="0023" n="23" rhead="PREF ACE."/> pour tous les angles poſſibles qu’elles peuvent <lb/>faire entr’elles. </s> <s xml:id="echoid-s131" xml:space="preserve">De-là je paſſai à une démonſ-<lb/>tration des poulies qui comprend toutes les <lb/>directions poſſibles des puiſſances, ou des <lb/>poids qui y ſont appliquez, ſoit que le centre <lb/>de ces poulies demeure fixe, ſoit qu’on le ſup-<lb/>poſe mobile. </s> <s xml:id="echoid-s132" xml:space="preserve">Enſuite au lieu de la démonſtra-<lb/>tion qu’on ne fait ordinairement que pour les <lb/>plans inclinez, j’en trouvai une qui s’étend <lb/>généralement à toutes ſortes de ſurfaces, & </s> <s xml:id="echoid-s133" xml:space="preserve">à <lb/>toutes les directions poſſibles des puiſſances, <lb/>ou des poids qui y ſont appliquez. </s> <s xml:id="echoid-s134" xml:space="preserve">Enfin d’une <lb/>ſeule démonſtration je découvris les proprié-<lb/>tez de toutes les eſpeces de leviers, de quel-<lb/>que figure, & </s> <s xml:id="echoid-s135" xml:space="preserve">dans quelque ſituation qu’ils <lb/>ſoient, & </s> <s xml:id="echoid-s136" xml:space="preserve">pour toutes les directions poſſibles <lb/>des puiſſances, ou des poids qui y ſont appli-<lb/>quez.</s> <s xml:id="echoid-s137" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s138" xml:space="preserve">Des vuës ſi étenduës me ſurprirent, & </s> <s xml:id="echoid-s139" xml:space="preserve">l’é-<lb/>vidence avec laquelle le détail de tout cela <lb/>me paroiſſoit, indépendamment même du <lb/>général, me confirma encore dans l’opinion <lb/>ou j’étois, qu’il faut entrer dans la génération <lb/>d e l’équilibre pour y voir en ſoi, & </s> <s xml:id="echoid-s140" xml:space="preserve">pour y <lb/>reconnoître les propriétez que tous les autres <pb file="0024" n="24" rhead="PREF ACE."/> principes ne prouvent, tout au plus, que par <lb/>néceſſité de conſéquence.</s> <s xml:id="echoid-s141" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s142" xml:space="preserve">Il y a encore un avantage dans la route <lb/>que je tiens, c’eſt qu’elle facilite extrémement <lb/>le calcul des forces, tant des poids, que des <lb/>puiſſances, en ce que leurs raports y ſont tou-<lb/>jours déterminez immédiatement par les ſinus <lb/>des angles que font leurs lignes de direction <lb/>avec celle de l’impreſſion qui réſulte de <lb/>leur concours d’action, & </s> <s xml:id="echoid-s143" xml:space="preserve">que cette méthode <lb/>détermine pour le point ou elles concourent. <lb/></s> <s xml:id="echoid-s144" xml:space="preserve">On y voit, que lors que deux puiſſances, <lb/>ou deux poids, ou bien une puiſſance & </s> <s xml:id="echoid-s145" xml:space="preserve">un <lb/>poids font équilibre, ſoit avec des cordes <lb/>ſeulement, ſoit à l’aide de quelque poulie, de <lb/>quelque ſurface, ou de quelque levier que ce <lb/>ſoit; </s> <s xml:id="echoid-s146" xml:space="preserve">ils ſont toujours entr’ eux en raiſon réci-<lb/>proque des ſinus de ces mêmes angles.</s> <s xml:id="echoid-s147" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s148" xml:space="preserve">J’avois deſſein d’expliquer avec cette mé-<lb/>thode les effets les plus ſurprenans, & </s> <s xml:id="echoid-s149" xml:space="preserve">les plus <lb/>difficiles des machines compoſées que l’on ren-<lb/>contre dans les arts, & </s> <s xml:id="echoid-s150" xml:space="preserve">dans la nature; </s> <s xml:id="echoid-s151" xml:space="preserve">mais cela <lb/>demandoit plus de loiſir, & </s> <s xml:id="echoid-s152" xml:space="preserve">même un plus <lb/>grand nombre d’expériences que l’état de ma <pb file="0025" n="25" rhead="PREF ACE."/> fortune ne me peut permetre: </s> <s xml:id="echoid-s153" xml:space="preserve">c’eſt pour cela <lb/>que je me ſuis déterminé à ne donner préſente-<lb/>ment que les Propoſitions fondamentales de <lb/>la Méchanique: </s> <s xml:id="echoid-s154" xml:space="preserve">peut-être que de plus habiles <lb/>gens que moy, & </s> <s xml:id="echoid-s155" xml:space="preserve">qui ſeront plus en état de <lb/>faire cette entrepriſe, voudront bien ſe donner <lb/>la peine d’en faire l’application à la Phyſique.</s> <s xml:id="echoid-s156" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s157" xml:space="preserve">Mais en attendant, je ne laiſſeray pas d’a-<lb/>maſſer tout ce que je pouray d’expériences <lb/>pour ce deſſein; </s> <s xml:id="echoid-s158" xml:space="preserve">c’eſt pourquoy je prie ceux <lb/>qui n’auront pas en vuë d’y travailler, de vou-<lb/>loir bien me communiquer celles qu’ils croi-<lb/>ront s’y pouvoir rapporter: </s> <s xml:id="echoid-s159" xml:space="preserve">& </s> <s xml:id="echoid-s160" xml:space="preserve">ſur tout de <lb/>me faire part de tout ce qui leur viendra de <lb/>difficultez ou de lumieres ſur les principes <lb/>qu’on propoſe icy, leur promettant d’en uſer <lb/>avec toute la docilité d’un homme qui ne <lb/>cherche que la vérité.</s> <s xml:id="echoid-s161" xml:space="preserve"/> </p> <pb file="0026" n="26"/> <figure> <image file="0026-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0026-01"/> </figure> </div> <div xml:id="echoid-div7" type="section" level="1" n="7"> <head xml:id="echoid-head7" xml:space="preserve">AVER TISSE MENT.</head> <p style="it"> <s xml:id="echoid-s162" xml:space="preserve">LES corollaires qu’on verra <lb/>citez dans la Solution de cba-<lb/>que Problême ſuivant, ſeront de la <lb/>Propoſition fondamentale qui le pré-<lb/>cède. </s> <s xml:id="echoid-s163" xml:space="preserve">C’eſt de peur d’ennuyer par <lb/>des rèpètitions trop frèquentes qu’on <lb/>ne la citera point.</s> <s xml:id="echoid-s164" xml:space="preserve"/> </p> <pb o="1" file="0027" n="27"/> <figure> <image file="0027-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0027-01"/> </figure> </div> <div xml:id="echoid-div8" type="section" level="1" n="8"> <head xml:id="echoid-head8" xml:space="preserve">PROJET <lb/>D’UNE <lb/>NOUVELLE MECHANIQUE.</head> <figure> <image file="0027-02" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0027-02"/> </figure> <p style="it"> <s xml:id="echoid-s165" xml:space="preserve">CE C 1<unsure/> n’ètant que pour ceux qui entendent aſſez <lb/>ces matières pour en pouvoir juger, on ne s’arrê-<lb/>tera point à répéter des Définitions, ny des Axio-<lb/>mes qui ſe trouvent par tout; </s> <s xml:id="echoid-s166" xml:space="preserve">en voici ſeulcment <lb/>un, avec une demande, & </s> <s xml:id="echoid-s167" xml:space="preserve">quelques Lemmes <lb/>particuliérement néceſſaires pour l’intelligence de ce Projet.</s> <s xml:id="echoid-s168" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div9" type="section" level="1" n="9"> <head xml:id="echoid-head9" xml:space="preserve"><emph style="sc">Axiome</emph>.</head> <p> <s xml:id="echoid-s169" xml:space="preserve">Les eſpaces que parcourt un même corps, ou des <lb/>corps égaux dans des tems égaux, ſont entre-eux <lb/>comme les forces qui les meuvent; </s> <s xml:id="echoid-s170" xml:space="preserve">& </s> <s xml:id="echoid-s171" xml:space="preserve">réciproque-<lb/>ment lorſque ces Eſpaces ſont entre-eux comme ces <pb o="2" file="0028" n="28" rhead="NOUVELLE"/> forces, elles les font parcourir au même corps, ou à <lb/> <anchor type="note" xlink:label="note-0028-01a" xlink:href="note-0028-01"/> des corps égaux en tems égaux.</s> <s xml:id="echoid-s172" xml:space="preserve"/> </p> <div xml:id="echoid-div9" type="float" level="2" n="1"> <note position="left" xlink:label="note-0028-01" xlink:href="note-0028-01a" xml:space="preserve">LEMMES.</note> </div> </div> <div xml:id="echoid-div11" type="section" level="1" n="10"> <head xml:id="echoid-head10" xml:space="preserve"><emph style="sc">Demande</emph>.</head> <p> <s xml:id="echoid-s173" xml:space="preserve">On ſuppoſe ici que dans tout corps qui ſe meut, <lb/>ou qui fait effort pour ſe mouvoir, il y a toûjours <lb/>un certain point qui ſurchargé de l’impreſſion de <lb/>tous les autres, détermine ce corps à ſuivre celle <lb/>qu’il a pour lors vers l’endroit où il tend. </s> <s xml:id="echoid-s174" xml:space="preserve">On ne ſe <lb/>met point en peine que ce point ſoit le même dans <lb/>toutes les ſituations poſſibles de ce corps: </s> <s xml:id="echoid-s175" xml:space="preserve">c’eſt aſſez <lb/>que dans chaque ſituation il y en ait un que l’on <lb/>appelle ici ſon Centre de gravité, ou, plus générale-<lb/>ment ſon Centre de direction, ou, d’équilibre, du moins <lb/>pour le tems qu’il détermine ainſi ce corps à ſuivre <lb/>ſon impreſſion; </s> <s xml:id="echoid-s176" xml:space="preserve">& </s> <s xml:id="echoid-s177" xml:space="preserve">la ligne qui joint ce point avec <lb/>celui où il tend, s’appelle ſa Ligne de direction.</s> <s xml:id="echoid-s178" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s179" xml:space="preserve">On ne met ceci en ſuppoſition que pour abréger; </s> <s xml:id="echoid-s180" xml:space="preserve">autrement <lb/>on le pourroit dèmontrer: </s> <s xml:id="echoid-s181" xml:space="preserve">car il n’y arien de plus évident que <lb/>de tous les points d’un corps, il y en a toûjours, & </s> <s xml:id="echoid-s182" xml:space="preserve">même <lb/>néceſſairement, quelqu’un autour duquel l’impreſſion qu’ils <lb/>ont tous vers le côtè ou ce corps tend, ſe trouve ſi ègalement <lb/>partagèe qu’ils demeureroient en èquilibre deſſus, ſi ſans cban-<lb/>ger la ſituation de ce corps par raport à l’endroit où il tend, <lb/>on le rendoit fixe; </s> <s xml:id="echoid-s183" xml:space="preserve">& </s> <s xml:id="echoid-s184" xml:space="preserve">par conſèquent l’impreſſion d’un tel <lb/>point ainſi ſurcbargè de celle de tous les autres, étant l<unsure/>a <lb/>même que s’il ètoit le ſeul qui en eût, il doit déterminer ce <lb/>corps à la ſuivre.</s> <s xml:id="echoid-s185" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s186" xml:space="preserve">Il n’y a rien-là, ce me ſemble, que de clair; </s> <s xml:id="echoid-s187" xml:space="preserve">cependant <lb/>s’il ſe trouvoit quelqu’un qui faute de le voir de même, fit <lb/>difficultè de l’accorder, il peut prendre dans la ſuite les corps <lb/>dont on parlera, pou@ des points qui ayent la peſanteur, ou <lb/>l’impreſſion qu’on y ſuppoſe; </s> <s xml:id="echoid-s188" xml:space="preserve">ou bien pour des puiſſances appli-<lb/>quées de même, & </s> <s xml:id="echoid-s189" xml:space="preserve">qui leur ſoient ègales: </s> <s xml:id="echoid-s190" xml:space="preserve">parce que les démonſ- <pb o="3" file="0029" n="29" rhead="ME’CHANIQUE."/> trations ſuivantes ſe peuvent également appliquor aux uns <lb/> <anchor type="note" xlink:label="note-0029-01a" xlink:href="note-0029-01"/> & </s> <s xml:id="echoid-s191" xml:space="preserve">aux autres.</s> <s xml:id="echoid-s192" xml:space="preserve"/> </p> <div xml:id="echoid-div11" type="float" level="2" n="1"> <note position="right" xlink:label="note-0029-01" xlink:href="note-0029-01a" xml:space="preserve">LEMMES.</note> </div> <figure> <image file="0029-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0029-01"/> </figure> </div> <div xml:id="echoid-div13" type="section" level="1" n="11"> <head xml:id="echoid-head11" xml:space="preserve">LEMME I.</head> <p style="it"> <s xml:id="echoid-s193" xml:space="preserve">LE poids A ètant ſuſpendu à une corde EH, ou à deux <lb/> <anchor type="note" xlink:label="note-0029-02a" xlink:href="note-0029-02"/> MH & </s> <s xml:id="echoid-s194" xml:space="preserve">NH attacbèes à un même point fixe H; <lb/></s> <s xml:id="echoid-s195" xml:space="preserve">ou bien ſoutenu ſur un pieu EH, ou ſur deux MH & </s> <s xml:id="echoid-s196" xml:space="preserve">NH <lb/>appuyez auſſi ſur un même point H: </s> <s xml:id="echoid-s197" xml:space="preserve">en ſorte que la li-<lb/>gne A H, qui joint ſon Centre de gravitè A avec ſon <lb/>point de ſuſpenſion, ou d’apuy, faſſe quelque angle que c@ <lb/>ſoit avec ſa ligne de direction AK: </s> <s xml:id="echoid-s198" xml:space="preserve">Ce poids tombera de A <lb/>vers B le long de l’arc A B dont H eſt le centre, juſqu’à <lb/>ce que la ligne AH ſoit dans la perpendiculaire, ou dans <lb/>le plan borizontal H B, & </s> <s xml:id="echoid-s199" xml:space="preserve">étant arrivé en B, ily demeu-<lb/>rera, ſi l’on n’y ſuppoſe à’autre cauſe que ſa peſanteur.</s> <s xml:id="echoid-s200" xml:space="preserve"/> </p> <div xml:id="echoid-div13" type="float" level="2" n="1"> <note position="right" xlink:label="note-0029-02" xlink:href="note-0029-02a" xml:space="preserve">fig. 1. <lb/>2. <lb/>3. <lb/>4.</note> </div> </div> <div xml:id="echoid-div15" type="section" level="1" n="12"> <head xml:id="echoid-head12" xml:space="preserve"><emph style="sc">Demonstr ation</emph>.</head> <p> <s xml:id="echoid-s201" xml:space="preserve">L’effet de la peſanteur du poids A, c’eſt de l’ap-<lb/>procher ( byp. </s> <s xml:id="echoid-s202" xml:space="preserve">) du centre de la terre, c’eſt-à-dire, <lb/>( dem. </s> <s xml:id="echoid-s203" xml:space="preserve">) d’en approcher ſon centre de gravité A, <lb/>tant que rien ne l’en empêche. </s> <s xml:id="echoid-s204" xml:space="preserve">Or dans la ſituation <lb/>preſente rien ne l’empêche de s’en approcher de la <lb/>longueur de A K, en tombant le long de l’arc A B <lb/>juſqu’en B; </s> <s xml:id="echoid-s205" xml:space="preserve">au contraire étant arrivé en B, la cor-<lb/>de E H, ou les cordes M H & </s> <s xml:id="echoid-s206" xml:space="preserve">N H; </s> <s xml:id="echoid-s207" xml:space="preserve">ou bien le plan <lb/>horizontal H B, le retiennent & </s> <s xml:id="echoid-s208" xml:space="preserve">l’empêchent de <lb/>deſcendre davantage; </s> <s xml:id="echoid-s209" xml:space="preserve">& </s> <s xml:id="echoid-s210" xml:space="preserve">par conſéquent ce poids <lb/>dans la ſituation preſente tombera le long de l’arc A B <lb/>juſqu’en B, & </s> <s xml:id="echoid-s211" xml:space="preserve">y étant arrivé, il y demeurera. </s> <s xml:id="echoid-s212" xml:space="preserve">Ce <lb/>qu’il faloit demontrer.</s> <s xml:id="echoid-s213" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div16" type="section" level="1" n="13"> <head xml:id="echoid-head13" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s214" xml:space="preserve">On prouvera de même que tout autre corps, de <pb o="4" file="0030" n="30" rhead="NOUVELLE"/> quelque côté qu’il tende, appuyé ſeulement ſur un <lb/> <anchor type="note" xlink:label="note-0030-01a" xlink:href="note-0030-01"/> de ſes points, ou même ſur une de ſes faces, de quel-<lb/>que largeur qu’elle ſoit, doit neceſſairement avancer <lb/>du côté ou il eſt pouſſé par ſa peſanteur, ou par <lb/>quelqu’autre force que ce ſoit, non-ſeulement tant <lb/>que ſa ligne de direction ne paſſe point par ce point <lb/>d’apuy, ny par aucun de la partie de cette face ſur <lb/>laquelle il s’appuye; </s> <s xml:id="echoid-s215" xml:space="preserve">mais encore tant qu’elle n’eſt <lb/>point perpendiculaire au plan, ou à la ſurface qui ſe <lb/>trouve à ſon paſſage; </s> <s xml:id="echoid-s216" xml:space="preserve">c’eſt-à-dire, tant que ſon centre <lb/>de direction n’eſt point appuyé: </s> <s xml:id="echoid-s217" xml:space="preserve">Car ce centre pou-<lb/>vant encore avancer du côté où il tend, la force qui <lb/>le pouſſe, ne manquera pas de l’y obliger; </s> <s xml:id="echoid-s218" xml:space="preserve">Mais auſſi <lb/>par une raiſon toute contraire, ſi-tôt que l’un, & </s> <s xml:id="echoid-s219" xml:space="preserve"><lb/>l’autre arrivera, ce corps demeurera néceſſairement <lb/>en cet état.</s> <s xml:id="echoid-s220" xml:space="preserve"/> </p> <div xml:id="echoid-div16" type="float" level="2" n="1"> <note position="left" xlink:label="note-0030-01" xlink:href="note-0030-01a" xml:space="preserve">LEMMES.</note> </div> <figure> <image file="0030-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0030-01"/> </figure> </div> <div xml:id="echoid-div18" type="section" level="1" n="14"> <head xml:id="echoid-head14" xml:space="preserve">LEMME II.</head> <p style="it"> <s xml:id="echoid-s221" xml:space="preserve">LE poids MN étant ſuſpendu par deux cordes P M & </s> <s xml:id="echoid-s222" xml:space="preserve"><lb/> <anchor type="note" xlink:label="note-0030-02a" xlink:href="note-0030-02"/> RN, attacbées aux clous P & </s> <s xml:id="echoid-s223" xml:space="preserve">R, & </s> <s xml:id="echoid-s224" xml:space="preserve">qui prolon-<lb/>gées concourent en H, ſa ligne de diréction A H paſſera <lb/>par ce point de concours.</s> <s xml:id="echoid-s225" xml:space="preserve"/> </p> <div xml:id="echoid-div18" type="float" level="2" n="1"> <note position="left" xlink:label="note-0030-02" xlink:href="note-0030-02a" xml:space="preserve">fig. 5. <lb/>6.</note> </div> </div> <div xml:id="echoid-div20" type="section" level="1" n="15"> <head xml:id="echoid-head15" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s226" xml:space="preserve">Premiérement ( fig. </s> <s xml:id="echoid-s227" xml:space="preserve">5. </s> <s xml:id="echoid-s228" xml:space="preserve">) l’effort que le poids M N <lb/>fait pour attirer le point P de la corde P M vers M, <lb/>étant le même qu’il feroit contre le point H, ſi cette <lb/>corde prolongée y étoit attachée, & </s> <s xml:id="echoid-s229" xml:space="preserve">non plus en P; <lb/></s> <s xml:id="echoid-s230" xml:space="preserve">ce corps, ou ce poids eſt ſoutenu par cette corde, <lb/>comme ſi elle n’étoit attachée qu’en H. </s> <s xml:id="echoid-s231" xml:space="preserve">Pour la <lb/>même raiſon il eſt auſſi ſoutenu par la corde R N, <lb/>comme ſi étant prolongée, elle n’étoit attachée qu’en <lb/>H: </s> <s xml:id="echoid-s232" xml:space="preserve">Il eſt donc ſoutenu par ces deux cordes enſem- <pb o="5" file="0031" n="31" rhead="ME’CHANIQUE."/> ble, comme ſi l’une, & </s> <s xml:id="echoid-s233" xml:space="preserve">l’autre étant prolongée, elles <lb/> <anchor type="note" xlink:label="note-0031-01a" xlink:href="note-0031-01"/> n’étoient attachées qu’en ce ſeul point: </s> <s xml:id="echoid-s234" xml:space="preserve">& </s> <s xml:id="echoid-s235" xml:space="preserve">par con-<lb/>ſéquent ( Lemm. </s> <s xml:id="echoid-s236" xml:space="preserve">1. </s> <s xml:id="echoid-s237" xml:space="preserve">) la ligne de direction A K de ce <lb/>poids ainſi ſuſpendu, paſſera par le point H où les <lb/>cordes P M & </s> <s xml:id="echoid-s238" xml:space="preserve">R N concourent. </s> <s xml:id="echoid-s239" xml:space="preserve">Ce qu’il F.</s> <s xml:id="echoid-s240" xml:space="preserve">D.</s> <s xml:id="echoid-s241" xml:space="preserve"/> </p> <div xml:id="echoid-div20" type="float" level="2" n="1"> <note position="right" xlink:label="note-0031-01" xlink:href="note-0031-01a" xml:space="preserve">LEMMES.</note> </div> <p> <s xml:id="echoid-s242" xml:space="preserve">Secondement ( fig. </s> <s xml:id="echoid-s243" xml:space="preserve">6. </s> <s xml:id="echoid-s244" xml:space="preserve">) l’effort que le poids M N fait <lb/>pour attirer le point P de ſa corde P M vers M, étant <lb/>le même que celui dont il preſſe ſon point C vers H, <lb/>il eſt encore ſoutenu par cette corde, comme il le <lb/>ſeroit par le pieu C H. </s> <s xml:id="echoid-s245" xml:space="preserve">Et pour la même raiſon il <lb/>eſt ſoutenu par la corde R N, comme il le ſeroit par <lb/>le pieu D H: </s> <s xml:id="echoid-s246" xml:space="preserve">Donc il eſt ſoutenu par ces deux cor-<lb/>des, comme il le ſeroit par ces deux pieux enſemble: <lb/></s> <s xml:id="echoid-s247" xml:space="preserve">& </s> <s xml:id="echoid-s248" xml:space="preserve">par conſéquent ( Lemm. </s> <s xml:id="echoid-s249" xml:space="preserve">1. </s> <s xml:id="echoid-s250" xml:space="preserve">) la ligne de direction <lb/>A K de ce poids ainſi ſuſpendu, paſſera encore par le <lb/>point H. </s> <s xml:id="echoid-s251" xml:space="preserve">Ce qu’il F. </s> <s xml:id="echoid-s252" xml:space="preserve">D.</s> <s xml:id="echoid-s253" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div22" type="section" level="1" n="16"> <head xml:id="echoid-head16" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s254" xml:space="preserve">De-là il eſt clair que de quelque côté que tende <lb/>le corps M N, & </s> <s xml:id="echoid-s255" xml:space="preserve">que ſa ligne de direction ſoit tour-<lb/>née, ſoit que cette impreſſion lui vienne de ſa pe-<lb/>ſanteur, ou de quelque autre force, cette ligne de <lb/>direction paſſera toûjours par le point H, ou les cor-<lb/>des qui le retiennent, doivent concourir, ou con-<lb/>courent en effet.</s> <s xml:id="echoid-s256" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div23" type="section" level="1" n="17"> <head xml:id="echoid-head17" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s257" xml:space="preserve">De ſorte que ſi ces cordes ne concourent qu’à une <lb/>diſtance infinie, c’eſt-à-dire qu’elles ſoient paralleles <lb/>entre-elles, cette ligne de direction ſera auſſi parallele <lb/>à l’une & </s> <s xml:id="echoid-s258" xml:space="preserve">à l’autre.</s> <s xml:id="echoid-s259" xml:space="preserve"/> </p> <figure> <image file="0031-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0031-01"/> </figure> <pb o="6" file="0032" n="32" rhead="NOUVELLE"/> <note position="left" xml:space="preserve">LEMMES.</note> <figure> <image file="0032-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0032-01"/> </figure> </div> <div xml:id="echoid-div24" type="section" level="1" n="18"> <head xml:id="echoid-head18" xml:space="preserve">LEMME III.</head> <p style="it"> <s xml:id="echoid-s260" xml:space="preserve">SI le point A ſans peſanteur eſt pouſſé en même tems, <lb/> <anchor type="note" xlink:label="note-0032-02a" xlink:href="note-0032-02"/> & </s> <s xml:id="echoid-s261" xml:space="preserve">uniformément par deux puiſſances E & </s> <s xml:id="echoid-s262" xml:space="preserve">F ſuivant <lb/>les lignes A C & </s> <s xml:id="echoid-s263" xml:space="preserve">A B, qui faſſent entre-elles quelque an-<lb/>gle C A B que ce ſoit, & </s> <s xml:id="echoid-s264" xml:space="preserve">que la force dont agit la puiſ-<lb/>ſance E, ſoit à celle dont agit la puiſſance F, comme A C <lb/>à A B. </s> <s xml:id="echoid-s265" xml:space="preserve">Ce point A ſuivra la Diagonale A D du paral-<lb/>lelogramme B C fait ſous ces deux lignes.</s> <s xml:id="echoid-s266" xml:space="preserve"/> </p> <div xml:id="echoid-div24" type="float" level="2" n="1"> <note position="left" xlink:label="note-0032-02" xlink:href="note-0032-02a" xml:space="preserve">fig. 7.</note> </div> </div> <div xml:id="echoid-div26" type="section" level="1" n="19"> <head xml:id="echoid-head19" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s267" xml:space="preserve">Le point A pouſſé par la puiſſance E vers C D, <lb/>l’eſt de même que s’il y étoit porté avec la ligne A B <lb/>toûjours parallele à elle-même, de la même viteſſe <lb/>qu’il y eſt pouſſé; </s> <s xml:id="echoid-s268" xml:space="preserve">Nous pouvons donc le regarder <lb/>comme pouſſé de cette maniére vers C D avec la <lb/>ligne A B toûjours parallelle à elle-même, ou à C D, <lb/>au même tems qu’il eſt pouſſé par la puiſſance F <lb/>le long de la même ligne A B. </s> <s xml:id="echoid-s269" xml:space="preserve">Or cela bien conçu, <lb/>il eſt clair qu’en quelque point, par exemple G, que <lb/>la ligne A B rencontre A D, le point A s’y trouvera <lb/>toûjours: </s> <s xml:id="echoid-s270" xml:space="preserve">parce que la force qui le porte avec A B <lb/>vers C D, eſt à celle qui le porte le long de la mê-<lb/>me A B, comme ( byp. </s> <s xml:id="echoid-s271" xml:space="preserve">) A C à A B; </s> <s xml:id="echoid-s272" xml:space="preserve">c’eſt-à-dire, en <lb/>tirant H K par le point G parallele à A B, comme <lb/>A K à K G: </s> <s xml:id="echoid-s273" xml:space="preserve">Donc ( ax. </s> <s xml:id="echoid-s274" xml:space="preserve">) au même tems que A B <lb/>parcourt A K, & </s> <s xml:id="echoid-s275" xml:space="preserve">qu’elle arrive avec le point A en <lb/>H K, ce même point parcourt une partie de A B <lb/>égale à K G; </s> <s xml:id="echoid-s276" xml:space="preserve">& </s> <s xml:id="echoid-s277" xml:space="preserve">par conſéquent il ſe trouve alors <lb/>en G. </s> <s xml:id="echoid-s278" xml:space="preserve">On démontrera de même qu’au même tems <lb/>que A B arrive en C D, le point A ſe trouve en D; <lb/></s> <s xml:id="echoid-s279" xml:space="preserve">& </s> <s xml:id="echoid-s280" xml:space="preserve">ainſi dans tous les autres points de la Diagona-<lb/>le A D: </s> <s xml:id="echoid-s281" xml:space="preserve">& </s> <s xml:id="echoid-s282" xml:space="preserve">par conſéquent ce point ainſi pouſſé ſe <pb o="7" file="0033" n="33" rhead="ME’CHANIQUE."/> meuvra exactement le long de cette ligne. </s> <s xml:id="echoid-s283" xml:space="preserve">Ce qu’il <lb/> <anchor type="note" xlink:label="note-0033-01a" xlink:href="note-0033-01"/> F. </s> <s xml:id="echoid-s284" xml:space="preserve">D.</s> <s xml:id="echoid-s285" xml:space="preserve"/> </p> <div xml:id="echoid-div26" type="float" level="2" n="1"> <note position="right" xlink:label="note-0033-01" xlink:href="note-0033-01a" xml:space="preserve">LEMMES.</note> </div> </div> <div xml:id="echoid-div28" type="section" level="1" n="20"> <head xml:id="echoid-head20" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s286" xml:space="preserve">Quand bien même les forces dont agiſſent les puiſ-<lb/>ſances E & </s> <s xml:id="echoid-s287" xml:space="preserve">F augmenteroient ou diminuëroient, <lb/>pourvu que ce fut ſuivant la même proportion de <lb/>part & </s> <s xml:id="echoid-s288" xml:space="preserve">d’autre; </s> <s xml:id="echoid-s289" xml:space="preserve">ce même point ſe meuvroit encore <lb/>exactement le long de A D: </s> <s xml:id="echoid-s290" xml:space="preserve">parce que ces forces <lb/>ſeroient encore entre-elles comme A C à A B, ou <lb/>comme A K à K G.</s> <s xml:id="echoid-s291" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div29" type="section" level="1" n="21"> <head xml:id="echoid-head21" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s292" xml:space="preserve">C’eſt la même choſe que le point A ſoit pouſſé le <lb/>long de A D par le concours d’action des puiſſances <lb/>E & </s> <s xml:id="echoid-s293" xml:space="preserve">F, ou qu’il y ſoit pouſſé par une ſeule puiſſance <lb/>d’une viteſſe égale à celle que lui cauſent ces deux <lb/>enſemble.</s> <s xml:id="echoid-s294" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div30" type="section" level="1" n="22"> <head xml:id="echoid-head22" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s295" xml:space="preserve">Puiſque le point A parcourt A D & </s> <s xml:id="echoid-s296" xml:space="preserve">A B en même <lb/>tems, la force, ou le compoſé de forces, qui le pouſ-<lb/>ſe le long de A D, eſt à celle qui le pouſſe le long <lb/>de A B, ( ax. </s> <s xml:id="echoid-s297" xml:space="preserve">) comme A D à A B. </s> <s xml:id="echoid-s298" xml:space="preserve">Et pour la même <lb/>raiſon, elle eſt à celle dont il eſt pouſſé le long de <lb/>A C, comme A D à A C.</s> <s xml:id="echoid-s299" xml:space="preserve"/> </p> <figure> <image file="0033-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0033-01"/> </figure> </div> <div xml:id="echoid-div31" type="section" level="1" n="23"> <head xml:id="echoid-head23" xml:space="preserve">LEMME IV.</head> <p style="it"> <s xml:id="echoid-s300" xml:space="preserve">PRèſentement que le point A ſoit le centre de diréction <lb/> <anchor type="note" xlink:label="note-0033-02a" xlink:href="note-0033-02"/> du corps E F, & </s> <s xml:id="echoid-s301" xml:space="preserve">que ce corps ſoit pouſſè en même tems <lb/>& </s> <s xml:id="echoid-s302" xml:space="preserve">uniformèment par deux puiſſances appliquées en E & </s> <s xml:id="echoid-s303" xml:space="preserve"><lb/>en F, ſuivant les lignes E C & </s> <s xml:id="echoid-s304" xml:space="preserve">F B, qui paſſent par le <lb/>point A, avec des forces qui ſoient entr’elles comme A C <pb o="8" file="0034" n="34" rhead="NOUVELLE"/> & </s> <s xml:id="echoid-s305" xml:space="preserve">A B: </s> <s xml:id="echoid-s306" xml:space="preserve">que l’on acbeve le parallelogramme B C, & </s> <s xml:id="echoid-s307" xml:space="preserve">que <lb/> <anchor type="note" xlink:label="note-0034-01a" xlink:href="note-0034-01"/> l’on regarde pour un moment ce corps comme s’il n’avoit au-<lb/>cune peſanteur. </s> <s xml:id="echoid-s308" xml:space="preserve">Quelque angle BAC que ces lignes faſſe<unsure/>nt <lb/>entr’elles, ce corps ainſi pouſſè ſuivra la diagonale A D.</s> <s xml:id="echoid-s309" xml:space="preserve"/> </p> <div xml:id="echoid-div31" type="float" level="2" n="1"> <note position="right" xlink:label="note-0033-02" xlink:href="note-0033-02a" xml:space="preserve">fig. 8.</note> <note position="left" xlink:label="note-0034-01" xlink:href="note-0034-01a" xml:space="preserve">LEMMES.</note> </div> </div> <div xml:id="echoid-div33" type="section" level="1" n="24"> <head xml:id="echoid-head24" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s310" xml:space="preserve">Les lignes EC & </s> <s xml:id="echoid-s311" xml:space="preserve">FB, ſuivant leſquelles les puiſ-<lb/>ſances E & </s> <s xml:id="echoid-s312" xml:space="preserve">F agiſſent, paſſant ( byp. </s> <s xml:id="echoid-s313" xml:space="preserve">) par le point A, <lb/>il eſt pouſſé en même-tems, & </s> <s xml:id="echoid-s314" xml:space="preserve">uniformement ſuivant <lb/>ces mêmes lignes vers CD & </s> <s xml:id="echoid-s315" xml:space="preserve">BD, par des forces <lb/>qui ſont entr’elles, ( byp. </s> <s xml:id="echoid-s316" xml:space="preserve">) comme A C à A B: </s> <s xml:id="echoid-s317" xml:space="preserve">Donc <lb/>( Lem. </s> <s xml:id="echoid-s318" xml:space="preserve">3. </s> <s xml:id="echoid-s319" xml:space="preserve">) le point A, & </s> <s xml:id="echoid-s320" xml:space="preserve">par conſéquent auſſi <lb/>( Dem. </s> <s xml:id="echoid-s321" xml:space="preserve">) le corps EF ſuivra la diagonale AD. </s> <s xml:id="echoid-s322" xml:space="preserve">Ce <lb/>qu’il F. </s> <s xml:id="echoid-s323" xml:space="preserve">D.</s> <s xml:id="echoid-s324" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div34" type="section" level="1" n="25"> <head xml:id="echoid-head25" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s325" xml:space="preserve">On voit que c’eſt la même choſe que le corps EF <lb/>ſoit pouſſé ſuivant AD par le concours d’action de <lb/>deux puiſſances appliquées en E, & </s> <s xml:id="echoid-s326" xml:space="preserve">en F, ſuivant <lb/>des lignes EC & </s> <s xml:id="echoid-s327" xml:space="preserve">FB qui paſſent par ſon centre de <lb/>gravité A, ou qu’il y ſoit pouſſé par une ſimple <lb/>puiſſance d’une viteſſe égale à celle que lui cauſent <lb/>ces deux enſemble: </s> <s xml:id="echoid-s328" xml:space="preserve">de ſorte que ſi AD étoit perpen-<lb/>diculaire à l’horizon, & </s> <s xml:id="echoid-s329" xml:space="preserve">que la peſanteur, que nous <lb/>concevons préſentement être dans ce corps, l’em-<lb/>portât le long de cette ligne avec une ſemblable vi-<lb/>teſſe, il la ſuivroit de même que s’il n’avoit en effet <lb/>aucune peſanteur, & </s> <s xml:id="echoid-s330" xml:space="preserve">qu’il fût pouſſé de la maniére <lb/>que nous venons de dire, par les deux puiſſances <lb/>E & </s> <s xml:id="echoid-s331" xml:space="preserve">F.</s> <s xml:id="echoid-s332" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div35" type="section" level="1" n="26"> <head xml:id="echoid-head26" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s333" xml:space="preserve">C’eſt pour cela auſſi que ce corps étant pouſſé par <lb/>ſa peſanteur ſuivant A D perpendiculaire à l’hori-<lb/>zon, s’il lui ſurvenoit quelqu’autre force qui le <pb o="9" file="0035" n="35" rhead="ME’CHANIQUE."/> pouſſât de même ſuivant quelqu’autre ligne qui ren-<lb/> <anchor type="note" xlink:label="note-0035-01a" xlink:href="note-0035-01"/> contrât celle-cy; </s> <s xml:id="echoid-s334" xml:space="preserve">Quelque angle que ces deux lignes <lb/>fiſſent entr’elles, ce corps ſuivroit la diagonale d’un <lb/>parallelogramme fait ſous des parties de ces lignes, <lb/>qui depuis le point de leur rencontre, fuſſent entre-<lb/>elles, comme la peſanteur de ce poids & </s> <s xml:id="echoid-s335" xml:space="preserve">cette nou-<lb/>velle force: </s> <s xml:id="echoid-s336" xml:space="preserve">Il la ſuivroit, dis-je, de la même <lb/>maniére que s’il n’étoit pouſſé que par une ſeule <lb/>puiſſance ſuivant cette diréction, & </s> <s xml:id="echoid-s337" xml:space="preserve">d’une viteſſe <lb/>égale à celle que lui pouroient cauſer ces deux en-<lb/>ſemble.</s> <s xml:id="echoid-s338" xml:space="preserve"/> </p> <div xml:id="echoid-div35" type="float" level="2" n="1"> <note position="right" xlink:label="note-0035-01" xlink:href="note-0035-01a" xml:space="preserve">LEMMES.</note> </div> </div> <div xml:id="echoid-div37" type="section" level="1" n="27"> <head xml:id="echoid-head27" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s339" xml:space="preserve">Il ſuit encore de cette propoſition, & </s> <s xml:id="echoid-s340" xml:space="preserve">du Corol-<lb/>laire 3. </s> <s xml:id="echoid-s341" xml:space="preserve">du Lemme 3.</s> <s xml:id="echoid-s342" xml:space="preserve">, que la force dont le corps E F <lb/>eſt pouſſé le long de A D, eſt à celle dont la puiſ-<lb/>ſance F le pouſſe le long de A B, comme A D à A B; <lb/></s> <s xml:id="echoid-s343" xml:space="preserve">& </s> <s xml:id="echoid-s344" xml:space="preserve">à celle dont la puiſſance E le pouſſe le long de A C, <lb/>comme A D à A C.</s> <s xml:id="echoid-s345" xml:space="preserve"/> </p> <figure> <image file="0035-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0035-01"/> </figure> </div> <div xml:id="echoid-div38" type="section" level="1" n="28"> <head xml:id="echoid-head28" xml:space="preserve">LEMME V.</head> <p style="it"> <s xml:id="echoid-s346" xml:space="preserve">L E S trois côtez d’un triangle réctiligne, quel qu’il ſoit, <lb/> <anchor type="note" xlink:label="note-0035-02a" xlink:href="note-0035-02"/> ſont entr’eux, comme les ſinus des angles auſquels ils <lb/>ſont ooppoſez.</s> <s xml:id="echoid-s347" xml:space="preserve"/> </p> <div xml:id="echoid-div38" type="float" level="2" n="1"> <note position="right" xlink:label="note-0035-02" xlink:href="note-0035-02a" xml:space="preserve">fig 9.</note> </div> </div> <div xml:id="echoid-div40" type="section" level="1" n="29"> <head xml:id="echoid-head29" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s348" xml:space="preserve">Soit ABD tel triangle réctiligne qu’on voudra, <lb/>inſcrit dans un cercle, dont C ſoit le centre; </s> <s xml:id="echoid-s349" xml:space="preserve">ſur <lb/>quelqu’un de ſes côtez, comme A B, ſoit menée du <lb/>centre C la perpendiculaire C E avec la ligne C A. <lb/></s> <s xml:id="echoid-s350" xml:space="preserve">Il eſt clair que A E eſt le ſinus de l’angle A C E, <lb/>qui eſt égal à l’angle D; </s> <s xml:id="echoid-s351" xml:space="preserve">Donc A B eſt le double du <pb o="10" file="0036" n="36" rhead="NOUVELLE"/> ſinus de l’angle D. </s> <s xml:id="echoid-s352" xml:space="preserve">Et pour la même raiſon, BD eſt <lb/> <anchor type="note" xlink:label="note-0036-01a" xlink:href="note-0036-01"/> le double du ſinus de l’angle BAD; </s> <s xml:id="echoid-s353" xml:space="preserve">& </s> <s xml:id="echoid-s354" xml:space="preserve">AD le dou-<lb/>ble auſſi de celui de l’angle B: </s> <s xml:id="echoid-s355" xml:space="preserve">Donc les trois côtez <lb/>de cetriangle ſont entr’eux, comme le double du ſinus <lb/>de chacun des angles auſquels ils ſont oppoſez: </s> <s xml:id="echoid-s356" xml:space="preserve">Donc <lb/>ils ſont auſſi entr’eux, comme ces mêmes ſinus. </s> <s xml:id="echoid-s357" xml:space="preserve">Ce <lb/>qu’il F. </s> <s xml:id="echoid-s358" xml:space="preserve">D.</s> <s xml:id="echoid-s359" xml:space="preserve"/> </p> <div xml:id="echoid-div40" type="float" level="2" n="1"> <note position="left" xlink:label="note-0036-01" xlink:href="note-0036-01a" xml:space="preserve">LEMMES.</note> </div> <figure> <image file="0036-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0036-01"/> </figure> <pb o="11" file="0037" n="37" rhead="MECHANIQUE."/> <figure> <image file="0037-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0037-01"/> </figure> </div> <div xml:id="echoid-div42" type="section" level="1" n="30"> <head xml:id="echoid-head30" xml:space="preserve">PROPOSITION <lb/>FONDAMENTALE <lb/>DES POIDS SUSPENDUS <lb/>AVEC DES CORDES</head> <p> <s xml:id="echoid-s360" xml:space="preserve">En quelque nombre qu’elles ſoient; </s> <s xml:id="echoid-s361" xml:space="preserve">& </s> <s xml:id="echoid-s362" xml:space="preserve">pour <lb/>tousles angles poſſibles qu’elles peuvent faire <lb/>entre-elles.</s> <s xml:id="echoid-s363" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s364" xml:space="preserve">LE Poids K ſoutenu avec les cordes P B & </s> <s xml:id="echoid-s365" xml:space="preserve">C R par <lb/> <anchor type="note" xlink:label="note-0037-01a" xlink:href="note-0037-01"/> les puiſſances P & </s> <s xml:id="echoid-s366" xml:space="preserve">R, & </s> <s xml:id="echoid-s367" xml:space="preserve">en équilibre avec elles, eſt <lb/>toujours à cbacune d’elles, comme le ſinus de l’angle P A R que <lb/>leurs cordes font entre-elles, à chacun des ſinus des angles <lb/>RAK & </s> <s xml:id="echoid-s368" xml:space="preserve">PAK que font avec la ligne de direction AK <lb/>de ce poids, chacune de ces cordes réciproquement priſes. </s> <s xml:id="echoid-s369" xml:space="preserve">Par <lb/>exemple, il eſt à la puiſſance P, comme le ſinus de R A P <lb/>au ſinus de R A K; </s> <s xml:id="echoid-s370" xml:space="preserve">& </s> <s xml:id="echoid-s371" xml:space="preserve">à la puiſſance R, comme le mème <lb/>ſinus de R A P au ſinus de P A K.</s> <s xml:id="echoid-s372" xml:space="preserve"/> </p> <div xml:id="echoid-div42" type="float" level="2" n="1"> <note position="right" xlink:label="note-0037-01" xlink:href="note-0037-01a" xml:space="preserve">fig. 10. <lb/>11. <lb/>12. <lb/>13. <lb/>14. <lb/>15. <lb/>16. <lb/>17.</note> </div> </div> <div xml:id="echoid-div44" type="section" level="1" n="31"> <head xml:id="echoid-head31" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s373" xml:space="preserve">Les impreſſions particuliéres que font les puiſſances <lb/>P & </s> <s xml:id="echoid-s374" xml:space="preserve">R, ſur le point A de ce corps, (fig. </s> <s xml:id="echoid-s375" xml:space="preserve">10. </s> <s xml:id="echoid-s376" xml:space="preserve">11. </s> <s xml:id="echoid-s377" xml:space="preserve">12. <lb/></s> <s xml:id="echoid-s378" xml:space="preserve">13. </s> <s xml:id="echoid-s379" xml:space="preserve">& </s> <s xml:id="echoid-s380" xml:space="preserve">14.) </s> <s xml:id="echoid-s381" xml:space="preserve">ou de ſa corde, (fig. </s> <s xml:id="echoid-s382" xml:space="preserve">15.) </s> <s xml:id="echoid-s383" xml:space="preserve">étant les mê-<lb/>mes qu’elles feroient, ſi elles le pouſſoient chacune <lb/>ſuivant ſa ligne de direction AP & </s> <s xml:id="echoid-s384" xml:space="preserve">AR: </s> <s xml:id="echoid-s385" xml:space="preserve">1°. </s> <s xml:id="echoid-s386" xml:space="preserve">Ce <pb o="12" file="0038" n="38" rhead="NOUVELLE"/> point regardé comme tiré ſeulement par ces deux <lb/> <anchor type="note" xlink:label="note-0038-01a" xlink:href="note-0038-01"/> puiſſances, doit tendre (Lemm. </s> <s xml:id="echoid-s387" xml:space="preserve">3.) </s> <s xml:id="echoid-s388" xml:space="preserve">le long de quelque <lb/>ligne AD, quiſoit la diagonale d’un parallelogramme <lb/>fait ſous des parties AB & </s> <s xml:id="echoid-s389" xml:space="preserve">AC des lignes de direction <lb/>des puiſſances P & </s> <s xml:id="echoid-s390" xml:space="preserve">R, qui ſoient entre-elles, comme <lb/>ces mêmes puiſſances. </s> <s xml:id="echoid-s391" xml:space="preserve">2°. </s> <s xml:id="echoid-s392" xml:space="preserve">Cette ligne AD doit être <lb/>la même que la ligne de direction A K de ce poids <lb/>prolongée du côté de D: </s> <s xml:id="echoid-s393" xml:space="preserve">autrement ces lignes A K <lb/>& </s> <s xml:id="echoid-s394" xml:space="preserve">AD faiſant en A quelque angle entre-elles, ce <lb/>point ainſi pouſſé, ou tiré ſuivant la ligne AD par <lb/>le concours d’action de ces puiſſances, & </s> <s xml:id="echoid-s395" xml:space="preserve">à même <lb/>tems ſuivant ſa ligne de direction A K par la peſan-<lb/>teur du poids K, devroit (Lemm. </s> <s xml:id="echoid-s396" xml:space="preserve">3.) </s> <s xml:id="echoid-s397" xml:space="preserve">ſuivre une <lb/>troiſiéme ligne qui fût la Diagonale d’un parallelo-<lb/>gramme fait ſous des parties de ces lignes priſes de-<lb/>puis A, & </s> <s xml:id="echoid-s398" xml:space="preserve">qui fuſſent entre-elles, comme la force <lb/>dont ce point eſt tiré par ces puiſſances ſuivant AD, <lb/>eſt à la peſanteur du poids K; </s> <s xml:id="echoid-s399" xml:space="preserve">ainſi ce poids ne ſeroit <lb/>plus en équilibre avec ces puiſſances, ce qui eſt contre <lb/>l’hypothêſe. </s> <s xml:id="echoid-s400" xml:space="preserve">3°. </s> <s xml:id="echoid-s401" xml:space="preserve">La force dont ce point eſt tiré ſuivant <lb/>AD, eſt auſſi égale à la peſanteur de ce poids; </s> <s xml:id="echoid-s402" xml:space="preserve">au-<lb/>trement cette ligne étant la même que la ligne de di-<lb/>rection de ce poids, il ſe meuvroit encore en haut, ou <lb/>en bas ſelon la différence de ces forces, ce qui eſt <lb/>encore contre l’hypothêſe: </s> <s xml:id="echoid-s403" xml:space="preserve">Donc ce point eſt tiré <lb/>de A vers D par le concours d’action des puiſſances <lb/>P & </s> <s xml:id="echoid-s404" xml:space="preserve">R ſuivant la ligne de direction de ce poids, & </s> <s xml:id="echoid-s405" xml:space="preserve"><lb/>d’une force êgale à ſa peſanteur. </s> <s xml:id="echoid-s406" xml:space="preserve">Or la force dont il <lb/>eſt ainſi tiré de A vers D, eſt à celle dont la puiſ-<lb/>ſance P le tire à elle, comme (Lemm. </s> <s xml:id="echoid-s407" xml:space="preserve">3. </s> <s xml:id="echoid-s408" xml:space="preserve">Cor. </s> <s xml:id="echoid-s409" xml:space="preserve">3.) </s> <s xml:id="echoid-s410" xml:space="preserve">A D <lb/>à A B; </s> <s xml:id="echoid-s411" xml:space="preserve">c’eſt-à-dire, (Lemm. </s> <s xml:id="echoid-s412" xml:space="preserve">5.) </s> <s xml:id="echoid-s413" xml:space="preserve">comme le ſinus de <lb/>l’angle D B A, ou de ſon complement PAR, (BC <lb/>eſt un parallelogramme) au ſinus de l’angle BDA, <lb/>ou de RAK égal à celui-ci, ou à ſon complement: <lb/></s> <s xml:id="echoid-s414" xml:space="preserve">Donc ce poids eſt à la puiſſance P, comme le ſinus <pb o="13" file="0039" n="39" rhead="MECHANIQUE."/> de l’angle PAR, au ſinus de l’angle RAK. </s> <s xml:id="echoid-s415" xml:space="preserve">On dé-<lb/> <anchor type="note" xlink:label="note-0039-01a" xlink:href="note-0039-01"/> montrera de même que ce même poids eſt à la puiſ-<lb/>ſance R, comme le même ſinus de l’angle PAR, au <lb/>ſinus de l’angle PAK.</s> <s xml:id="echoid-s416" xml:space="preserve"/> </p> <div xml:id="echoid-div44" type="float" level="2" n="1"> <note position="left" xlink:label="note-0038-01" xlink:href="note-0038-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> <note position="right" xlink:label="note-0039-01" xlink:href="note-0039-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> <p> <s xml:id="echoid-s417" xml:space="preserve">Voilà pour les ſix premiers cas. </s> <s xml:id="echoid-s418" xml:space="preserve">(fig. </s> <s xml:id="echoid-s419" xml:space="preserve">10. </s> <s xml:id="echoid-s420" xml:space="preserve">11. </s> <s xml:id="echoid-s421" xml:space="preserve">12. </s> <s xml:id="echoid-s422" xml:space="preserve">13. <lb/></s> <s xml:id="echoid-s423" xml:space="preserve">14. </s> <s xml:id="echoid-s424" xml:space="preserve">& </s> <s xml:id="echoid-s425" xml:space="preserve">15.) </s> <s xml:id="echoid-s426" xml:space="preserve">Pour les deux autres, (fig. </s> <s xml:id="echoid-s427" xml:space="preserve">16. </s> <s xml:id="echoid-s428" xml:space="preserve">& </s> <s xml:id="echoid-s429" xml:space="preserve">17.) </s> <s xml:id="echoid-s430" xml:space="preserve"><lb/>il ne faut que concevoir le poids K de telle figure <lb/>que le point A lui appartienne; </s> <s xml:id="echoid-s431" xml:space="preserve">c’eſt-à-dire qu’il lui <lb/>ſoit continu, ſans àugmenter ny diminuër ſa peſan-<lb/>teur, ni ſans changer ſon centre de gravité, non <lb/>plus que les angles RAK & </s> <s xml:id="echoid-s432" xml:space="preserve">PAK que font les <lb/>cordes que tiennent les puiſſances P & </s> <s xml:id="echoid-s433" xml:space="preserve">R, avec ſa <lb/>ligne de direction A K. </s> <s xml:id="echoid-s434" xml:space="preserve">Or cela bien conçu, il eſt <lb/>clair par ce qui vient d’être dit, que la peſanteur <lb/>de tout le poids K eſt égale à la force dont le point A <lb/>en ce cas ſeroit pouſſé, ou tiré de A vers D: </s> <s xml:id="echoid-s435" xml:space="preserve">ainſi <lb/>une telle force étant (Lemm. </s> <s xml:id="echoid-s436" xml:space="preserve">3. </s> <s xml:id="echoid-s437" xml:space="preserve">Cor. </s> <s xml:id="echoid-s438" xml:space="preserve">3.) </s> <s xml:id="echoid-s439" xml:space="preserve">à celle de la <lb/>puiſſance P, comme AD à AB; </s> <s xml:id="echoid-s440" xml:space="preserve">le poids K, qui eſt le <lb/>même (byp.) </s> <s xml:id="echoid-s441" xml:space="preserve">que s’il avoit une telle figure, eſt encore <lb/>ici à la puiſſance P, qui eſt encore auſſi (byp) la même, <lb/>comme AD à AB; </s> <s xml:id="echoid-s442" xml:space="preserve">c’eſt-à-dire, (Lemm. </s> <s xml:id="echoid-s443" xml:space="preserve">5.) </s> <s xml:id="echoid-s444" xml:space="preserve">comme <lb/>le ſinus de l’angle DBA, ou de ſon complement PAR, <lb/>au ſinus de l’angle BDA, ou de RAK qui lui eſt <lb/>égal; </s> <s xml:id="echoid-s445" xml:space="preserve">& </s> <s xml:id="echoid-s446" xml:space="preserve">de même à la puiſſance R, comme le ſinus <lb/>du même angle PAR au ſinus de l’angle PAK.</s> <s xml:id="echoid-s447" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s448" xml:space="preserve">Et par conſéquent, quelque ſoit le poids K, & </s> <s xml:id="echoid-s449" xml:space="preserve">de <lb/>quelque maniére qu’il ſoit ſoutenu avec les cordes <lb/>BP, & </s> <s xml:id="echoid-s450" xml:space="preserve">CR par les puiſſances P & </s> <s xml:id="echoid-s451" xml:space="preserve">R, il eſt toujours <lb/>à chacune de ces puiſſances, comme le ſinus de l’an-<lb/>gle que leurs cordes font entre-elles, à chacun des <lb/>ſinus des angles que font chacune de ces cordes ré-<lb/>ciproquement priſes avec ſa ligne de direction. </s> <s xml:id="echoid-s452" xml:space="preserve">Ce <lb/>qu’il F. </s> <s xml:id="echoid-s453" xml:space="preserve">D.</s> <s xml:id="echoid-s454" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s455" xml:space="preserve">Ce ſeroit ici le lieu de répondre à la digreſſion que Monſieur <pb o="14" file="0040" n="40" rhead="MECHANIQUE."/> Borelli à faite contre ce ſentiment dans ſon Traité du mo u-<lb/> <anchor type="note" xlink:label="note-0040-01a" xlink:href="note-0040-01"/> vement des Animaux, tom. </s> <s xml:id="echoid-s456" xml:space="preserve">1. </s> <s xml:id="echoid-s457" xml:space="preserve">chap. </s> <s xml:id="echoid-s458" xml:space="preserve">13. </s> <s xml:id="echoid-s459" xml:space="preserve">Mais la bréveté <lb/>qu’on s’eſt propoſée dans cet eſſay, ne permet pas de s’étendre <lb/>ici autant qu’il faudroit pour cela: </s> <s xml:id="echoid-s460" xml:space="preserve">On en verra dans peu <lb/>un examen particulier, où l’on fera voir en quoy cet Illuſtre <lb/>Autbeur s’eſt mépris.</s> <s xml:id="echoid-s461" xml:space="preserve"/> </p> <div xml:id="echoid-div45" type="float" level="2" n="2"> <note position="left" xlink:label="note-0040-01" xlink:href="note-0040-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> </div> <div xml:id="echoid-div47" type="section" level="1" n="32"> <head xml:id="echoid-head32" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s462" xml:space="preserve">Il ſuit de cette propoſition que les puiſſances P & </s> <s xml:id="echoid-s463" xml:space="preserve"><lb/>R ſont entr’elles en raiſon réciproque des ſinus des <lb/>angles que font leurs lignes de direction avec celle <lb/>du poid qu’elles ſoutiennent, c’eſt-à-dire, en raiſon <lb/>réciproque des diſtances de leurs lignes de direction <lb/>à celle de ce poids.</s> <s xml:id="echoid-s464" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div48" type="section" level="1" n="33"> <head xml:id="echoid-head33" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s465" xml:space="preserve">D’où l’on voit qu’en mettant une troiſiéme puiſ-<lb/>ſance, dont le nom ſoit K, à la place de ce poids, <lb/>qui faſſe équilibre, comme lui, avec les deux P & </s> <s xml:id="echoid-s466" xml:space="preserve">R <lb/>qui le ſoutiennent: </s> <s xml:id="echoid-s467" xml:space="preserve">ces trois puiſſances K, P, & </s> <s xml:id="echoid-s468" xml:space="preserve">R <lb/>ſeront entr’elles, comme les ſinus des angles P A R, <lb/>RAK, & </s> <s xml:id="echoid-s469" xml:space="preserve">PAK.</s> <s xml:id="echoid-s470" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div49" type="section" level="1" n="34"> <head xml:id="echoid-head34" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s471" xml:space="preserve">D’où il ſuit, non ſeulement que chacune de ces <lb/>puiſſances, de quelque maniére qu’on les prenne, <lb/>eſt toûjours plus petite que la ſomme des deux autres, <lb/>de même que chacun de ces ſinus eſt toujours moin-<lb/>dre que la ſomme des deux autres.</s> <s xml:id="echoid-s472" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div50" type="section" level="1" n="35"> <head xml:id="echoid-head35" xml:space="preserve"><emph style="sc">Corollaire</emph> IV.</head> <p> <s xml:id="echoid-s473" xml:space="preserve">Mais encore qu’elles ſont toutes trois priſes à <lb/>diſcrétion deux à deux, en raiſon réciproque des ſinus <lb/>des angles que font leurs cordes, ou leurs lignes de <lb/>diréction avec celle de la troiſiéme; </s> <s xml:id="echoid-s474" xml:space="preserve">C’eſt-à-dire, <pb o="15" file="0041" n="41" rhead="NOUVELLE"/> en raiſon réciproque des diſtances de leurs lignes de <lb/> <anchor type="note" xlink:label="note-0041-01a" xlink:href="note-0041-01"/> diréction à quelque point que ce ſoit de celle de la <lb/>troiſiéme.</s> <s xml:id="echoid-s475" xml:space="preserve"/> </p> <div xml:id="echoid-div50" type="float" level="2" n="1"> <note position="right" xlink:label="note-0041-01" xlink:href="note-0041-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> </div> <div xml:id="echoid-div52" type="section" level="1" n="36"> <head xml:id="echoid-head36" xml:space="preserve"><emph style="sc">Crollaire</emph> V.</head> <p> <s xml:id="echoid-s476" xml:space="preserve">Il ſuit encore de cette propoſition, que de quel-<lb/>que côté que tende le corps K, ſoit que cette impreſ-<lb/>ſion lui vienne de ſa peſanteur, ou de quelqu’autre <lb/>force; </s> <s xml:id="echoid-s477" xml:space="preserve">celle de cette impreſſion, tant que ce corps <lb/>ſera en équilibre avec les puiſſances P & </s> <s xml:id="echoid-s478" xml:space="preserve">R, ſera <lb/>toujours à celle dont chacune d’elles le retient, com-<lb/>me le ſinus de l’angle P A R que font leurs cordes <lb/>entr’elles, à chacun des ſinus des angles R A K & </s> <s xml:id="echoid-s479" xml:space="preserve"><lb/>PAK, que font chacune de ces cordes réciproque-<lb/>ment priſes, avec la ligne de direction A K de ce <lb/>corps, qui paſſe toûjours (Lem. </s> <s xml:id="echoid-s480" xml:space="preserve">2. </s> <s xml:id="echoid-s481" xml:space="preserve">Cor. </s> <s xml:id="echoid-s482" xml:space="preserve">1.) </s> <s xml:id="echoid-s483" xml:space="preserve">par le <lb/>point A de leur concours.</s> <s xml:id="echoid-s484" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div53" type="section" level="1" n="37"> <head xml:id="echoid-head37" xml:space="preserve"><emph style="sc">Corollaire</emph> VI.</head> <p> <s xml:id="echoid-s485" xml:space="preserve">Ce qui fait voir que ſi l’angle de ces cordes étoit infi-<lb/>niment aigu, c’eſt-à-dire, qu’elles fuſſent paralleles <lb/>entre-elles, la force, quelle qu’elle fût, dont ce corps <lb/>agiroit contre ces puiſſances, ſeroit égale à la ſomme <lb/>de celles dont elles lui reſiſtent: </s> <s xml:id="echoid-s486" xml:space="preserve">De ſorte que s’il <lb/>agiſſoit ſeulement comme poids, il ſeroit alors égal <lb/>à la ſomme de ces puiſſances; </s> <s xml:id="echoid-s487" xml:space="preserve">& </s> <s xml:id="echoid-s488" xml:space="preserve">chacune d’elle ſeroit <lb/>à l’autre en raiſon réciproque des diſtances de leurs <lb/>lignes de direction à celle de ce corps, ou de ce poids, <lb/>qui pour lors leur ſeroit (Lem. </s> <s xml:id="echoid-s489" xml:space="preserve">2. </s> <s xml:id="echoid-s490" xml:space="preserve">Cor. </s> <s xml:id="echoid-s491" xml:space="preserve">2.) </s> <s xml:id="echoid-s492" xml:space="preserve">auſſi parallele.</s> <s xml:id="echoid-s493" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div54" type="section" level="1" n="38"> <head xml:id="echoid-head38" xml:space="preserve"><emph style="sc">Corollaire</emph> VII.</head> <p> <s xml:id="echoid-s494" xml:space="preserve">Il ſuit de plus, que ſi le poids K eſt à chacune des <lb/>puiſſances P & </s> <s xml:id="echoid-s495" xml:space="preserve">R, comme les ſinus de l’angle R A P, <lb/>à chacun des ſinus des angles R A K & </s> <s xml:id="echoid-s496" xml:space="preserve">P A K, elles <lb/>le ſoutiendront en cet état, & </s> <s xml:id="echoid-s497" xml:space="preserve">feront équilibre avec <pb o="16" file="0042" n="42" rhead="MECHANIQUE."/> lui: </s> <s xml:id="echoid-s498" xml:space="preserve">Car ces deux puiſſances étant alors entr’elles, <lb/> <anchor type="note" xlink:label="note-0042-01a" xlink:href="note-0042-01"/> comme les ſinus des angles BDA & </s> <s xml:id="echoid-s499" xml:space="preserve">BAD, c’eſt-à-<lb/>dire, (Lem. </s> <s xml:id="echoid-s500" xml:space="preserve">5.) </s> <s xml:id="echoid-s501" xml:space="preserve">comme AB à BD, ou à CA qui lui <lb/>eſt égale; </s> <s xml:id="echoid-s502" xml:space="preserve">l’impreſſion compoſée qui réſulte du <lb/>concours d’action de ces deux puiſſances ſur le point <lb/>A, doit (Lem. </s> <s xml:id="echoid-s503" xml:space="preserve">3.) </s> <s xml:id="echoid-s504" xml:space="preserve">le faire tendre de A vers D ſui-<lb/>vant AD, d’une force qui ſoit à celle de chacune <lb/>de ces puiſſances P & </s> <s xml:id="echoid-s505" xml:space="preserve">R, (Lem. </s> <s xml:id="echoid-s506" xml:space="preserve">3. </s> <s xml:id="echoid-s507" xml:space="preserve">Cor. </s> <s xml:id="echoid-s508" xml:space="preserve">3.) </s> <s xml:id="echoid-s509" xml:space="preserve">comme <lb/>AD eſt à chacun des côtez AB & </s> <s xml:id="echoid-s510" xml:space="preserve">AC, ou BD, du <lb/>parallelogramme BC; </s> <s xml:id="echoid-s511" xml:space="preserve">ou bien (Lemm. </s> <s xml:id="echoid-s512" xml:space="preserve">5.) </s> <s xml:id="echoid-s513" xml:space="preserve">comme <lb/>le ſinus de l’angle DBA, ou de ſon complement <lb/>PAR, à chacun des ſinus des angles BDA & </s> <s xml:id="echoid-s514" xml:space="preserve">BAD, <lb/>ou de RAK & </s> <s xml:id="echoid-s515" xml:space="preserve">de PAK, quileur ſontégaux, ou qui <lb/>ſont leurs complemens; </s> <s xml:id="echoid-s516" xml:space="preserve">c’eſt-à-dire, (byp.) </s> <s xml:id="echoid-s517" xml:space="preserve">comme <lb/>le poids K à chacune de ces puiſſances: </s> <s xml:id="echoid-s518" xml:space="preserve">& </s> <s xml:id="echoid-s519" xml:space="preserve">par conſé-<lb/>quent la force dont ces puiſſances tirent ou pouſſent <lb/>le point A de ce corps, ou de ſa corde vers D, eſt <lb/>égale à celle, dont il eſt tiré vers K ſuivant la même <lb/>ligne DK par la peſanteur de ce poids: </s> <s xml:id="echoid-s520" xml:space="preserve">ainſi elles <lb/>le doivent ſoutenir en cet état, & </s> <s xml:id="echoid-s521" xml:space="preserve">demeurer en équi-<lb/>libre avec lui.</s> <s xml:id="echoid-s522" xml:space="preserve"/> </p> <div xml:id="echoid-div54" type="float" level="2" n="1"> <note position="left" xlink:label="note-0042-01" xlink:href="note-0042-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> </div> <div xml:id="echoid-div56" type="section" level="1" n="39"> <head xml:id="echoid-head39" xml:space="preserve"><emph style="sc">Corollaire</emph> VIII.</head> <p> <s xml:id="echoid-s523" xml:space="preserve">Ce qui fait voir que l’on peut faire ſoutenir quel-<lb/> <anchor type="note" xlink:label="note-0042-02a" xlink:href="note-0042-02"/> que grand poids K, que ce ſoit, à quelque puiſſance R <lb/>que ce puiſſe être par le moyen d’une corde ſeulement. <lb/></s> <s xml:id="echoid-s524" xml:space="preserve">Il ne faut pour cela que de quelque point D comme <lb/>centre, avec le rayon A D perpendiculaire à l’hori-<lb/>zon, & </s> <s xml:id="echoid-s525" xml:space="preserve">pris à diſcrétion, décrire l’arc A C B; </s> <s xml:id="echoid-s526" xml:space="preserve">& </s> <s xml:id="echoid-s527" xml:space="preserve">y <lb/>ayant inſcrit A C qui ſoit à A D, comme la puiſ-<lb/>ſance R au poids K, joignez D C, & </s> <s xml:id="echoid-s528" xml:space="preserve">apres avoir <lb/>dirigé par le point A la corde A P de ce poids pa-<lb/>rallelement à C D, attachez-la au crochet P, & </s> <s xml:id="echoid-s529" xml:space="preserve">luy <lb/>appliquez en A la puiſſance R ſuivant A C. </s> <s xml:id="echoid-s530" xml:space="preserve">Il eſt <lb/>clair par le Corollaire précédent que cette puiſſance <pb o="17" file="0043" n="43" rhead="MECHANIQUE."/> étant (byp.) </s> <s xml:id="echoid-s531" xml:space="preserve">à ce poids comme A C à A D; </s> <s xml:id="echoid-s532" xml:space="preserve">c’eſt-à-<lb/> <anchor type="note" xlink:label="note-0043-01a" xlink:href="note-0043-01"/> dire, (Lemm. </s> <s xml:id="echoid-s533" xml:space="preserve">5.) </s> <s xml:id="echoid-s534" xml:space="preserve">comme le ſinus de l’angle D, ou <lb/>de ſon égal D A P, au ſinus de l’angle D C A, ou deſon <lb/>complement PAC; </s> <s xml:id="echoid-s535" xml:space="preserve">elle ſe ſoutiendraen cet état.</s> <s xml:id="echoid-s536" xml:space="preserve"/> </p> <div xml:id="echoid-div56" type="float" level="2" n="1"> <note position="left" xlink:label="note-0042-02" xlink:href="note-0042-02a" xml:space="preserve">fig. 18.</note> <note position="right" xlink:label="note-0043-01" xlink:href="note-0043-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> </div> <div xml:id="echoid-div58" type="section" level="1" n="40"> <head xml:id="echoid-head40" xml:space="preserve"><emph style="sc">Corollaire</emph> IX.</head> <p> <s xml:id="echoid-s537" xml:space="preserve">Ainſi il n’y a point de force R, quelque petite qu’on <lb/>l’imagine, qui ne ſoit capable de mouvoir quelque <lb/>grand poids K que ce ſoit, ſuſpendu à une corde, <lb/>& </s> <s xml:id="echoid-s538" xml:space="preserve">de le faire ſortir de la ligne PF qui tombe à plomb <lb/>de ſon point de ſuſpenſion: </s> <s xml:id="echoid-s539" xml:space="preserve">& </s> <s xml:id="echoid-s540" xml:space="preserve">cela, juſqu’à ce que <lb/>les ſinus des angles que font leurs lignes de direction <lb/>AC & </s> <s xml:id="echoid-s541" xml:space="preserve">AD avec AP, qui va du point où elles <lb/>concourent, à ce point de ſuſpenſion, ſoient en raiſon <lb/>réciproque de ce poids & </s> <s xml:id="echoid-s542" xml:space="preserve">de cette puiſſance.</s> <s xml:id="echoid-s543" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div59" type="section" level="1" n="41"> <head xml:id="echoid-head41" xml:space="preserve"><emph style="sc">Corollaire</emph> X.</head> <p> <s xml:id="echoid-s544" xml:space="preserve">Et parce que ce mouvement eſt impoſſible, à moins <lb/>que ce poids ne monte de même que le point A de ſa <lb/>corde, du moins de la hauteur du ſinus verſe H E <lb/>de l’angle APE fait par la partie AP de ſa corde, <lb/>avec la ligne AF, qui tombe à plomb de ſon point A <lb/>de ſuſpenſion: </s> <s xml:id="echoid-s545" xml:space="preserve">il ſuit évidemment qu’il n’y a point <lb/>de force, quelque petite qu’on l’imagine, qui ne ſoit <lb/>capable de faire monter du moins à cette hauteur <lb/>quelque grand poids que ce ſoit à l’aide ſeulement <lb/>d’une corde attachée à quelque point fixe.</s> <s xml:id="echoid-s546" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s547" xml:space="preserve">La raiſon pour laquelle on dit ici du moins, c’eſt pour <lb/>s’ accommoder à toute bypothêſe: </s> <s xml:id="echoid-s548" xml:space="preserve">car ſi l’on regarde les lignes <lb/>de direction des poids comme paralleles entre-elles, ce poids <lb/>monte-ici juſtement de cette hauteur; </s> <s xml:id="echoid-s549" xml:space="preserve">mais ſi elles concourent <lb/>en quelque endroit du monde, il doit néceſſairement monter <lb/>plus baut, & </s> <s xml:id="echoid-s550" xml:space="preserve">ce d’autant plus (quoi. </s> <s xml:id="echoid-s551" xml:space="preserve">qu’en proportion diffé-<lb/>rente) que l’angle que fait ſa ligne de direction avec celle qui <pb o="18" file="0044" n="44" rhead="NOUVELLE"/> tombe à plomb de ſon point de ſuſpenſion, lorſque ce poids <lb/> <anchor type="note" xlink:label="note-0044-01a" xlink:href="note-0044-01"/> n’eſt plus dans cette même ligne, eſt plus obtus. </s> <s xml:id="echoid-s552" xml:space="preserve">Tout cela eſt <lb/>clair; </s> <s xml:id="echoid-s553" xml:space="preserve">c’eſt pourquoi on ne s’y arrête pas davantage.</s> <s xml:id="echoid-s554" xml:space="preserve"/> </p> <div xml:id="echoid-div59" type="float" level="2" n="1"> <note position="left" xlink:label="note-0044-01" xlink:href="note-0044-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> </div> <div xml:id="echoid-div61" type="section" level="1" n="42"> <head xml:id="echoid-head42" xml:space="preserve"><emph style="sc">Corollaire</emph> XI.</head> <p> <s xml:id="echoid-s555" xml:space="preserve">Il ſuit encore du Corollaire 7. </s> <s xml:id="echoid-s556" xml:space="preserve">que ſi au lieu du <lb/> <anchor type="note" xlink:label="note-0044-02a" xlink:href="note-0044-02"/> poid K, on mettoit quelque nouvelle puiſſance, (quel-<lb/>le ſoit auſſi appellée K) qui fûtaux deux premiéres P <lb/>& </s> <s xml:id="echoid-s557" xml:space="preserve">R, commele ſinus de l’angle P A R, aux ſinus des <lb/>angles RAK, & </s> <s xml:id="echoid-s558" xml:space="preserve">PAK; </s> <s xml:id="echoid-s559" xml:space="preserve">c’eſt-à-dire, que ces trois <lb/>puiſſances K, P, & </s> <s xml:id="echoid-s560" xml:space="preserve">R fuſſent entr’elles, comme les <lb/>ſinus de ces trois angles PAR, RAK, & </s> <s xml:id="echoid-s561" xml:space="preserve">PAK; <lb/></s> <s xml:id="echoid-s562" xml:space="preserve">elles demeureroient en équilibre entr’elles, de ſorte <lb/>qu’aucune ne l’emporteroit ſur aucune autre.</s> <s xml:id="echoid-s563" xml:space="preserve"/> </p> <div xml:id="echoid-div61" type="float" level="2" n="1"> <note position="left" xlink:label="note-0044-02" xlink:href="note-0044-02a" xml:space="preserve">fig. 10. <lb/>11. <lb/>12. <lb/>13. <lb/>14. <lb/>15. <lb/>16. <lb/>17.</note> </div> </div> <div xml:id="echoid-div63" type="section" level="1" n="43"> <head xml:id="echoid-head43" xml:space="preserve"><emph style="sc">Corollaire</emph> XII.</head> <p> <s xml:id="echoid-s564" xml:space="preserve">Ce qui fait encore voir, que ſans rien changer à <lb/>l’inclinaiſon des cordes B P & </s> <s xml:id="echoid-s565" xml:space="preserve">C R de ces puiſſances, <lb/>une infinité d’autres miſes en leurs places, pouront <lb/>demeurer en équilibre trois à trois, pourvu qu’elles <lb/>ſoient entr’elles, comme ces trois premiéres.</s> <s xml:id="echoid-s566" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div64" type="section" level="1" n="44"> <head xml:id="echoid-head44" xml:space="preserve"><emph style="sc">Corollaire</emph> XIII.</head> <p> <s xml:id="echoid-s567" xml:space="preserve">On peut auſſi en changeant l’inclinaiſon de ces <lb/>cordes conſerver l’équilibre de ces trois premiéres <lb/>puiſſances K, P, & </s> <s xml:id="echoid-s568" xml:space="preserve">R, dans ſix variations différentes <lb/>de leurs angles, pourvu que ces puiſſances faſſent <lb/>échange entr’elles de leurs cordes, juſqu’à ce que <lb/>chacune d’elles ſe trouve appliquée ſucceſſivement <lb/>à chacune de ces cordes pendant deux ſituations dif-<lb/>férentes des deux autres. </s> <s xml:id="echoid-s569" xml:space="preserve">Pour l’appercevoir il ne faut <lb/>que s’imaginer que lorſque deux de ces puiſſances, <lb/>par exemple P & </s> <s xml:id="echoid-s570" xml:space="preserve">R, font échange de leurs cordes, il <lb/>ſe fait en même-tems une échange des angles que <lb/>ces mêmes cordes faiſoient auparavant avec celle de <pb o="19" file="0045" n="45" rhead="MECHANIQUE."/> la puiſſance K, qui n’en change point alors, ſans qu’il <lb/> <anchor type="note" xlink:label="note-0045-01a" xlink:href="note-0045-01"/> arrive encore aucun changement à celui qu’elles font <lb/>entr’elles: </s> <s xml:id="echoid-s571" xml:space="preserve">de cette maniére l’on aura deux des cas <lb/>dont il eſt ici queſtion. </s> <s xml:id="echoid-s572" xml:space="preserve">On en trouvera encore deux <lb/>en concevant de même l’échange des angles qui ſe <lb/>fait dans l’échange des cordes de P & </s> <s xml:id="echoid-s573" xml:space="preserve">de K; </s> <s xml:id="echoid-s574" xml:space="preserve">& </s> <s xml:id="echoid-s575" xml:space="preserve">encore <lb/>deux pour l’échange de celles de K & </s> <s xml:id="echoid-s576" xml:space="preserve">de R: </s> <s xml:id="echoid-s577" xml:space="preserve">c’eſt ainſi <lb/>qu’on les aura tous ſix.</s> <s xml:id="echoid-s578" xml:space="preserve"/> </p> <div xml:id="echoid-div64" type="float" level="2" n="1"> <note position="right" xlink:label="note-0045-01" xlink:href="note-0045-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> </div> <div xml:id="echoid-div66" type="section" level="1" n="45"> <head xml:id="echoid-head45" xml:space="preserve"><emph style="sc">Corollaire</emph> XIV.</head> <p> <s xml:id="echoid-s579" xml:space="preserve">Mais tant que chacune de ces puiſſances demeure <lb/>appliquée à la même corde, l’on ne peut en changer <lb/>la direction, c’eſt-à-dire, l’inclinaiſon de ces cordes, <lb/>ſans en rompre l’équilibre; </s> <s xml:id="echoid-s580" xml:space="preserve">parce qu’il n’eſt pas poſ-<lb/>ſible de trouver ſeulement deux ſituations de cordes, <lb/>ou les ſinus de ces trois angles R A P, R A K, & </s> <s xml:id="echoid-s581" xml:space="preserve">P A K, <lb/>ayent le même raport entr’eux.</s> <s xml:id="echoid-s582" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div67" type="section" level="1" n="46"> <head xml:id="echoid-head46" xml:space="preserve"><emph style="sc">Corollaire</emph> XV.</head> <p> <s xml:id="echoid-s583" xml:space="preserve">C’eſt ce qui fait que tout autant de fois que les <lb/>angles RAP, que font entr’elles les cordes des puiſ-<lb/>ſances P & </s> <s xml:id="echoid-s584" xml:space="preserve">R, ſont différens, les poids qu’elles ſou-<lb/>tiennent ſont différens auſſi. </s> <s xml:id="echoid-s585" xml:space="preserve">En effet plus cet angle <lb/>eſt obtus, plus le poids qu elles ſoutiennent, doit être <lb/>petit, quoi-qu’en proportion différente: </s> <s xml:id="echoid-s586" xml:space="preserve">puiſque plus <lb/>cet angle eſt obtus, moins eſt grande la raiſon de AD <lb/>à AB, & </s> <s xml:id="echoid-s587" xml:space="preserve">à BD; </s> <s xml:id="echoid-s588" xml:space="preserve">c’eſt-à-dire, (Lemm. </s> <s xml:id="echoid-s589" xml:space="preserve">5.) </s> <s xml:id="echoid-s590" xml:space="preserve">celle du <lb/>ſinus de l’angle DBA, ou de ſon complement PAR, <lb/>à chacun des ſinus de l’angle BDA, ou de RAK qui <lb/>lui eſt égal, ou ſon complement; </s> <s xml:id="echoid-s591" xml:space="preserve">& </s> <s xml:id="echoid-s592" xml:space="preserve">de l’angle BAD, <lb/>ou PAK encore égal à celui-ci, ou ſon complement: <lb/></s> <s xml:id="echoid-s593" xml:space="preserve">qui eſt juſtement celle que ce poids doit alors avoir à <lb/>chacune de ces puiſſances pour demeurer en équilibre <lb/>avec elles; </s> <s xml:id="echoid-s594" xml:space="preserve">De ſorte que cet angle peut être ſi obtus <lb/>que ces mêmes puiſſances pourront faire équilibre <pb o="20" file="0046" n="46" rhead="NOUVELLE"/> avec ce poids, quelque petit qu’il ſoit: </s> <s xml:id="echoid-s595" xml:space="preserve">c’eſt ainſi qu’il <lb/> <anchor type="note" xlink:label="note-0046-01a" xlink:href="note-0046-01"/> peut diminuer à l’infini, & </s> <s xml:id="echoid-s596" xml:space="preserve">faire cependant tou-<lb/>jours équilibre avec elles, que@ques grandes qu’on les <lb/>ſuppoſe.</s> <s xml:id="echoid-s597" xml:space="preserve"/> </p> <div xml:id="echoid-div67" type="float" level="2" n="1"> <note position="left" xlink:label="note-0046-01" xlink:href="note-0046-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> </div> <div xml:id="echoid-div69" type="section" level="1" n="47"> <head xml:id="echoid-head47" xml:space="preserve"><emph style="sc">Corollaire</emph> XVI.</head> <p> <s xml:id="echoid-s598" xml:space="preserve">Voilà ce qui peut arriver en changeant la direction <lb/>de l’une, & </s> <s xml:id="echoid-s599" xml:space="preserve">de l’autre de ces deux puiſſances; </s> <s xml:id="echoid-s600" xml:space="preserve">Mais à <lb/>n’en déplacer qu’une: </s> <s xml:id="echoid-s601" xml:space="preserve">1°. </s> <s xml:id="echoid-s602" xml:space="preserve">Si elles ſont égales, ou ſi <lb/>étant inégales, il s’en trouve une qui ait une direc-<lb/>tion horizontale, il eſt clair qu’on changeroit le ra-<lb/>port des ſinus des angles que leurs cordes font avec <lb/>la ligne de direction du poids qu’elles ſoutiennent: <lb/></s> <s xml:id="echoid-s603" xml:space="preserve">ainſi (Cor. </s> <s xml:id="echoid-s604" xml:space="preserve">I.) </s> <s xml:id="echoid-s605" xml:space="preserve">elles ne pouroient plus ſoutenir n’y ce <lb/>poids, ny aucun autre. </s> <s xml:id="echoid-s606" xml:space="preserve">2°. </s> <s xml:id="echoid-s607" xml:space="preserve">Au contraire ſi étant inéga-<lb/>les, elles n’ont aucune de leurs cordes qui ſoit hori-<lb/>zontale; </s> <s xml:id="echoid-s608" xml:space="preserve">au lieu du poids qu’elles ſoutiennent, on <lb/>poura encore leur en faire ſoutenir un autre, pourvu <lb/>qu’on change tellement celle de leurs directions qui <lb/>fait le plus grand angle avec celle du poids qu’elles <lb/>ſoutiennent, qu’on lui en faſſe faire un autre encore <lb/>avec elle, qui ſoit le complement du premier: </s> <s xml:id="echoid-s609" xml:space="preserve">Car <lb/>les ſinus des angles de leurs cordes avec la direction <lb/>du poids appliqué à leur point de concours, étant <lb/>encore les mêmes qu’auparavant, elles en pouront <lb/>encore ſoutenir un (Cor. </s> <s xml:id="echoid-s610" xml:space="preserve">II.) </s> <s xml:id="echoid-s611" xml:space="preserve">à qui elles ſeront com-<lb/>me ces mêmes ſinus réciproquement pris, à celui de <lb/>l’angle que feront alors leurs cordes entr’elles; </s> <s xml:id="echoid-s612" xml:space="preserve">c’eſt-<lb/>à-dire, qui ſera à celui qu’elles ſoutenoient aupara-<lb/>vant, comme ce dernier ſinus à celui de l’angle qu’el-<lb/>les faiſoient alors entr’elles. </s> <s xml:id="echoid-s613" xml:space="preserve">Mais auſſi par une raiſon <lb/>toute contraire en tout autre changement de cette <lb/>lignede direction ces deux puiſſances ne peuvent plus <lb/>rien ſoutenir.</s> <s xml:id="echoid-s614" xml:space="preserve"/> </p> <pb o="21" file="0047" n="47" rhead="MECHANIQUE."/> <p style="it"> <s xml:id="echoid-s615" xml:space="preserve">La raiſon pour laquelle on vient de demander que ce <lb/> <anchor type="note" xlink:label="note-0047-01a" xlink:href="note-0047-01"/> changement ſe fit dans celle des directions de ces puiſſances, qui <lb/>fait le plus grand angle avec celle du poids qu’elles ſoutien-<lb/>nent; </s> <s xml:id="echoid-s616" xml:space="preserve">c’eſt que ſi l’on faiſoit un tel cbangement à l’autre, l’an-<lb/>gle des cordes ſe tourneroit en deſſous, ce qui détermineroit <lb/>l’action de ces deux puiſſances à ſeconder ſa peſanteur, plutôt <lb/>qu’àla ſoutenir.</s> <s xml:id="echoid-s617" xml:space="preserve"/> </p> <div xml:id="echoid-div69" type="float" level="2" n="1"> <note position="right" xlink:label="note-0047-01" xlink:href="note-0047-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>d s cordes ſeu-<lb/>lement.</note> </div> </div> <div xml:id="echoid-div71" type="section" level="1" n="48"> <head xml:id="echoid-head48" xml:space="preserve"><emph style="sc">Corollaire</emph> XVII.</head> <p> <s xml:id="echoid-s618" xml:space="preserve">Puiſque ce poids pour faire équilibre avec les <lb/>puiſſances P & </s> <s xml:id="echoid-s619" xml:space="preserve">R, doit toujours être à chacune d’elles, <lb/>comme le ſinus de l’angle PAR de leurs cordes, à <lb/>chacun des ſinus des angles RAK & </s> <s xml:id="echoid-s620" xml:space="preserve">PAK: </s> <s xml:id="echoid-s621" xml:space="preserve">Il ſuit <lb/>évidemment que ſi enfin cet angle, à force de deve-<lb/>nir obtus, devenoit à rien; </s> <s xml:id="echoid-s622" xml:space="preserve">c’eſt-à-dire, que R A & </s> <s xml:id="echoid-s623" xml:space="preserve"><lb/>AP ne fiſſent plus qu’une même ligne droite, ce poids <lb/>ſeroit auſſi réduit à rien, & </s> <s xml:id="echoid-s624" xml:space="preserve">ces deux puiſſances agi-<lb/>roient ſeulement alors l’une contre l’autre: </s> <s xml:id="echoid-s625" xml:space="preserve">ainſi tant <lb/>qu’il reſte quelque poids attaché à ces cordes entre ces <lb/>puiſſances, elles font toujours quelque angle PAR <lb/>que ce ſoit.</s> <s xml:id="echoid-s626" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div72" type="section" level="1" n="49"> <head xml:id="echoid-head49" xml:space="preserve"><emph style="sc">Corollaire</emph> XVIII.</head> <p> <s xml:id="echoid-s627" xml:space="preserve">De-là on voit qu’il n’y a point de force imaginable, <lb/>n’y de poids, quelques grands qu’on les conçoive, qui <lb/>appliquez aux extrémitez d’une corde parfaitement <lb/>fléxible, la puiſſent tellement bander qu’elle devienne <lb/>parfaitement droite, pour peu de peſanteur qu’on y <lb/>ſuppoſe: </s> <s xml:id="echoid-s628" xml:space="preserve">car quelque prodigieuſe que ſoit cette force, <lb/>& </s> <s xml:id="echoid-s629" xml:space="preserve">quelques grands que ſoient ces poids, ils auront <lb/>toujours quelque raport à la peſanteur de cette corde: <lb/></s> <s xml:id="echoid-s630" xml:space="preserve">& </s> <s xml:id="echoid-s631" xml:space="preserve">par conſéquent elle ſe courbera toujours.</s> <s xml:id="echoid-s632" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s633" xml:space="preserve">Iuſqu’ici nous n’avons eu aucun égard à cette peſan-<lb/>@eur de cordes, c’eſt pour cela que nous n’y avons conſidéré <pb o="22" file="0048" n="48" rhead="NOUVELLE"/> que l’angle qu’elles font entr’elles à l’endroit ou le poids <lb/> <anchor type="note" xlink:label="note-0048-01a" xlink:href="note-0048-01"/> qu’elles ſoutiennent, leur eſt attacbé: </s> <s xml:id="echoid-s634" xml:space="preserve">ce ſera en les conſidé-<lb/>rant de même que nous prendrons quelque jour leur peſanteur <lb/>pourune infinité de poids, qui leur étant appliquez dans toute <lb/>leur longueur, leur font faire une infinité d’angles quinous ſer-<lb/>viront à en déterminer la courbure pour toutes ſortes d’bypo-<lb/>theſes.</s> <s xml:id="echoid-s635" xml:space="preserve"/> </p> <div xml:id="echoid-div72" type="float" level="2" n="1"> <note position="left" xlink:label="note-0048-01" xlink:href="note-0048-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> </div> <div xml:id="echoid-div74" type="section" level="1" n="50"> <head xml:id="echoid-head50" xml:space="preserve"><emph style="sc">Corollaire</emph> XIX.</head> <p> <s xml:id="echoid-s636" xml:space="preserve">On pourroit encore de même démontrer le raport <lb/>du poids K ſuſpendu partrois cordes, ou par davanta-<lb/>ge, même juſqu’à l’infini, aux puiſſances qui le ſou-<lb/>tiendroient; </s> <s xml:id="echoid-s637" xml:space="preserve">en prenant pour une ſeule, l’impreſſion <lb/>compoſée de chacun des points, ou deux de ces cor-<lb/>des concourent, laquelle réſulte du concours d’action <lb/>des deux puiſſances qui y ſont appliquées; </s> <s xml:id="echoid-s638" xml:space="preserve">& </s> <s xml:id="echoid-s639" xml:space="preserve">ainſi <lb/>toujours de même, juſqu’à ce qu’enfin toutes ces <lb/>puiſſances, en quelque nombre qu’elles ſoient, fuſſent <lb/>réduites à deux ſeulement: </s> <s xml:id="echoid-s640" xml:space="preserve">ce qui réduiroit toujours <lb/>tous ces cas, quclques différens qu’ils fuſſent, à celui <lb/>de la propoſition préſente, qui pour cela a été appel-<lb/>lée Fondamentale.</s> <s xml:id="echoid-s641" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div75" type="section" level="1" n="51"> <head xml:id="echoid-head51" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s642" xml:space="preserve">Lorſque de chacun des nocuds qui lient enſemble les <lb/>cordes avec leſquelles un poids eſt ſoutenu par plu-<lb/>ſieurs puiſſances, ſuivant chacune ſa direction, il <lb/>ne s’en répand point plus de trois dans un même <lb/>plan; </s> <s xml:id="echoid-s643" xml:space="preserve">alors cette réduction ſe peut faire ſans con-<lb/>noître aucun raport entre ces puiſſances, n’y d’au-<lb/>cune d’elles à ce poids. </s> <s xml:id="echoid-s644" xml:space="preserve">Mais lorſque le contraire <lb/>arrive, c’eſt-à-dire, lorſque d’un même næud il ſe <lb/>répand plus de trois cordes dans un même plan; </s> <s xml:id="echoid-s645" xml:space="preserve">alors <lb/>pour chacun de ces plans, il faut connoître entre leſ-<lb/>quelles on voudra des forces qui y ſont appliquées.</s> <s xml:id="echoid-s646" xml:space="preserve"> <pb o="23" file="0049" n="49" rhead="MECHANIQUE."/> autant de raports, moins trois, qu’il y aura de telles <lb/> <anchor type="note" xlink:label="note-0049-01a" xlink:href="note-0049-01"/> forces, ou qu’un même noeud y répandra de cordes. <lb/></s> <s xml:id="echoid-s647" xml:space="preserve">Par exemple, pour 4. </s> <s xml:id="echoid-s648" xml:space="preserve">cordes, il ne faudra connoître <lb/>qu’un tel raport. </s> <s xml:id="echoid-s649" xml:space="preserve">Pour 5. </s> <s xml:id="echoid-s650" xml:space="preserve">il en faudra 2. </s> <s xml:id="echoid-s651" xml:space="preserve">Pour 6. </s> <s xml:id="echoid-s652" xml:space="preserve">3. </s> <s xml:id="echoid-s653" xml:space="preserve"><lb/>Pour 7. </s> <s xml:id="echoid-s654" xml:space="preserve">4. </s> <s xml:id="echoid-s655" xml:space="preserve">& </s> <s xml:id="echoid-s656" xml:space="preserve">ainſi toujours 3. </s> <s xml:id="echoid-s657" xml:space="preserve">moins que le nombre <lb/>de ces cordes.</s> <s xml:id="echoid-s658" xml:space="preserve"/> </p> <div xml:id="echoid-div75" type="float" level="2" n="1"> <note position="right" xlink:label="note-0049-01" xlink:href="note-0049-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſ@u-<lb/>lement.</note> </div> <p style="it"> <s xml:id="echoid-s659" xml:space="preserve">On n’entre point dans le détail de tout ceci, de peur de ſe <lb/>trop éloigner de la breveté qu’on s’eſt propoſée; </s> <s xml:id="echoid-s660" xml:space="preserve">outre qu’en voi-<lb/>la, ce me ſemble, aſſez pour juger de l’étenduë & </s> <s xml:id="echoid-s661" xml:space="preserve">de la fécondité <lb/>de ces principes: </s> <s xml:id="echoid-s662" xml:space="preserve">on le verra démontré dans un autre ouvrage.</s> <s xml:id="echoid-s663" xml:space="preserve"/> </p> <figure> <image file="0049-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0049-01"/> </figure> </div> <div xml:id="echoid-div77" type="section" level="1" n="52"> <head xml:id="echoid-head52" xml:space="preserve">PROBLEME.</head> <p style="it"> <s xml:id="echoid-s664" xml:space="preserve">LEs forces de trois bommes K, P, & </s> <s xml:id="echoid-s665" xml:space="preserve">R, étant données, <lb/>ou même ſeulement le raport qu’elles ont entr’elles, les <lb/>appliquer tellement à trois cordes A K, A P, & </s> <s xml:id="echoid-s666" xml:space="preserve">A R, qui <lb/>aboutiſſent à un même næud A, qu’aucun des trois ne l’emporte <lb/>ſur aucun des deux autres.</s> <s xml:id="echoid-s667" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div78" type="section" level="1" n="53"> <head xml:id="echoid-head53" xml:space="preserve"><emph style="sc">Solution</emph>.</head> <p> <s xml:id="echoid-s668" xml:space="preserve">Premièremènt, s’il y a quelqu’une des forces de <lb/>ces trois perſonnes, qui ſoit égale, ou plus grande que <lb/>la ſomme de celles des deux autres, ce Problême par <lb/>le Corollaire 3. </s> <s xml:id="echoid-s669" xml:space="preserve">de la propoſition préſente eſt impoſ-<lb/>ſible.</s> <s xml:id="echoid-s670" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s671" xml:space="preserve">Secondement ſi chacune de ces forces, de quelque <lb/>maniére qu’on les prenne, eſt en effet moindre que <lb/>la ſomme des deux autres; </s> <s xml:id="echoid-s672" xml:space="preserve">placez à diſcretion une de <lb/>ces trois perſonnes, par exemple K, & </s> <s xml:id="echoid-s673" xml:space="preserve">lui donnez la <lb/>direction AK qu’il vous plaira. </s> <s xml:id="echoid-s674" xml:space="preserve">Ayant prolongé à <lb/>diſcretion AK du côté de E, faite ſur AE le triangle <lb/>AEF, tel que ſes trois côtez AE, EF, & </s> <s xml:id="echoid-s675" xml:space="preserve">AF, ſoient <lb/>entr’eux, comme les forces de ces trois perſonnes K, <pb o="24" file="0050" n="50" rhead="NOUVELLE"/> P, & </s> <s xml:id="echoid-s676" xml:space="preserve">R. </s> <s xml:id="echoid-s677" xml:space="preserve">Placez enſuite la perſonne R ſuivant le côté <lb/> <anchor type="note" xlink:label="note-0050-01a" xlink:href="note-0050-01"/> AF de ce triangle, & </s> <s xml:id="echoid-s678" xml:space="preserve">la perſonne P ſuivant AP paral-<lb/>lele à EF. </s> <s xml:id="echoid-s679" xml:space="preserve">Cela fait, ces trois hommes ſeront dans la <lb/>ſituation requiſe pour demeurer en équilibre.</s> <s xml:id="echoid-s680" xml:space="preserve"/> </p> <div xml:id="echoid-div78" type="float" level="2" n="1"> <note position="left" xlink:label="note-0050-01" xlink:href="note-0050-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> </div> <div xml:id="echoid-div80" type="section" level="1" n="54"> <head xml:id="echoid-head54" xml:space="preserve"><emph style="sc">Demonstration</emph></head> <p> <s xml:id="echoid-s681" xml:space="preserve">Leurs forces K, P, & </s> <s xml:id="echoid-s682" xml:space="preserve">R ſont (Hyp.) </s> <s xml:id="echoid-s683" xml:space="preserve">entr’elles, <lb/>comme AE, EF, & </s> <s xml:id="echoid-s684" xml:space="preserve">AF; </s> <s xml:id="echoid-s685" xml:space="preserve">C’eſt-à-dire, (Lemm. </s> <s xml:id="echoid-s686" xml:space="preserve">5.) <lb/></s> <s xml:id="echoid-s687" xml:space="preserve">comme les ſinus des angles AFE, EAF, & </s> <s xml:id="echoid-s688" xml:space="preserve">AEF, <lb/>qui ſont les mêmes que ceux desangles PAR, RAK, <lb/>& </s> <s xml:id="echoid-s689" xml:space="preserve">PAK: </s> <s xml:id="echoid-s690" xml:space="preserve">Donc (Corol. </s> <s xml:id="echoid-s691" xml:space="preserve">II.) </s> <s xml:id="echoid-s692" xml:space="preserve">ces trois hommes demeu-<lb/>reront ainſien équilibre, ſans qu’aucun d’eux le puiſſe <lb/>emporter ſur lequel que ce ſoit de deux autres. </s> <s xml:id="echoid-s693" xml:space="preserve">Ce <lb/>qu’il faloit faire.</s> <s xml:id="echoid-s694" xml:space="preserve"/> </p> <figure> <image file="0050-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0050-01"/> </figure> <pb o="25" file="0051" n="51" rhead="MECHANIQUE."/> <figure> <image file="0051-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0051-01"/> </figure> </div> <div xml:id="echoid-div81" type="section" level="1" n="55"> <head xml:id="echoid-head55" xml:space="preserve">PROPOSITION <lb/>FONDAMENTALE <lb/>DES POULIES,</head> <p> <s xml:id="echoid-s695" xml:space="preserve">Soit que le centre en demeure fixe, ſoit qu’on <lb/>le ſuppoſe mobile; </s> <s xml:id="echoid-s696" xml:space="preserve">& </s> <s xml:id="echoid-s697" xml:space="preserve">pour toutes les <lb/>directions poſſibles des puiſſances, ou des <lb/>poids qui y ſont appliquez.</s> <s xml:id="echoid-s698" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s699" xml:space="preserve">SOit la puiſſance, ou le poids D appliquè, ou ſuſpendu <lb/> <anchor type="note" xlink:label="note-0051-01a" xlink:href="note-0051-01"/> au centre mobile d’une poulie A, autour de laquelle paſſe <lb/>la corde PMNR, dont les extremitez ſont retenuës par <lb/>les puiſſances P & </s> <s xml:id="echoid-s700" xml:space="preserve">R: </s> <s xml:id="echoid-s701" xml:space="preserve">quelque angle MHN que faſ-<lb/>ſent entre-elles les parties MP, & </s> <s xml:id="echoid-s702" xml:space="preserve">NR de cette corde, <lb/>prolongées juſqu’àce qu’elles concourent en H; </s> <s xml:id="echoid-s703" xml:space="preserve">le poids, ou la <lb/>puiſſance D, ſera toujours à cbacune des puiſſances P & </s> <s xml:id="echoid-s704" xml:space="preserve">R, <lb/>comme le ſinus de cet angle, au ſinus de ſa moitié.</s> <s xml:id="echoid-s705" xml:space="preserve"/> </p> <div xml:id="echoid-div81" type="float" level="2" n="1"> <note position="right" xlink:label="note-0051-01" xlink:href="note-0051-01a" xml:space="preserve">fig. 20 <lb/>21. <lb/>22. <lb/>23.</note> </div> </div> <div xml:id="echoid-div83" type="section" level="1" n="56"> <head xml:id="echoid-head56" xml:space="preserve"><emph style="sc">Demonstration.</emph></head> <p> <s xml:id="echoid-s706" xml:space="preserve">Il eſt clair que tant quele poids, ou la puiſſance D, <lb/>demeure ainſi en équilibre avec les puiſſances P & </s> <s xml:id="echoid-s707" xml:space="preserve">R <lb/>ſur la poulie A, non-ſeulement cette poulie demeure <lb/>ſans mouvement; </s> <s xml:id="echoid-s708" xml:space="preserve">mais encore la corde PMNR de-<lb/>meure fixe deſſus ſans gliſſer, n’y ſe mouvoir non plus <lb/>que ſi elley étoit colée, ou phyſiquement unie depuis</s> </p> <pb o="26" file="0052" n="52" rhead="NOUVELLE"/> <p> <s xml:id="echoid-s709" xml:space="preserve">M juſqu’à N; </s> <s xml:id="echoid-s710" xml:space="preserve">& </s> <s xml:id="echoid-s711" xml:space="preserve">les points M, & </s> <s xml:id="echoid-s712" xml:space="preserve">N de cette corde <lb/> <anchor type="note" xlink:label="note-0052-01a" xlink:href="note-0052-01"/> ſont auſſi fixes ſur cette poulie, tant que dure cet <lb/>équilibre, que ſi en effet PM, & </s> <s xml:id="echoid-s713" xml:space="preserve">RN étoient deux <lb/>cordes, qui y fuſſent ſéparément attachées: </s> <s xml:id="echoid-s714" xml:space="preserve">Les puiſ-<lb/>ſances P, & </s> <s xml:id="echoid-s715" xml:space="preserve">R reſi ſtent donc au poids, ou à la puiſſan-<lb/>ce D, de même que ſi MP, & </s> <s xml:id="echoid-s716" xml:space="preserve">RN étoient deux cor-<lb/>des différentes & </s> <s xml:id="echoid-s717" xml:space="preserve">attachées aux points M & </s> <s xml:id="echoid-s718" xml:space="preserve">N de la <lb/>poulie A, ſuivant les tangentes en cespoints: </s> <s xml:id="echoid-s719" xml:space="preserve">De ſorte <lb/>que l’on peut regarder cette poulie comme un corps <lb/>qui tend vers D, ſuivant AD, d’une force égale à <lb/>celle du poids, ou de la puiſſance D; </s> <s xml:id="echoid-s720" xml:space="preserve">mais qui eſt re-<lb/>tenu avec les cordes PM & </s> <s xml:id="echoid-s721" xml:space="preserve">RN par les puiſſances <lb/>P & </s> <s xml:id="echoid-s722" xml:space="preserve">R. </s> <s xml:id="echoid-s723" xml:space="preserve">Or en ce cas, non-ſeulement ſa ligne de di-<lb/>rection AD paſſeroit (Lemm. </s> <s xml:id="echoid-s724" xml:space="preserve">2. </s> <s xml:id="echoid-s725" xml:space="preserve">Cor. </s> <s xml:id="echoid-s726" xml:space="preserve">1.) </s> <s xml:id="echoid-s727" xml:space="preserve">par le point H, <lb/>ou concourent ces cordes prolongées; </s> <s xml:id="echoid-s728" xml:space="preserve">mais encore <lb/>cette poulie regardée avec une telle impreſſion, ſeroit <lb/>(prop. </s> <s xml:id="echoid-s729" xml:space="preserve">fond. </s> <s xml:id="echoid-s730" xml:space="preserve">des cordes. </s> <s xml:id="echoid-s731" xml:space="preserve">Cor. </s> <s xml:id="echoid-s732" xml:space="preserve">5.) </s> <s xml:id="echoid-s733" xml:space="preserve">aux puiſſances P, & </s> <s xml:id="echoid-s734" xml:space="preserve">R, <lb/>qui la retiennent, comme le finus de l’angle MHN, <lb/>aux ſinus des angles NHA, & </s> <s xml:id="echoid-s735" xml:space="preserve">MHA: </s> <s xml:id="echoid-s736" xml:space="preserve">Donc en effet <lb/>la ligne de direction AD de la poulie A ainſi tirée <lb/>par le poids, ou la puiſſance D, contre les puiſſances <lb/>P & </s> <s xml:id="echoid-s737" xml:space="preserve">R, paſſe toujours par le point H, ou leurs cordes <lb/>prolongées concourent; </s> <s xml:id="echoid-s738" xml:space="preserve">& </s> <s xml:id="echoid-s739" xml:space="preserve">le poids, ou la puiſſance <lb/>D, eſt auſſi à chacune de ces puiſſances, comme le <lb/>ſinus de l’angle MHN à chacun des ſinus des angles <lb/>NHA, & </s> <s xml:id="echoid-s740" xml:space="preserve">MHA. </s> <s xml:id="echoid-s741" xml:space="preserve">Or à cauſe que HA, paſſe auſſi <lb/>parle centre de cette poulie, & </s> <s xml:id="echoid-s742" xml:space="preserve">que MH, & </s> <s xml:id="echoid-s743" xml:space="preserve">NH ſont <lb/>deux tangentes, les angles NHA, & </s> <s xml:id="echoid-s744" xml:space="preserve">MHA, ſont <lb/>chacun la moitié de l’angle MHN: </s> <s xml:id="echoid-s745" xml:space="preserve">Donc le poids, <lb/>ou la puiſſance D, eſt toujours à chacune de puiſſan-<lb/>ces P & </s> <s xml:id="echoid-s746" xml:space="preserve">R, comme le ſinus de l’angle MHN, que <lb/>ſont leurs cordes entr’elles, au ſinus de ſa moitié. </s> <s xml:id="echoid-s747" xml:space="preserve">Ce <lb/>qu’il faloit démontrer.</s> <s xml:id="echoid-s748" xml:space="preserve"/> </p> <div xml:id="echoid-div83" type="float" level="2" n="1"> <note position="left" xlink:label="note-0052-01" xlink:href="note-0052-01a" xml:space="preserve">DES <lb/>POULIES.</note> </div> <figure> <image file="0052-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0052-01"/> </figure> <pb o="27" file="0053" n="53" rhead="MECHANIQUE."/> </div> <div xml:id="echoid-div85" type="section" level="1" n="57"> <head xml:id="echoid-head57" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <note position="right" xml:space="preserve">DES <lb/>POULIES.</note> <p> <s xml:id="echoid-s749" xml:space="preserve">Il ſuit de-là que le poids, ou la puiſſance D, tant <lb/>que dure cet équilibre, eſt toujours à chacune des <lb/>puiſſances P, & </s> <s xml:id="echoid-s750" xml:space="preserve">R, en raiſon réciproque des ſinus des <lb/>angles que font les lignes de direction de ce poids, & </s> <s xml:id="echoid-s751" xml:space="preserve"><lb/>de cette puiſſance avec celle de l’autre; </s> <s xml:id="echoid-s752" xml:space="preserve">c’eſt-à-dire, <lb/>en raiſon réciproque des diſtances des lignes de di-<lb/>rection de ce poids & </s> <s xml:id="echoid-s753" xml:space="preserve">de cette puiſſance, à quelque <lb/>point que ce ſoit de celle de l’autre. </s> <s xml:id="echoid-s754" xml:space="preserve">Et pour la même <lb/>raiſon les puiſſances P & </s> <s xml:id="echoid-s755" xml:space="preserve">R ſont auſſi entr’elles en <lb/>raiſon rèciproque des diſtances de leurs lignes de di-<lb/>rection à quelque point que ce ſoit de celle du poids, <lb/>ou de la puiſſance D, qu’elles ſoutiennent.</s> <s xml:id="echoid-s756" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div86" type="section" level="1" n="58"> <head xml:id="echoid-head58" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s757" xml:space="preserve">Si la puiſſance, ou le poids D, (le tout étant <lb/>appliqué à la poulie A comme cy-deſſus) eſt à chacune <lb/>des puiſſances P, & </s> <s xml:id="echoid-s758" xml:space="preserve">R, comme le ſinus de l’angle <lb/>MHN, que font leurs cordes entr’elles, au ſinus de <lb/>ſa moitié; </s> <s xml:id="echoid-s759" xml:space="preserve">elles le ſoutiendront en cet état: </s> <s xml:id="echoid-s760" xml:space="preserve">parce <lb/>que cette raiſon étant la même qu’il doit avoir aux <lb/>puiſſances qui l’y pourroient ſoutenir; </s> <s xml:id="echoid-s761" xml:space="preserve">les puiſſances <lb/>p, & </s> <s xml:id="echoid-s762" xml:space="preserve">R leur ſont néceſſairement égales; </s> <s xml:id="echoid-s763" xml:space="preserve">& </s> <s xml:id="echoid-s764" xml:space="preserve">par conſé-<lb/>quent étant appliquées de même, elles l’y doivent <lb/>auſſi ſoutenir.</s> <s xml:id="echoid-s765" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div87" type="section" level="1" n="59"> <head xml:id="echoid-head59" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s766" xml:space="preserve">Ce qui fait voir que lorſque lepoids D, eſt à cha-<lb/>cune des puiſſances P, & </s> <s xml:id="echoid-s767" xml:space="preserve">R, en raiſon réciproque des <lb/>diſtances des lignes de direction de ce poids, & </s> <s xml:id="echoid-s768" xml:space="preserve">de <lb/>cette puiſſance, à quelque point que ce ſoit de celle <lb/>de l’autre, il demeure en équilibre avec elles. </s> <s xml:id="echoid-s769" xml:space="preserve">Ce <lb/>qu’on dit du poids D ſe doit auſſi de la puiſſance D.</s> <s xml:id="echoid-s770" xml:space="preserve"/> </p> <pb o="28" file="0054" n="54" rhead="NOUVELLE"/> </div> <div xml:id="echoid-div88" type="section" level="1" n="60"> <head xml:id="echoid-head60" xml:space="preserve"><emph style="sc">Corollaire</emph> IV.</head> <note position="left" xml:space="preserve">DES <lb/>POULIES.</note> <p> <s xml:id="echoid-s771" xml:space="preserve">Il ſuit de plus de cette propoſition que les puiſ-<lb/>ſances P, & </s> <s xml:id="echoid-s772" xml:space="preserve">R agiſſent contre la puiſſance, ou le poids <lb/>D; </s> <s xml:id="echoid-s773" xml:space="preserve">& </s> <s xml:id="echoid-s774" xml:space="preserve">ce poids, ou cette puiſſance réciproquement <lb/>auſſi contr’elles, de la même maniére que ſi leurs cor-<lb/>des PM, & </s> <s xml:id="echoid-s775" xml:space="preserve">RN, prolongées juſqu’en H, y etoient <lb/>noiiées avec la corde AD du poids, ou de la puiſſance <lb/>D; </s> <s xml:id="echoid-s776" xml:space="preserve">& </s> <s xml:id="echoid-s777" xml:space="preserve">que la poulie A étant ôtée, elles tiraſſent contre <lb/>ce poids, ou contre cette puiſſance ſuivant les mêmes <lb/>angles MHA, & </s> <s xml:id="echoid-s778" xml:space="preserve">NHA: </s> <s xml:id="echoid-s779" xml:space="preserve">puis qu’alors le poids, ou la <lb/>puiſſance D ſeroit encore (prop. </s> <s xml:id="echoid-s780" xml:space="preserve">fond. </s> <s xml:id="echoid-s781" xml:space="preserve">des cordes. </s> <s xml:id="echoid-s782" xml:space="preserve">Cor. </s> <s xml:id="echoid-s783" xml:space="preserve">2. </s> <s xml:id="echoid-s784" xml:space="preserve">& </s> <s xml:id="echoid-s785" xml:space="preserve"><lb/>5.) </s> <s xml:id="echoid-s786" xml:space="preserve">à chacune d’elles, comme le ſinus de l’angle MHN <lb/>de leurs cordes, au ſinus de la moitié de cet angle. <lb/></s> <s xml:id="echoid-s787" xml:space="preserve">Ainſi que les puiſſances P & </s> <s xml:id="echoid-s788" xml:space="preserve">R ſoutiennent le poids, <lb/>ou la puiſſance D, par le moyen de la poulie A, avec <lb/>la même corde PMNR; </s> <s xml:id="echoid-s789" xml:space="preserve">ou qu’elles ſoutiennent l’un <lb/>ou l’autre ſans poulie, mais avec chacune une corde <lb/>PH, & </s> <s xml:id="echoid-s790" xml:space="preserve">RH, noiiée avec AD en H; </s> <s xml:id="echoid-s791" xml:space="preserve">C’eſt toujours <lb/>la même choſe, pourvu que les angles de ces cordes <lb/>avec la ligne de direction du poids, ou de la puiſſan-<lb/>ce D, ſoient les mêmes dans l’un & </s> <s xml:id="echoid-s792" xml:space="preserve">l’autre cas.</s> <s xml:id="echoid-s793" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div89" type="section" level="1" n="61"> <head xml:id="echoid-head61" xml:space="preserve"><emph style="sc">Corollaire</emph> V.</head> <p> <s xml:id="echoid-s794" xml:space="preserve">On voit encore que ſi au lieu du poids, ou de la <lb/>puiſſance D, on attachoit la corde AD à quelque <lb/>clou, comme en D, ou ailleurs; </s> <s xml:id="echoid-s795" xml:space="preserve">ce clou feroit la même <lb/>réſiſtance contre les puiſſances P, & </s> <s xml:id="echoid-s796" xml:space="preserve">R, que fait pré-<lb/>ſentement le poids, ou la puiſſance D: </s> <s xml:id="echoid-s797" xml:space="preserve">ainſi n’étant <lb/>arrivé aucun changement du côté de ces puiſſances, <lb/>la direction de la corde AD ſeroit non ſeulement en-<lb/>core la même qu’auparavant, c’eſt-à-dire, qu’elle <lb/>diviſeroit encore l’angle MHN en deux parties éga-<lb/>les; </s> <s xml:id="echoid-s798" xml:space="preserve">mais encore la réſiſtance de ce clou ſeroit égale <lb/>à celle du poids, ou de la puiſſance D: </s> <s xml:id="echoid-s799" xml:space="preserve">D’où il ſuit</s> </p> <pb o="29" file="0055" n="55" rhead="MECHANIQUE."/> <p> <s xml:id="echoid-s800" xml:space="preserve">I°. </s> <s xml:id="echoid-s801" xml:space="preserve">Que lors qu’une poulie ſur laquelle deux puiſſan-<lb/> <anchor type="note" xlink:label="note-0055-01a" xlink:href="note-0055-01"/> ces font équilibre, comme ici P & </s> <s xml:id="echoid-s802" xml:space="preserve">R, eſt ſuſpenduë <lb/>ou retenuë par une corde telle qu’eſt ici AD, cette <lb/>corde ſe dirige toujours en ſorte qu’elle diviſe en <lb/>deux également l’angle des tangentes de cette poulie <lb/>aux points ou les cordes de ces puiſſances la touchent. <lb/></s> <s xml:id="echoid-s803" xml:space="preserve">(Il eſt clair que la même choſe arriveroit, ſi au lieu <lb/>de ces puiſſances on appliquoit à leurs cordes des poids <lb/>qui demeuraſſent de même en équilibre ſur cette pou-<lb/>lie) 2°. </s> <s xml:id="echoid-s804" xml:space="preserve">La charge du clou, ou la corde qui retient cette <lb/>poulie, eſt attachée, étant égale à la réſiſtance qu’il fait <lb/>contre chacune de ces puiſſances, ou des poids mis en <lb/>leurs places; </s> <s xml:id="echoid-s805" xml:space="preserve">eſt à chacun d’eux, ou d’elles, comme le <lb/>ſinus de l’angle de ces tangentes, au ſinus de ſa moitié.</s> <s xml:id="echoid-s806" xml:space="preserve"/> </p> <div xml:id="echoid-div89" type="float" level="2" n="1"> <note position="right" xlink:label="note-0055-01" xlink:href="note-0055-01a" xml:space="preserve">DES <lb/>POULIES.</note> </div> </div> <div xml:id="echoid-div91" type="section" level="1" n="62"> <head xml:id="echoid-head62" xml:space="preserve"><emph style="sc">Corollaire</emph> VI.</head> <p> <s xml:id="echoid-s807" xml:space="preserve">Préſentement ſi la poulie A, au lieu d’être arrêtée <lb/>en D, ou en quelqu’autre point de ſa corde AD, <lb/>étoit ſeulement fixe en ſon centre A; </s> <s xml:id="echoid-s808" xml:space="preserve">les puiſſances <lb/>P & </s> <s xml:id="echoid-s809" xml:space="preserve">R agiſſant encore deſſus de la même maniére <lb/>qu’auparavant, & </s> <s xml:id="echoid-s810" xml:space="preserve">cette poulie leur faiſant auſſi en-<lb/>core la même réſiſtance qu’elle leur faiſoit, lors qu’elle <lb/>étoit retenuë parla corde AH; </s> <s xml:id="echoid-s811" xml:space="preserve">elle doit en recevoir <lb/>encore la même impreſſion, & </s> <s xml:id="echoid-s812" xml:space="preserve">ſuivant la même di-<lb/>rection qu’auparavant: </s> <s xml:id="echoid-s813" xml:space="preserve">Ainſi l’effort commun des <lb/>puiſſances P & </s> <s xml:id="echoid-s814" xml:space="preserve">R ſur cette poulie, ne tend encore <lb/>qu’à la mouvoir ſuivant HA avec la même force dont <lb/>elles tiroient auparavant contre le poids, ou la puiſ-<lb/>ſance D, ou contre le clou qu’on vient de ſuſpoſer en <lb/>D: </s> <s xml:id="echoid-s815" xml:space="preserve">de ſorte que la charge de cette poulie, lors qu’elle eſt <lb/>fixe, eſt toujours à chacune des puiſſances P & </s> <s xml:id="echoid-s816" xml:space="preserve">R, ou <lb/>à chacun des poids qui leur étant égaux, ſeroient mis <lb/>en leurs places, comme le ſinus de l’angle de leurs par-<lb/>ties de cordes PM & </s> <s xml:id="echoid-s817" xml:space="preserve">RN prolongées juſqu’à ce <lb/>qu’elles concourent, au ſinus de ſa moitié.</s> <s xml:id="echoid-s818" xml:space="preserve"/> </p> <pb o="30" file="0056" n="56" rhead="NOUVELLE"/> </div> <div xml:id="echoid-div92" type="section" level="1" n="63"> <head xml:id="echoid-head63" xml:space="preserve"><emph style="sc">Corollaire</emph> VII.</head> <note position="left" xml:space="preserve">DES <lb/>POULIES.</note> <p> <s xml:id="echoid-s819" xml:space="preserve">D’où il ſuit que plus l’angle MHN, que ces par-<lb/>ties de corde prolongées du côté de H, font entr’elles, <lb/>ſera obtus, moins ſera grande la charge de la poulie A, <lb/>ſoit que le centre en ſoit fixe, ou qu’il ſoit mobile: <lb/></s> <s xml:id="echoid-s820" xml:space="preserve">parce que plus cet angle ſera obtus, moins ſera grande <lb/>la raiſon de ſon ſinus au ſinus de ſa moitié, quoi qu’en <lb/>proportion difféentes: </s> <s xml:id="echoid-s821" xml:space="preserve">De ſorte que cet angle peut <lb/>être ſi obtus, que la poulie A ne ſera chargée que ſi <lb/>peu qu’on voudra des puiſſances P & </s> <s xml:id="echoid-s822" xml:space="preserve">R: </s> <s xml:id="echoid-s823" xml:space="preserve">juſques-là <lb/>même qu’elle pourroit être ſoutenuë contre elles <lb/>par une puiſſance, ou un poids D indéfiniment petit, <lb/>c’eſt-à-dire, moindre que quelque poids donné, que <lb/>ce ſoit: </s> <s xml:id="echoid-s824" xml:space="preserve">Car il ne faut pour cela qu’ouvrir l’angle <lb/>MHN, que font entr’elles les parties de corde PM <lb/>& </s> <s xml:id="echoid-s825" xml:space="preserve">RN, juſqu’à ce qu’enfin ſon ſinus ſoit au ſinus de <lb/>ſa moitié en moindre raiſon que le poids donné, à <lb/>chacune des puiſſances P & </s> <s xml:id="echoid-s826" xml:space="preserve">R.</s> <s xml:id="echoid-s827" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div93" type="section" level="1" n="64"> <head xml:id="echoid-head64" xml:space="preserve"><emph style="sc">Corollaire</emph> VIII.</head> <p> <s xml:id="echoid-s828" xml:space="preserve">Au contraire on peut rendre cet angle MHN <lb/>ſiaigu que la puiſſance, ou le poids D, ou quel-<lb/>qu’autre en ſa place, devra être plus grand que <lb/>chacune des puiſſances P & </s> <s xml:id="echoid-s829" xml:space="preserve">R pour être en équilibre <lb/>avec elles; </s> <s xml:id="echoid-s830" xml:space="preserve">mais ce poids ne peut pas ainſi augmèn-<lb/>ter à l’infini, de même que nous venons de dire qu’il <lb/>peut diminuer; </s> <s xml:id="echoid-s831" xml:space="preserve">Car ne pouvant jamais être plus <lb/>grand que lors que cet angle eſt infiniment aigu, <lb/>c’eſt-à-dire, lors que les parties de corde MP & </s> <s xml:id="echoid-s832" xml:space="preserve">NR <lb/>ſont paralleles; </s> <s xml:id="echoid-s833" xml:space="preserve">& </s> <s xml:id="echoid-s834" xml:space="preserve">le ſinus de cet angle n’étant encore <lb/>alors que double du ſinus de ſa moitié, ce poids ne <lb/>peut être tout au plus que double de chacune des puiſ-<lb/>ſances P & </s> <s xml:id="echoid-s835" xml:space="preserve">R.</s> <s xml:id="echoid-s836" xml:space="preserve"/> </p> <pb o="31" file="0057" n="57" rhead="MECHANIQUE."/> </div> <div xml:id="echoid-div94" type="section" level="1" n="65"> <head xml:id="echoid-head65" xml:space="preserve"><emph style="sc">Corollaire</emph> IX.</head> <note position="right" xml:space="preserve">DES <lb/>POULIES.</note> <p> <s xml:id="echoid-s837" xml:space="preserve">D’ou l’on voit en général que lors qu’un poids fait <lb/>ainfi équilibre avec deux puiſſances, ou avec deux au-<lb/>tres poids par le moyen de quelque poulie à moufle, <lb/>il ne peut être tout au plus que double de chacune de <lb/>ces puiſſances, ou de ces autres poids; </s> <s xml:id="echoid-s838" xml:space="preserve">mais au con-<lb/>traire il peut diminuer à l’infini, & </s> <s xml:id="echoid-s839" xml:space="preserve">faire cependant <lb/>toujours équilibre avec elles, ou avec ces poids, quoi <lb/>qu’ils demeurent toujours les mêmes.</s> <s xml:id="echoid-s840" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div95" type="section" level="1" n="66"> <head xml:id="echoid-head66" xml:space="preserve"><emph style="sc">Corollaire</emph> X.</head> <p> <s xml:id="echoid-s841" xml:space="preserve">Ce qui fait voir que ſur une infinité de cas diffé-<lb/>rens ou cet équilibre peut arriver, il n’y en a qu’un <lb/>feul ou le poids, ou bien la puiſſance D, puiſſe être <lb/>double de chacune des puiſſances P & </s> <s xml:id="echoid-s842" xml:space="preserve">R; </s> <s xml:id="echoid-s843" xml:space="preserve">& </s> <s xml:id="echoid-s844" xml:space="preserve">que dans <lb/>tousles autres il eſt toujours moindre que double, & </s> <s xml:id="echoid-s845" xml:space="preserve"><lb/>même moindre à l’infini que chacune d’elles.</s> <s xml:id="echoid-s846" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s847" xml:space="preserve">Tous ceux qwi ſe mêlent de Mècbanique, ſcavent aſſez <lb/>que l’on regarde ordinairement comme génèrale & </s> <s xml:id="echoid-s848" xml:space="preserve">comme <lb/>abſolument vraye cette propoſition: </s> <s xml:id="echoid-s849" xml:space="preserve">qu’un poids attaché, <lb/>ou ſuſpendu au centre mobile d’une poulie à moufle, <lb/>& </s> <s xml:id="echoid-s850" xml:space="preserve">en équilibre avec une puiſſance appliquée à l’ex-<lb/>trémitè d’une corde qui embraſſant cette poulie, à <lb/>ſon autre extrémité retenuë par quelque clou, ou <lb/>autrement, eſt double de cette puiſſance. </s> <s xml:id="echoid-s851" xml:space="preserve">Cependant <lb/>on voit par ce dernier Corollaire que ſur une infinité de cas <lb/>diffèrens, ou cet équilibre peut arriver, cette propoſition n’eſt <lb/>vraye que dans un ſeul, qui eſt lors que les parties de corde <lb/>gui touchent cette poulie ſont paralleles; </s> <s xml:id="echoid-s852" xml:space="preserve">& </s> <s xml:id="echoid-s853" xml:space="preserve">fauſſe dans tous <lb/>les autres. </s> <s xml:id="echoid-s854" xml:space="preserve">Il eſt vrai que dans la dèmonſtration que les Au-<lb/>teurs qui l’ont avancèe, en donnent, ils ſuppoſent tous que ces <lb/>parties de corde touchent cette poulie aux extrèmitez d’un <lb/>même diametre, & </s> <s xml:id="echoid-s855" xml:space="preserve">conſèquemment qu’elles ſont paralleles;</s> <s xml:id="echoid-s856" xml:space="preserve"> <pb o="32" file="0058" n="58" rhead="NOUVELLE"/> mais outre qu’il eſt rare qu’elles le ſoient, c’eſt que n’ayant <lb/> <anchor type="note" xlink:label="note-0058-01a" xlink:href="note-0058-01"/> point fait cette reſtriction dans leur propoſition, ils la regar-<lb/>dent dans la ſuite comme générale, & </s> <s xml:id="echoid-s857" xml:space="preserve">l’appliquent indiffé-<lb/>remment à toutes les machines ou l’on ſe ſert de moufles, ſans <lb/>avoir égard à la ſituation de leurs cordes; </s> <s xml:id="echoid-s858" xml:space="preserve">Ce qui les a jettez <lb/>dans des mépriſes conſidérables, comme on le verra dans les <lb/>Corollaires 15. </s> <s xml:id="echoid-s859" xml:space="preserve">& </s> <s xml:id="echoid-s860" xml:space="preserve">17.</s> <s xml:id="echoid-s861" xml:space="preserve"/> </p> <div xml:id="echoid-div95" type="float" level="2" n="1"> <note position="left" xlink:label="note-0058-01" xlink:href="note-0058-01a" xml:space="preserve">DES <lb/>POULIES.</note> </div> </div> <div xml:id="echoid-div97" type="section" level="1" n="67"> <head xml:id="echoid-head67" xml:space="preserve"><emph style="sc">Corollaire</emph> XI.</head> <p> <s xml:id="echoid-s862" xml:space="preserve">Soit que le centre de la poulie A demeure fixe, <lb/>ſoit qu’on le ſuppoſe mobile, tant que ces puiſſances <lb/>demeurent ainſi en équilibre aux extrémitez P & </s> <s xml:id="echoid-s863" xml:space="preserve">R <lb/>d’une même corde paſſée par deſſus, ou par deſſous <lb/>cette poulie, elles ſont toujours égales: </s> <s xml:id="echoid-s864" xml:space="preserve">puiſque quel-<lb/>que ſoit la charge de cette poulie, ſi elle eſt fixe, ou <lb/>bien le poids qui y eſt attaché, ſi elle eſt mobile; </s> <s xml:id="echoid-s865" xml:space="preserve">cette <lb/>charge, ou ce poids, tant que dure cet équilibre, eſt <lb/>toujours à chacune d’elles, comme le ſinus de l’angle <lb/>de leurs parties de corde, au ſinus de ſa moitié.</s> <s xml:id="echoid-s866" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div98" type="section" level="1" n="68"> <head xml:id="echoid-head68" xml:space="preserve"><emph style="sc">Corollaire</emph> XII.</head> <p> <s xml:id="echoid-s867" xml:space="preserve">Si ces parties de corde des puiſſances P, & </s> <s xml:id="echoid-s868" xml:space="preserve">R, <lb/>lorſqu’elles ſoutiennent la puiſſance, ou le poids D, <lb/>ne ſont point paralleles, ces deux mêmes puiſſances <lb/>le pourront ſoutenir dans deux ſituations différentes <lb/>de leurs cordes; </s> <s xml:id="echoid-s869" xml:space="preserve">parce que ces cordes peuvent faire <lb/>des angles égaux en H de part & </s> <s xml:id="echoid-s870" xml:space="preserve">d’autre de la poulie <lb/>A, avec ſa ligne de direction AH, ſoit en s’écartant <lb/>l’une de l’autre, (fig. </s> <s xml:id="echoid-s871" xml:space="preserve">20. </s> <s xml:id="echoid-s872" xml:space="preserve">& </s> <s xml:id="echoid-s873" xml:space="preserve">22.) </s> <s xml:id="echoid-s874" xml:space="preserve">ſoit en s’approchant; <lb/></s> <s xml:id="echoid-s875" xml:space="preserve">(fig. </s> <s xml:id="echoid-s876" xml:space="preserve">21. </s> <s xml:id="echoid-s877" xml:space="preserve">& </s> <s xml:id="echoid-s878" xml:space="preserve">23.) </s> <s xml:id="echoid-s879" xml:space="preserve">& </s> <s xml:id="echoid-s880" xml:space="preserve">par conſéquent les mêmes puiſſan-<lb/>ces P, & </s> <s xml:id="echoid-s881" xml:space="preserve">R, qui dans l’une de ces ſituations de cor-<lb/>des, ſont capables de ſoutenir la puiſſance, ou le <lb/>poids D, le pourront encore ſoutenir dans l’autre.</s> <s xml:id="echoid-s882" xml:space="preserve"/> </p> <pb o="33" file="0059" n="59" rhead="MECHANIQUE."/> </div> <div xml:id="echoid-div99" type="section" level="1" n="69"> <head xml:id="echoid-head69" xml:space="preserve"><emph style="sc">Corollaire</emph> XIII.</head> <note position="right" xml:space="preserve">DES <lb/>POULIES.</note> <p> <s xml:id="echoid-s883" xml:space="preserve">Mais ſi les parties de corde des puiſſances P & </s> <s xml:id="echoid-s884" xml:space="preserve">R <lb/>ſont paralleles, elles ne pourront ſoutenir la puiſſan-<lb/>ce, ou le poids D qu’en cette ſeule ſituation; </s> <s xml:id="echoid-s885" xml:space="preserve">parce <lb/>qu’il n’eſt pas poſſible d’en trouver d’autre, ou cette <lb/>puiſſance, ou bien ce poids, ſoit double de chacune <lb/>des puiſſances P & </s> <s xml:id="echoid-s886" xml:space="preserve">R qui le ſoutiennent.</s> <s xml:id="echoid-s887" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div100" type="section" level="1" n="70"> <head xml:id="echoid-head70" xml:space="preserve"><emph style="sc">Corollaire</emph> XIV.</head> <p> <s xml:id="echoid-s888" xml:space="preserve">On voit encore de cette propoſition que le poids <lb/> <anchor type="note" xlink:label="note-0059-02a" xlink:href="note-0059-02"/> D étant en équilibre avec la puiſſance R, par le <lb/>moyen de pluſieurs moufles A, B, C, &</s> <s xml:id="echoid-s889" xml:space="preserve">c. </s> <s xml:id="echoid-s890" xml:space="preserve">Séparez, <lb/>& </s> <s xml:id="echoid-s891" xml:space="preserve">appliquez le long du poteau EG, ou de quelqu’au-<lb/>tre corps que ce ſoit, de la maniére qu’on les voit <lb/>ici; </s> <s xml:id="echoid-s892" xml:space="preserve">eſt à cette puiſſance, comme le produit des ſinus <lb/>des angles H, K, L, &</s> <s xml:id="echoid-s893" xml:space="preserve">c. </s> <s xml:id="echoid-s894" xml:space="preserve">que font, lors qu’on les pro-<lb/>longe, les parties des cordes EK, FL, GR, &</s> <s xml:id="echoid-s895" xml:space="preserve">c. <lb/></s> <s xml:id="echoid-s896" xml:space="preserve">qui touchent ces poulies; </s> <s xml:id="echoid-s897" xml:space="preserve">au produit des ſinus de <lb/>chacun leur moitié: </s> <s xml:id="echoid-s898" xml:space="preserve">Car ſelon la propoſition préſen-<lb/>te, la réſiſtance de la poulie A, ou bien le poids D eſt <lb/>à la réſiſtance de la poulie B, comme le ſinus de l’angle <lb/>H, au ſinus de ſa moitié. </s> <s xml:id="echoid-s899" xml:space="preserve">La réſiſtance de la poulie <lb/>B eſt auſſi à celle de la poulie C, comme le ſinus de <lb/>l’angle K, au ſinus de ſa moitié. </s> <s xml:id="echoid-s900" xml:space="preserve">Enfin la réſiſtance <lb/>de la poulie C eſt encore à celle de la puiſſance R, <lb/>comme le ſinus de l’angle L, au ſinus de ſa moitié; </s> <s xml:id="echoid-s901" xml:space="preserve">& </s> <s xml:id="echoid-s902" xml:space="preserve"><lb/>ainſi de même à l’infini: </s> <s xml:id="echoid-s903" xml:space="preserve">Donc en multipliant par <lb/>ordre les termes de toutes ces proportions, c’eſt-à-<lb/>dire, les antécédens par les antécédens, & </s> <s xml:id="echoid-s904" xml:space="preserve">les conſé-<lb/>quens par les conſéquens, on aura le poids D à la <lb/>puiſſance R, comme le produit des ſinus des angles <lb/>H, K, L, &</s> <s xml:id="echoid-s905" xml:space="preserve">c. </s> <s xml:id="echoid-s906" xml:space="preserve">au produit des ſinus de chacun leur <lb/>moitié.</s> <s xml:id="echoid-s907" xml:space="preserve"/> </p> <div xml:id="echoid-div100" type="float" level="2" n="1"> <note position="right" xlink:label="note-0059-02" xlink:href="note-0059-02a" xml:space="preserve">fig. 24. <lb/>205.</note> </div> <pb o="34" file="0060" n="60" rhead="NOUVELLE"/> </div> <div xml:id="echoid-div102" type="section" level="1" n="71"> <head xml:id="echoid-head71" xml:space="preserve"><emph style="sc">Corollaire</emph> XV.</head> <note position="left" xml:space="preserve">DES <lb/>POULIES.</note> <p> <s xml:id="echoid-s908" xml:space="preserve">Il ſuit de ce dernier Corollaire que ſur une in-<lb/>finité de cas différens, ou le poids D peut être en <lb/>équilibre avec la puiſſance R, à l’aide de pluſieurs <lb/>moufles, de la maniére qu’ils ſont ici; </s> <s xml:id="echoid-s909" xml:space="preserve">il n’y en a qu’un <lb/>ſeul, ou ce poids ſoit à cette puiſſance comme le der-<lb/>nier terme d’une progreſſion double, qui en a autant <lb/>qu’il y a de moufles, plus un, eſt au premier; </s> <s xml:id="echoid-s910" xml:space="preserve">c’eſt-à-<lb/>dire, ici, comme 8. </s> <s xml:id="echoid-s911" xml:space="preserve">à 1. </s> <s xml:id="echoid-s912" xml:space="preserve">C’eſt lors que les parties de <lb/>corde qui ſont tangentes de ces moufles, ſont paralle-<lb/>les entr’elles: </s> <s xml:id="echoid-s913" xml:space="preserve">parce que les angles H, K, L, &</s> <s xml:id="echoid-s914" xml:space="preserve">c. </s> <s xml:id="echoid-s915" xml:space="preserve">étant <lb/>alors infiniment aigus, leurs ſinus ſont doubles des <lb/>ſinus de leurs moitiez. </s> <s xml:id="echoid-s916" xml:space="preserve">Pour dans tous les autres, il <lb/>eſt toujours à cette puiſſance en moindre raiſon que <lb/>ce dernier terme au premier, & </s> <s xml:id="echoid-s917" xml:space="preserve">même, en moindre à <lb/>l’infini: </s> <s xml:id="echoid-s918" xml:space="preserve">parce que les angles H, K, L, &</s> <s xml:id="echoid-s919" xml:space="preserve">c. </s> <s xml:id="echoid-s920" xml:space="preserve">ne pou-<lb/>vant devenir plus aigus que lorſque ces parties de <lb/>cordes ſont paralleles, les raiſons de leurs ſinus aux <lb/>ſinus de leurs moitiez, ne peuvent auſſi jamais être <lb/>chacune plus grande que celle de 2. </s> <s xml:id="echoid-s921" xml:space="preserve">à I. </s> <s xml:id="echoid-s922" xml:space="preserve">au contraire <lb/>ces angles pouvant devenir toujours plus obtus juſqu’à <lb/>l’infini, les raiſons de leurs ſinus aux ſinus de leurs <lb/>moitiez, peuvent auſſi diminuer à l’infini.</s> <s xml:id="echoid-s923" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s924" xml:space="preserve">On voit aſſez la mépriſe de ceux qui dans cet uſage des <lb/>poulies à moufle, ont avancé comme propoſition générale que le <lb/>poids D eſt à la puiſſance R, comme le dernier terme <lb/>d’une progreſſion double, qui en à autant qu’il y a <lb/>de moufles, plus un, eſt au premier. </s> <s xml:id="echoid-s925" xml:space="preserve">Ce qui les a trom-<lb/>pez c’eſt l’uſage trop étendu qu’ils ont donné à la propoſition <lb/>raportée dans la réfléxion qui ſuit le Corollaire 10. </s> <s xml:id="echoid-s926" xml:space="preserve">de la <lb/>propoſition préſente.</s> <s xml:id="echoid-s927" xml:space="preserve"/> </p> <figure> <image file="0060-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0060-01"/> </figure> <pb o="35" file="0061" n="61" rhead="MECHANIQUE."/> </div> <div xml:id="echoid-div103" type="section" level="1" n="72"> <head xml:id="echoid-head72" xml:space="preserve"><emph style="sc">Corollaire</emph> XVI.</head> <note position="right" xml:space="preserve">DES <lb/>POULIES.</note> <p> <s xml:id="echoid-s928" xml:space="preserve">Préſentement ſi l’on ſe ſert de pluſieurs poulies <lb/>liées enſemble, comme on les voit dans les figures 26. <lb/></s> <s xml:id="echoid-s929" xml:space="preserve"> <anchor type="note" xlink:label="note-0061-02a" xlink:href="note-0061-02"/> & </s> <s xml:id="echoid-s930" xml:space="preserve">27. </s> <s xml:id="echoid-s931" xml:space="preserve">Il ſuit encore de cette propoſition que la puiſ-<lb/>ſance R eſt au poids D qu’elle ſoutient à l’aide de ces <lb/>poulies, comme le produit des ſinus des moitiez de <lb/>chacun des angles que font, ſi on les prolonge, les <lb/>cordes tangentes des poulies mobiles L, K, & </s> <s xml:id="echoid-s932" xml:space="preserve">H, à la <lb/>ſomme des produits de chacun des ſinus de ces mêmes <lb/>angles par les ſinus des moitiez de chacun des autres. <lb/></s> <s xml:id="echoid-s933" xml:space="preserve">Par exemple, ſoit le ſinus de l’angle de ces cordes fait <lb/>en A, appellé a; </s> <s xml:id="echoid-s934" xml:space="preserve">& </s> <s xml:id="echoid-s935" xml:space="preserve">celui de ſa moitié appellé b. </s> <s xml:id="echoid-s936" xml:space="preserve">Celui de <lb/>l’angle C, appellé c; </s> <s xml:id="echoid-s937" xml:space="preserve">& </s> <s xml:id="echoid-s938" xml:space="preserve">celui de ſa moitié appellé d. </s> <s xml:id="echoid-s939" xml:space="preserve"><lb/>Enfin celui de l’angle E, appellé e; </s> <s xml:id="echoid-s940" xml:space="preserve">& </s> <s xml:id="echoid-s941" xml:space="preserve">celui de ſa moi-<lb/>tié appellé f. </s> <s xml:id="echoid-s942" xml:space="preserve">Cela ſuppoſé il ſuit, dî-je, de cette pro-<lb/>poſition que la puiſſance R en ce cas eſt au poids D, <lb/>comme bàf à adf + cbf + ebd. </s> <s xml:id="echoid-s943" xml:space="preserve">Parceque la corde <lb/>RRORNRM étant également bandée dans tou-<lb/>tes ſes parties, & </s> <s xml:id="echoid-s944" xml:space="preserve">d’une force égale à celle de la puiſ-<lb/>ſance R; </s> <s xml:id="echoid-s945" xml:space="preserve">on la peut regarder, tant que dure cet équi-<lb/>libre, comme diviſée en autant de cordes RRO, <lb/>RN, & </s> <s xml:id="echoid-s946" xml:space="preserve">RM, qu’il y a de poulies L, K, & </s> <s xml:id="echoid-s947" xml:space="preserve">H, dans <lb/>l’écharpe LH; </s> <s xml:id="echoid-s948" xml:space="preserve">& </s> <s xml:id="echoid-s949" xml:space="preserve">chacune de ces cordes comme fixe <lb/>en O, N, & </s> <s xml:id="echoid-s950" xml:space="preserve">M, & </s> <s xml:id="echoid-s951" xml:space="preserve">tirée du côté de R, R, & </s> <s xml:id="echoid-s952" xml:space="preserve">R par la <lb/>puiſſance R, ou par d’autres qui lui ſoient égales. </s> <s xml:id="echoid-s953" xml:space="preserve"><lb/>Et parce que les poulies L, K, & </s> <s xml:id="echoid-s954" xml:space="preserve">H, portent chacune <lb/>quelque choſe du poids D, regardons-le auſſi comme <lb/>diviſé en autant de parties x, y, & </s> <s xml:id="echoid-s955" xml:space="preserve">z, dont la partie x <lb/>ſoit portée par la poulie L; </s> <s xml:id="echoid-s956" xml:space="preserve">la partie y, par la poulie <lb/>K; </s> <s xml:id="echoid-s957" xml:space="preserve">& </s> <s xml:id="echoid-s958" xml:space="preserve">la partie z, par la poulie H.</s> <s xml:id="echoid-s959" xml:space="preserve"/> </p> <div xml:id="echoid-div103" type="float" level="2" n="1"> <note position="right" xlink:label="note-0061-02" xlink:href="note-0061-02a" xml:space="preserve">fig. 26.</note> </div> <pb o="36" file="0062" n="62" rhead="NOUVELLE"/> <p> <s xml:id="echoid-s960" xml:space="preserve">Cela conçcû il eſt clair par la propoſition qu’on vient <lb/> <anchor type="note" xlink:label="note-0062-01a" xlink:href="note-0062-01"/> de démontrer</s> </p> <div xml:id="echoid-div104" type="float" level="2" n="2"> <note position="left" xlink:label="note-0062-01" xlink:href="note-0062-01a" xml:space="preserve">DES <lb/>POULIES.</note> </div> <note position="right" xml:space="preserve"> <lb/>que{ # {x. R :: a. b. \\ R. y :: d. c.} # Donc {x. y :: ad. bc. \\ & \\ x. x + y :: ad. ad + bc. # }Donc <lb/># {z. R :: e. f. \\ R. x :: b. a.} # Donc z. x :: eb. fa. <lb/></note> <p> <s xml:id="echoid-s961" xml:space="preserve">z. </s> <s xml:id="echoid-s962" xml:space="preserve">x + y :</s> <s xml:id="echoid-s963" xml:space="preserve">: adeb. </s> <s xml:id="echoid-s964" xml:space="preserve">aadf + abcf. </s> <s xml:id="echoid-s965" xml:space="preserve">Donc z. </s> <s xml:id="echoid-s966" xml:space="preserve">z + x + y :</s> <s xml:id="echoid-s967" xml:space="preserve">: <lb/>adeb. </s> <s xml:id="echoid-s968" xml:space="preserve">adeb + aadf + abcf. </s> <s xml:id="echoid-s969" xml:space="preserve">Or R. </s> <s xml:id="echoid-s970" xml:space="preserve">z :</s> <s xml:id="echoid-s971" xml:space="preserve">: f. </s> <s xml:id="echoid-s972" xml:space="preserve">e. </s> <s xml:id="echoid-s973" xml:space="preserve">Donc R. <lb/></s> <s xml:id="echoid-s974" xml:space="preserve">z + x + y = D :</s> <s xml:id="echoid-s975" xml:space="preserve">: adefb. </s> <s xml:id="echoid-s976" xml:space="preserve">adeeb + aadef + aebcf. </s> <s xml:id="echoid-s977" xml:space="preserve">Et en <lb/>diviſant les deux terniers termes de cette derniére <lb/>proportion, par ae, l’on aura R. </s> <s xml:id="echoid-s978" xml:space="preserve">D :</s> <s xml:id="echoid-s979" xml:space="preserve">: dfb. </s> <s xml:id="echoid-s980" xml:space="preserve">edb + <lb/>adf + bcf. </s> <s xml:id="echoid-s981" xml:space="preserve">Ce qu’il faloit démontrer.</s> <s xml:id="echoid-s982" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div106" type="section" level="1" n="73"> <head xml:id="echoid-head73" xml:space="preserve"><emph style="sc">Corollaire</emph> XVII.</head> <p> <s xml:id="echoid-s983" xml:space="preserve">D’où il ſuit que dans cet uſage des poulies, lors <lb/>que les parties de corde, qui touchent celles de l’é-<lb/>charpe LH, ſont paralleles, la puiſſance R eſt au <lb/>poids D, comme l’unité au double du nombre des <lb/>poulies ſuſpenduës; </s> <s xml:id="echoid-s984" xml:space="preserve">mais que dans tout autre cas, <lb/>elle lui eſt toujours en plus grande raiſon; </s> <s xml:id="echoid-s985" xml:space="preserve">& </s> <s xml:id="echoid-s986" xml:space="preserve">même, <lb/>cette raiſon augmente, quoi qu’en proportion diffé-<lb/>rente, à meſure que les angles A, C & </s> <s xml:id="echoid-s987" xml:space="preserve">E deviennent <lb/>moins aigus, ou plus obtus.</s> <s xml:id="echoid-s988" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s989" xml:space="preserve">On voit aſſez que tous ces Corollaires avec une infinité <lb/>d’autres qu’on pourroit encore tirer de cette propoſition, dé-<lb/>pendent abſolument de ſon univerſalité, & </s> <s xml:id="echoid-s990" xml:space="preserve">que ſans cela <lb/>il ſeroit impoſſible de réſoudre une infinité de Problêmes <lb/>qu’on peut faire ſur cette matiére. </s> <s xml:id="echoid-s991" xml:space="preserve">Par exemple celui-ci.</s> <s xml:id="echoid-s992" xml:space="preserve"/> </p> <figure> <image file="0062-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0062-01"/> </figure> <pb o="37" file="0063" n="63" rhead="MECHANIQUE."/> <figure> <image file="0063-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0063-01"/> </figure> <note position="right" xml:space="preserve">DES <lb/>POULIES.</note> </div> <div xml:id="echoid-div107" type="section" level="1" n="74"> <head xml:id="echoid-head74" xml:space="preserve">PROBLEME.</head> <p style="it"> <s xml:id="echoid-s993" xml:space="preserve">F Aire ſoutenir quelque poids D que ce ſoit, à quelque <lb/> <anchor type="note" xlink:label="note-0063-02a" xlink:href="note-0063-02"/> puiſſance R que ce puiſſe être, par le moyen d’une ſeule <lb/>poulie à moufle.</s> <s xml:id="echoid-s994" xml:space="preserve"/> </p> <div xml:id="echoid-div107" type="float" level="2" n="1"> <note position="right" xlink:label="note-0063-02" xlink:href="note-0063-02a" xml:space="preserve">fig. 28.</note> </div> </div> <div xml:id="echoid-div109" type="section" level="1" n="75"> <head xml:id="echoid-head75" xml:space="preserve"><emph style="sc">Solution</emph></head> <p> <s xml:id="echoid-s995" xml:space="preserve">Premiérement ſi cette puiſſance eſt plus grande <lb/>que ce poids, faite ſur EH perpendiculaire à l’hori-<lb/>zon, & </s> <s xml:id="echoid-s996" xml:space="preserve">priſe à diſcretion, le triangle Iſocelle EFH, <lb/>dont chacun des côtez EF, & </s> <s xml:id="echoid-s997" xml:space="preserve">FH, ſoit en même rai-<lb/>ſon à ſa baſe EH, que la puiſſance R, dont il eſt <lb/>queſtion, eſt au poids D. </s> <s xml:id="echoid-s998" xml:space="preserve">Du point H tirez HG per-<lb/>pendiculaire à HF, & </s> <s xml:id="echoid-s999" xml:space="preserve">égale au rayon de la poulie <lb/>A, dont on ſe veut ſervir; </s> <s xml:id="echoid-s1000" xml:space="preserve">du point G faite encore <lb/>GM parallele à HE, & </s> <s xml:id="echoid-s1001" xml:space="preserve">qui rencontre HF en M; </s> <s xml:id="echoid-s1002" xml:space="preserve">& </s> <s xml:id="echoid-s1003" xml:space="preserve"><lb/>de ce même point M faite auſſi MA parallele à HG, <lb/>& </s> <s xml:id="echoid-s1004" xml:space="preserve">qui rencontre auſſi EH en A; </s> <s xml:id="echoid-s1005" xml:space="preserve">placez enſuite en <lb/>ce point le centre de la poulie A, & </s> <s xml:id="echoid-s1006" xml:space="preserve">ſuſpendez le poids <lb/>D à cette poulie ſuivant AH. </s> <s xml:id="echoid-s1007" xml:space="preserve">Enfin aprés avoir di-<lb/>rigé la partie MR dè ſa corde ſuivant HF, & </s> <s xml:id="echoid-s1008" xml:space="preserve">ſon <lb/>autre partie PN parallelement à EF; </s> <s xml:id="echoid-s1009" xml:space="preserve">arrêtez cette <lb/>corde par un bout au clou P, & </s> <s xml:id="echoid-s1010" xml:space="preserve">appliquez à l’autre <lb/>la puiſſance R: </s> <s xml:id="echoid-s1011" xml:space="preserve">cela fait elle ſoutiendra ce poids en <lb/>cet état.</s> <s xml:id="echoid-s1012" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div110" type="section" level="1" n="76"> <head xml:id="echoid-head76" xml:space="preserve"><emph style="sc">Demonstration.</emph></head> <p> <s xml:id="echoid-s1013" xml:space="preserve">Le poids D peut être ſoutenu en cet état (Cor. </s> <s xml:id="echoid-s1014" xml:space="preserve">2.) <lb/></s> <s xml:id="echoid-s1015" xml:space="preserve">par deux puiſſances, à chacune deſquelles il ſoit <lb/>comme le ſinus de l’angle RHP, au ſinus de ſa moitié <lb/>EHR, ou EHP; </s> <s xml:id="echoid-s1016" xml:space="preserve">c’eſt-à-dire, (Lemm. </s> <s xml:id="echoid-s1017" xml:space="preserve">5.) </s> <s xml:id="echoid-s1018" xml:space="preserve">comme <lb/>EH à EF, ou bien (hyp.) </s> <s xml:id="echoid-s1019" xml:space="preserve">comme il eſt à la puiſſance <lb/>R: </s> <s xml:id="echoid-s1020" xml:space="preserve">Donc le clou P faiſant la fonction d’une de ces <pb o="38" file="0064" n="64" rhead="NOUVELLE"/> puiſſances, celle-ci ſoutiendra ce poids en cet état. </s> <s xml:id="echoid-s1021" xml:space="preserve">Ce <lb/> <anchor type="note" xlink:label="note-0064-01a" xlink:href="note-0064-01"/> qu’il faloit faire.</s> <s xml:id="echoid-s1022" xml:space="preserve"/> </p> <div xml:id="echoid-div110" type="float" level="2" n="1"> <note position="left" xlink:label="note-0064-01" xlink:href="note-0064-01a" xml:space="preserve">DES <lb/>POULIES.</note> </div> <p> <s xml:id="echoid-s1023" xml:space="preserve">Secondement, ſi c’eſt le poids D qui ſoit plus <lb/>grand que la puiſſance R, quelque grand qu’il ſoit, <lb/>& </s> <s xml:id="echoid-s1024" xml:space="preserve">quelque petite elle, qu’elle puiſſe être, on le lui <lb/> <anchor type="note" xlink:label="note-0064-02a" xlink:href="note-0064-02"/> pourra encore faire ſoutenir: </s> <s xml:id="echoid-s1025" xml:space="preserve">voici comment. </s> <s xml:id="echoid-s1026" xml:space="preserve">Ayant <lb/>décrit du centre F avec le rayon HF perpendicu-<lb/>laire à l’horizon, & </s> <s xml:id="echoid-s1027" xml:space="preserve">pris à diſcretion, l’arc HEB; </s> <s xml:id="echoid-s1028" xml:space="preserve">& </s> <s xml:id="echoid-s1029" xml:space="preserve"><lb/>y ayant inſcrit une ligne HE, qui ſoit à HF, comme <lb/>la puiſſance R au poids D; </s> <s xml:id="echoid-s1030" xml:space="preserve">joignez FE, & </s> <s xml:id="echoid-s1031" xml:space="preserve">faite en-<lb/>ſuite HG perpendiculaire à HF, & </s> <s xml:id="echoid-s1032" xml:space="preserve">égale au rayon de <lb/>la poulie A, dont on ſe veut ſervir; </s> <s xml:id="echoid-s1033" xml:space="preserve">Du point G faite <lb/>encore GM parallele à HE, & </s> <s xml:id="echoid-s1034" xml:space="preserve">qui rencontre HF en <lb/>M, d’où il faut faire encore MA parallele à HG, & </s> <s xml:id="echoid-s1035" xml:space="preserve">qui <lb/>rencontre HE en A, où vous placerez le centre de <lb/>la poulie A, à laquelle vous appliquerez la puiſſance <lb/>R ſuivant AH, & </s> <s xml:id="echoid-s1036" xml:space="preserve">le poids D ſuivant HF, l’autre <lb/>bout de ſa corde étant attaché en P fuivant NP pa-<lb/>rallele à EF. </s> <s xml:id="echoid-s1037" xml:space="preserve">Cela fait la puiſſance R ſoutiendra en-<lb/>core le poids D en cet état.</s> <s xml:id="echoid-s1038" xml:space="preserve"/> </p> <div xml:id="echoid-div111" type="float" level="2" n="2"> <note position="left" xlink:label="note-0064-02" xlink:href="note-0064-02a" xml:space="preserve">fig. 29.</note> </div> </div> <div xml:id="echoid-div113" type="section" level="1" n="77"> <head xml:id="echoid-head77" xml:space="preserve"><emph style="sc">Demonstration.</emph></head> <p> <s xml:id="echoid-s1039" xml:space="preserve">Cette puiſſance peut faire équilibre en cet état avec <lb/>deux autres, pourvu qu’elle ſoit à chacune d’elles, <lb/>(Cor. </s> <s xml:id="echoid-s1040" xml:space="preserve">2.) </s> <s xml:id="echoid-s1041" xml:space="preserve">comme le ſinus de l’angle PHF, au ſinus <lb/>de ſa moitié PHE, ou à EF; </s> <s xml:id="echoid-s1042" xml:space="preserve">c’eſt-à-dire, (Lemm. </s> <s xml:id="echoid-s1043" xml:space="preserve">5.) <lb/></s> <s xml:id="echoid-s1044" xml:space="preserve">Comme HE à HF, ou EHF; </s> <s xml:id="echoid-s1045" xml:space="preserve">ou bien (hyp.) </s> <s xml:id="echoid-s1046" xml:space="preserve">comme <lb/>elle eſt au poids D: </s> <s xml:id="echoid-s1047" xml:space="preserve">Donc le clou P faiſant la fonc-<lb/>tion d’une de ces puiſſances, & </s> <s xml:id="echoid-s1048" xml:space="preserve">le poids D celle de <lb/>l’autre, la puiſſance R doit le ſoutenir en cet état. </s> <s xml:id="echoid-s1049" xml:space="preserve"><lb/>Ce qu’il F. </s> <s xml:id="echoid-s1050" xml:space="preserve">D.</s> <s xml:id="echoid-s1051" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div114" type="section" level="1" n="78"> <head xml:id="echoid-head78" xml:space="preserve"><emph style="sc">Corollaire.</emph></head> <p> <s xml:id="echoid-s1052" xml:space="preserve">On voit que l’on peut toujours faire ſoutenir quel- <pb o="39" file="0065" n="65" rhead="MECHANIQUE."/> que poids que ce ſoit, à quelque puiſſance que ce <lb/> <anchor type="note" xlink:label="note-0065-01a" xlink:href="note-0065-01"/> puiſſe être, par le moyen d’une ſeule poulie à mouf-<lb/>fle: </s> <s xml:id="echoid-s1053" xml:space="preserve">en appliquant le moindre des deux au centre <lb/>mobile de cette poulie, & </s> <s xml:id="echoid-s1054" xml:space="preserve">le plus grand au bout libre <lb/>de la corde qui l’embraſſe. </s> <s xml:id="echoid-s1055" xml:space="preserve">La raiſon en eſt évidente <lb/>par les Corollaires 7. </s> <s xml:id="echoid-s1056" xml:space="preserve">& </s> <s xml:id="echoid-s1057" xml:space="preserve">8.</s> <s xml:id="echoid-s1058" xml:space="preserve"/> </p> <div xml:id="echoid-div114" type="float" level="2" n="1"> <note position="right" xlink:label="note-0065-01" xlink:href="note-0065-01a" xml:space="preserve">DES <lb/>POULIES.</note> </div> </div> <div xml:id="echoid-div116" type="section" level="1" n="79"> <head xml:id="echoid-head79" xml:space="preserve"><emph style="sc">Remarque.</emph></head> <p> <s xml:id="echoid-s1059" xml:space="preserve">Il n’eſt pas cependant toujours néceſſaire de les <lb/>diſpoſer ainſi: </s> <s xml:id="echoid-s1060" xml:space="preserve">car s’il arrive que le plus grand des <lb/>deux ne ſoit point plus que double de l’autre, on ſera <lb/>libre d’appliquer celuy qu’on voudra au centre mo-<lb/>bile de la poulie A: </s> <s xml:id="echoid-s1061" xml:space="preserve">puiſque, qu’il y ſoit, (Cor. </s> <s xml:id="echoid-s1062" xml:space="preserve">8.) <lb/></s> <s xml:id="echoid-s1063" xml:space="preserve">ou non, (Cor. </s> <s xml:id="echoid-s1064" xml:space="preserve">7.) </s> <s xml:id="echoid-s1065" xml:space="preserve">il peut toujours monter juſqu’à <lb/>être double de l’autre, mais jamais davantage: </s> <s xml:id="echoid-s1066" xml:space="preserve">(Cor. </s> <s xml:id="echoid-s1067" xml:space="preserve">8.) </s> <s xml:id="echoid-s1068" xml:space="preserve"><lb/>ainſi lorſque l’un des deux (la puiſſance ou le poids) <lb/>fera plus que double de l’autre, il faudra néceſſai-<lb/>rement les appliquer ſuivant le Corollaire ci-deſſus; </s> <s xml:id="echoid-s1069" xml:space="preserve"><lb/>c’eſt-à-dire, appliquer le moindre au centre du mouf-<lb/>fle dont on ſe ſert, & </s> <s xml:id="echoid-s1070" xml:space="preserve">le plus grand ou bout libre <lb/>de la corde qui embraſſe cette poulie à moufle.</s> <s xml:id="echoid-s1071" xml:space="preserve"/> </p> <figure> <image file="0065-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0065-01"/> </figure> <pb o="40" file="0066" n="66" rhead="NOUVELLE"/> <figure> <image file="0066-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0066-01"/> </figure> </div> <div xml:id="echoid-div117" type="section" level="1" n="80"> <head xml:id="echoid-head80" xml:space="preserve">PROPOSITION <lb/>FONDAMENTALE <lb/>DES POIDS SOUTENUS</head> <p> <s xml:id="echoid-s1072" xml:space="preserve">Sur quelque eſpéce de ſurfaces que ceſoit; <lb/></s> <s xml:id="echoid-s1073" xml:space="preserve">& </s> <s xml:id="echoid-s1074" xml:space="preserve">pour toutes les directions poſſibles <lb/>des puiſſances qui y ſont appliquées.</s> <s xml:id="echoid-s1075" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s1076" xml:space="preserve">TELLE que ſoit la ſurface GH; </s> <s xml:id="echoid-s1077" xml:space="preserve">le poids EO, & </s> <s xml:id="echoid-s1078" xml:space="preserve"><lb/> <anchor type="note" xlink:label="note-0066-01a" xlink:href="note-0066-01"/> la puiſſance R qui le ſoutient deſſus, ſont toujours en-<lb/>tr’ eux en raiſon réciproque des ſinus des angles que font leurs <lb/>lignes de direction avec AD tirée perpendiculairement du point <lb/>A de leur concours, ſur la ſurface GH.</s> <s xml:id="echoid-s1079" xml:space="preserve"/> </p> <div xml:id="echoid-div117" type="float" level="2" n="1"> <note position="left" xlink:label="note-0066-01" xlink:href="note-0066-01a" xml:space="preserve">fig. 30. <lb/>31. <lb/>32. <lb/>33. <lb/>34. <lb/>35. <lb/>36.</note> </div> </div> <div xml:id="echoid-div119" type="section" level="1" n="81"> <head xml:id="echoid-head81" xml:space="preserve"><emph style="sc">Demonstration.</emph></head> <p> <s xml:id="echoid-s1080" xml:space="preserve">A fin que ce poids & </s> <s xml:id="echoid-s1081" xml:space="preserve">cette puiſſance demeurent ainſi <lb/>en equilibre ſur la ſurface GH, telle qu’elle ſoit, <lb/>il faut 1°. </s> <s xml:id="echoid-s1082" xml:space="preserve">que leurs lignes de direction EB & </s> <s xml:id="echoid-s1083" xml:space="preserve">FC, <lb/>ſe rencontrent en quelque point A: </s> <s xml:id="echoid-s1084" xml:space="preserve">car ſi elles étoient <lb/>paralleles, il eſt clair que le centre de gravité de ce <lb/>poids pourroit encore deſcendre d’une longueur éga-<lb/>le à la diſtance qu’elles auroient entr’elles; </s> <s xml:id="echoid-s1085" xml:space="preserve">ce qui <lb/>eſt contre l’hypothêſe. </s> <s xml:id="echoid-s1086" xml:space="preserve">2°. </s> <s xml:id="echoid-s1087" xml:space="preserve">Les impreſſions particu-<lb/>liéres que font ſur le point A, & </s> <s xml:id="echoid-s1088" xml:space="preserve">la peſanteur de ce <lb/>poids, & </s> <s xml:id="echoid-s1089" xml:space="preserve">la puiſſance R qui le retient; </s> <s xml:id="echoid-s1090" xml:space="preserve">étant les mê-<lb/>mes que ſi ce point étoit pouſſé en même temps par <pb o="41" file="0067" n="67" rhead="MECHANIQUE."/> deux forces qui leur fuſſent égales, ſuivant leurs lignes <lb/> <anchor type="note" xlink:label="note-0067-01a" xlink:href="note-0067-01"/> de direction AC & </s> <s xml:id="echoid-s1091" xml:space="preserve">AB:</s> <s xml:id="echoid-s1092" xml:space="preserve">le poids EO ainſi ſoutenu ſur <lb/>la ſurface GH par le concours d’action de ſa peſan-<lb/>teur & </s> <s xml:id="echoid-s1093" xml:space="preserve">de la puiſſance R, doittendre (Lem. </s> <s xml:id="echoid-s1094" xml:space="preserve">4. </s> <s xml:id="echoid-s1095" xml:space="preserve">Cor. </s> <s xml:id="echoid-s1096" xml:space="preserve">2.) <lb/></s> <s xml:id="echoid-s1097" xml:space="preserve">ſuivant quelque ligne AD qui ſoit la diagonale d’un <lb/>parallelogramme fait ſous des parties AC & </s> <s xml:id="echoid-s1098" xml:space="preserve">AB de <lb/>leurs lignes de direction, qui ſoient entr’elles, com-<lb/>me ce poids eſt à cette puiſſance. </s> <s xml:id="echoid-s1099" xml:space="preserve">3°. </s> <s xml:id="echoid-s1100" xml:space="preserve">Cette ligne AD <lb/>doit toujours être perpendiculaire à la ſurface GH, <lb/>telle qu’elle ſoit; </s> <s xml:id="echoid-s1101" xml:space="preserve">par exemple, au point O: </s> <s xml:id="echoid-s1102" xml:space="preserve">autrement <lb/>AD devenant alors la ligne dedirection de ce corps, il <lb/>rouleroit ou couleroit (Cor. </s> <s xml:id="echoid-s1103" xml:space="preserve">Lemm. </s> <s xml:id="echoid-s1104" xml:space="preserve">1.) </s> <s xml:id="echoid-s1105" xml:space="preserve">vers H, ou vers <lb/>G, ſelon que cette ligne ſe trouveroit au-deſſus ou <lb/>au-deſſous du point O. </s> <s xml:id="echoid-s1106" xml:space="preserve">4°. </s> <s xml:id="echoid-s1107" xml:space="preserve">Cette même ligne doit <lb/>encore paſſer par quelqu’un des points où ce poids <lb/>touche cette ſurface, c’eſt-à-dire, par quelqu’un des <lb/>points de ſa baze: </s> <s xml:id="echoid-s1108" xml:space="preserve">autrement, & </s> <s xml:id="echoid-s1109" xml:space="preserve">pour la même rai-<lb/>ſon, il rouleroit, ou couleroit encore du côté qu’elle <lb/>s’en écarteroit. </s> <s xml:id="echoid-s1110" xml:space="preserve">5°. </s> <s xml:id="echoid-s1111" xml:space="preserve">La force dont ce poids eſt ainſi <lb/>pouſſé ou tiré ſuivant AD par le concours d’action <lb/>de ſa peſanteur & </s> <s xml:id="echoid-s1112" xml:space="preserve">de la puiſſance R, eſt à celle de <lb/>cette puiſſance, (Lemm. </s> <s xml:id="echoid-s1113" xml:space="preserve">4. </s> <s xml:id="echoid-s1114" xml:space="preserve">Cor. </s> <s xml:id="echoid-s1115" xml:space="preserve">3.) </s> <s xml:id="echoid-s1116" xml:space="preserve">comme AD à AB; </s> <s xml:id="echoid-s1117" xml:space="preserve"><lb/>c’eſt-à-dire, (Lem. </s> <s xml:id="echoid-s1118" xml:space="preserve">5.) </s> <s xml:id="echoid-s1119" xml:space="preserve">commele ſinus de l’angle DBA, <lb/>ou de ſon complement BAC, au ſinus de l’an-<lb/>gle BDA, ou de DAC qui luy eſt égal. </s> <s xml:id="echoid-s1120" xml:space="preserve">Elle <lb/>eſt auſſi pour la même raiſon à la peſanteur de ce <lb/>poids, comme le ſinus du même angle BAC, au ſinus <lb/>de BAD: </s> <s xml:id="echoid-s1121" xml:space="preserve">Ainſi la peſanteur de ce poids eſt à la <lb/>force de cette puiſſance, comme le ſinus de l’angle <lb/>BAD au ſinus de l’angle DAC; </s> <s xml:id="echoid-s1122" xml:space="preserve">& </s> <s xml:id="echoid-s1123" xml:space="preserve">par conſéquent <lb/>ce poids, & </s> <s xml:id="echoid-s1124" xml:space="preserve">cette puiſſance ſont entr’eux en raiſon <lb/>réciproque des ſinus des angles que font leurs li-<lb/>gnes de diréction avec la ligne tirée de leur point de <lb/>concours perpendiculairement à la ſurfacc GH, <lb/>telle quelle ſoit. </s> <s xml:id="echoid-s1125" xml:space="preserve">Ce qu’il F.</s> <s xml:id="echoid-s1126" xml:space="preserve">D.</s> <s xml:id="echoid-s1127" xml:space="preserve"/> </p> <div xml:id="echoid-div119" type="float" level="2" n="1"> <note position="right" xlink:label="note-0067-01" xlink:href="note-0067-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus ſur <lb/>des ſurfaces.</note> </div> <pb o="42" file="0068" n="68" rhead="NOUVELLE"/> <note position="left" xml:space="preserve">DES POIDS <lb/>ſoutenus ſur <lb/>des ſurfaces.</note> <p style="it"> <s xml:id="echoid-s1128" xml:space="preserve">Les cas ou le point de concours de ces lignes de direction <lb/>ſe trouve encore dans ce poids, mais au-deſſous de ſon centre <lb/>de gravité, ſe réſoudront comme ceux des figures 33. </s> <s xml:id="echoid-s1129" xml:space="preserve">34. </s> <s xml:id="echoid-s1130" xml:space="preserve">& </s> <s xml:id="echoid-s1131" xml:space="preserve"><lb/>36. </s> <s xml:id="echoid-s1132" xml:space="preserve">& </s> <s xml:id="echoid-s1133" xml:space="preserve">ceux où il ſe trouvera debors, ſe réſoudront auſſi de <lb/>même, en regardant ſeulement ce point comme apparte-<lb/>nant à ce poids, de la maniére que nous avons fait en traitant <lb/>des poids ſoutenus avec des cordes ſeulement, figure 16. </s> <s xml:id="echoid-s1134" xml:space="preserve">& </s> <s xml:id="echoid-s1135" xml:space="preserve">17. <lb/></s> <s xml:id="echoid-s1136" xml:space="preserve">Tout cela eſt aiſé; </s> <s xml:id="echoid-s1137" xml:space="preserve">c’eſt pourquoy on n’exprime point ici les <lb/>figures de tous ces cas.</s> <s xml:id="echoid-s1138" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s1139" xml:space="preserve">On n’exprime point non plus la figure d’aucune ſurface <lb/>borizontale: </s> <s xml:id="echoid-s1140" xml:space="preserve">parce que la ligne de direction de quelque poids <lb/>que ce ſoit, lui étant toujours perpendiculaire, il s’y ſoutient <lb/>de lui-même, & </s> <s xml:id="echoid-s1141" xml:space="preserve">ſans le ſecours d’aucune puiſſance, par la <lb/>même raiſon qu’il en a beſoin, comme l’on vient de voir, pour <lb/>demeurer ſur quelque autre ſurface que ce ſoit. </s> <s xml:id="echoid-s1142" xml:space="preserve">Cette propo-<lb/>ſition ne laiſſe pas cependant de s’étendre encore juſques-là, <lb/>comme on le verra dans les Corollaires 9. </s> <s xml:id="echoid-s1143" xml:space="preserve">& </s> <s xml:id="echoid-s1144" xml:space="preserve">10. </s> <s xml:id="echoid-s1145" xml:space="preserve">ainſi on <lb/>n’en peut pas concevoir une plus générale.</s> <s xml:id="echoid-s1146" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div121" type="section" level="1" n="82"> <head xml:id="echoid-head82" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s1147" xml:space="preserve">On voit des articles 3. </s> <s xml:id="echoid-s1148" xml:space="preserve">& </s> <s xml:id="echoid-s1149" xml:space="preserve">4. </s> <s xml:id="echoid-s1150" xml:space="preserve">de cette démonſtra-<lb/>tion que le poids EO ne peut-être ſoutenu par quel-<lb/>que puiſſance R que ce ſoit, ſur quelque ſurface <lb/>quece puiſſe être, à moins que la ligne AD ne tombe <lb/>perpendiculairement ſur cette ſurface, & </s> <s xml:id="echoid-s1151" xml:space="preserve">qu’elle ne <lb/>paſſe en même temps par quelqu’un des points ou <lb/>ce poids touche cette même ſurface; </s> <s xml:id="echoid-s1152" xml:space="preserve">c’eſt-à-dire, <lb/>par quelqu’un des points de la baze de ce même <lb/>poids.</s> <s xml:id="echoid-s1153" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div122" type="section" level="1" n="83"> <head xml:id="echoid-head83" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s1154" xml:space="preserve">Mais auſſi pour la même raiſon dés que l’un &</s> <s xml:id="echoid-s1155" xml:space="preserve"> <pb o="43" file="0069" n="69" rhead="MECHANIQUE."/> l’autre arrivera, la puiſſance qui y ſera alors appli-<lb/> <anchor type="note" xlink:label="note-0069-01a" xlink:href="note-0069-01"/> quée ne manquera pas de l’y ſoutenir.</s> <s xml:id="echoid-s1156" xml:space="preserve"/> </p> <div xml:id="echoid-div122" type="float" level="2" n="1"> <note position="right" xlink:label="note-0069-01" xlink:href="note-0069-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus ſur <lb/>des ſutfaces.</note> </div> </div> <div xml:id="echoid-div124" type="section" level="1" n="84"> <head xml:id="echoid-head84" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s1157" xml:space="preserve">De-là on voit qu’il n’y a point non plus de puiſſan-<lb/>ce capable de ſoutenir un poids rond, ou quelque <lb/>ſphére que ce ſoit, ſur quelque ſurface que ce puiſſe <lb/>être, à moins que le concours de leurs lignes de di-<lb/>rection ne ſe faſſe dans ſon centre de grandeur; </s> <s xml:id="echoid-s1158" xml:space="preserve">parce <lb/>que c’eſt le ſeul point de ce corps, d’où l’on puiſſe <lb/>tirer une ligne qui paſſe par ſa baze, & </s> <s xml:id="echoid-s1159" xml:space="preserve">qui ſoit en <lb/>même temps perpendiculaire à la ſurface ſur laquelle <lb/>il eſt ſoutenu. </s> <s xml:id="echoid-s1160" xml:space="preserve">Dans tout autre configuration de poids, <lb/>il en va tout autrement: </s> <s xml:id="echoid-s1161" xml:space="preserve">parce que l’on y peut trou-<lb/>ver pluſieurs points, d’où il eſt poſſible de tirer de <lb/>telles perpendiculaires.</s> <s xml:id="echoid-s1162" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div125" type="section" level="1" n="85"> <head xml:id="echoid-head85" xml:space="preserve"><emph style="sc">Corollaire</emph> IV.</head> <p> <s xml:id="echoid-s1163" xml:space="preserve">Il n’y a point encore de puiſſance R, telle quelle <lb/>ſoit, qui puiſſe ſoutenir aucun poids ſur quelque ſur-<lb/>face que ce puiſſe être, à moins que ſa ligne de di-<lb/>rection AR ne ſe trouve dans le complement NAO <lb/>de l’angle CAO, que fait la ligne de direction AC <lb/>de ce poids avec AO tirée du point A perpendiculai-<lb/>rement à cette ſurface GH. </s> <s xml:id="echoid-s1164" xml:space="preserve">Car 1°. </s> <s xml:id="echoid-s1165" xml:space="preserve">ſi elle concou-<lb/>roit avec AN elle ne feroit plus aueun angle avec <lb/>AC: </s> <s xml:id="echoid-s1166" xml:space="preserve">ainſi cette puiſſance porteroit ſeule tout ce poids, <lb/>& </s> <s xml:id="echoid-s1167" xml:space="preserve">non avec ce plan; </s> <s xml:id="echoid-s1168" xml:space="preserve">ce qui eſt contre l’hypothêſe. <lb/></s> <s xml:id="echoid-s1169" xml:space="preserve">2°. </s> <s xml:id="echoid-s1170" xml:space="preserve">Si cette ligne de direction concouroit avec AO, <lb/>ou ſi elle ſortoit de l’angle NAO, la diagonale AD <lb/>du parallelogramme BC ſe tourneroit vers G; </s> <s xml:id="echoid-s1171" xml:space="preserve">ce <lb/>qui feroit néceſſairement (Corol. </s> <s xml:id="echoid-s1172" xml:space="preserve">Lemm. </s> <s xml:id="echoid-s1173" xml:space="preserve">I.) </s> <s xml:id="echoid-s1174" xml:space="preserve">tomber ce <lb/>poids de ce côté-là, ce qui eſt encore contre l’hypo-<lb/>thêſe. </s> <s xml:id="echoid-s1175" xml:space="preserve">Donc la ligne de direction AB de la puiſſance <lb/>R, doit toujours ſe trouver dans le complement NAO <pb o="44" file="0070" n="70" rhead="NOUVELLE"/> <anchor type="note" xlink:label="note-0070-01a" xlink:href="note-0070-01"/> de l’angle CAO: </s> <s xml:id="echoid-s1176" xml:space="preserve">de ſorte que c’eſt-là tout l’eſpace <lb/>du mouvement qu’elle peut avoir.</s> <s xml:id="echoid-s1177" xml:space="preserve"/> </p> <div xml:id="echoid-div125" type="float" level="2" n="1"> <note position="left" xlink:label="note-0070-01" xlink:href="note-0070-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus ſur <lb/>des ſurfaces.</note> </div> </div> <div xml:id="echoid-div127" type="section" level="1" n="86"> <head xml:id="echoid-head86" xml:space="preserve"><emph style="sc">Corollaire</emph> V.</head> <p> <s xml:id="echoid-s1178" xml:space="preserve">Le plan BAC, ou le parallelogramme BC de <lb/>cette ligne avec AC ligne de direction du poids <lb/>EO, eſt auſſi toujours perpendiculaire à la ſur-<lb/>face que la ligne GH repreſente, & </s> <s xml:id="echoid-s1179" xml:space="preserve">ſur laquelle ce <lb/>poids eſt ſoutenu; </s> <s xml:id="echoid-s1180" xml:space="preserve">puiſque la diagonale AD de ce pa-<lb/>rallelogramme BC, l’eſt toujours à cette même ſur-<lb/>face.</s> <s xml:id="echoid-s1181" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div128" type="section" level="1" n="87"> <head xml:id="echoid-head87" xml:space="preserve"><emph style="sc">Corollaire</emph> VI.</head> <p> <s xml:id="echoid-s1182" xml:space="preserve">On voit auſſi de l’article cinquiéme de la demonſ-<lb/>tration ci-deſſus, que le poids EO eſt à la charge de <lb/>cette ſurface, c’eſt-à-dire, à l’impreſſion que lui & </s> <s xml:id="echoid-s1183" xml:space="preserve">la <lb/>puiſſance R font enſemble deſſus, ou à la réſiſtance <lb/>qu’elle leur fait, comme le ſinus de l’angle BAD, au <lb/>ſinus de l’angle BAC.</s> <s xml:id="echoid-s1184" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div129" type="section" level="1" n="88"> <head xml:id="echoid-head88" xml:space="preserve"><emph style="sc">Corollaire</emph> VII.</head> <p> <s xml:id="echoid-s1185" xml:space="preserve">D’où il ſuit que le poids EO, la charge de la ſur-<lb/>face GH, & </s> <s xml:id="echoid-s1186" xml:space="preserve">la puiſſance R, ſont entr’eux comme <lb/>les ſinus des angles BAD, BAC, ou ABD ſon com-<lb/>plement, & </s> <s xml:id="echoid-s1187" xml:space="preserve">DAC, ou ADB qui lui eſt égal; </s> <s xml:id="echoid-s1188" xml:space="preserve">c’eſt-à-<lb/>dire, (Lemm. </s> <s xml:id="echoid-s1189" xml:space="preserve">5.) </s> <s xml:id="echoid-s1190" xml:space="preserve">comme les lignes BD, AD, & </s> <s xml:id="echoid-s1191" xml:space="preserve"><lb/>AB.</s> <s xml:id="echoid-s1192" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div130" type="section" level="1" n="89"> <head xml:id="echoid-head89" xml:space="preserve"><emph style="sc">Corollaire</emph> VIII.</head> <p> <s xml:id="echoid-s1193" xml:space="preserve">D’où il ſuit encore que plus l’angle BAC eſt obtus, <lb/>moins la charge de cette ſurface eſt grande: </s> <s xml:id="echoid-s1194" xml:space="preserve">de ſorte <lb/>qu’il le peut devenir juſqu’à un tel point, quelle ſera <lb/>ſi petite qu’on voudra; </s> <s xml:id="echoid-s1195" xml:space="preserve">c’eſt ainſi qu’elle peut dimi-<lb/>nuer à l’infini.</s> <s xml:id="echoid-s1196" xml:space="preserve"/> </p> <pb o="45" file="0071" n="71" rhead="MECHANIQUE."/> </div> <div xml:id="echoid-div131" type="section" level="1" n="90"> <head xml:id="echoid-head90" xml:space="preserve"><emph style="sc">Corollaire</emph> IX.</head> <note position="right" xml:space="preserve">DES POIDS <lb/>ſoutenus ſur <lb/>des ſurfaces.</note> <p> <s xml:id="echoid-s1197" xml:space="preserve">Mais elle ne peut pas augmenter de même; </s> <s xml:id="echoid-s1198" xml:space="preserve">parce <lb/>que ne pouvant jamais être plus grande, que lorſque <lb/>cet angle eſt infiniment aigu; </s> <s xml:id="echoid-s1199" xml:space="preserve">c’eſt-à-dire, lorſque <lb/>les lignes AC & </s> <s xml:id="echoid-s1200" xml:space="preserve">AB concourent avec AO: </s> <s xml:id="echoid-s1201" xml:space="preserve">& </s> <s xml:id="echoid-s1202" xml:space="preserve">AD <lb/>n’étant encore alors qu’égale à la ſomme de AB & </s> <s xml:id="echoid-s1203" xml:space="preserve">de <lb/>BD; </s> <s xml:id="echoid-s1204" xml:space="preserve">la charge de cette ſurface, qui eſt alors hori-<lb/>zontale ne peut jamais être plus grande que la ſomme <lb/>de ce poids, & </s> <s xml:id="echoid-s1205" xml:space="preserve">de cette puiſſance.</s> <s xml:id="echoid-s1206" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s1207" xml:space="preserve">On entend ici par ſurface horizontale unplan qui le ſoit, <lb/>ou bien un point d’une ſurface courbe dont toutes les tangentes <lb/>ſoient auſſi borizontales.</s> <s xml:id="echoid-s1208" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div132" type="section" level="1" n="91"> <head xml:id="echoid-head91" xml:space="preserve"><emph style="sc">Corollaire</emph> X.</head> <p> <s xml:id="echoid-s1209" xml:space="preserve">On voit encore qu’il faut d’autant moins de force <lb/> <anchor type="note" xlink:label="note-0071-02a" xlink:href="note-0071-02"/> pour ſoutenir ainſi un poids ſuivant la même direc-<lb/>tion AB ſur un même point de quelque ſurface <lb/>que ce ſoit, que cette ſurface, ſi elle eſt droite <lb/>(fig. </s> <s xml:id="echoid-s1210" xml:space="preserve">30. </s> <s xml:id="echoid-s1211" xml:space="preserve">& </s> <s xml:id="echoid-s1212" xml:space="preserve">33.) </s> <s xml:id="echoid-s1213" xml:space="preserve">ou bien ſi elle eſt courbe; </s> <s xml:id="echoid-s1214" xml:space="preserve">(fig. </s> <s xml:id="echoid-s1215" xml:space="preserve">32. <lb/></s> <s xml:id="echoid-s1216" xml:space="preserve">36.)</s> <s xml:id="echoid-s1217" xml:space="preserve">, que ſa tangente au point ou la perpendiculai-<lb/>re AO la rencontre, eſt plus inclinée, quoi qu’en <lb/>proportion différente: </s> <s xml:id="echoid-s1218" xml:space="preserve">parce que la raiſon du ſinus <lb/>de l’angle CAD, au ſinus de l’angle BAD, en eſt <lb/>toujours moindre; </s> <s xml:id="echoid-s1219" xml:space="preserve">& </s> <s xml:id="echoid-s1220" xml:space="preserve">comme cette inclinaiſon avec <lb/>l’horizon peut diminuer à l’infini, la force qu’il faut <lb/>pour ſoutenir quelque poids ſuivant la même direc-<lb/>tion ſur quelqu’une de ces ſurfaces, ſoit droite, ſoit <lb/>courbe, peut auſſi diminuer à l’infini: </s> <s xml:id="echoid-s1221" xml:space="preserve">De ſorte que <lb/>lorſqu’elle ſera infiniment inclinée; </s> <s xml:id="echoid-s1222" xml:space="preserve">c’eſt-à-dire, ho-<lb/>rizontale, du moins dans le point où la perpendicu-<lb/>laire AO la rencontre, cette force ſera nulle, & </s> <s xml:id="echoid-s1223" xml:space="preserve">ré-<lb/>duite à zéro; </s> <s xml:id="echoid-s1224" xml:space="preserve">c’eſt-à-dire, qu’il n’en faudra plus du <lb/>tout pour l’y ſoutenir.</s> <s xml:id="echoid-s1225" xml:space="preserve"/> </p> <div xml:id="echoid-div132" type="float" level="2" n="1"> <note position="right" xlink:label="note-0071-02" xlink:href="note-0071-02a" xml:space="preserve">fig 30. <lb/>32. <lb/>33. <lb/>36.</note> </div> <pb o="46" file="0072" n="72" rhead="NOUVELLE"/> <note position="left" xml:space="preserve">DES POIDS <lb/>ſoutenus ſur <lb/>des ſurfaces.</note> </div> <div xml:id="echoid-div134" type="section" level="1" n="92"> <head xml:id="echoid-head92" xml:space="preserve"><emph style="sc">Corollaire</emph> XI.</head> <p> <s xml:id="echoid-s1226" xml:space="preserve">Au contraire pour ſoutenir ce poids ſur le même <lb/>point d’une ſurface toujours également inclinée, telle <lb/>quelle ſoit, mais ſuivant différentes directions de <lb/>puiſſance; </s> <s xml:id="echoid-s1227" xml:space="preserve">il faut d’autant plus de forces que l’angle <lb/>DAB fait dans l’eſpace NAD par la perpendiculaire <lb/>AO avec la ligne de direction AB de la puiſſance qui <lb/>ſoutient ce poids, differe davantage de l’angle droit: <lb/></s> <s xml:id="echoid-s1228" xml:space="preserve">parce qu’alors la raiſon du ſinus de l’angle CAD, au ſi-<lb/>nus de l’angle BAD, en ſera d’autant plus grande, quoi <lb/>qu’en proportion différente: </s> <s xml:id="echoid-s1229" xml:space="preserve">& </s> <s xml:id="echoid-s1230" xml:space="preserve">conime cet angle peut-<lb/>être plus ou moins grand qu’un angle droit, & </s> <s xml:id="echoid-s1231" xml:space="preserve">en diffé-<lb/>rer de plus en plus juſqu’au concours de AB avec AN, <lb/>ou avec AO, ſans que AB ſorte de l’eſpace NAO; </s> <s xml:id="echoid-s1232" xml:space="preserve">la <lb/>puiſſance qui ſoutient ce poids peut auſſi augmenter <lb/>juſques-là: </s> <s xml:id="echoid-s1233" xml:space="preserve">mais différemment ſelon que AB s’appro-<lb/>che de l’une, ou de l’autre de ces deux lignes. </s> <s xml:id="echoid-s1234" xml:space="preserve">Car <lb/>1°. </s> <s xml:id="echoid-s1235" xml:space="preserve">ne pouvant jamais être plus grande par l’appro-<lb/>che de AB vers AN, que lorſque l’angle NAB, que <lb/>ces lignes font entr’elles, eſt le plus petit qu’elles <lb/>puiſſent faire; </s> <s xml:id="echoid-s1236" xml:space="preserve">c’eſt-à-dire, celui qu’elles font immé-<lb/>diatement avant que de concourir; </s> <s xml:id="echoid-s1237" xml:space="preserve">& </s> <s xml:id="echoid-s1238" xml:space="preserve">le ſinus de CAD <lb/>étant encore alors un peu moindre que celui de <lb/>BAO, quoique d’une difference infiniment petite: </s> <s xml:id="echoid-s1239" xml:space="preserve"><lb/>Cette puiſſance ne peut être tout au plus de ce côté-là, <lb/>qu’un peu moindre que ce poids d’une différence qui <lb/>ſoit auſſi infiniment petite. </s> <s xml:id="echoid-s1240" xml:space="preserve">2°. </s> <s xml:id="echoid-s1241" xml:space="preserve">Au contraire du côté de <lb/>AO elle pent augmenter à l’infini: </s> <s xml:id="echoid-s1242" xml:space="preserve">parce que la raiſon <lb/>du ſinus de CAD, à celuy de BAD, augmentant à <lb/>meſure quela ligne AB s’approche de AO, en s’éloi-<lb/>gnant de la ſituation qu’elle auroit ſi elle faiſoit un an-<lb/>gle droit avec AO; </s> <s xml:id="echoid-s1243" xml:space="preserve">cette force peut auſſi augmenter <lb/>de ce côté-là juſqu’à ce que AB, & </s> <s xml:id="echoid-s1244" xml:space="preserve">AO concourent. </s> <s xml:id="echoid-s1245" xml:space="preserve"><lb/>Or en ce cas l’angle BAD étant infiniment petit, la <pb o="47" file="0073" n="73" rhead="MECHANIQUE."/> raiſon du ſinus de CAD à celui de cet angle, ſera auſſi <lb/> <anchor type="note" xlink:label="note-0073-01a" xlink:href="note-0073-01"/> infinie; </s> <s xml:id="echoid-s1246" xml:space="preserve">& </s> <s xml:id="echoid-s1247" xml:space="preserve">par conſéquent auſſi celle de cette puiſ-<lb/>ſance à ce poids. </s> <s xml:id="echoid-s1248" xml:space="preserve">Ce qui fait voir que cette puiſ-<lb/>ſance peut augmenter à l’infini dans le mouvement <lb/>qu’elle peut avoir depuis la ſituation où ſa ligne de <lb/>direction feroit un angle droit avec la perpendiculaire <lb/>AO, juſqu’au concours de ces deux mêmes lignes, & </s> <s xml:id="echoid-s1249" xml:space="preserve"><lb/>demeurer cependant toûjours en équilibre avec le <lb/>même poids, & </s> <s xml:id="echoid-s1250" xml:space="preserve">ſur le même point d’une ſurface in-<lb/>clinée, telle quelle ſoit.</s> <s xml:id="echoid-s1251" xml:space="preserve"/> </p> <div xml:id="echoid-div134" type="float" level="2" n="1"> <note position="right" xlink:label="note-0073-01" xlink:href="note-0073-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus ſur <lb/>des ſurfaces.</note> </div> </div> <div xml:id="echoid-div136" type="section" level="1" n="93"> <head xml:id="echoid-head93" xml:space="preserve"><emph style="sc">Corollaire</emph> XII.</head> <p> <s xml:id="echoid-s1252" xml:space="preserve">Pour les plans perpendiculaires à l’horizon, ou plu-<lb/> <anchor type="note" xlink:label="note-0073-02a" xlink:href="note-0073-02"/> tôt paralleles à la ligne de direction de ce poids; </s> <s xml:id="echoid-s1253" xml:space="preserve">& </s> <s xml:id="echoid-s1254" xml:space="preserve"><lb/>pour les points des ſurfaces courbes, d’où l’on peut <lb/>tirer des tangentes qui ſoient auſſi perpendiculai-<lb/>res à l’horizon: </s> <s xml:id="echoid-s1255" xml:space="preserve">la ligne de direction AB de la puiſ-<lb/>ſance qui ſoutient ce poids ſur, ou contre ces plans, <lb/>ou ces points de ſurfaces courbes, ne pouvant (Co-<lb/>rol. </s> <s xml:id="echoid-s1256" xml:space="preserve">4.) </s> <s xml:id="echoid-s1257" xml:space="preserve">s’éloigner de la ſituation où elle ſeroit, ſi <lb/>elle faiſoit un angle droit avec AO, qu’en s’appro-<lb/>chant de AO, puiſque l’angle NAO en ce cas eſt <lb/>droit; </s> <s xml:id="echoid-s1258" xml:space="preserve">cette puiſſance ne peut auſſi augmenter que <lb/>de ce côté-là: </s> <s xml:id="echoid-s1259" xml:space="preserve">De ſorte qu’elle peut à la vérité aug-<lb/>menter à l’infini de même que ſur les ſurfaces incli-<lb/>nées: </s> <s xml:id="echoid-s1260" xml:space="preserve">mais elle ne peut jamais être moins grande <lb/>que lorſque ſa ligne de direction AB fait le plus petit <lb/>angle qu’elle puiſſe faire avec AN; </s> <s xml:id="echoid-s1261" xml:space="preserve">c’eſt-à-dire, <lb/>qu’en ne ſurpaſſant ce poids que d’une différence in-<lb/>finiment petite.</s> <s xml:id="echoid-s1262" xml:space="preserve"/> </p> <div xml:id="echoid-div136" type="float" level="2" n="1"> <note position="right" xlink:label="note-0073-02" xlink:href="note-0073-02a" xml:space="preserve">fig. 31. <lb/>34. <lb/>35.</note> </div> </div> <div xml:id="echoid-div138" type="section" level="1" n="94"> <head xml:id="echoid-head94" xml:space="preserve"><emph style="sc">Corollaire</emph> XIII.</head> <note position="right" xml:space="preserve">fig. 30. <lb/>31. <lb/>32. <lb/>33. <lb/>34. <lb/>35. <lb/>36.</note> <p> <s xml:id="echoid-s1263" xml:space="preserve">De ſorte que dés le moment que AB & </s> <s xml:id="echoid-s1264" xml:space="preserve">AN vien-<lb/>nent à concourir enſemble, le ſinus de CAD deve-<lb/>nant alors égal à celui de BAD, cette puiſſance de- <pb o="48" file="0074" n="74" rhead="NOUVELLE"/> <anchor type="note" xlink:label="note-0074-01a" xlink:href="note-0074-01"/> vient auſſi égale à ce poids, qui ceſſe auſſi-tôt de s’ap-<lb/>puyer ſur, ou contre la ſurface GH, telle qu’elle <lb/>ſoit.</s> <s xml:id="echoid-s1265" xml:space="preserve"/> </p> <div xml:id="echoid-div138" type="float" level="2" n="1"> <note position="left" xlink:label="note-0074-01" xlink:href="note-0074-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus ſur <lb/>des ſurfaces.</note> </div> <p style="it"> <s xml:id="echoid-s1266" xml:space="preserve">Ce n’eſt que pour ne pas embaraſſer l’imagination de ceux <lb/>qui ſont accoutumez à regarder une ſurface perpendiculaire à <lb/>l’horizon, comme parallele à la ligne de direction d’un poids, & </s> <s xml:id="echoid-s1267" xml:space="preserve"><lb/>une horizontale, comme lui étant perpendiculaire; </s> <s xml:id="echoid-s1268" xml:space="preserve">que l’on s’eſt <lb/>accommodé à cette bypotbêſe dans ces trois derniers Corollaires: <lb/></s> <s xml:id="echoid-s1269" xml:space="preserve">car pour les rendre auſſi généraux qu’on les puiſſe imaginer, & </s> <s xml:id="echoid-s1270" xml:space="preserve"><lb/>pour toutes ſortes d’bypotbêſes, il ne faut que regarder les plans, <lb/>ou les tangentes des ſurfaces courbes, que l’on dit ici perpen-<lb/>diculaires à l’borizon, comme paralleles ſeulement à la ligne <lb/>de direction de ce poids, ſans avoir égard à l’angle qu’elles <lb/>font, ou qu’elles peuvent faire avec l’borizon. </s> <s xml:id="echoid-s1271" xml:space="preserve">De même celles <lb/>que l’on appelle ici borizontales, ſe doivent ſeulement regar-<lb/>der comme perpendiculaires à cette ligne. </s> <s xml:id="echoid-s1272" xml:space="preserve">De cette maniére ces <lb/>Corollaires ſeront ſi généraux, qu’ils ſe pourront appliquer à <lb/>toutes les directions poſſibles d’un corps ſoutenu ſur ou contre <lb/>quelque ſurface que ce puiſſe être.</s> <s xml:id="echoid-s1273" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s1274" xml:space="preserve">Il eſt encore bon de remarquer que lorſqu’on dit ici qu’un <lb/>poids eſt ſoutenu ſur le même point d’une ſurface, l’on ne pré-<lb/>tend pas dire qu’il ne la rencontre jamais qu’en un ſeul point; <lb/></s> <s xml:id="echoid-s1275" xml:space="preserve">mais on entend ſeulement que la ligne AD, qui tombe du <lb/>point A perpendiculairement deſſus, la rencontre toujours dans <lb/>le même point O, tant que ce poids eſt ſoutenu deſſus, quoi que <lb/>ce ſoit ſuivant différentes directions de puiſſances. </s> <s xml:id="echoid-s1276" xml:space="preserve">La raiſon <lb/>de cette précaution eſt évidente du côté des ſurfaces courbes, <lb/>dont tous les points ont cbacun une tangente d’une direction <lb/>particuliére. </s> <s xml:id="echoid-s1277" xml:space="preserve">Pour du côté des plans, on la reconnoîtra dans <lb/>le Corollaire 23. </s> <s xml:id="echoid-s1278" xml:space="preserve">où l’on verra que dans l’bypotbêſe du con-<lb/>cours des lignes de direction des poids en quelque points de la <lb/>terre que ce ſoit, ils ne péſent pas toujours également deſſus, <lb/>quoique la ligne de direction de la puiſſance qui leur eſt appli- <pb o="49" file="0075" n="75" rhead="MECHANIQUE"/> quée demeure toujours la même. </s> <s xml:id="echoid-s1279" xml:space="preserve">Au contraire, ils péſent tou-<lb/> <anchor type="note" xlink:label="note-0075-01a" xlink:href="note-0075-01"/> jours également ſur le même point de quelque ſurface que ce <lb/>ſoit, à moins qu’on ne change la ligne de direction de cotte <lb/>puiſſance, ou la ſi@uation de cette ſurface. </s> <s xml:id="echoid-s1280" xml:space="preserve">C’eſt pour cela <lb/>que dans les trois Corollaires précédens, où l’on examine ſépa-<lb/>rément le changement que peut cauſer dans l’action d’un poids <lb/>les différentes inclinaiſons de la même, ou des différentes ſurfaces <lb/>ſur leſquelles il eſt ſoutenu, & </s> <s xml:id="echoid-s1281" xml:space="preserve">les différentes lignes de direction des <lb/>puiſſances qui l’yſoutiennent;</s> <s xml:id="echoid-s1282" xml:space="preserve">on l’aregardé comme appliqué non-<lb/>ſeulement à la même ſurface, mais auſſi toujours au même point.</s> <s xml:id="echoid-s1283" xml:space="preserve"/> </p> <div xml:id="echoid-div139" type="float" level="2" n="2"> <note position="right" xlink:label="note-0075-01" xlink:href="note-0075-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus ſur <lb/>des ſurfaces.</note> </div> </div> <div xml:id="echoid-div141" type="section" level="1" n="95"> <head xml:id="echoid-head95" xml:space="preserve"><emph style="sc">Corollaire</emph> XIV.</head> <p> <s xml:id="echoid-s1284" xml:space="preserve">Puiſque la puiſſance qui ſoutient quelque poids <lb/>que ce ſoit ſur le même point de quelque ſurface <lb/>que ce puiſſe être, eſt d’autant plus grande que <lb/>ſa ligne de direction AB s’éloigne davantage de la <lb/>ſituaton où elle feroit un angle droit avec AO, ſans <lb/>cependant ſortir de l’eſpace NAO: </s> <s xml:id="echoid-s1285" xml:space="preserve">Il s’enſuit qu’elle <lb/>n’eſt jamais moindre que lorſqu’elle eſt parallele au <lb/>plan, ou à quelqu’une des tangentes au point de la ſur-<lb/>face courbe, ſur lequel ce poids eſt ſoutenu.</s> <s xml:id="echoid-s1286" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div142" type="section" level="1" n="96"> <head xml:id="echoid-head96" xml:space="preserve"><emph style="sc">Corollaire</emph> XV.</head> <p> <s xml:id="echoid-s1287" xml:space="preserve">D’où l’on voit dans l’hypothêſe ordinaire, ou l’on <lb/> <anchor type="note" xlink:label="note-0075-02a" xlink:href="note-0075-02"/> regarde HK comme parallele à AC, que les triangles <lb/>BAD, HKG étant alors ſemblables, cette puiſſance qui <lb/>eſt à ce poids (Cor. </s> <s xml:id="echoid-s1288" xml:space="preserve">7.) </s> <s xml:id="echoid-s1289" xml:space="preserve">comme AB à BD, lui ſera auſſi <lb/>comme HK à HG: </s> <s xml:id="echoid-s1290" xml:space="preserve">& </s> <s xml:id="echoid-s1291" xml:space="preserve">comme elle eſt alors la moin-<lb/>dre qu’elle puiſſe jamais être ſelon le Corollaire précé-<lb/>dent; </s> <s xml:id="echoid-s1292" xml:space="preserve">il s’enſuit qu’elle ne peut jamais être en moindre <lb/>raiſon au poids qu’elle ſoutient ſur un plan incliné, <lb/>qu’eſt celle de la hauteur de ce plan à ſa longueur.</s> <s xml:id="echoid-s1293" xml:space="preserve"/> </p> <div xml:id="echoid-div142" type="float" level="2" n="1"> <note position="right" xlink:label="note-0075-02" xlink:href="note-0075-02a" xml:space="preserve">fig. 30.</note> </div> </div> <div xml:id="echoid-div144" type="section" level="1" n="97"> <head xml:id="echoid-head97" xml:space="preserve"><emph style="sc">Corollaire</emph> XVI.</head> <p> <s xml:id="echoid-s1294" xml:space="preserve">On voit encore que toute puiſſance qui peut ſoute- <pb o="50" file="0076" n="76" rhead="NOUVELLE"/> <anchor type="note" xlink:label="note-0076-01a" xlink:href="note-0076-01"/> nir un poids ſur un plan incliné ſuivant une ligne <lb/>de direction, qui faſſe avec une perpendiculaire faite <lb/>ſur AD au point A, un angle moindre que celui de <lb/>cette perpendiculaire avec AN; </s> <s xml:id="echoid-s1295" xml:space="preserve">l’y peut ſoutenir <lb/> <anchor type="note" xlink:label="note-0076-02a" xlink:href="note-0076-02"/> encore, & </s> <s xml:id="echoid-s1296" xml:space="preserve">ſur le même point, ſuivant une autre ligne <lb/>de direction, qui paſſant de l’autre coté de cette per-<lb/>pendiculaire, faſſe avec elle un angle égal au pre-<lb/>mier: </s> <s xml:id="echoid-s1297" xml:space="preserve">car les deux angles que font alors ces deux li-<lb/>gnes de direction avec AD, étant complemens à deux <lb/>droits, l’un de l’autre, leurs ſinus ſeront égaux; </s> <s xml:id="echoid-s1298" xml:space="preserve">& </s> <s xml:id="echoid-s1299" xml:space="preserve"><lb/>par conſéquent ils ſeront en même raiſon au ſinus de <lb/>l’angle CAD; </s> <s xml:id="echoid-s1300" xml:space="preserve">& </s> <s xml:id="echoid-s1301" xml:space="preserve">par conſéquent auſſi ce même poids <lb/>ſeroit alors en même raiſon aux puiſſances, qui pla-<lb/>cées ſuivant ces différentes directions, le ſoutien-<lb/>droient l’une aprés l’autre; </s> <s xml:id="echoid-s1302" xml:space="preserve">ainſi elles ſeroient égales <lb/>entr’elles: </s> <s xml:id="echoid-s1303" xml:space="preserve">Donc la même puiſſance qui ſoutient ce <lb/>poids ſuivant une de ces directions, le peut encore <lb/>ſoutenir ſuivant l’autre ſur le même plan incliné <lb/>GH.</s> <s xml:id="echoid-s1304" xml:space="preserve"/> </p> <div xml:id="echoid-div144" type="float" level="2" n="1"> <note position="left" xlink:label="note-0076-01" xlink:href="note-0076-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus ſur <lb/>des ſurfaces.</note> <note position="left" xlink:label="note-0076-02" xlink:href="note-0076-02a" xml:space="preserve">fig 30. <lb/>33.</note> </div> <p style="it"> <s xml:id="echoid-s1305" xml:space="preserve">On verra par le Corollaire 23. </s> <s xml:id="echoid-s1306" xml:space="preserve">que pour rendre ce dernier <lb/>Corollaire général pour toutes ſortes d’hypotbéſes, il faut que <lb/>ce poids ſe trouve alors ſur le même point d’un plan toujours <lb/>également incliné.</s> <s xml:id="echoid-s1307" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div146" type="section" level="1" n="98"> <head xml:id="echoid-head98" xml:space="preserve"><emph style="sc">Corollaire</emph> XVII.</head> <p> <s xml:id="echoid-s1308" xml:space="preserve">1°. </s> <s xml:id="echoid-s1309" xml:space="preserve">Si AB ne concourt point avec la perpendiculaire <lb/>faite ſur AD au point A, la puiſſance R qui ſoutiet le <lb/>poid EO ſuivant cette même ligne AB, eſt à ce même <lb/>poids en plus grande raiſon que le ſinus de CAD au ſi-<lb/>nus total; </s> <s xml:id="echoid-s1310" xml:space="preserve">c’eſt-à-dire, dans l’hypoth êſe ordinaire, en <lb/>plus grande raiſon que la hauteur dé ce plan (Cor. <lb/></s> <s xml:id="echoid-s1311" xml:space="preserve">14. </s> <s xml:id="echoid-s1312" xml:space="preserve">& </s> <s xml:id="echoid-s1313" xml:space="preserve">15.) </s> <s xml:id="echoid-s1314" xml:space="preserve">à ſa longueur. </s> <s xml:id="echoid-s1315" xml:space="preserve">2°. </s> <s xml:id="echoid-s1316" xml:space="preserve">Si l’angle de cette per-<lb/>pendiculaire avec la ligne de direction AB de cette <lb/>puiſſance, eſt moindre que l’angle de cette même per- <pb o="51" file="0077" n="77" rhead="MECHANIQUE."/> pendiculaire avec AN, cette puiſſance eſt auſſi moin-<lb/> <anchor type="note" xlink:label="note-0077-01a" xlink:href="note-0077-01"/> dre que ce poids: </s> <s xml:id="echoid-s1317" xml:space="preserve">puiſque de quelque côté que AB <lb/>faſſe cet angle avec cette pendiculaire, la puiſſance qui <lb/>ſoutiendra ce poids, (Cor. </s> <s xml:id="echoid-s1318" xml:space="preserve">16.) </s> <s xml:id="echoid-s1319" xml:space="preserve">ſera la même; </s> <s xml:id="echoid-s1320" xml:space="preserve">il s’enſuit <lb/>(Cor. </s> <s xml:id="echoid-s1321" xml:space="preserve">II. </s> <s xml:id="echoid-s1322" xml:space="preserve">n. </s> <s xml:id="echoid-s1323" xml:space="preserve">I.) </s> <s xml:id="echoid-s1324" xml:space="preserve">qu’elle ſera moindre que ce poids dans <lb/>l’un & </s> <s xml:id="echoid-s1325" xml:space="preserve">l’autrecas: </s> <s xml:id="echoid-s1326" xml:space="preserve">ainſi l’on peut conclure en général <lb/>qu’une même puiſſance peut ſoutenir un même poids <lb/>ſur un même plan incliné ſuivant deux différentes di-<lb/>rections: </s> <s xml:id="echoid-s1327" xml:space="preserve">pourvu qu’elle ſoit moindre que ce poids, & </s> <s xml:id="echoid-s1328" xml:space="preserve"><lb/>qu’elle lui ſoit cependant en plus grande raiſon que le <lb/>ſinus de l’angle CAD au ſinus total; </s> <s xml:id="echoid-s1329" xml:space="preserve">c’eſt-à-dire, dans <lb/>l’hypothêſe ordinaire, où l’on regarde la ligne de di-<lb/>rection de ce poids comme parallele à la hauteur de ce <lb/>plan: </s> <s xml:id="echoid-s1330" xml:space="preserve">pourvu que cette puiſſance moindre que ce <lb/>poids, lui ſoit cependant en plus grande raiſon que la <lb/>hauteur de ce plan à ſa longueur.</s> <s xml:id="echoid-s1331" xml:space="preserve"/> </p> <div xml:id="echoid-div146" type="float" level="2" n="1"> <note position="right" xlink:label="note-0077-01" xlink:href="note-0077-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus ſur <lb/>des ſurfaces.</note> </div> </div> <div xml:id="echoid-div148" type="section" level="1" n="99"> <head xml:id="echoid-head99" xml:space="preserve"><emph style="sc">Corollaire</emph> XVIII.</head> <p> <s xml:id="echoid-s1332" xml:space="preserve">En tout autre cas; </s> <s xml:id="echoid-s1333" xml:space="preserve">c’eſt-à-dire, lorſque cette puiſ-<lb/>ſance eſt plus grande que ce poids, ou du moins qu’el-<lb/>le lui eſt égale, ou bien lorſqu’elle lui eſt en même <lb/>raiſon que le ſinus de l’angle CAD au ſinus total, <lb/>elle ne le peut ſoutenir ſur le même point de ce plan <lb/>que ſuivant une ſeule direction. </s> <s xml:id="echoid-s1334" xml:space="preserve">Tout cela eſt mani-<lb/>feſte par une raiſon toute contraire à celle du Corol-<lb/>laire précédent.</s> <s xml:id="echoid-s1335" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s1336" xml:space="preserve">On voit aſſez comment ces quatre derniers Corollaires ſe <lb/>peuvent appliquer à toutes ſortes de ſurfaces courbes, pour les <lb/>points ou elles peuvent être touchées par des plans inclinez. <lb/></s> <s xml:id="echoid-s1337" xml:space="preserve">Il n’eſt pas difficile non plus de reconnoitre ce qui leur peut <lb/>convenir de ce que nous allons encore dire des ſurfaces planes; </s> <s xml:id="echoid-s1338" xml:space="preserve"><lb/>c’eſt pourquoi nous ne parlerons plus d’orénavant que de <lb/>celles-ci.</s> <s xml:id="echoid-s1339" xml:space="preserve"/> </p> <pb o="52" file="0078" n="78" rhead="NOUVELLE"/> <note position="left" xml:space="preserve">DES POIDS <lb/>ſoutenus ſur <lb/>des ſurfaces.</note> </div> <div xml:id="echoid-div149" type="section" level="1" n="100"> <head xml:id="echoid-head100" xml:space="preserve"><emph style="sc">Corollaire</emph> XIX.</head> <p> <s xml:id="echoid-s1340" xml:space="preserve">Puiſque (Cor. </s> <s xml:id="echoid-s1341" xml:space="preserve">7.) </s> <s xml:id="echoid-s1342" xml:space="preserve">le poids EO, la puiſſance R, & </s> <s xml:id="echoid-s1343" xml:space="preserve"><lb/>la charge du plan GH, ſont entr’eux comme les li-<lb/> <anchor type="note" xlink:label="note-0078-02a" xlink:href="note-0078-02"/> gnes BD, BA, & </s> <s xml:id="echoid-s1344" xml:space="preserve">AD; </s> <s xml:id="echoid-s1345" xml:space="preserve">ſi la ligne de direction de la <lb/>puiſſance R eſt parallele au plan GH, & </s> <s xml:id="echoid-s1346" xml:space="preserve">AC celle de <lb/>ce poids parallele auſſi à HK hauteur de ce plan: </s> <s xml:id="echoid-s1347" xml:space="preserve">ce <lb/>poids, cette puiſſance, & </s> <s xml:id="echoid-s1348" xml:space="preserve">la charge de ce même plan, <lb/>ſeront entr’eux comme la longueur de ce plan, ſa hau-<lb/>teur, & </s> <s xml:id="echoid-s1349" xml:space="preserve">ſa baſe; </s> <s xml:id="echoid-s1350" xml:space="preserve">c’eſt-à-dire, comme GH, HK, & </s> <s xml:id="echoid-s1351" xml:space="preserve"><lb/>KG; </s> <s xml:id="echoid-s1352" xml:space="preserve">parce qu’alors les triangles GHK, & </s> <s xml:id="echoid-s1353" xml:space="preserve">DBA ſont <lb/>ſemblables.</s> <s xml:id="echoid-s1354" xml:space="preserve"/> </p> <div xml:id="echoid-div149" type="float" level="2" n="1"> <note position="left" xlink:label="note-0078-02" xlink:href="note-0078-02a" xml:space="preserve">fig. 30.</note> </div> </div> <div xml:id="echoid-div151" type="section" level="1" n="101"> <head xml:id="echoid-head101" xml:space="preserve"><emph style="sc">Corollaire</emph> XX.</head> <p> <s xml:id="echoid-s1355" xml:space="preserve">Et pour la même raiſon, ſi cette ligne de direction <lb/> <anchor type="note" xlink:label="note-0078-03a" xlink:href="note-0078-03"/> AB eſt parallele à l’horizon; </s> <s xml:id="echoid-s1356" xml:space="preserve">c’eſt-à-dire, à la baze <lb/>GK de ce plan; </s> <s xml:id="echoid-s1357" xml:space="preserve">& </s> <s xml:id="echoid-s1358" xml:space="preserve">que AC ſoit encore parallele à ſa <lb/>hauteur HK: </s> <s xml:id="echoid-s1359" xml:space="preserve">Ce poids, cette puiſſance, & </s> <s xml:id="echoid-s1360" xml:space="preserve">la charge <lb/>de ce plan, ſeront alors entr’eux comme la baze de <lb/>de ce plan, ſa hauteur, & </s> <s xml:id="echoid-s1361" xml:space="preserve">ſa longueur; </s> <s xml:id="echoid-s1362" xml:space="preserve">c’eſt-à-dire, <lb/>comme GK, KH, HG; </s> <s xml:id="echoid-s1363" xml:space="preserve">parce qu’alors les triangles <lb/>GKH, & </s> <s xml:id="echoid-s1364" xml:space="preserve">DBA ſont encore ſemblables.</s> <s xml:id="echoid-s1365" xml:space="preserve"/> </p> <div xml:id="echoid-div151" type="float" level="2" n="1"> <note position="left" xlink:label="note-0078-03" xlink:href="note-0078-03a" xml:space="preserve">fig. 33.</note> </div> <p style="it"> <s xml:id="echoid-s1366" xml:space="preserve">Fuſqu’ici nous n’avons regardé le même poids que comme <lb/>appliqué au même endroit d’un plan toujours également incli-<lb/>né: </s> <s xml:id="echoid-s1367" xml:space="preserve">Mais s’il ſe trouvoit ſucceſſivement en différens points, <lb/>qu’arriveroit-il? </s> <s xml:id="echoid-s1368" xml:space="preserve">Le voici.</s> <s xml:id="echoid-s1369" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div153" type="section" level="1" n="102"> <head xml:id="echoid-head102" xml:space="preserve"><emph style="sc">Corollaire</emph> XXI.</head> <p> <s xml:id="echoid-s1370" xml:space="preserve">Il ſuit encore de cette propoſition que les puiſſances <lb/> <anchor type="note" xlink:label="note-0078-04a" xlink:href="note-0078-04"/> P & </s> <s xml:id="echoid-s1371" xml:space="preserve">R qui ſoutiennent ſucceſſivement le même poids <lb/>A, ou des poids égaux ſur les points O, & </s> <s xml:id="echoid-s1372" xml:space="preserve">Q d’un <lb/>même plan HG, ſont entr’elles en raiſon compoſée <lb/>de celles des ſinus des angles QAD, & </s> <s xml:id="echoid-s1373" xml:space="preserve">QAP; </s> <s xml:id="echoid-s1374" xml:space="preserve">OAR <lb/>& </s> <s xml:id="echoid-s1375" xml:space="preserve">OAD: </s> <s xml:id="echoid-s1376" xml:space="preserve">car puiſque la puiſſance P eſt au poids A <pb o="53" file="0079" n="79" rhead="MECHANIQUE"/> appliqué en Q, comme le ſinus de l’angle QAD, au <lb/> <anchor type="note" xlink:label="note-0079-01a" xlink:href="note-0079-01"/> ſinus de l’angle QAP; </s> <s xml:id="echoid-s1377" xml:space="preserve">& </s> <s xml:id="echoid-s1378" xml:space="preserve">quele même poids appliqué <lb/>en O, eſt à la puiſſance R, comme le ſinus de l’angle <lb/>OAR, à celui de l’angle OAD: </s> <s xml:id="echoid-s1379" xml:space="preserve">Il ſuit en multipliant <lb/>par ordre ces deux rangées de proportionnelles; </s> <s xml:id="echoid-s1380" xml:space="preserve">que <lb/>la puiſſance P eſt à la puiſſance R, comme le produit <lb/>des finus des angles QAD, & </s> <s xml:id="echoid-s1381" xml:space="preserve">OAR, au produit de <lb/>ceux des angles QAP, & </s> <s xml:id="echoid-s1382" xml:space="preserve">OAD; </s> <s xml:id="echoid-s1383" xml:space="preserve">c’eſt-à-dire, en <lb/>raiſon compoſée de celles des ſinus des angles QAD, <lb/>& </s> <s xml:id="echoid-s1384" xml:space="preserve">QAP; </s> <s xml:id="echoid-s1385" xml:space="preserve">OAR, & </s> <s xml:id="echoid-s1386" xml:space="preserve">OAD.</s> <s xml:id="echoid-s1387" xml:space="preserve"/> </p> <div xml:id="echoid-div153" type="float" level="2" n="1"> <note position="left" xlink:label="note-0078-04" xlink:href="note-0078-04a" xml:space="preserve">fig. 37. <lb/>38.</note> <note position="right" xlink:label="note-0079-01" xlink:href="note-0079-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus ſur <lb/>des ſurfaces.</note> </div> </div> <div xml:id="echoid-div155" type="section" level="1" n="103"> <head xml:id="echoid-head103" xml:space="preserve"><emph style="sc">Corollaire</emph> XXII.</head> <p> <s xml:id="echoid-s1388" xml:space="preserve">D’où l’on voit que tant que les lignes de direction <lb/>AP, & </s> <s xml:id="echoid-s1389" xml:space="preserve">AR ſont la même, ou paralleles entr’elles, <lb/>& </s> <s xml:id="echoid-s1390" xml:space="preserve">que celles du poids A appliqué en Q, & </s> <s xml:id="echoid-s1391" xml:space="preserve">en O, ſont <lb/>auſſi paralleles; </s> <s xml:id="echoid-s1392" xml:space="preserve">les angles QAP, & </s> <s xml:id="echoid-s1393" xml:space="preserve">OAR étant <lb/>alors égaux entr’eux, de même que les angles QAD, <lb/>& </s> <s xml:id="echoid-s1394" xml:space="preserve">OAD, les puiſſances P & </s> <s xml:id="echoid-s1395" xml:space="preserve">R ſont auſſi pour lors <lb/>égales; </s> <s xml:id="echoid-s1396" xml:space="preserve">ce qui fait voir que ce même poids A péſe <lb/>alors également ſur quelque point O, ou Q de ce <lb/>plan, qu’il ſoit appliqué.</s> <s xml:id="echoid-s1397" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div156" type="section" level="1" n="104"> <head xml:id="echoid-head104" xml:space="preserve"><emph style="sc">Corollaire</emph> XXIII.</head> <p> <s xml:id="echoid-s1398" xml:space="preserve">Au contraire, s’il n’y a que les lignes de de direc-<lb/>tion AP, & </s> <s xml:id="echoid-s1399" xml:space="preserve">AR, de ces puiſſances qui ſoient paral-<lb/>leles entr’elles, & </s> <s xml:id="echoid-s1400" xml:space="preserve">que celles de ce même poids A ap-<lb/>pliqué ſucceſſivement en O, & </s> <s xml:id="echoid-s1401" xml:space="preserve">en Q, concourent en <lb/>quelque point D, que ce ſoit; </s> <s xml:id="echoid-s1402" xml:space="preserve">par exemple, au cen-<lb/>tre de la terre: </s> <s xml:id="echoid-s1403" xml:space="preserve">les angles QAP, & </s> <s xml:id="echoid-s1404" xml:space="preserve">OAR étant en-<lb/>core égaux, les puiſſances P & </s> <s xml:id="echoid-s1405" xml:space="preserve">R ſeront entr’elles <lb/>comme les ſinus des angles QAD, & </s> <s xml:id="echoid-s1406" xml:space="preserve">OAD; </s> <s xml:id="echoid-s1407" xml:space="preserve">ou bien, <lb/>à cauſe des paralleles QA, AO, ces puiſſances ſeront <lb/>entr’elles comme les ſinus des angles SAD, & </s> <s xml:id="echoid-s1408" xml:space="preserve">ASA, <lb/>ou ASD ſon complement; </s> <s xml:id="echoid-s1409" xml:space="preserve">c’eſt-à-dire, (Lem. </s> <s xml:id="echoid-s1410" xml:space="preserve">5.) </s> <s xml:id="echoid-s1411" xml:space="preserve">com-<lb/>me DS à AD. </s> <s xml:id="echoid-s1412" xml:space="preserve">Ce qui fait voir qu’en ce cas plus le poids</s> </p> <pb o="54" file="0080" n="80" rhead="NOUVELLE"/> <note position="left" xml:space="preserve">DES POIDS <lb/>ſoutenus ſur <lb/>des ſurfaces.</note> <p> <s xml:id="echoid-s1413" xml:space="preserve">A eſt haut ſur le plan HG, plus auſſi la puiſſance qui l’y <lb/>doit ſoutenir ſuivant une certaine direction, doit être <lb/>grande.</s> <s xml:id="echoid-s1414" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s1415" xml:space="preserve">On voit préſentement qu’il peut y avoir bien de la diffé-<lb/>rence entre un poids ſoutenu ſur un même plan, & </s> <s xml:id="echoid-s1416" xml:space="preserve">un poids <lb/>ſoutenu ſur le même point du plan; </s> <s xml:id="echoid-s1417" xml:space="preserve">c’eſt auſſi pour cela qu’on <lb/>à pris ſoin ci-deſſus de ne les pas confondre, & </s> <s xml:id="echoid-s1418" xml:space="preserve">de faire re-<lb/>marquer cette différence dans la ſeconde réfléxion qui ſuit le <lb/>Corollaire 13.</s> <s xml:id="echoid-s1419" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div157" type="section" level="1" n="105"> <head xml:id="echoid-head105" xml:space="preserve"><emph style="sc">Corollaire</emph> XXIV.</head> <p> <s xml:id="echoid-s1420" xml:space="preserve">Mais s’il n’y a que les lignes de direction du poids <lb/>A placé tantôt en O, & </s> <s xml:id="echoid-s1421" xml:space="preserve">tantôt en Q, qui ſoient pa-<lb/>ralleles entr’elles; </s> <s xml:id="echoid-s1422" xml:space="preserve">les puiſſances P & </s> <s xml:id="echoid-s1423" xml:space="preserve">R ſeront alors <lb/>entr’elles comme les ſinus des angles OAR, & </s> <s xml:id="echoid-s1424" xml:space="preserve">QAP; <lb/></s> <s xml:id="echoid-s1425" xml:space="preserve">c’eſt-à-dire, en raiſon réciproque des ſinus des angles <lb/>que font leurs lignes de direction avec AO, & </s> <s xml:id="echoid-s1426" xml:space="preserve">AQ, <lb/>tirées des points A ou ces lignes de direction con-<lb/>courent avec celles de ce poids, perpendiculairement <lb/>au plan GH.</s> <s xml:id="echoid-s1427" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div158" type="section" level="1" n="106"> <head xml:id="echoid-head106" xml:space="preserve"><emph style="sc">Corollaire</emph> XXV.</head> <p> <s xml:id="echoid-s1428" xml:space="preserve">Si préſentement on conçoit que les puiſſances P & </s> <s xml:id="echoid-s1429" xml:space="preserve">R <lb/>ſoient égales, & </s> <s xml:id="echoid-s1430" xml:space="preserve">que les poids A & </s> <s xml:id="echoid-s1431" xml:space="preserve">A ſoient différens; <lb/></s> <s xml:id="echoid-s1432" xml:space="preserve">on trouvera de méme que ces poids, qu’elles ſoutien-<lb/>nent ſur les points Q & </s> <s xml:id="echoid-s1433" xml:space="preserve">O du plan GH, ſeront entr’-<lb/>eux en raiſon compoſée de celles des ſinus des angles <lb/>QAP, & </s> <s xml:id="echoid-s1434" xml:space="preserve">QAD; </s> <s xml:id="echoid-s1435" xml:space="preserve">OAD, & </s> <s xml:id="echoid-s1436" xml:space="preserve">OAR; </s> <s xml:id="echoid-s1437" xml:space="preserve">c’eſt-à-dire, com-<lb/>me le produit des ſinus des angles QAP & </s> <s xml:id="echoid-s1438" xml:space="preserve">OAD, au <lb/>produit de ceux des angles QAD & </s> <s xml:id="echoid-s1439" xml:space="preserve">OAR: </s> <s xml:id="echoid-s1440" xml:space="preserve">De ſorte <lb/>que 1°. </s> <s xml:id="echoid-s1441" xml:space="preserve">Lorſque les lignes de direction de ces puiſ-<lb/>ſances ſont paralleles entr’elles, & </s> <s xml:id="echoid-s1442" xml:space="preserve">celles de ces poids <lb/>paralleles auſſi; </s> <s xml:id="echoid-s1443" xml:space="preserve">ces poids ſont égaux. </s> <s xml:id="echoid-s1444" xml:space="preserve">2°. </s> <s xml:id="echoid-s1445" xml:space="preserve">Mais s’il <lb/>n’y a que celles de ces puiſſances qui le ſoient; </s> <s xml:id="echoid-s1446" xml:space="preserve">ces <pb o="55" file="0081" n="81" rhead="MECHANIQUE."/> poids ſont entr’eux en raiſon réciproque des <lb/> <anchor type="note" xlink:label="note-0081-01a" xlink:href="note-0081-01"/> ſinus des angles que font leurs lignes de direction <lb/>avec les perpendiculaires au plan GH en Q, & </s> <s xml:id="echoid-s1447" xml:space="preserve">en O. <lb/></s> <s xml:id="echoid-s1448" xml:space="preserve">3°. </s> <s xml:id="echoid-s1449" xml:space="preserve">Au contraire s’il n’y a que les lignes de direction <lb/>de ces poids qui ſoient paralleles entr’elles; </s> <s xml:id="echoid-s1450" xml:space="preserve">ils ſeront <lb/>entr’eux en même raiſon que les ſinus des angles que <lb/>font ces mêmes perpendiculaires avec les lignes de di-<lb/>rection des puiſſances qui les ſoutiennent.</s> <s xml:id="echoid-s1451" xml:space="preserve"/> </p> <div xml:id="echoid-div158" type="float" level="2" n="1"> <note position="right" xlink:label="note-0081-01" xlink:href="note-0081-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus ſur <lb/>des ſurfaces.</note> </div> <p style="it"> <s xml:id="echoid-s1452" xml:space="preserve">Il y auroit encore bien des Corollaires à tirer de cette pro-<lb/>poſition non-ſeulement par raport aux plans, mais auſſi par <lb/>raport aux poli-plans: </s> <s xml:id="echoid-s1453" xml:space="preserve">en voilà aſſés pour en pouvoir juger.</s> <s xml:id="echoid-s1454" xml:space="preserve"/> </p> <figure> <image file="0081-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0081-01"/> </figure> </div> <div xml:id="echoid-div160" type="section" level="1" n="107"> <head xml:id="echoid-head107" xml:space="preserve">PROBLEME.</head> <p style="it"> <s xml:id="echoid-s1455" xml:space="preserve">L Apaiſſance R étant donnée, la diſpoſer tellement qu’elle <lb/> <anchor type="note" xlink:label="note-0081-02a" xlink:href="note-0081-02"/> puiſſe ſoutenir le poids EO auſſi donné, ſur le <lb/>plan GHMN incliné, & </s> <s xml:id="echoid-s1456" xml:space="preserve">ſuivant ſa longueur GH, de la <lb/>hauteur HK; </s> <s xml:id="echoid-s1457" xml:space="preserve">& </s> <s xml:id="echoid-s1458" xml:space="preserve">ſuivant ſa largeur GN, de la hauteur <lb/>NL.</s> <s xml:id="echoid-s1459" xml:space="preserve"/> </p> <div xml:id="echoid-div160" type="float" level="2" n="1"> <note position="right" xlink:label="note-0081-02" xlink:href="note-0081-02a" xml:space="preserve">fig. 39. <lb/>40.</note> </div> </div> <div xml:id="echoid-div162" type="section" level="1" n="108"> <head xml:id="echoid-head108" xml:space="preserve"><emph style="sc">Solution.</emph></head> <p> <s xml:id="echoid-s1460" xml:space="preserve">Aiant placé ce poids en quelque point, ou partie <lb/>de ce plan qu’il vous plaira, de quelqu’un des points <lb/>ou ſa baze le touche; </s> <s xml:id="echoid-s1461" xml:space="preserve">par exemple, du point O, tirez <lb/>OD perpendiculaire à ce plan, marquez auſſi la ligne <lb/>de direction FC de ce poids, qui rencontre auſſi ce <lb/>plan en L, & </s> <s xml:id="echoid-s1462" xml:space="preserve">OD prolongée en A. </s> <s xml:id="echoid-s1463" xml:space="preserve">Enſuite de quel-<lb/>que point C qu’il vous plaira de la ligne AC, faite <lb/>AC à CD, comme le poids EO à la puiſſance R. <lb/></s> <s xml:id="echoid-s1464" xml:space="preserve">1°. </s> <s xml:id="echoid-s1465" xml:space="preserve">Si CD n’eſt pas aſſez longue pour atteindre de C <lb/>juſqu’en quelque point D de la ligne AD, ce Pro-<lb/>blême eſt impoſſible: </s> <s xml:id="echoid-s1466" xml:space="preserve">parce que la diagonale tirée du <pb o="56" file="0082" n="82" rhead="NOUVELLE"/> <anchor type="note" xlink:label="note-0082-01a" xlink:href="note-0082-01"/> point A dans le parallelogramme fait ſous ces deux <lb/>lignes; </s> <s xml:id="echoid-s1467" xml:space="preserve">c’eft-à-dire, ſous AC & </s> <s xml:id="echoid-s1468" xml:space="preserve">CD, étant alors dif-<lb/>férente de AD, elle ne ſeroit pas perpendiculaire à <lb/>ce plan: </s> <s xml:id="echoid-s1469" xml:space="preserve">ainſi ce poids rouleroit alors (n. </s> <s xml:id="echoid-s1470" xml:space="preserve">3. </s> <s xml:id="echoid-s1471" xml:space="preserve">Demonſt.) <lb/></s> <s xml:id="echoid-s1472" xml:space="preserve">du côté de L. </s> <s xml:id="echoid-s1473" xml:space="preserve">2°. </s> <s xml:id="echoid-s1474" xml:space="preserve">Mais ſi CD peut atteindre juſqu’en <lb/>quelque point D de la ligne AD, achevez le paral-<lb/>lelogramme BC, & </s> <s xml:id="echoid-s1475" xml:space="preserve">placez la puiſſance R ſuivant <lb/>AB: </s> <s xml:id="echoid-s1476" xml:space="preserve">alors elle ſoutiendra ce poids ſur ce plan.</s> <s xml:id="echoid-s1477" xml:space="preserve"/> </p> <div xml:id="echoid-div162" type="float" level="2" n="1"> <note position="left" xlink:label="note-0082-01" xlink:href="note-0082-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus ſur <lb/>des ſurfaces.</note> </div> </div> <div xml:id="echoid-div164" type="section" level="1" n="109"> <head xml:id="echoid-head109" xml:space="preserve"><emph style="sc">Demonstration.</emph></head> <p> <s xml:id="echoid-s1478" xml:space="preserve">Puiſque (Hyp.) </s> <s xml:id="echoid-s1479" xml:space="preserve">cette puiſſance eſt à ce poids com-<lb/>me CD, ou AB qui lui eſt égale, eſt à AC; </s> <s xml:id="echoid-s1480" xml:space="preserve">leur con-<lb/>cours d’action doit le pouſſer (Lemm. </s> <s xml:id="echoid-s1481" xml:space="preserve">4. </s> <s xml:id="echoid-s1482" xml:space="preserve">Cor. </s> <s xml:id="echoid-s1483" xml:space="preserve">2.) <lb/></s> <s xml:id="echoid-s1484" xml:space="preserve">ſuivant AD perpendiculaire (Hyp.) </s> <s xml:id="echoid-s1485" xml:space="preserve">au plan GM, <lb/>& </s> <s xml:id="echoid-s1486" xml:space="preserve">qui paſſe auſſi (Hyp.) </s> <s xml:id="echoid-s1487" xml:space="preserve">par la baſe de ce poids: </s> <s xml:id="echoid-s1488" xml:space="preserve"><lb/>Donc (Cor. </s> <s xml:id="echoid-s1489" xml:space="preserve">2.) </s> <s xml:id="echoid-s1490" xml:space="preserve">il doit demeurer deſſus en équilibre <lb/>avec cette puiſſance. </s> <s xml:id="echoid-s1491" xml:space="preserve">Ce qu’il F. </s> <s xml:id="echoid-s1492" xml:space="preserve">D.</s> <s xml:id="echoid-s1493" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div165" type="section" level="1" n="110"> <head xml:id="echoid-head110" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s1494" xml:space="preserve">Il eſt clair que ſi la puiſſance R ceſſoit de retenir le <lb/>poids EO, il couleroit le long de OL.</s> <s xml:id="echoid-s1495" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div166" type="section" level="1" n="111"> <head xml:id="echoid-head111" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s1496" xml:space="preserve">Si CD eſt la plus petite ligne qui puiſſe atteindre <lb/>de C juſqu’en AD; </s> <s xml:id="echoid-s1497" xml:space="preserve">c’eſt-à-dire, ſi l’angle ADC eſt <lb/>droit, l’angle BAD le ſera auſſi; </s> <s xml:id="echoid-s1498" xml:space="preserve">& </s> <s xml:id="echoid-s1499" xml:space="preserve">par conſéquent <lb/>cette puiſſance eſt la plus petite (Cor. </s> <s xml:id="echoid-s1500" xml:space="preserve">14.) </s> <s xml:id="echoid-s1501" xml:space="preserve">qui puiſ-<lb/>ſe ſoutenir ce poids ſur ce plan, & </s> <s xml:id="echoid-s1502" xml:space="preserve">elle ne l’y poura <lb/>ſoutenir non plus que (Cor. </s> <s xml:id="echoid-s1503" xml:space="preserve">16.) </s> <s xml:id="echoid-s1504" xml:space="preserve">ſuivant cette ſeule <lb/>direction.</s> <s xml:id="echoid-s1505" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div167" type="section" level="1" n="112"> <head xml:id="echoid-head112" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s1506" xml:space="preserve">Si CD n’eſt pas la plus petite qui puiſſe atteindre <lb/>depuis C juſqu’en AD, mais qu’elle ſoit cependant <lb/>cncore moindre que AC; </s> <s xml:id="echoid-s1507" xml:space="preserve">cette même puiſſance poura <pb o="57" file="0083" n="83" rhead="MECHANIQUE."/> ſoutenir ce poids ſuivant deux directions différentes; <lb/></s> <s xml:id="echoid-s1508" xml:space="preserve"> <anchor type="note" xlink:label="note-0083-01a" xlink:href="note-0083-01"/> parce qu’en ce cas CD poura rencontrer AD en deux <lb/>points également éloignez de la perpendiculaire tirée <lb/>du point C ſur AD. </s> <s xml:id="echoid-s1509" xml:space="preserve">Ce qui revient au Corollaire 16. <lb/></s> <s xml:id="echoid-s1510" xml:space="preserve">de la propoſition précédente.</s> <s xml:id="echoid-s1511" xml:space="preserve"/> </p> <div xml:id="echoid-div167" type="float" level="2" n="1"> <note position="right" xlink:label="note-0083-01" xlink:href="note-0083-01a" xml:space="preserve">DESPOIDS <lb/>ſoutenus ſur <lb/>des ſurfaces.</note> </div> </div> <div xml:id="echoid-div169" type="section" level="1" n="113"> <head xml:id="echoid-head113" xml:space="preserve"><emph style="sc">Corollaire</emph> IV.</head> <p> <s xml:id="echoid-s1512" xml:space="preserve">Enfin ſi CD eſt plus grande que AC, elle ne poura <lb/>encore ſoutenir ce poids que ſuivant cette ſeule di-<lb/>rection. </s> <s xml:id="echoid-s1513" xml:space="preserve">Ce qui revient auſſi au Corollaire 18.</s> <s xml:id="echoid-s1514" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div170" type="section" level="1" n="114"> <head xml:id="echoid-head114" xml:space="preserve"><emph style="sc">Corollaire</emph> V.</head> <p> <s xml:id="echoid-s1515" xml:space="preserve">Mais auſſi CD pouvant être infiniment plus gran-<lb/>de que AC, cette puiſſance peut auſſi être infiniment <lb/>plus grande que ce poids, & </s> <s xml:id="echoid-s1516" xml:space="preserve">demeurer cependant tou-<lb/>jours en équilibre avec lui. </s> <s xml:id="echoid-s1517" xml:space="preserve">Ce qui revient encore au <lb/>nombre 2. </s> <s xml:id="echoid-s1518" xml:space="preserve">du Corollaire 11.</s> <s xml:id="echoid-s1519" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s1520" xml:space="preserve">Tout cela ſe peut aiſément appliquer à toutes ſortes de <lb/>ſurfaces courbes: </s> <s xml:id="echoid-s1521" xml:space="preserve">ceux pour qui ce Projet eſt écrit le voyent <lb/>aſſez.</s> <s xml:id="echoid-s1522" xml:space="preserve"/> </p> <figure> <image file="0083-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0083-01"/> </figure> <pb o="58" file="0084" n="84" rhead="NOUVELLE"/> <figure> <image file="0084-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0084-01"/> </figure> </div> <div xml:id="echoid-div171" type="section" level="1" n="115"> <head xml:id="echoid-head115" xml:space="preserve">PROPOSITION <lb/>FONDAMENTALE <lb/>POUR <lb/>TOUTES SOR TES DE LEVIERS,</head> <p> <s xml:id="echoid-s1523" xml:space="preserve">De quelque eſpéce, & </s> <s xml:id="echoid-s1524" xml:space="preserve">dans quelque ſituation <lb/>qu’ils ſoient; </s> <s xml:id="echoid-s1525" xml:space="preserve">& </s> <s xml:id="echoid-s1526" xml:space="preserve">pour toutes les directions <lb/>poſſibles des puiſſances, ou des poids qui y <lb/>ſont appliquez.</s> <s xml:id="echoid-s1527" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s1528" xml:space="preserve">SOIENT les puißances E & </s> <s xml:id="echoid-s1529" xml:space="preserve">F appliquées aux points <lb/> <anchor type="note" xlink:label="note-0084-01a" xlink:href="note-0084-01"/> O & </s> <s xml:id="echoid-s1530" xml:space="preserve">X du Levier MN, de quelque eſpéce, & </s> <s xml:id="echoid-s1531" xml:space="preserve">en <lb/>quelque ſituation qu’il ſoit; </s> <s xml:id="echoid-s1532" xml:space="preserve">quelque angle OAX que faſ-<lb/>ſent auſſi entr’elles, les lignes de direction de ces puiſſances, <lb/>indéfiniment prolongées vers A: </s> <s xml:id="echoid-s1533" xml:space="preserve">Ces deux puißances feroni <lb/>équilibre ſur le point fixe B de ce levier, par ou paſſe la dia-<lb/>gonale AG du parallelogramme RS, dont les côtez AS & </s> <s xml:id="echoid-s1534" xml:space="preserve">AR <lb/>ſont entr’eux, comme les puiſſances E, & </s> <s xml:id="echoid-s1535" xml:space="preserve">F.</s> <s xml:id="echoid-s1536" xml:space="preserve"/> </p> <div xml:id="echoid-div171" type="float" level="2" n="1"> <note position="left" xlink:label="note-0084-01" xlink:href="note-0084-01a" xml:space="preserve">fig 41. <lb/>42. <lb/>43. <lb/>44. <lb/>45. <lb/>46.</note> </div> </div> <div xml:id="echoid-div173" type="section" level="1" n="116"> <head xml:id="echoid-head116" xml:space="preserve"><emph style="sc">Demonstration.</emph></head> <p> <s xml:id="echoid-s1537" xml:space="preserve">Concevons pour un moment que MAN eſt la <lb/>figure de ce Levier, & </s> <s xml:id="echoid-s1538" xml:space="preserve">qu’au lieu d’être tiré, ou <lb/>pouſſé par les puiſſances E & </s> <s xml:id="echoid-s1539" xml:space="preserve">F, ſon point A ſoit ſeu- <pb o="59" file="0085" n="85" rhead="MECHANIQUE."/> lement pouſſé, mais en même-tems, vers R ſuivant <lb/> <anchor type="note" xlink:label="note-0085-01a" xlink:href="note-0085-01"/> AR par une puiſſance égale à F, & </s> <s xml:id="echoid-s1540" xml:space="preserve">vers S ſuivant AS <lb/>par une autre puiſſance égale à E. </s> <s xml:id="echoid-s1541" xml:space="preserve">Il eſt conſtant <lb/>(Lemm. </s> <s xml:id="echoid-s1542" xml:space="preserve">3.) </s> <s xml:id="echoid-s1543" xml:space="preserve">que l’impreſſion compoſée que ce point <lb/>recevroit alors du concours d’action de ces deux <lb/>puiſſances, s’il étoit ainſi pouſſé, ne tendroit qu’à le <lb/>mouvoir ſuivant la diagonale AG du parallelogram-<lb/>me RS; </s> <s xml:id="echoid-s1544" xml:space="preserve">ainſi ne faiſant d’impreſſion ſur tout le corps <lb/>MAN, qu’autant que le point A, à raiſon de l’ob-<lb/>ſtacle qu’il leur fait, lui en communique, toute l’ac-<lb/>tion de ces deux puiſſances ſur ce corps, ſe trouveroit <lb/>réünie dans la ſeule ligne AG, ou AB; </s> <s xml:id="echoid-s1545" xml:space="preserve">ſi donc il <lb/>ſe trouvoit quelque point fixe dans cette ligne, par <lb/>exemple B, ce point ſoutiendroit, lui ſeul, toute <lb/>l’impreſſion de ces deux puiſſances ſur ce corps, ou <lb/>ce Levier; </s> <s xml:id="echoid-s1546" xml:space="preserve">& </s> <s xml:id="echoid-s1547" xml:space="preserve">par conſéquent ce corps n’en ayant <lb/>alors que ce qu’elles lui en communiquent, demeu-<lb/>reroit ainſi en équilibre ſur ce point, ſans qu’aucune <lb/>d’elles l’emportât ſur l’autre. </s> <s xml:id="echoid-s1548" xml:space="preserve">Oſtons préſentement <lb/>ces deux puiſſances, & </s> <s xml:id="echoid-s1549" xml:space="preserve">remettons les deux premiéres <lb/>E, & </s> <s xml:id="echoid-s1550" xml:space="preserve">F en action comme auparavant, prenant encore <lb/>MAN pour la figure du Levier, ou elles ſont appli-<lb/>quées. </s> <s xml:id="echoid-s1551" xml:space="preserve">Il eſt clair que ces deux puiſſances faiſant <lb/>toute la même impreſſion ſur le point A de ce Levier, <lb/>& </s> <s xml:id="echoid-s1552" xml:space="preserve">ſuivant la même direction AG, que faiſoient <lb/>auparavant celles que nous leur avions ſuppoſées <lb/>égales; </s> <s xml:id="echoid-s1553" xml:space="preserve">leur action doit ſe trouver de même que <lb/>celle de ces deux ſuppoſées, toute réünie dans la <lb/>ſeule ligne AB: </s> <s xml:id="echoid-s1554" xml:space="preserve">& </s> <s xml:id="echoid-s1555" xml:space="preserve">par conſéquent le point fixe B la <lb/>ſoutenant, lui ſeul, toute entiére, elles doivent encore <lb/>demeurer en équilibre ſur ce point. </s> <s xml:id="echoid-s1556" xml:space="preserve">Et ſi enfin on <lb/>retranche la partie MAN du corps AMN, en ſorte <lb/>qu’il n’en reſte plus que le Levier MN: </s> <s xml:id="echoid-s1557" xml:space="preserve">Il eſt encore <lb/>clair que les puiſſances E & </s> <s xml:id="echoid-s1558" xml:space="preserve">F agiſſant encore ſur ce <lb/>levier de même que lors qu’il étoit joint à la partie <pb o="60" file="0086" n="86" rhead="NOUVELLE"/> <anchor type="note" xlink:label="note-0086-01a" xlink:href="note-0086-01"/> retranchée, toute leur action doit encore ſe trouver <lb/>réünie dans le ſeul point B de ce levier, par ou paſſe <lb/>la diagonale AG du parallelogramme RS; </s> <s xml:id="echoid-s1559" xml:space="preserve">& </s> <s xml:id="echoid-s1560" xml:space="preserve">par <lb/>conſéquent, elles doivent encore demeurer en équi-<lb/>libre ſur ce point. </s> <s xml:id="echoid-s1561" xml:space="preserve">Ce qu’il faloit démontrer.</s> <s xml:id="echoid-s1562" xml:space="preserve"/> </p> <div xml:id="echoid-div173" type="float" level="2" n="1"> <note position="right" xlink:label="note-0085-01" xlink:href="note-0085-01a" xml:space="preserve">DES <lb/>LEVIERS.</note> <note position="left" xlink:label="note-0086-01" xlink:href="note-0086-01a" xml:space="preserve">DES <lb/>LEVIERS.</note> </div> </div> <div xml:id="echoid-div175" type="section" level="1" n="117"> <head xml:id="echoid-head117" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s1563" xml:space="preserve">On voit aſſez qu’en quelque ſituation, de quel-<lb/>que eſpéce, & </s> <s xml:id="echoid-s1564" xml:space="preserve">de quelque figure, que ſoit le levier <lb/>auquel deux puiſſances, telles qu’elles ſoient, ſont <lb/>appliquées, elles ſeront toujours en équilibre, quelque <lb/>angle que faſſent entr’elles leurs lignes de direction, <lb/>tant qu’il y aura un point fixe de ce levier dans la <lb/>diagonale AG; </s> <s xml:id="echoid-s1565" xml:space="preserve">c’eſt pourquoi on ne s’arrête point <lb/>ici à exprimer les figures detous les cas de cette pro-<lb/>poſition, chacun le pourra faire à ſon loiſir.</s> <s xml:id="echoid-s1566" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div176" type="section" level="1" n="118"> <head xml:id="echoid-head118" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s1567" xml:space="preserve">On voit encore que les deux mêmes puiſſances E <lb/>& </s> <s xml:id="echoid-s1568" xml:space="preserve">F peuvent faire ſucceſſivement équilibre ſur une <lb/>infinité de points B de ce même levier en changeant <lb/>ſeulement leurs directions: </s> <s xml:id="echoid-s1569" xml:space="preserve">puis qu’on les peut varier <lb/>en tant de maniéres que AG paſſera ſucceſſivement <lb/>par tous les points imaginables de ce levier, excepté <lb/>par les points O & </s> <s xml:id="echoid-s1570" xml:space="preserve">X, où ces deux puiſſances ſont <lb/>appliquées.</s> <s xml:id="echoid-s1571" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div177" type="section" level="1" n="119"> <head xml:id="echoid-head119" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s1572" xml:space="preserve">Ce qui fait voir que dans la ſuppoſition du concours <lb/>des lignes de direction des poids au centre de la terre, <lb/>leurs centres de gravité, ou de direction peuvent chan-<lb/>ger inceſſamment à meſure qu’ils s’en approchent, ou <lb/>qu’ils s’en éloignent, ſelon la différente ſituation qu’ils <lb/>peuvent avoir par raport à lui, excepté dans les ſphéri-<lb/>ques. </s> <s xml:id="echoid-s1573" xml:space="preserve">On nes’arrête point à démontrer cela ſur des fi- <pb o="61" file="0087" n="87" rhead="MECHANIQUE."/> gures particuliéres, parce qu’il ſuit ſi naturellement <lb/> <anchor type="note" xlink:label="note-0087-01a" xlink:href="note-0087-01"/> du Corollaire 2. </s> <s xml:id="echoid-s1574" xml:space="preserve">qu’il n’y a perſonne qui ne le puiſſe <lb/>faire de ſoi-même.</s> <s xml:id="echoid-s1575" xml:space="preserve"/> </p> <div xml:id="echoid-div177" type="float" level="2" n="1"> <note position="right" xlink:label="note-0087-01" xlink:href="note-0087-01a" xml:space="preserve">DES <lb/>LEVIERS.</note> </div> </div> <div xml:id="echoid-div179" type="section" level="1" n="120"> <head xml:id="echoid-head120" xml:space="preserve"><emph style="sc">Corollaire</emph> IV.</head> <p> <s xml:id="echoid-s1576" xml:space="preserve">On voit de plus de cette propoſition que le point fixe <lb/>de ce levier, où bien ſon appui B eſt pouſſé de A vers <lb/>G ſuivant AG par l’impreſſion compoſée qu’il reçoit <lb/>du concours d’action de ces deux puiſſances, & </s> <s xml:id="echoid-s1577" xml:space="preserve">que <lb/>la réſiſtance qu’il leur fait, étant égale à cette même <lb/>impreſſion, eſt à la force de chacune d’elles (Lemm. <lb/></s> <s xml:id="echoid-s1578" xml:space="preserve">3. </s> <s xml:id="echoid-s1579" xml:space="preserve">Cor. </s> <s xml:id="echoid-s1580" xml:space="preserve">3.) </s> <s xml:id="echoid-s1581" xml:space="preserve">Comme la diagonale AG à chacun des <lb/>côtez du parallelogramme RS, qui les répréſen-<lb/>tent; </s> <s xml:id="echoid-s1582" xml:space="preserve">c’eſt-à-dire, que cette réſiſtance du point fixe, <lb/>où bien de l’appui B de ce levier, eſt à la force de la <lb/>puiſſance E, comme AG à AS; </s> <s xml:id="echoid-s1583" xml:space="preserve">& </s> <s xml:id="echoid-s1584" xml:space="preserve">à celle de la puiſ-<lb/>ſance F, comme la même AG à AR.</s> <s xml:id="echoid-s1585" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div180" type="section" level="1" n="121"> <head xml:id="echoid-head121" xml:space="preserve"><emph style="sc">Corollaire</emph> V.</head> <p> <s xml:id="echoid-s1586" xml:space="preserve">D’où il ſuit que lors que l’angle OAX eſt infini-<lb/>ment aigu, c’eſt-à-dire, lors que les lignes de direc-<lb/>tion EX & </s> <s xml:id="echoid-s1587" xml:space="preserve">OF des puiſſances E & </s> <s xml:id="echoid-s1588" xml:space="preserve">F ſont paralleles, <lb/>le point A étant alors infiniment éloigné de B, la <lb/>ligne AG, c’eſt-à-dire, (Cor. </s> <s xml:id="echoid-s1589" xml:space="preserve">4.) </s> <s xml:id="echoid-s1590" xml:space="preserve">la direction de <lb/>l’appui B leur eſt auſſi parallele, & </s> <s xml:id="echoid-s1591" xml:space="preserve">toujours vers <lb/>l’endroit, où tendent ces deux puiſſances, lors qu’elles <lb/>tirent du même côté; </s> <s xml:id="echoid-s1592" xml:space="preserve">ou bien vers celui, ou tend la <lb/>plus proche de cet appui, lors qu’elles tirent de diffé-<lb/>rens cótez.</s> <s xml:id="echoid-s1593" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div181" type="section" level="1" n="122"> <head xml:id="echoid-head122" xml:space="preserve"><emph style="sc">Corollaire</emph> VI.</head> <p> <s xml:id="echoid-s1594" xml:space="preserve">Il ſuit encore du Corollaire 4. </s> <s xml:id="echoid-s1595" xml:space="preserve">pour tous les leviers <lb/> <anchor type="note" xlink:label="note-0087-02a" xlink:href="note-0087-02"/> de l’eſpéce exprimée dans les figures 41. </s> <s xml:id="echoid-s1596" xml:space="preserve">& </s> <s xml:id="echoid-s1597" xml:space="preserve">42. </s> <s xml:id="echoid-s1598" xml:space="preserve">que <lb/>plus l’angle OAX, que les lignes de direction des <lb/>puiſſances E & </s> <s xml:id="echoid-s1599" xml:space="preserve">F, font entr’elles, ſera obtus, moins <pb o="62" file="0088" n="88" rhead="NOUVELLE"/> <anchor type="note" xlink:label="note-0088-01a" xlink:href="note-0088-01"/> ſera grande la charge de l’appui B de chacun de ces le-<lb/>viers; </s> <s xml:id="echoid-s1600" xml:space="preserve">parce que plus cet angle ſera obtus, moins ſera <lb/>grande la raiſon de AG aux côtez du parallelogramme <lb/>RS, dont elle eſt diagonale, quoi qu’en proportion <lb/>différente: </s> <s xml:id="echoid-s1601" xml:space="preserve">De ſorte que l’on peut faire cet angle <lb/>ſi obtus que la charge de l’appui B de ce levier ſera <lb/>ſi petite qu’on voudra, quoi que les deux mêmes <lb/>puiſſances demeurent toujours en équilibre deſſus; <lb/></s> <s xml:id="echoid-s1602" xml:space="preserve">juſque-là même qu’elle peut devenir indéfiniment <lb/>petite, c’eſt-à-dire, moindre que quelque poids don-<lb/>né que ce ſoit. </s> <s xml:id="echoid-s1603" xml:space="preserve">Il ne faut pour cela qu’ouvrir l’angle <lb/>OAX juſqu’à ce que la diagonale AG du paralle-<lb/>logramme RS, ſoit à chacun de ſes côtez AR & </s> <s xml:id="echoid-s1604" xml:space="preserve"><lb/>AS en moindre raiſon que celle qui eſt entre ce poids <lb/>donné, & </s> <s xml:id="echoid-s1605" xml:space="preserve">chacune des puiſſances F & </s> <s xml:id="echoid-s1606" xml:space="preserve">E que l’on ſup-<lb/>poſe appliquées à ce levier.</s> <s xml:id="echoid-s1607" xml:space="preserve"/> </p> <div xml:id="echoid-div181" type="float" level="2" n="1"> <note position="right" xlink:label="note-0087-02" xlink:href="note-0087-02a" xml:space="preserve">fig. 41. <lb/>42.</note> <note position="left" xlink:label="note-0088-01" xlink:href="note-0088-01a" xml:space="preserve">DES <lb/>LEVIERS.</note> </div> </div> <div xml:id="echoid-div183" type="section" level="1" n="123"> <head xml:id="echoid-head123" xml:space="preserve"><emph style="sc">Corollaire</emph> VII.</head> <p> <s xml:id="echoid-s1608" xml:space="preserve">La charge de ce point ne peut pas de même <lb/>augmenter à l’infini: </s> <s xml:id="echoid-s1609" xml:space="preserve">Car ne pouvant jamais être <lb/>plus grande que lors que l’angle OAX eſt infini-<lb/>ment aigu, c’eſt-à-dire, lors que les lignes de direc-<lb/>tion de ces deux puiſſances ſont paralleles; </s> <s xml:id="echoid-s1610" xml:space="preserve">& </s> <s xml:id="echoid-s1611" xml:space="preserve">la <lb/>diagonale AG n’étant encore alors qu’égale à la <lb/>ſomme des côtez AR & </s> <s xml:id="echoid-s1612" xml:space="preserve">RG du parallelogramme <lb/>RS alors infiniment long; </s> <s xml:id="echoid-s1613" xml:space="preserve">la charge de cet appui ne <lb/>peut être, tout au plus, qu’égale à la ſomme des puiſ-<lb/>ſances E & </s> <s xml:id="echoid-s1614" xml:space="preserve">F.</s> <s xml:id="echoid-s1615" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div184" type="section" level="1" n="124"> <head xml:id="echoid-head124" xml:space="preserve"><emph style="sc">Corollaire</emph> VIII.</head> <note position="left" xml:space="preserve">fig. 43. <lb/>44. <lb/>45. <lb/>46.</note> <p> <s xml:id="echoid-s1616" xml:space="preserve">Pour les leviers des eſpeces exprimées dans les <lb/>figures 43. </s> <s xml:id="echoid-s1617" xml:space="preserve">44. </s> <s xml:id="echoid-s1618" xml:space="preserve">45. </s> <s xml:id="echoid-s1619" xml:space="preserve">& </s> <s xml:id="echoid-s1620" xml:space="preserve">46. </s> <s xml:id="echoid-s1621" xml:space="preserve">Il en va tout autrement: <lb/></s> <s xml:id="echoid-s1622" xml:space="preserve">car plus l’angle OAX eſt obtus, plus la charge, <lb/>ou la réſiſtance de l’appui B eſt grande; </s> <s xml:id="echoid-s1623" xml:space="preserve">parce que <lb/>la raiſon de la diagonale AG à chacun des côtez du <pb o="63" file="0089" n="89" rhead="MECHANIQUE."/> parallelogramme RS, en eſt plus grande, quoi qu’en <lb/> <anchor type="note" xlink:label="note-0089-01a" xlink:href="note-0089-01"/> proportion différente; </s> <s xml:id="echoid-s1624" xml:space="preserve">mais elle ne peut pas pour <lb/>cela augmenter à l’infini: </s> <s xml:id="echoid-s1625" xml:space="preserve">car ne pouvant jamais être <lb/>plus grande que lors que cet angle eſt infiniment <lb/>obtus, c’eſt-à-dire, lors que les lignes de direction <lb/>de ces puiſſances concourent en une ſeule ligne <lb/>droite; </s> <s xml:id="echoid-s1626" xml:space="preserve">& </s> <s xml:id="echoid-s1627" xml:space="preserve">la diagonale AG n’étant alors qu’égale à <lb/>la ſomme des côtez AR & </s> <s xml:id="echoid-s1628" xml:space="preserve">GR du parallelogramme <lb/>RS alors encore infiniment long; </s> <s xml:id="echoid-s1629" xml:space="preserve">la réſiſtance de cet <lb/>appui ne peut par conféquent être encore, toutau plus, <lb/>qu’égale à la ſomme des forces de ces deux puiſſances.</s> <s xml:id="echoid-s1630" xml:space="preserve"/> </p> <div xml:id="echoid-div184" type="float" level="2" n="1"> <note position="right" xlink:label="note-0089-01" xlink:href="note-0089-01a" xml:space="preserve">DES <lb/>LEVIERS.</note> </div> </div> <div xml:id="echoid-div186" type="section" level="1" n="125"> <head xml:id="echoid-head125" xml:space="preserve"><emph style="sc">Corollaire</emph> IX.</head> <p> <s xml:id="echoid-s1631" xml:space="preserve">Au contraire, plus cet angle OAX eſt aigu, <lb/>moins eſt grande la charge, ou la réſiſtance de l’ap-<lb/>pui B dece levier: </s> <s xml:id="echoid-s1632" xml:space="preserve">car plus cet angle eſt aigu, moins <lb/>eſt grande la raiſon de la diagonale AG aux côtez <lb/>du parallelogramme RS, quoi qu’en proportion dif-<lb/>férente; </s> <s xml:id="echoid-s1633" xml:space="preserve">mais elle ne peut pas non plus ainſi diminuer <lb/>à l’infini de même que nous venons de dire (Cor. </s> <s xml:id="echoid-s1634" xml:space="preserve">6.) <lb/></s> <s xml:id="echoid-s1635" xml:space="preserve">qu’elle le peut dans les leviers de l’eſpece exprimée <lb/>dans les figures 41. </s> <s xml:id="echoid-s1636" xml:space="preserve">& </s> <s xml:id="echoid-s1637" xml:space="preserve">42. </s> <s xml:id="echoid-s1638" xml:space="preserve">Car ne pouvant jamais être <lb/>moindre que lors que cet angle eſt infiniment aigu, <lb/>c’eſt-à-dire, lors que les lignes de direction de ces <lb/>puiſſances deviennent parallel’es; </s> <s xml:id="echoid-s1639" xml:space="preserve">& </s> <s xml:id="echoid-s1640" xml:space="preserve">le point G, qui <lb/>à meſure que cet angle devient plus aigu, s’approche <lb/>de plus en plus de la ligne AR, (fig. </s> <s xml:id="echoid-s1641" xml:space="preserve">43. </s> <s xml:id="echoid-s1642" xml:space="preserve">& </s> <s xml:id="echoid-s1643" xml:space="preserve">44.) </s> <s xml:id="echoid-s1644" xml:space="preserve"><lb/>ou AS (fig. </s> <s xml:id="echoid-s1645" xml:space="preserve">45. </s> <s xml:id="echoid-s1646" xml:space="preserve">& </s> <s xml:id="echoid-s1647" xml:space="preserve">46.) </s> <s xml:id="echoid-s1648" xml:space="preserve">entrant alors dans cette <lb/>ligne, AG demeure encore égale à la différence de <lb/>AR à AS; </s> <s xml:id="echoid-s1649" xml:space="preserve">& </s> <s xml:id="echoid-s1650" xml:space="preserve">par conſéquent la charge, ou la réſiſ-<lb/>tance d’e l’appui B ne peut jamais être moindre que <lb/>la différence des forces des puiſſances E & </s> <s xml:id="echoid-s1651" xml:space="preserve">F.</s> <s xml:id="echoid-s1652" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div187" type="section" level="1" n="126"> <head xml:id="echoid-head126" xml:space="preserve"><emph style="sc">Corollaire</emph> X.</head> <p> <s xml:id="echoid-s1653" xml:space="preserve">D’où l’on voit en général 1°. </s> <s xml:id="echoid-s1654" xml:space="preserve">que dans toutes ſor- <pb o="64" file="0090" n="90" rhead="NOUVELLE"/> <anchor type="note" xlink:label="note-0090-01a" xlink:href="note-0090-01"/> tes de Leviers, de quelque figure, en quelque ſitua-<lb/>tion, & </s> <s xml:id="echoid-s1655" xml:space="preserve">de quelque eſpéce qu’ils ſoient, quelles que <lb/>ſoient auſſi les lignes de direction de puiſſances, ou <lb/>des poids qui y ſont appliquez, la charge de leur apui <lb/> <anchor type="note" xlink:label="note-0090-02a" xlink:href="note-0090-02"/> ne peut être jamais plus grande (Corol. </s> <s xml:id="echoid-s1656" xml:space="preserve">7. </s> <s xml:id="echoid-s1657" xml:space="preserve">& </s> <s xml:id="echoid-s1658" xml:space="preserve">8.) <lb/></s> <s xml:id="echoid-s1659" xml:space="preserve">que la ſomme de ces poids, ou de ces puiſſances. </s> <s xml:id="echoid-s1660" xml:space="preserve"><lb/>2°. </s> <s xml:id="echoid-s1661" xml:space="preserve">que dans les leviers de l’eſpéce exprimée par les <lb/>figures 41. </s> <s xml:id="echoid-s1662" xml:space="preserve">& </s> <s xml:id="echoid-s1663" xml:space="preserve">42. </s> <s xml:id="echoid-s1664" xml:space="preserve">elle peut (Cor. </s> <s xml:id="echoid-s1665" xml:space="preserve">6.) </s> <s xml:id="echoid-s1666" xml:space="preserve">diminuer à l’infi-<lb/>ni. </s> <s xml:id="echoid-s1667" xml:space="preserve">3°. </s> <s xml:id="echoid-s1668" xml:space="preserve">Mais que dans toute autre eſpéce (fig. </s> <s xml:id="echoid-s1669" xml:space="preserve">43. </s> <s xml:id="echoid-s1670" xml:space="preserve">44. </s> <s xml:id="echoid-s1671" xml:space="preserve"><lb/>45. </s> <s xml:id="echoid-s1672" xml:space="preserve">& </s> <s xml:id="echoid-s1673" xml:space="preserve">46.) </s> <s xml:id="echoid-s1674" xml:space="preserve">elle ne peut jamais être moindre (Cor. </s> <s xml:id="echoid-s1675" xml:space="preserve">9.) </s> <s xml:id="echoid-s1676" xml:space="preserve"><lb/>que la différence des deux puiſſances, ou des deux <lb/>poids qui y ſont appliquez.</s> <s xml:id="echoid-s1677" xml:space="preserve"/> </p> <div xml:id="echoid-div187" type="float" level="2" n="1"> <note position="left" xlink:label="note-0090-01" xlink:href="note-0090-01a" xml:space="preserve">DES <lb/>LEVIERS.</note> <note position="left" xlink:label="note-0090-02" xlink:href="note-0090-02a" xml:space="preserve">fig. 41. <lb/>42. <lb/>43. <lb/>44. <lb/>45. <lb/>46.</note> </div> <p> <s xml:id="echoid-s1678" xml:space="preserve">Perſonne que je fçache n’avoit encore démontré la charge, <lb/>n’y la direction des points d’apuy des leviers: </s> <s xml:id="echoid-s1679" xml:space="preserve">il ne paroit p@ls <lb/>même qu’il ſoit aiſé de le faire par les principes ordinaires; <lb/></s> <s xml:id="echoid-s1680" xml:space="preserve">ſans cela cependant il y à bien des Problêmes qu’on ne ſçau-<lb/>roit réſoudre. </s> <s xml:id="echoid-s1681" xml:space="preserve">Par exemple, ſans la connoiſſance de la direc-<lb/>tion des apuis, il n’eſt pas poſſible de démontrer qu’elles doi-<lb/>vent être les directions de deux puiſſances, ſoit égales, <lb/>ſoit inégales, pour qu’elles puiſſent faire équilibre ſur <lb/>quelque levier que ce ſoit, dont l’apui eſt une ſphére; </s> <s xml:id="echoid-s1682" xml:space="preserve">n’y <lb/>ſur combien de points de ce levier ainſi apuié, il eſt poſſi-<lb/>ble qu’elles faſſent équilibre en changeant ſeulement <lb/>leurs directions. </s> <s xml:id="echoid-s1683" xml:space="preserve">Il n’eſt pas poſſible non plus ſans la connoiſ-<lb/>ſance & </s> <s xml:id="echoid-s1684" xml:space="preserve">de la direction, & </s> <s xml:id="echoid-s1685" xml:space="preserve">de la charge des apuis des leviers de <lb/>trouver le point d’apui de celui auquel tant de puiſ-<lb/>ſances qu’on voudra, ſoient appliquées, pour toutes <lb/>les directions poſſibles dans leſquelles on les peut ſup-<lb/>poſer; </s> <s xml:id="echoid-s1686" xml:space="preserve">ny deux puiſſances étant données avec leurs <lb/>directions & </s> <s xml:id="echoid-s1687" xml:space="preserve">leurs points d’application à un levier, <lb/>de trouver quelle doit être la direction, & </s> <s xml:id="echoid-s1688" xml:space="preserve">le point <lb/>d’application d’une troiſiéme puiſſance auſſi donnée, <lb/>pour que toutes trois enſemble faſſent équilibre ſur <lb/>quelque point donné, que ce ſoit, de ce levier. </s> <s xml:id="echoid-s1689" xml:space="preserve">Ilfaut <pb o="65" file="0091" n="91" rhead="MECHANIQUE."/> penſer la même choſe de tout autre Problême ſur les leviers <lb/> <anchor type="note" xlink:label="note-0091-01a" xlink:href="note-0091-01"/> dont la ſolution dépend de la détermination de la charge, <lb/>& </s> <s xml:id="echoid-s1690" xml:space="preserve">de la direction de leurs apuis.</s> <s xml:id="echoid-s1691" xml:space="preserve"/> </p> <div xml:id="echoid-div188" type="float" level="2" n="2"> <note position="right" xlink:label="note-0091-01" xlink:href="note-0091-01a" xml:space="preserve">DES <lb/>LEVIERS.</note> </div> </div> <div xml:id="echoid-div190" type="section" level="1" n="127"> <head xml:id="echoid-head127" xml:space="preserve"><emph style="sc">Corollaire</emph> XI.</head> <p> <s xml:id="echoid-s1692" xml:space="preserve">Il ſuit encore de cette propoſition que les puiſſan-<lb/>ces E & </s> <s xml:id="echoid-s1693" xml:space="preserve">F, qui demeurent ainſi en équilibre ſur le <lb/>levier MN, ſont en raiſon réciproque des lignes BD <lb/>& </s> <s xml:id="echoid-s1694" xml:space="preserve">BP tirées de ſon point d’apui B perpendiculaire-<lb/>ment ſur leurs lignes de direction, qu’elles quelles <lb/>ſoient: </s> <s xml:id="echoid-s1695" xml:space="preserve">Car puiſque la puiſſance E eſt à la puiſſance <lb/>F, comme AS, ou GR à AR; </s> <s xml:id="echoid-s1696" xml:space="preserve">c’eſt-à-dire, (Lemm. <lb/></s> <s xml:id="echoid-s1697" xml:space="preserve">5.) </s> <s xml:id="echoid-s1698" xml:space="preserve">comme le ſinus de l’angle GAR, ou de OAB ſon <lb/>(fig. </s> <s xml:id="echoid-s1699" xml:space="preserve">41. </s> <s xml:id="echoid-s1700" xml:space="preserve">42. </s> <s xml:id="echoid-s1701" xml:space="preserve">43. </s> <s xml:id="echoid-s1702" xml:space="preserve">& </s> <s xml:id="echoid-s1703" xml:space="preserve">44.) </s> <s xml:id="echoid-s1704" xml:space="preserve">égal, ou (fig. </s> <s xml:id="echoid-s1705" xml:space="preserve">45. </s> <s xml:id="echoid-s1706" xml:space="preserve">& </s> <s xml:id="echoid-s1707" xml:space="preserve">46.) </s> <s xml:id="echoid-s1708" xml:space="preserve"><lb/>ſon complement, eſt au ſinus de l’angle RGA, ou de <lb/>XAB ſon (fig. </s> <s xml:id="echoid-s1709" xml:space="preserve">41. </s> <s xml:id="echoid-s1710" xml:space="preserve">42. </s> <s xml:id="echoid-s1711" xml:space="preserve">43. </s> <s xml:id="echoid-s1712" xml:space="preserve">45. </s> <s xml:id="echoid-s1713" xml:space="preserve">& </s> <s xml:id="echoid-s1714" xml:space="preserve">46.) </s> <s xml:id="echoid-s1715" xml:space="preserve">égal, ou (fig. </s> <s xml:id="echoid-s1716" xml:space="preserve"><lb/>44.) </s> <s xml:id="echoid-s1717" xml:space="preserve">ſon complement; </s> <s xml:id="echoid-s1718" xml:space="preserve">& </s> <s xml:id="echoid-s1719" xml:space="preserve">que d’ailleurs BP eſt le <lb/>ſinus de l’angle OAB, & </s> <s xml:id="echoid-s1720" xml:space="preserve">BD celui de l’angle XAB: </s> <s xml:id="echoid-s1721" xml:space="preserve">Il <lb/>ſuit quela puiſſance E eſt à la puiſſance F, comme BP <lb/>à BD; </s> <s xml:id="echoid-s1722" xml:space="preserve">c’eſt-à-dire, en raiſon réciproque des diſtances <lb/>de leurs lignes de direction, au point d’appui de leur <lb/>levier. </s> <s xml:id="echoid-s1723" xml:space="preserve">Ce qui fait que lors que l’angle OAX eſt <lb/>infiniment aigu, c’eſt-à-dire, lors que les lignes de <lb/>direction de ces deux puiſſances ſont paralleles, & </s> <s xml:id="echoid-s1724" xml:space="preserve"><lb/>que ce levier eſt droit, quelque ſituation qu’il ait, <lb/>ces deux puiſſances ſont toujours entr’elles, tant <lb/>que dure leur équilibre, en raiſon réciproque des <lb/>bras de ce levier pris depuis ſon point d’appui, juſ-<lb/>qu’aux points où elles lui ſont appliquées.</s> <s xml:id="echoid-s1725" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div191" type="section" level="1" n="128"> <head xml:id="echoid-head128" xml:space="preserve"><emph style="sc">Corollaire</emph> XII.</head> <p> <s xml:id="echoid-s1726" xml:space="preserve">L’inverſe de ce dernier Corollaire ſuit encore ma-<lb/>nifeſtement de cette propoſition: </s> <s xml:id="echoid-s1727" xml:space="preserve">ſçavoir que lors <pb o="66" file="0092" n="92" rhead="NOUVELLE"/> que ces deux puiſſances ſont en raiſon réciproque <lb/> <anchor type="note" xlink:label="note-0092-01a" xlink:href="note-0092-01"/> des diſtances de leurs lignes de direction au point <lb/>d’appui de leur levier, de quelque eſpece qu’il ſoit, <lb/>elles demeurent toûjours en équilibre; </s> <s xml:id="echoid-s1728" xml:space="preserve">puis - qu’un tel <lb/>point ne ſe peut trouver ailleurs que dans la diagonale <lb/>AG du parallelogramme RS: </s> <s xml:id="echoid-s1729" xml:space="preserve">ainſi lorſque, leurs lignes <lb/>de direction étant paralleles, & </s> <s xml:id="echoid-s1730" xml:space="preserve">leur levier droit, <lb/>elles ſont en raiſon réciproque des bras de ce levier, <lb/>elles demeurent toujours enéquilibre deſſus.</s> <s xml:id="echoid-s1731" xml:space="preserve"/> </p> <div xml:id="echoid-div191" type="float" level="2" n="1"> <note position="left" xlink:label="note-0092-01" xlink:href="note-0092-01a" xml:space="preserve">DES <lb/>LEVIERS.</note> </div> </div> <div xml:id="echoid-div193" type="section" level="1" n="129"> <head xml:id="echoid-head129" xml:space="preserve"><emph style="sc">Corollaire</emph> XIII.</head> <p style="it"> <s xml:id="echoid-s1732" xml:space="preserve">D’où l’on voit en général que lors que deux puiſſances <lb/>font équilibre ſur un levier, de quelque eſpece, & </s> <s xml:id="echoid-s1733" xml:space="preserve">en quelque <lb/>ſituation qu’il ſoit, elles ſont toujours entr’elles en raiſon <lb/>réciproque des diſtances de leurs lignes de direction au point <lb/>d’appui de ce levier. </s> <s xml:id="echoid-s1734" xml:space="preserve">Et réciproquement auſſi, lors qu’elles <lb/>ſont en raiſon réciproque des diſtances de ces mémes lignes au <lb/>point d’appui de ce levier, de quelque eſpece, de quelque figure, <lb/>& </s> <s xml:id="echoid-s1735" xml:space="preserve">en quelque ſituation qu’il ſoit, elles font toujours équilibre <lb/>deſſus.</s> <s xml:id="echoid-s1736" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s1737" xml:space="preserve">Voila ce qu’on appelle ordinairement le premier principe <lb/>de Méchanique, & </s> <s xml:id="echoid-s1738" xml:space="preserve">ce que perſonne, du moins que je con-<lb/>noiſſe, n’avoit encore démontré de cette maniére, n’y ſi gé-<lb/>néralement. </s> <s xml:id="echoid-s1739" xml:space="preserve">Le voici encore d’une autre façon.</s> <s xml:id="echoid-s1740" xml:space="preserve"/> </p> <figure> <image file="0092-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0092-01"/> </figure> </div> <div xml:id="echoid-div194" type="section" level="1" n="130"> <head xml:id="echoid-head130" xml:space="preserve">LEMME VI.</head> <p style="it"> <s xml:id="echoid-s1741" xml:space="preserve">S I deux puiſſances D & </s> <s xml:id="echoid-s1742" xml:space="preserve">E appliquées au même point A <lb/> <anchor type="note" xlink:label="note-0092-02a" xlink:href="note-0092-02"/> d’un levier AB, dont le point fixe ſoit B, ſont entr’elles <lb/>en raiſon réciproque des ſinus des angles que font leurs lignes <lb/>de direction, qu’elles quelles ſoient, avec ce levier; </s> <s xml:id="echoid-s1743" xml:space="preserve">elles <lb/>demeureront en équilibre.</s> <s xml:id="echoid-s1744" xml:space="preserve"/> </p> <div xml:id="echoid-div194" type="float" level="2" n="1"> <note position="left" xlink:label="note-0092-02" xlink:href="note-0092-02a" xml:space="preserve">fig. 47. <lb/>48. <lb/>49.</note> </div> <pb o="67" file="0093" n="93" rhead="MECHANIQUE."/> </div> <div xml:id="echoid-div196" type="section" level="1" n="131"> <head xml:id="echoid-head131" xml:space="preserve"><emph style="sc">Demonstration.</emph></head> <note position="right" xml:space="preserve">DES <lb/>LEVIERS.</note> <p> <s xml:id="echoid-s1745" xml:space="preserve">Le point A ainſi tiré par les puiſſances D & </s> <s xml:id="echoid-s1746" xml:space="preserve">E, <lb/>doit en recevoir une impreſſion ſuivant quelque <lb/>ligne AG qui ſoit la diagonale d’un parallelogramme <lb/>RS fait ſous des parties de leurs lignes de direction <lb/>AS & </s> <s xml:id="echoid-s1747" xml:space="preserve">AR qui ſoient entr’elles, (Lemm. </s> <s xml:id="echoid-s1748" xml:space="preserve">3.) </s> <s xml:id="echoid-s1749" xml:space="preserve">comme <lb/>ces mêmes puiſſances; </s> <s xml:id="echoid-s1750" xml:space="preserve">& </s> <s xml:id="echoid-s1751" xml:space="preserve">la force de cette impreſſion <lb/>doit être à chacune des puiſſances D & </s> <s xml:id="echoid-s1752" xml:space="preserve">E, comme <lb/>(Lem. </s> <s xml:id="echoid-s1753" xml:space="preserve">3. </s> <s xml:id="echoid-s1754" xml:space="preserve">Cor. </s> <s xml:id="echoid-s1755" xml:space="preserve">3.) </s> <s xml:id="echoid-s1756" xml:space="preserve">AG à chacune des lignes AS & </s> <s xml:id="echoid-s1757" xml:space="preserve">AR, <lb/>ou SG qui lui eſt égale; </s> <s xml:id="echoid-s1758" xml:space="preserve">c’eſt-à-dire, (Lem. </s> <s xml:id="echoid-s1759" xml:space="preserve">5.) </s> <s xml:id="echoid-s1760" xml:space="preserve">comme <lb/>le ſinus de l’angle ASG, à chacun des ſinus des angles <lb/>AGS, ou GAR qui lui eſt égal, & </s> <s xml:id="echoid-s1761" xml:space="preserve">GAS; </s> <s xml:id="echoid-s1762" xml:space="preserve">& </s> <s xml:id="echoid-s1763" xml:space="preserve">par conſé-<lb/>quent les puiſſances D & </s> <s xml:id="echoid-s1764" xml:space="preserve">E ſont entr’elles, commeles ſi-<lb/>nus GAR & </s> <s xml:id="echoid-s1765" xml:space="preserve">GAS. </s> <s xml:id="echoid-s1766" xml:space="preserve">Elles ſont auſſi (hyp.) </s> <s xml:id="echoid-s1767" xml:space="preserve">entr’elles, com-<lb/>me les ſinus des angles RAB & </s> <s xml:id="echoid-s1768" xml:space="preserve">SAB; </s> <s xml:id="echoid-s1769" xml:space="preserve">& </s> <s xml:id="echoid-s1770" xml:space="preserve">par conſé-<lb/>quent les ſinus de GAR & </s> <s xml:id="echoid-s1771" xml:space="preserve">de RAB, de même que <lb/>ceux de GAS & </s> <s xml:id="echoid-s1772" xml:space="preserve">SAB ſont égaux entr’eux: </s> <s xml:id="echoid-s1773" xml:space="preserve">donc les an-<lb/>gles GAR & </s> <s xml:id="echoid-s1774" xml:space="preserve">RAB, auſſi-bien que GAS & </s> <s xml:id="echoid-s1775" xml:space="preserve">SAB, ſont <lb/>auſſi égaux, ou, du moins, complemens l’un de l’autre à <lb/>deux droits. </s> <s xml:id="echoid-s1776" xml:space="preserve">D’où il ſuit que AG eſt en ligne droite avec <lb/>AB; </s> <s xml:id="echoid-s1777" xml:space="preserve">& </s> <s xml:id="echoid-s1778" xml:space="preserve">par conſéquent le point fixe B dulevier AB ſou-<lb/>tient, lui ſeul, l’impreſſion toute entiére que le point A <lb/>reçoit du concours d’action des puiſſances D & </s> <s xml:id="echoid-s1779" xml:space="preserve">E; </s> <s xml:id="echoid-s1780" xml:space="preserve">& </s> <s xml:id="echoid-s1781" xml:space="preserve">ce <lb/>point demeurant ainſi en repos, ces puiſſances doivent <lb/>auſſi par conſéquent demeurer en équilibre. </s> <s xml:id="echoid-s1782" xml:space="preserve">Ce qu’il <lb/>faloit démontrer.</s> <s xml:id="echoid-s1783" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div197" type="section" level="1" n="132"> <head xml:id="echoid-head132" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s1784" xml:space="preserve">On voit de-là qu’une même puiſſance peut ainſi <lb/>faire ſucceſſivement équilibre avec pluſieurs autres, <lb/>quelques grandes, ou quelques petites qu’on les ſuppo-<lb/>ſe, pourvu qu’elle ſoit à chacune d’elles en raiſon ré- <pb o="68" file="0094" n="94" rhead="NOUVELLE"/> <anchor type="note" xlink:label="note-0094-01a" xlink:href="note-0094-01"/> ciproque des ſinus des angles que font avec ce levier <lb/>ſa ligne de direction, & </s> <s xml:id="echoid-s1785" xml:space="preserve">celle de la puiſſance avec <lb/>laquelle on la compare; </s> <s xml:id="echoid-s1786" xml:space="preserve">c’eſt-à-dire, pourvû que les <lb/>produits de ces deux puiſſances par les perpendicu-<lb/>laires tirées du point d’appui de ce levier ſur leurs <lb/>lignes de direction, ſoient égaux.</s> <s xml:id="echoid-s1787" xml:space="preserve"/> </p> <div xml:id="echoid-div197" type="float" level="2" n="1"> <note position="left" xlink:label="note-0094-01" xlink:href="note-0094-01a" xml:space="preserve">DES <lb/>LEVIERS.</note> </div> </div> <div xml:id="echoid-div199" type="section" level="1" n="133"> <head xml:id="echoid-head133" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s1788" xml:space="preserve">D’où l’on voit que l’action d’une puiſſance ne ſe <lb/>prend pas ſeulement de la grandeur de ſa force; </s> <s xml:id="echoid-s1789" xml:space="preserve">mais <lb/>auſſi de la diſtance de ſa ligne de direction au point <lb/>d’appui du levier ſur lequel elle agit: </s> <s xml:id="echoid-s1790" xml:space="preserve">De ſorte que <lb/>le produit de cette diſtance par la force de cette puiſ-<lb/>ſance, eſt la meſure juſte de ſon action, ou de l’im-<lb/>preſſion qu’elle fait ſur ce levier.</s> <s xml:id="echoid-s1791" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div200" type="section" level="1" n="134"> <head xml:id="echoid-head134" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s1792" xml:space="preserve">Ainſi en quelque point que ce ſoit d’un levier <lb/>quelle ſoit appliquée, pourvû que la diſtance du <lb/>point d’appui de ce levier à ſa ligne de direction ſoit <lb/>toujours la même, ſon action ſur ce levier ſera auſſi <lb/>toujours la même.</s> <s xml:id="echoid-s1793" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div201" type="section" level="1" n="135"> <head xml:id="echoid-head135" xml:space="preserve"><emph style="sc">Corollaire</emph> IV.</head> <p> <s xml:id="echoid-s1794" xml:space="preserve">Et par la même raiſon ſi différentes puiſſances éga-<lb/>les agiſſoient ſucceſſivement ſuivant la même direc-<lb/>tion, ou ſuivant des directions également diſtantes <lb/>du point d’appui du levier auquel elles ſeroient appli-<lb/>quées; </s> <s xml:id="echoid-s1795" xml:space="preserve">leurs actions ſur ce levier ſeroient auſſi-<lb/>égales.</s> <s xml:id="echoid-s1796" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div202" type="section" level="1" n="136"> <head xml:id="echoid-head136" xml:space="preserve"><emph style="sc">Corollaire</emph> V.</head> <p> <s xml:id="echoid-s1797" xml:space="preserve">Et réciproquement ſi ces puiſſances ainſi appliquées <lb/>à ce levier agiſſent également ſur lui, elles ſeront <pb o="69" file="0095" n="95" rhead="MECHANIQUE."/> auſſi égales entr’elles: </s> <s xml:id="echoid-s1798" xml:space="preserve">puis que les diſtances de leurs <lb/> <anchor type="note" xlink:label="note-0095-01a" xlink:href="note-0095-01"/> lignes de direction au point d’appui du levier auquel <lb/>elles ſont appliquées, ſont (hyp.) </s> <s xml:id="echoid-s1799" xml:space="preserve">égales, & </s> <s xml:id="echoid-s1800" xml:space="preserve">que les <lb/>produits de chacune d’elles par la force de chacune <lb/>de ces puiſſances, ſont (Cor. </s> <s xml:id="echoid-s1801" xml:space="preserve">2.) </s> <s xml:id="echoid-s1802" xml:space="preserve">auſſi égaux.</s> <s xml:id="echoid-s1803" xml:space="preserve"/> </p> <div xml:id="echoid-div202" type="float" level="2" n="1"> <note position="right" xlink:label="note-0095-01" xlink:href="note-0095-01a" xml:space="preserve">DES <lb/>LEVIERS.</note> </div> <figure> <image file="0095-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0095-01"/> </figure> <pb o="70" file="0096" n="96" rhead="NOUVELLE"/> <figure> <image file="0096-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0096-01"/> </figure> </div> <div xml:id="echoid-div204" type="section" level="1" n="137"> <head xml:id="echoid-head137" xml:space="preserve">AUTRE <lb/>PROPOSITION <lb/>DES LEVIERS, <lb/>Pour tous les cas poſſibles de la fondamentale <lb/>précédente.</head> <p style="it"> <s xml:id="echoid-s1804" xml:space="preserve">SI les puiſſances E & </s> <s xml:id="echoid-s1805" xml:space="preserve">F appliquées à diſſérens points X <lb/> <anchor type="note" xlink:label="note-0096-01a" xlink:href="note-0096-01"/> & </s> <s xml:id="echoid-s1806" xml:space="preserve">O du levier BO, de quelque ſigure, de quelque eſpéce, <lb/>& </s> <s xml:id="echoid-s1807" xml:space="preserve">en quelque ſituation qu’il ſoit; </s> <s xml:id="echoid-s1808" xml:space="preserve">ſont entr’elles en raiſon <lb/>rèciproque des diſtances BD & </s> <s xml:id="echoid-s1809" xml:space="preserve">BP de leurs lignes de direction <lb/>XE & </s> <s xml:id="echoid-s1810" xml:space="preserve">OF, qu’elles quelles ſoient, au point ſixe B de ce <lb/>levier; </s> <s xml:id="echoid-s1811" xml:space="preserve">c’eſt-à-dire, ſi E. </s> <s xml:id="echoid-s1812" xml:space="preserve">F. </s> <s xml:id="echoid-s1813" xml:space="preserve">:</s> <s xml:id="echoid-s1814" xml:space="preserve">: BP. </s> <s xml:id="echoid-s1815" xml:space="preserve">BD. </s> <s xml:id="echoid-s1816" xml:space="preserve">Ces deux <lb/>puiſſances ainſi appliquées demeureront en équilibre ſur ce <lb/>levier.</s> <s xml:id="echoid-s1817" xml:space="preserve"/> </p> <div xml:id="echoid-div204" type="float" level="2" n="1"> <note position="left" xlink:label="note-0096-01" xlink:href="note-0096-01a" xml:space="preserve">ſig 50. <lb/>51. <lb/>52.</note> </div> </div> <div xml:id="echoid-div206" type="section" level="1" n="138"> <head xml:id="echoid-head138" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s1818" xml:space="preserve">Si la puiſſance E étoit appliquée en O ſuivant une <lb/>ligne de direction OQ, dont la diſtance BQ au point <lb/>d’appui B de ce levier, fût égale à BD qui eſt celle <lb/>de ce même point à la ligne de direction XE qu’elle <lb/>a préſentement, cette puiſſance (Lemm. </s> <s xml:id="echoid-s1819" xml:space="preserve">6.) </s> <s xml:id="echoid-s1820" xml:space="preserve">demeu-<lb/>reroit en équilibre avec la puiſſance F. </s> <s xml:id="echoid-s1821" xml:space="preserve">Or l’action <lb/>de cette puiſſance appliquée en X ſuivant XE, eſt <lb/>la même (Lemm. </s> <s xml:id="echoid-s1822" xml:space="preserve">6. </s> <s xml:id="echoid-s1823" xml:space="preserve">Corol. </s> <s xml:id="echoid-s1824" xml:space="preserve">3.) </s> <s xml:id="echoid-s1825" xml:space="preserve">qu’elle ſeroit alors ſur <lb/>ce levier: </s> <s xml:id="echoid-s1826" xml:space="preserve">Donc ces deux puiſſances ainſi appliquées <pb o="71" file="0097" n="97" rhead="MECHANIQUE."/> en X & </s> <s xml:id="echoid-s1827" xml:space="preserve">en O, doivent demeurer en équilibre. </s> <s xml:id="echoid-s1828" xml:space="preserve">Ce qu’il <lb/> <anchor type="note" xlink:label="note-0097-01a" xlink:href="note-0097-01"/> faloit démontrer.</s> <s xml:id="echoid-s1829" xml:space="preserve"/> </p> <div xml:id="echoid-div206" type="float" level="2" n="1"> <note position="right" xlink:label="note-0097-01" xlink:href="note-0097-01a" xml:space="preserve">DES <lb/>LEVIERS.</note> </div> </div> <div xml:id="echoid-div208" type="section" level="1" n="139"> <head xml:id="echoid-head139" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s1830" xml:space="preserve">Il ſuit réciproquement de cette propoſition que ſi <lb/>les puiſſances E & </s> <s xml:id="echoid-s1831" xml:space="preserve">F font équilibre ſur quelque levier <lb/>que ce ſoit, elles ſeront entr’elles en raiſon récipro-<lb/>que des diſtances de ſon point d’appui à leurs lignes de <lb/>direction: </s> <s xml:id="echoid-s1832" xml:space="preserve">puis que ſi quelque nouvelle puiſſance miſe <lb/>en la place, par exemple, de la puiſſance E, & </s> <s xml:id="echoid-s1833" xml:space="preserve">appli-<lb/>quée comme elle au levier BO, étoit en telle raiſon à la <lb/>puiſſance F, elle feroit équilibre avec elle. </s> <s xml:id="echoid-s1834" xml:space="preserve">Orune telle <lb/>puiſſance (Lemm. </s> <s xml:id="echoid-s1835" xml:space="preserve">6. </s> <s xml:id="echoid-s1836" xml:space="preserve">Cor. </s> <s xml:id="echoid-s1837" xml:space="preserve">5.) </s> <s xml:id="echoid-s1838" xml:space="preserve">ſeroit égale à la puiſſance <lb/>E: </s> <s xml:id="echoid-s1839" xml:space="preserve">Donc la puiſſance E ſeroit auſſi alors à la puiſſance <lb/>F en raiſon réciproque des diſtances de leurs lignes <lb/>de direction au point d’appui du levier ſur lequel <lb/>elles feroient équilibre.</s> <s xml:id="echoid-s1840" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s1841" xml:space="preserve">On ne met point ici tous les Corollaires qu’on pourroit tirer <lb/>tant de cette propoſition, que de la précédente, par raport à la <lb/>Balance, à la Romaine, au Tour, aux Rouës à dent, aux Ci-<lb/>ſeaux, aux Tenailles, aux Etocs, &</s> <s xml:id="echoid-s1842" xml:space="preserve">c. </s> <s xml:id="echoid-s1843" xml:space="preserve">ce ſera lors qu’on en <lb/>fera l’application à la phyſique; </s> <s xml:id="echoid-s1844" xml:space="preserve">D’ailleurs tres-peu d’atten-<lb/>tion ſuſſit préſentement pour en déduire beaucoup plus que la <lb/>breveté, qu’on s’eſt propoſée dans ce projet, ne permet de ſaire.</s> <s xml:id="echoid-s1845" xml:space="preserve"/> </p> <figure> <image file="0097-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0097-01"/> </figure> </div> <div xml:id="echoid-div209" type="section" level="1" n="140"> <head xml:id="echoid-head140" xml:space="preserve">PROBLEME.</head> <p style="it"> <s xml:id="echoid-s1846" xml:space="preserve">TANT de puiſſances qu’on voudra appliquées à un <lb/> <anchor type="note" xlink:label="note-0097-02a" xlink:href="note-0097-02"/> même levier; </s> <s xml:id="echoid-s1847" xml:space="preserve">par exemple, les cinq que voici M, N, <lb/>O, P, Q, étant données avec leurs lignes de direction AM, <lb/>CN, EO, HP, & </s> <s xml:id="echoid-s1848" xml:space="preserve">GQ, quelles quelles ſoient; </s> <s xml:id="echoid-s1849" xml:space="preserve">trouver le <lb/>point de ce levier ſur lequel elles peuvent ainſi appliquées de-<lb/>meurer toutes enſemble en équilibre.</s> <s xml:id="echoid-s1850" xml:space="preserve"/> </p> <div xml:id="echoid-div209" type="float" level="2" n="1"> <note position="right" xlink:label="note-0097-02" xlink:href="note-0097-02a" xml:space="preserve">ſig. 53. <lb/>54. <lb/>55. <lb/>56. <lb/>57. <lb/>58. <lb/>59.</note> </div> <pb o="72" file="0098" n="98" rhead="NOUVELLE"/> </div> <div xml:id="echoid-div211" type="section" level="1" n="141"> <head xml:id="echoid-head141" xml:space="preserve"><emph style="sc">Solution</emph>.</head> <note position="left" xml:space="preserve">DES <lb/>LEVIERS.</note> <p> <s xml:id="echoid-s1851" xml:space="preserve">Premiérement ſi ces lignes de direction ſont toutes <lb/> <anchor type="note" xlink:label="note-0098-02a" xlink:href="note-0098-02"/> paralleles entr’elles, de quelque côté que ces puiſ-<lb/>ſances tirent, il faut prendre ſur le levier AH pro-<lb/>longé s’il en eſt beſoin, H λ à G λ, commela puiſſance <lb/>Q eſt à la puiſſance P, & </s> <s xml:id="echoid-s1852" xml:space="preserve">I’on aura le point λ ſur le-<lb/>quel ces deux puiſſances ainſi appliquées feroient <lb/>é<unsure/>quilibre, (Corol. </s> <s xml:id="echoid-s1853" xml:space="preserve">12.) </s> <s xml:id="echoid-s1854" xml:space="preserve">ſi elles étoient ſeules. </s> <s xml:id="echoid-s1855" xml:space="preserve">La <lb/>direction de ce point (Cor. </s> <s xml:id="echoid-s1856" xml:space="preserve">5.) </s> <s xml:id="echoid-s1857" xml:space="preserve">étant vers Q paralle-<lb/>lement à HP, c’eſt-à-dire, (hyp.) </s> <s xml:id="echoid-s1858" xml:space="preserve">à EO; </s> <s xml:id="echoid-s1859" xml:space="preserve">& </s> <s xml:id="echoid-s1860" xml:space="preserve">ſa charge <lb/>étant (Cor. </s> <s xml:id="echoid-s1861" xml:space="preserve">7. </s> <s xml:id="echoid-s1862" xml:space="preserve">ſig. </s> <s xml:id="echoid-s1863" xml:space="preserve">53.) </s> <s xml:id="echoid-s1864" xml:space="preserve">égale à la ſomme des forces de <lb/>ces deux puiſſances, ou (Cor. </s> <s xml:id="echoid-s1865" xml:space="preserve">9. </s> <s xml:id="echoid-s1866" xml:space="preserve">ſig. </s> <s xml:id="echoid-s1867" xml:space="preserve">54. </s> <s xml:id="echoid-s1868" xml:space="preserve">& </s> <s xml:id="echoid-s1869" xml:space="preserve">55.) </s> <s xml:id="echoid-s1870" xml:space="preserve">à leur <lb/>diffèrence: </s> <s xml:id="echoid-s1871" xml:space="preserve">au lieu d’elles on en peut ſuppoſer une <lb/>nouvelle de cette valeur appliquée en ce point λ, & </s> <s xml:id="echoid-s1872" xml:space="preserve"><lb/>dirigée du côté de la puiſſance Q parallelement à EO. <lb/></s> <s xml:id="echoid-s1873" xml:space="preserve">Il eſt clair que cette nouvelle puiſſance faiſant la <lb/>même impreſſion ſur ce levier que les puiſſances P & </s> <s xml:id="echoid-s1874" xml:space="preserve"><lb/>Q y en faiſoient auparavant, ſon centre d’équilibre <lb/>avec la puiſſance O, ſera le point lequel les trois <lb/>puiſſances O, P, & </s> <s xml:id="echoid-s1875" xml:space="preserve">Q feroient équilibre, ſi elles <lb/>étoient ſeules, & </s> <s xml:id="echoid-s1876" xml:space="preserve">appliquées comme elles ſont: </s> <s xml:id="echoid-s1877" xml:space="preserve"><lb/>l’ayant donc trouvé comme l’on vient de faire le point <lb/>λ; </s> <s xml:id="echoid-s1878" xml:space="preserve">c’eſt-à-dire, ayant fait λF à FE, comme la puiſ-<lb/>ſance O, à la charge qui réſulte au point λ du concours <lb/>d’action des puiſſances P & </s> <s xml:id="echoid-s1879" xml:space="preserve">Q; </s> <s xml:id="echoid-s1880" xml:space="preserve">on aura le point F <lb/>dont la direction ſera encore (Cor. </s> <s xml:id="echoid-s1881" xml:space="preserve">5.) </s> <s xml:id="echoid-s1882" xml:space="preserve">parallele à celle <lb/>de ces puiſſances, & </s> <s xml:id="echoid-s1883" xml:space="preserve">du côté de Q; </s> <s xml:id="echoid-s1884" xml:space="preserve">& </s> <s xml:id="echoid-s1885" xml:space="preserve">la charge en <lb/>ſera égale (Cor. </s> <s xml:id="echoid-s1886" xml:space="preserve">7. </s> <s xml:id="echoid-s1887" xml:space="preserve">ſig. </s> <s xml:id="echoid-s1888" xml:space="preserve">53.) </s> <s xml:id="echoid-s1889" xml:space="preserve">à la ſomme de ces trois-ci <lb/>O, P, & </s> <s xml:id="echoid-s1890" xml:space="preserve">Q; </s> <s xml:id="echoid-s1891" xml:space="preserve">ou (ſig. </s> <s xml:id="echoid-s1892" xml:space="preserve">54.) </s> <s xml:id="echoid-s1893" xml:space="preserve">à la ſomme de la puiſſance <lb/>O, & </s> <s xml:id="echoid-s1894" xml:space="preserve">de la diſſérence qui eſt entre les puiſſances P & </s> <s xml:id="echoid-s1895" xml:space="preserve"><lb/>Q; </s> <s xml:id="echoid-s1896" xml:space="preserve">ou enſin (Cor. </s> <s xml:id="echoid-s1897" xml:space="preserve">9. </s> <s xml:id="echoid-s1898" xml:space="preserve">ſig. </s> <s xml:id="echoid-s1899" xml:space="preserve">55.) </s> <s xml:id="echoid-s1900" xml:space="preserve">à la diſſérence qui eſt <lb/>entre la puiſſance O, & </s> <s xml:id="echoid-s1901" xml:space="preserve">la diſſèrence des puiſſances <lb/>P & </s> <s xml:id="echoid-s1902" xml:space="preserve">Q: </s> <s xml:id="echoid-s1903" xml:space="preserve">ainſi au lieu de la puiſſance O, & </s> <s xml:id="echoid-s1904" xml:space="preserve">de celle que <lb/>nous avons ſuppoſée en λ, ſi nous en concevons en- <pb o="73" file="0099" n="99" rhead="MECHANIQUE."/> core un autre de la valeur de la charge du point F, <lb/> <anchor type="note" xlink:label="note-0099-01a" xlink:href="note-0099-01"/> appliquée en ce point, & </s> <s xml:id="echoid-s1905" xml:space="preserve">dirigée du côté de la puiſ-<lb/>ſance Q parallelement à NC; </s> <s xml:id="echoid-s1906" xml:space="preserve">nous trouverons ſon <lb/>centre d’equilibre D avec la puiſſance N, de même <lb/>que nous avons déja trouvé les points F, & </s> <s xml:id="echoid-s1907" xml:space="preserve">λ. </s> <s xml:id="echoid-s1908" xml:space="preserve">Enſin <lb/>ôtant encore la puiſſance N avec celle qui étoit appli-<lb/>quée en F, & </s> <s xml:id="echoid-s1909" xml:space="preserve">mettant à leur défaut au point D une <lb/>autre puiſſance égale à ſa charge, & </s> <s xml:id="echoid-s1910" xml:space="preserve">dirigée du côté <lb/>de la puiſſance O parallelement à AM; </s> <s xml:id="echoid-s1911" xml:space="preserve">on trouvera <lb/>encore le centre de direction B qui lui eſt commun <lb/>avec la puiſſance M, de même que l’on vient de faire <lb/>les points λ, F, & </s> <s xml:id="echoid-s1912" xml:space="preserve">D; </s> <s xml:id="echoid-s1913" xml:space="preserve">& </s> <s xml:id="echoid-s1914" xml:space="preserve">ce point B ſera (Cor. </s> <s xml:id="echoid-s1915" xml:space="preserve">12.) <lb/></s> <s xml:id="echoid-s1916" xml:space="preserve">celui ſur lequel les cinq puiſſances M, N, O, P, & </s> <s xml:id="echoid-s1917" xml:space="preserve">Q, <lb/>ainſi dirigées demeureront en équilibre. </s> <s xml:id="echoid-s1918" xml:space="preserve">Ce qu’il <lb/>faloit premiérement trouver.</s> <s xml:id="echoid-s1919" xml:space="preserve"/> </p> <div xml:id="echoid-div211" type="float" level="2" n="1"> <note position="left" xlink:label="note-0098-02" xlink:href="note-0098-02a" xml:space="preserve">ſig. 53. <lb/>54. <lb/>55.</note> <note position="right" xlink:label="note-0099-01" xlink:href="note-0099-01a" xml:space="preserve">DES <lb/>LEVIERS</note> </div> </div> <div xml:id="echoid-div213" type="section" level="1" n="142"> <head xml:id="echoid-head142" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s1920" xml:space="preserve">On voit aſſez que ſi ce levier AH, (ſig. </s> <s xml:id="echoid-s1921" xml:space="preserve">55.)</s> <s xml:id="echoid-s1922" xml:space="preserve">ſe fû@t <lb/>terminé en H, ce Problême en ce cas auroit été im-<lb/>poſſible.</s> <s xml:id="echoid-s1923" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div214" type="section" level="1" n="143"> <head xml:id="echoid-head143" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s1924" xml:space="preserve">Ce que nous venons de démontrer des leviers <lb/> <anchor type="note" xlink:label="note-0099-02a" xlink:href="note-0099-02"/> droits, ſe peut fort aiſément appliquer à toutes ſor-<lb/>tes de leviers courbes; </s> <s xml:id="echoid-s1925" xml:space="preserve">par exemple, à celui de la fig. <lb/></s> <s xml:id="echoid-s1926" xml:space="preserve">56. </s> <s xml:id="echoid-s1927" xml:space="preserve">en faiſant ſeulement à diſcretion quelque ligne <lb/>droite ah, qui rencontre (il n’importe comment) <lb/>toutes les lignes de direction des puiſſances M, N, <lb/>O, P, & </s> <s xml:id="echoid-s1928" xml:space="preserve">Q, en a, c, e, h, & </s> <s xml:id="echoid-s1929" xml:space="preserve">g: </s> <s xml:id="echoid-s1930" xml:space="preserve">car la regardant com-<lb/>me un levier auquel ces puiſſances ſont appliquées <lb/>en ces mêmes points, on en trouvera, comme l’on <lb/>vient de faire, le point d’appui b, avec ſa ligne de <lb/>direction bB, qui rencontrant le levier AH, donnera <lb/>le point B qui ſera ſon point d’appui: </s> <s xml:id="echoid-s1931" xml:space="preserve">Puis que les <lb/>rapports de diſtances de ce point aux lignes de di- <pb o="74" file="0100" n="100" rhead="NOUVELLE"/> rection de ces puiſſ nces, qui ſont (hyp & </s> <s xml:id="echoid-s1932" xml:space="preserve">Cor. </s> <s xml:id="echoid-s1933" xml:space="preserve">5. <lb/></s> <s xml:id="echoid-s1934" xml:space="preserve"> <anchor type="note" xlink:label="note-0100-01a" xlink:href="note-0100-01"/> paralleles à b B, ſont les mêmes que ceux des diſtances <lb/>du point b à ces mêmes lignes.</s> <s xml:id="echoid-s1935" xml:space="preserve"/> </p> <div xml:id="echoid-div214" type="float" level="2" n="1"> <note position="right" xlink:label="note-0099-02" xlink:href="note-0099-02a" xml:space="preserve">ſig. 56.</note> <note position="left" xlink:label="note-0100-01" xlink:href="note-0100-01a" xml:space="preserve">DES <lb/>LEVIERS.</note> </div> <p> <s xml:id="echoid-s1936" xml:space="preserve">Secondement, s’il ſe trouve quelques -unes des <lb/> <anchor type="note" xlink:label="note-0100-02a" xlink:href="note-0100-02"/> lignes de direction des puiſſances données, qui ne ſoient <lb/>point paralleles entr’elles, quelles que ſoient les autres, <lb/>& </s> <s xml:id="echoid-s1937" xml:space="preserve">de quelque côté que ces puiſſances tirent encore; <lb/></s> <s xml:id="echoid-s1938" xml:space="preserve">Voici comment on pourra trouver le point d’appui <lb/>du levier AH auquel elles ſont appliquées. </s> <s xml:id="echoid-s1939" xml:space="preserve">Soient, ſi <lb/>l’on veut, les directions HP & </s> <s xml:id="echoid-s1940" xml:space="preserve">GQ des puiſſan-<lb/>ces P & </s> <s xml:id="echoid-s1941" xml:space="preserve">Q, non paralleles, & </s> <s xml:id="echoid-s1942" xml:space="preserve">qu’elles ſe rencon-<lb/>trent en V: </s> <s xml:id="echoid-s1943" xml:space="preserve">faite VR à VS, comme la puiſſance P à <lb/>la puiſſance Q; </s> <s xml:id="echoid-s1944" xml:space="preserve">achevez le parallelogramme RS, <lb/>& </s> <s xml:id="echoid-s1945" xml:space="preserve">faite la diagonale VK qui rencontre en λ le <lb/>levier AH prolongé juſqu’où il en ſera beſoin: </s> <s xml:id="echoid-s1946" xml:space="preserve">ce <lb/>point ſera (prop. </s> <s xml:id="echoid-s1947" xml:space="preserve">ſond.) </s> <s xml:id="echoid-s1948" xml:space="preserve">celui ſur lequel ces deux puiſ-<lb/>ſances ainſi appliquées feroient équilibre, ſi elles <lb/>étoient ſeules; </s> <s xml:id="echoid-s1949" xml:space="preserve">& </s> <s xml:id="echoid-s1950" xml:space="preserve">ſa charge, dont la direction eſt <lb/>de V vers K ſuivant VK, (Cor. </s> <s xml:id="echoid-s1951" xml:space="preserve">4.) </s> <s xml:id="echoid-s1952" xml:space="preserve">ſera à chacune <lb/>des puiſſances P, & </s> <s xml:id="echoid-s1953" xml:space="preserve">Q, comme VK à chacune des <lb/>lignes VR & </s> <s xml:id="echoid-s1954" xml:space="preserve">VS qui les répréſentent: </s> <s xml:id="echoid-s1955" xml:space="preserve">De ſorte que <lb/>ſi au lieu de ces deux puiſſances, on en appliquoit <lb/>quelqu’autre au point λ de ce levier, ſuivant cette <lb/>même direction VK, & </s> <s xml:id="echoid-s1956" xml:space="preserve">qui eût ce même raport à <lb/>chacune d’elles, c’eſt-à-dire, qui ſût égale à la <lb/>charge de ce point; </s> <s xml:id="echoid-s1957" xml:space="preserve">elle feroit ſeule ainſi appliquée <lb/>la même impreſſion ſur ce levier que font préſente-<lb/>ment ces deux enſemble; </s> <s xml:id="echoid-s1958" xml:space="preserve">& </s> <s xml:id="echoid-s1959" xml:space="preserve">par conſéquent ſon <lb/>centre d’équilibre avec la puiſſance O, ſeroit celui <lb/>des trois puiſſances, O, P, & </s> <s xml:id="echoid-s1960" xml:space="preserve">Q; </s> <s xml:id="echoid-s1961" xml:space="preserve">c’eſt-à-dire, le <lb/>point ſur lequel elles feroient equilibre, ſi elles <lb/>étoient ſeules, & </s> <s xml:id="echoid-s1962" xml:space="preserve">ainſi appliquées. </s> <s xml:id="echoid-s1963" xml:space="preserve">S’il arrive que <lb/>VK & </s> <s xml:id="echoid-s1964" xml:space="preserve">OE ſoient paralleles, on trouvera ce point <lb/>comme l’on a fait dans l’hypothêſe des directions <pb o="75" file="0101" n="101" rhead="MECHANIQUE."/> paralleles; </s> <s xml:id="echoid-s1965" xml:space="preserve">mais ſi elles ſe rencontrent en quelque <lb/> <anchor type="note" xlink:label="note-0101-01a" xlink:href="note-0101-01"/> point T, il faut, comme l’on vient de faire, prendre <lb/>TY à TZ, comme la puiſſance ſuppoſée en λ, eſt à <lb/>la puiſſance O; </s> <s xml:id="echoid-s1966" xml:space="preserve">& </s> <s xml:id="echoid-s1967" xml:space="preserve">ayant achevé le parallelogramme <lb/>YZ, faite ſa diagonale TX qui rencontre en F le <lb/>levier AH: </s> <s xml:id="echoid-s1968" xml:space="preserve">ce point ſera encore (Prop. </s> <s xml:id="echoid-s1969" xml:space="preserve">fond.) </s> <s xml:id="echoid-s1970" xml:space="preserve">celui <lb/>ſur lequel ces deux puiſſances, ou bien ces trois-ici O, <lb/>P, & </s> <s xml:id="echoid-s1971" xml:space="preserve">Q, feroient équilibre, ſi elles étoient ſeules, & </s> <s xml:id="echoid-s1972" xml:space="preserve"><lb/>appliquees ſuivant les conditions de ce Problême; </s> <s xml:id="echoid-s1973" xml:space="preserve">& </s> <s xml:id="echoid-s1974" xml:space="preserve"><lb/>ſa charge, dont la direction eſt de T vers X, (Cor. </s> <s xml:id="echoid-s1975" xml:space="preserve">4.) <lb/></s> <s xml:id="echoid-s1976" xml:space="preserve">eſt en ce cas à la puiſſance O, comme TX à TZ: </s> <s xml:id="echoid-s1977" xml:space="preserve">ôtant <lb/>donc encore cette puiſſance avec celle que nous <lb/>avions ſuppoſée en λ au lieu des puiſſances P & </s> <s xml:id="echoid-s1978" xml:space="preserve">Q; </s> <s xml:id="echoid-s1979" xml:space="preserve"><lb/>& </s> <s xml:id="echoid-s1980" xml:space="preserve">en mettant une autre en F, dirigée de T vers X <lb/>ſuivant TX, & </s> <s xml:id="echoid-s1981" xml:space="preserve">qui ſoit à la puiſſance O, comme <lb/>TX à TZ; </s> <s xml:id="echoid-s1982" xml:space="preserve">elle fera encore ſeule ainſi appliquée la <lb/>même impreſſion ſur le leuier AH, que lui faiſoient <lb/>auparavant les trois puiſſances O, P, & </s> <s xml:id="echoid-s1983" xml:space="preserve">Q; </s> <s xml:id="echoid-s1984" xml:space="preserve">& </s> <s xml:id="echoid-s1985" xml:space="preserve">par <lb/>conſéquent ſon centre d’équilibre avec la puiſſance <lb/>N, ſera celui des quatres puiſſances N, O, P, & </s> <s xml:id="echoid-s1986" xml:space="preserve">Q. </s> <s xml:id="echoid-s1987" xml:space="preserve"><lb/>S’il arrive que TX & </s> <s xml:id="echoid-s1988" xml:space="preserve">NC ſoient paralleles, on trou-<lb/>vera encore ce point comme dans la première partie <lb/>de cette ſolution; </s> <s xml:id="echoid-s1989" xml:space="preserve">mais ſi elles ſe rencontrent en quel-<lb/>que point β, il faut encore prendre βγ à βε, comme <lb/>la puiſſance ſuppoſée en F, eſt à la puiſſance N; </s> <s xml:id="echoid-s1990" xml:space="preserve">& </s> <s xml:id="echoid-s1991" xml:space="preserve"><lb/>ayant achevé le parallelogramme εγ, il faut prolon-<lb/>ger ſa diagonale βδ juſqu’à ce qu’elle rencontre le <lb/>levier AH en D, & </s> <s xml:id="echoid-s1992" xml:space="preserve">ce point ſera (prop. </s> <s xml:id="echoid-s1993" xml:space="preserve">fond.) </s> <s xml:id="echoid-s1994" xml:space="preserve">celui <lb/>ſur lequel ces deux puiſſances, ou bien ces quatre-ci <lb/>N, O, P, & </s> <s xml:id="echoid-s1995" xml:space="preserve">Q, feroient équilibre, ſi elles étoient <lb/>ſeules, & </s> <s xml:id="echoid-s1996" xml:space="preserve">appliquées ſuivant les directions données <lb/>de ce problême. </s> <s xml:id="echoid-s1997" xml:space="preserve">La direction de ce point ſera encore <lb/>de β vers δ ſuivant βδ, (Cor. </s> <s xml:id="echoid-s1998" xml:space="preserve">4.) </s> <s xml:id="echoid-s1999" xml:space="preserve">& </s> <s xml:id="echoid-s2000" xml:space="preserve">ſa charge ſera <lb/>auſſi à la puiſſance N, comme βδ à βε: </s> <s xml:id="echoid-s2001" xml:space="preserve">ainſi au lieu <lb/>de la puiſſance N, & </s> <s xml:id="echoid-s2002" xml:space="preserve">de celle que nous venons de <pb o="76" file="0102" n="102" rhead="NOUVELLE"/> ſuppoſer en F, ſi l’on en ſuppoſe encore quelqu’autre <lb/> <anchor type="note" xlink:label="note-0102-01a" xlink:href="note-0102-01"/> en D, dirigée de β vers δ ſuivant βδ, & </s> <s xml:id="echoid-s2003" xml:space="preserve">qui ſoit à la <lb/>puiſſance N, comme βδ à βε; </s> <s xml:id="echoid-s2004" xml:space="preserve">cette nouvelle puiſ-<lb/>ſance ainſi appliquée au levier AH, fera encore <lb/>feule ſur lui la même impreſſion que faiſoient aupa-<lb/>ravant les quatre puiſſances N, O, P, & </s> <s xml:id="echoid-s2005" xml:space="preserve">Q; </s> <s xml:id="echoid-s2006" xml:space="preserve">& </s> <s xml:id="echoid-s2007" xml:space="preserve">par <lb/>conſéquent ſon centre d’équilibre avec la puiſſance <lb/>M, ſera celui des cinq puiſſances données dans ce <lb/>problême. </s> <s xml:id="echoid-s2008" xml:space="preserve">S’il arrive que βδ ſoit parallele à AM, on <lb/>trouvera encore ce point comme l’'on a fait dans le cas <lb/>des directions paralleles; </s> <s xml:id="echoid-s2009" xml:space="preserve">mais ſi elles ſe rencontrent <lb/>en quelque point L, il faut encore prendre L θ à Lω, <lb/>comme la puiſſance qu’on ſuppoſe en D, eſt à la puiſ-<lb/>ſance M; </s> <s xml:id="echoid-s2010" xml:space="preserve">& </s> <s xml:id="echoid-s2011" xml:space="preserve">ayant achevé le parallelogramme ωθ, il <lb/>faut tirer la diagonale LI, qui rencontrant le levier <lb/>AH, donnera le point B pour centre d’equilibre <lb/>(Prop. </s> <s xml:id="echoid-s2012" xml:space="preserve">fond.) </s> <s xml:id="echoid-s2013" xml:space="preserve">de ces deux puiſſances; </s> <s xml:id="echoid-s2014" xml:space="preserve">c’eſt-à-dire en <lb/>remettant à la place de celle qu’on ſuppoſe en D, les <lb/>quatre N, O, P, & </s> <s xml:id="echoid-s2015" xml:space="preserve">Q, qu’elle ſuppléoit; </s> <s xml:id="echoid-s2016" xml:space="preserve">ſur lequel <lb/>les cinq puiſſances données, & </s> <s xml:id="echoid-s2017" xml:space="preserve">appliquées ſuivant <lb/>les conditions de ce Problême, feront équilibre. </s> <s xml:id="echoid-s2018" xml:space="preserve">Ce <lb/>qu’il faloit encore trouver.</s> <s xml:id="echoid-s2019" xml:space="preserve"/> </p> <div xml:id="echoid-div215" type="float" level="2" n="2"> <note position="left" xlink:label="note-0100-02" xlink:href="note-0100-02a" xml:space="preserve">ſig 57. <lb/>58. <lb/>59.</note> <note position="right" xlink:label="note-0101-01" xlink:href="note-0101-01a" xml:space="preserve">DES <lb/>LEVIERS.</note> <note position="left" xlink:label="note-0102-01" xlink:href="note-0102-01a" xml:space="preserve">DES <lb/>LEVIERS.</note> </div> <p style="it"> <s xml:id="echoid-s2020" xml:space="preserve">Ce Problème, tout facile qu’il paroit à réſoudre par les <lb/>principes q@’on ſuit ici, eſt peut - é<unsure/>tre un des plus diſſiciles <lb/>qu’on puiſſe propoſer à ceux qui ſont dans les principes ordi-<lb/>naires; </s> <s xml:id="echoid-s2021" xml:space="preserve">non ſeulement quant à la maniére de déterminer les <lb/>points d’appui des levi<unsure/>ers pour toutes les directions poſſibles <lb/>des puiſſances, ou des poids qui y ſont appliquez; </s> <s xml:id="echoid-s2022" xml:space="preserve">mais auſſi <lb/>quant à celle de reconnoitre leur direction, & </s> <s xml:id="echoid-s2023" xml:space="preserve">leur charge, <lb/>de quelque eſpece, & </s> <s xml:id="echoid-s2024" xml:space="preserve">dans quelque ſituation qu’ils ſoient: <lb/></s> <s xml:id="echoid-s2025" xml:space="preserve">C’eſt peut ètre auſſi ce qui a fait que les Mécbaniciens, qui <lb/>ont entrepris de démontrer le cas de la figu<unsure/>re 53. </s> <s xml:id="echoid-s2026" xml:space="preserve">n’ont oſé tou-<lb/>ch@r aux autres.</s> <s xml:id="echoid-s2027" xml:space="preserve"/> </p> <pb o="77" file="0103" n="103" rhead="MECHANIQUE."/> <p style="it"> <s xml:id="echoid-s2028" xml:space="preserve">Outre les Poulies, les Surfaces, & </s> <s xml:id="echoid-s2029" xml:space="preserve">les Leviers, on met <lb/> <anchor type="note" xlink:label="note-0103-01a" xlink:href="note-0103-01"/> encore la Vis au rang des machines qu’on regarde comme <lb/>capitales entre celles qui ſont propres à faciliter les mouve-<lb/>mens; </s> <s xml:id="echoid-s2030" xml:space="preserve">mais parce que cette dernière ſe rapporte toute au plan <lb/>incliné & </s> <s xml:id="echoid-s2031" xml:space="preserve">au levier, dont on va voir qu’elle eſt compoſée, <lb/>nous n’en dirons ici que tres@ peu de choſes.</s> <s xml:id="echoid-s2032" xml:space="preserve"/> </p> <div xml:id="echoid-div216" type="float" level="2" n="3"> <note position="right" xlink:label="note-0103-01" xlink:href="note-0103-01a" xml:space="preserve">DES <lb/>LEVIERS.</note> </div> <figure> <image file="0103-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0103-01"/> </figure> <pb o="78" file="0104" n="104" rhead="NOUVELLE"/> <figure> <image file="0104-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0104-01"/> </figure> </div> <div xml:id="echoid-div218" type="section" level="1" n="144"> <head xml:id="echoid-head144" xml:space="preserve">DE LA VIS <lb/>REMARQUES. <lb/>I.</head> <p> <s xml:id="echoid-s2033" xml:space="preserve">ON voit aſſez que tout l’uſage de la Vis eſt de <lb/>tirer, ou de repouſſer ſuivantla direction de ſon <lb/>axe tout ce qui lui fait quelque réſiſtance; </s> <s xml:id="echoid-s2034" xml:space="preserve">& </s> <s xml:id="echoid-s2035" xml:space="preserve">ſi elle <lb/>eſtſixe, la force, ou le corps contre lequel on s’en ſert, <lb/>doit tirer, ou preſſer ſon écrouë du côté diametra-<lb/>lement oppoſé à celui vers lequel il eſt forcé d’avan-<lb/>cer. </s> <s xml:id="echoid-s2036" xml:space="preserve">Au contraire ſi c’eſt l’écrouë qui ſoit ſixe, cette <lb/>force, où ce corps doit tirer, ou preſſer la vis elle-<lb/>même de ce même côté.</s> <s xml:id="echoid-s2037" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div219" type="section" level="1" n="145"> <head xml:id="echoid-head145" xml:space="preserve">II.</head> <p> <s xml:id="echoid-s2038" xml:space="preserve">De ſorte, que dans l’uſage de la vis, lors qu’elle <lb/>eſt ſixe, l’on doit regarder tous les points de ſon <lb/>écrouë, comme tirez ou preſſez parallelement à ſon <lb/>axe du côté que cette écrouë eſt preſſée, ou tirée par <lb/>la force, ou par le poids dont elle eſt chargée.</s> <s xml:id="echoid-s2039" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div220" type="section" level="1" n="146"> <head xml:id="echoid-head146" xml:space="preserve">III.</head> <p> <s xml:id="echoid-s2040" xml:space="preserve">D’où l’on voit que les lignes de direction de tous <lb/>ces points, ſont toutes obliques à celui des pas de <lb/>cette vis ſur lequel ceux de ces points qui le tou-<lb/>chent, s’appuyent; </s> <s xml:id="echoid-s2041" xml:space="preserve">& </s> <s xml:id="echoid-s2042" xml:space="preserve">par conſéquent (démonſt. </s> <s xml:id="echoid-s2043" xml:space="preserve">des <pb o="79" file="0105" n="105" rhead="MECHANIQUE."/> ſurf. </s> <s xml:id="echoid-s2044" xml:space="preserve">n. </s> <s xml:id="echoid-s2045" xml:space="preserve">3.) </s> <s xml:id="echoid-s2046" xml:space="preserve">ſi cette vis, & </s> <s xml:id="echoid-s2047" xml:space="preserve">ſon écrouë étoient Mathé-<lb/> <anchor type="note" xlink:label="note-0105-01a" xlink:href="note-0105-01"/> matiquement juſtes, chacun d’eux tendroit à couler <lb/>du côté que ſa ligne de direction s’écarteroit de la <lb/>perpendiculaire tirée de lui à ce pas; </s> <s xml:id="echoid-s2048" xml:space="preserve">& </s> <s xml:id="echoid-s2049" xml:space="preserve">parce que <lb/>cet écartement ſe feroit pour tous du même côté, à <lb/>cauſe du paralleliſme de toutes ces lignes, & </s> <s xml:id="echoid-s2050" xml:space="preserve">de la <lb/>pente uniforme du cordon de cette vis dans toute <lb/>ſa longueur: </s> <s xml:id="echoid-s2051" xml:space="preserve">Il ſuit évidemment que tous ces points <lb/>devroient s’accorder dans un même mouvement qui <lb/>emportât cette écrouë ſuivant le fil de ce cordon, <lb/>c’eſt-à-dire, en tournoyant du côté de cet écartement; <lb/></s> <s xml:id="echoid-s2052" xml:space="preserve">ſi dans ſon frotement avec la vis, l’inégalité de leurs <lb/>parties ne les acrochoient point enſemble.</s> <s xml:id="echoid-s2053" xml:space="preserve"/> </p> <div xml:id="echoid-div220" type="float" level="2" n="1"> <note position="right" xlink:label="note-0105-01" xlink:href="note-0105-01a" xml:space="preserve">DELA VIS.</note> </div> </div> <div xml:id="echoid-div222" type="section" level="1" n="147"> <head xml:id="echoid-head147" xml:space="preserve">IV.</head> <p> <s xml:id="echoid-s2054" xml:space="preserve">La même choſe ſe doit entendre de tous les points <lb/>de la vis, ſi c’eſt l’écrouë qui ſoit fixe.</s> <s xml:id="echoid-s2055" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div223" type="section" level="1" n="148"> <head xml:id="echoid-head148" xml:space="preserve">V.</head> <p> <s xml:id="echoid-s2056" xml:space="preserve">Ainſi à regarder l’une & </s> <s xml:id="echoid-s2057" xml:space="preserve">l’autre dans une juſteſſe <lb/>Mathématique, il faut néceſſairement quelque force <lb/>pour retenir celle des deux qui eſt mobile, contre <lb/>l’impreſſion de la force, ou du poids qui la charge. <lb/></s> <s xml:id="echoid-s2058" xml:space="preserve">La voici.</s> <s xml:id="echoid-s2059" xml:space="preserve"/> </p> <figure> <image file="0105-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0105-01"/> </figure> </div> <div xml:id="echoid-div224" type="section" level="1" n="149"> <head xml:id="echoid-head149" xml:space="preserve">PROPOSITION <lb/>DE LA VIS.</head> <p style="it"> <s xml:id="echoid-s2060" xml:space="preserve">LOrſqu’une puiſſance ſoutient quelque poids, ou l’action <lb/>de quelque autre force à l’aide d’une vis, ſoit que<unsure/> <lb/>cette vis ſoit fixe, ou que ce ſoit ſon écrouë cette puiſſance <pb o="80" file="0106" n="106" rhead="NOUVELLE"/> eſt toujours à ce poids, ou a cette force, quelle qu’elle ſoit, <lb/> <anchor type="note" xlink:label="note-0106-01a" xlink:href="note-0106-01"/> comme la diſtance qui eſt entre deux des pas de cette vis, à <lb/>la circonfèrence a’un cercle, dont le rayon eſt égal à la diſtance <lb/>qui eſt entre cette puiſſance & </s> <s xml:id="echoid-s2061" xml:space="preserve">l’axe de cette mème vis.</s> <s xml:id="echoid-s2062" xml:space="preserve"/> </p> <div xml:id="echoid-div224" type="float" level="2" n="1"> <note position="left" xlink:label="note-0106-01" xlink:href="note-0106-01a" xml:space="preserve">DELAVIS</note> </div> </div> <div xml:id="echoid-div226" type="section" level="1" n="150"> <head xml:id="echoid-head150" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s2063" xml:space="preserve">Premiérement ſi la vis VXYZ eſt fixe, concevons que <lb/> <anchor type="note" xlink:label="note-0106-02a" xlink:href="note-0106-02"/> le point A de ſon écrouë PQ ſoit retenu ſur un de ſes <lb/>pas GH par quelque puiſſance R dont la direction AB <lb/>ſoit dans le plan de cette écrouë, & </s> <s xml:id="echoid-s2064" xml:space="preserve">perpendiculaire à <lb/>EP qui y eſt auſſi, & </s> <s xml:id="echoid-s2065" xml:space="preserve">qui paſſe par le point A. </s> <s xml:id="echoid-s2066" xml:space="preserve">Il eſt clair <lb/>que cette puiſſance retenant par ce moyen toute l’é-<lb/>crouë PQ, ce point A fait ſur elle la même impreſſion <lb/>que s’il ſoutenoit lui ſeul toute l’action du poids, ou de <lb/>la force, quelle qu’elle ſoit, qui pouſſe cette écrouë, <lb/>ou qui la tire (Remarq. </s> <s xml:id="echoid-s2067" xml:space="preserve">1.) </s> <s xml:id="echoid-s2068" xml:space="preserve">vers ZY parallelement à <lb/>l’axe MS de cette vis: </s> <s xml:id="echoid-s2069" xml:space="preserve">Ainſi l’on peut regarder le <lb/>point A de cette écrouë, comme ayant lui ſeul ſuivant <lb/>AC perpendiculaire au plan de cette écrouë, & </s> <s xml:id="echoid-s2070" xml:space="preserve">pa-<lb/>rallele à MS, toute l’impreſſion qu’elle reçoit de ſa <lb/>charge; </s> <s xml:id="echoid-s2071" xml:space="preserve">de ſorte que ſi l’on fait AD perpendiculaire <lb/>au pas GH de cette vis, & </s> <s xml:id="echoid-s2072" xml:space="preserve">que de quelque point D <lb/>de cette ligne, l’on acheve le parallelogramme BC, <lb/>on verra (prop. </s> <s xml:id="echoid-s2073" xml:space="preserve">fond. </s> <s xml:id="echoid-s2074" xml:space="preserve">des ſurf.) </s> <s xml:id="echoid-s2075" xml:space="preserve">que la puiſſance R ſera à <lb/>la charge de l’écrouë PQ, comme AB, à AC, ou à BD <lb/>qui lui eſt égale; </s> <s xml:id="echoid-s2076" xml:space="preserve">c’eſt-à-dire, à cauſe que le triangle <lb/>HFG roulé ſur la vis VXYZ, eſt ſemblable au <lb/>triangle ABD; </s> <s xml:id="echoid-s2077" xml:space="preserve">comme HF au demi tour FG de cette <lb/>vis, ou comme 2HF, c’eſt-à-dire, HK à un tour <lb/>entier. </s> <s xml:id="echoid-s2078" xml:space="preserve">Or regardant EAP comme un levier dont <lb/>l’appui eſt le point E de l’axe MS de cette vis, & </s> <s xml:id="echoid-s2079" xml:space="preserve">qui <lb/>ſe trouve dans le plan de ſon écrouë; </s> <s xml:id="echoid-s2080" xml:space="preserve">la puiſſance P, <lb/>dont la direction eſt auſſi dans ce même plan, & </s> <s xml:id="echoid-s2081" xml:space="preserve">per-<lb/>pendiculaire à EP, & </s> <s xml:id="echoid-s2082" xml:space="preserve">conſéquemment parallele à AB, <pb o="81" file="0107" n="107" rhead="MECHANIQUE."/> ſoutenant ainſi (Hyp.) </s> <s xml:id="echoid-s2083" xml:space="preserve">le point A, ou la charge de <lb/> <anchor type="note" xlink:label="note-0107-01a" xlink:href="note-0107-01"/> l’écrouë PQ au lieu de la puiſſance R, eſt à cette der-<lb/>niére puiſſance, (Lemm 6. </s> <s xml:id="echoid-s2084" xml:space="preserve">Cor. </s> <s xml:id="echoid-s2085" xml:space="preserve">2.) </s> <s xml:id="echoid-s2086" xml:space="preserve">comme EA à EP, <lb/>ou comme le tour entier de cette vis à la circonfé-<lb/>rence d’un cercle dont le rayon ſoit EP: </s> <s xml:id="echoid-s2087" xml:space="preserve">Ainſi en <lb/>multipliant par ordre ces deux rangées de proportio-<lb/>nelles, l’on aura la puiſſance P à la charge de l’écrouë <lb/>PQ, comme la diſtance HK qui eſt entre deux des <lb/>pas de cette vis, à la circonference d’un cercle dont <lb/>le rayon eſt égal à la diſtance EP qui eſt entre cette <lb/>fance & </s> <s xml:id="echoid-s2088" xml:space="preserve">l’axe de cette même vis.</s> <s xml:id="echoid-s2089" xml:space="preserve"/> </p> <div xml:id="echoid-div226" type="float" level="2" n="1"> <note position="left" xlink:label="note-0106-02" xlink:href="note-0106-02a" xml:space="preserve">ſig. 60.</note> <note position="right" xlink:label="note-0107-01" xlink:href="note-0107-01a" xml:space="preserve">DELA VIS.</note> </div> <p> <s xml:id="echoid-s2090" xml:space="preserve">Secondement, ſi c’eſt l’écrouë PQ qui ſoit fixe, <lb/>concevons que le point A appartient à la vis VXYZ, & </s> <s xml:id="echoid-s2091" xml:space="preserve"><lb/>qu’il eſt retenu ſur le pas de cette écrouë par quelque <lb/>puiſſance R dont la direction ſoit encore ſuivant AB <lb/>qu’on ſuppoſe dans le plan de cetteécrouë, & </s> <s xml:id="echoid-s2092" xml:space="preserve">perpen-<lb/>diculaire à EP qui y eſt auſſi. </s> <s xml:id="echoid-s2093" xml:space="preserve">Il eſt encore clair que <lb/>cette puiſſance retenant par ce moyen toute la vis <lb/>VXYZ, ce point A fait encore ſur elle la même <lb/>impreſſion ſuivant AC parallele à MS, que s’il ſou-<lb/>tenoit, lui ſeul, toute l’action de ce qui pouſſe, (re-<lb/>marq. </s> <s xml:id="echoid-s2094" xml:space="preserve">1.) </s> <s xml:id="echoid-s2095" xml:space="preserve">ou de ce qui tire cette vis vers ZY: </s> <s xml:id="echoid-s2096" xml:space="preserve">ainſi pour <lb/>la même raiſon que ci-deſſus, la puiſſance R ſera <lb/>encore à la charge de cette vis, comme AB à BD; <lb/></s> <s xml:id="echoid-s2097" xml:space="preserve">c’eſt-à-dire, comme HF à FG, ou bien comme HK <lb/>au circuit de cette vis; </s> <s xml:id="echoid-s2098" xml:space="preserve">& </s> <s xml:id="echoid-s2099" xml:space="preserve">la puiſſance T, qui au lieu de <lb/>la puiſſance R retient cette vis, eſt auſſi à cette dernié-<lb/>re puiſſance, (Lemm. </s> <s xml:id="echoid-s2100" xml:space="preserve">6. </s> <s xml:id="echoid-s2101" xml:space="preserve">Cor. </s> <s xml:id="echoid-s2102" xml:space="preserve">2.) </s> <s xml:id="echoid-s2103" xml:space="preserve">comme EA à ST, ou <lb/>comme un tour entier de cette vis à la circonférence <lb/>d’un cercle dont ST ſoit le rayon: </s> <s xml:id="echoid-s2104" xml:space="preserve">ainſi en multi-<lb/>pliant encore par ordre ces deux rangées de propor-<lb/>tionnelles, on aura encore la puiſſance T à la charge <lb/>de cette vis, comme la diſtance qui eſt entre deux de <lb/>ſes pas, à la circonférence d’un cercle dont le rayon eſt <lb/>égal à la diſtance qui eſt entre cette puiſſance & </s> <s xml:id="echoid-s2105" xml:space="preserve">l’axe de <lb/>cette même vis.</s> <s xml:id="echoid-s2106" xml:space="preserve"/> </p> <pb o="82" file="0108" n="108" rhead="NOUVELLE"/> <p> <s xml:id="echoid-s2107" xml:space="preserve">Donc lors qu’une puiſſance ſoutient un poids, ou <lb/> <anchor type="note" xlink:label="note-0108-01a" xlink:href="note-0108-01"/> l’action de quelqu’autre force que ce ſoit, à l’aide <lb/>d’une vis, ſoit que cette vis ſoit fixe, ou que ce ſoit <lb/>ſon écrouë, cette puiſſance eſt toujours à ce poids, ou <lb/>à cette force, commela diſtance qui eſt entre deux des <lb/>pas de cette vis, à la circonférence d’un cercle, dont <lb/>le rayon eſt égal à la diſtance qui eſt entre cette puiſ-<lb/>ſance & </s> <s xml:id="echoid-s2108" xml:space="preserve">l’axe de cette même vis. </s> <s xml:id="echoid-s2109" xml:space="preserve">Ce qu’il faloit dé-<lb/>montrer.</s> <s xml:id="echoid-s2110" xml:space="preserve"/> </p> <div xml:id="echoid-div227" type="float" level="2" n="2"> <note position="left" xlink:label="note-0108-01" xlink:href="note-0108-01a" xml:space="preserve">DE LA VIS.</note> </div> </div> <div xml:id="echoid-div229" type="section" level="1" n="151"> <head xml:id="echoid-head151" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s2111" xml:space="preserve">D’où l’on voit que pour peu que la raiſon d’une <lb/>puiſſance à un poids, ou à quelqu’autre force, ſur-<lb/>paſſe celle de la diſtance, qui eſt entre deux des pas <lb/>d’une vis, à la circonférence d’un cercle, dont le <lb/>rayon ſoit la diſtance de l’axe de cette vis à cette <lb/>puiſſance, elle pourra ainſi appliquée ſurmonter ce <lb/>poids, ou cette force à l’aide de cette vis, & </s> <s xml:id="echoid-s2112" xml:space="preserve">elle le <lb/>fera dautant plus aiſément que cette raiſon ſera plus <lb/>grande.</s> <s xml:id="echoid-s2113" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s2114" xml:space="preserve">L’obſtacle que le frotement de la vis avec ſon écrouë <lb/>fait à ce mouvement, doit être compté comme faiſant partie <lb/>de ſa charge: </s> <s xml:id="echoid-s2115" xml:space="preserve">c’eſt ainſi qu’on la peut réduire à une juſteſſe <lb/>Mathématique, de meme que toute autre machine.</s> <s xml:id="echoid-s2116" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div230" type="section" level="1" n="152"> <head xml:id="echoid-head152" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s2117" xml:space="preserve">D’où il ſuit que plus les pas d’une vis ſont ſerrez, <lb/>& </s> <s xml:id="echoid-s2118" xml:space="preserve">que la diſtance de ſon axe à la puiſſance qui y eſt <lb/>appliquée, eſt grande, plus il eſt facile à cette puiſ-<lb/>ſance de ſurmonter le poids, ou la force qui agit <lb/>contr’elle.</s> <s xml:id="echoid-s2119" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div231" type="section" level="1" n="153"> <head xml:id="echoid-head153" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s2120" xml:space="preserve">Il ſuit encore de cette propoſition qu’une même <pb o="83" file="0109" n="109" rhead="MECHANIQUE."/> puiſſance peut également mouvoir un même poids, <lb/> <anchor type="note" xlink:label="note-0109-01a" xlink:href="note-0109-01"/> ou ſurmonter une même force, à l’aide d’une même <lb/>vis, ſoit qu’on la ſuppoſe appliquée à cette vis, où <lb/>qu’elle le ſoit à ſon écrouë pourvû qu’elle ſoit éga-<lb/>lement diſtante de ſon axe dans l’un & </s> <s xml:id="echoid-s2121" xml:space="preserve">l’autre cas.</s> <s xml:id="echoid-s2122" xml:space="preserve"/> </p> <div xml:id="echoid-div231" type="float" level="2" n="1"> <note position="right" xlink:label="note-0109-01" xlink:href="note-0109-01a" xml:space="preserve">DELA VIS.</note> </div> <p style="it"> <s xml:id="echoid-s2123" xml:space="preserve">On ne dit rien ici du Coin, de la Scie, du Mortier, &</s> <s xml:id="echoid-s2124" xml:space="preserve">c. <lb/></s> <s xml:id="echoid-s2125" xml:space="preserve">ce ſont plûtôt des inſtrumens, ou des outils propres à faciliter <lb/>la diviſion d’un corps, que des machines qui puiſſent en fa-<lb/>ciliter le mouvement; </s> <s xml:id="echoid-s2126" xml:space="preserve">on aura peut-ê<unsure/>tre un jour occaſion d’en <lb/>parler.</s> <s xml:id="echoid-s2127" xml:space="preserve"/> </p> <pb file="0110" n="110"/> <pb file="0111" n="111"/> </div> <div xml:id="echoid-div233" type="section" level="1" n="154"> <head xml:id="echoid-head154" xml:space="preserve">EXAMEN <lb/>DE L’OPINION <lb/>DE M BORELLI <lb/>SUR LES PROPRIETEZ DES POIDS <lb/>ſuſpendus par des cordes.</head> <pb file="0112" n="112"/> <pb file="0113" n="113"/> <figure> <image file="0113-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0113-01"/> </figure> </div> <div xml:id="echoid-div234" type="section" level="1" n="155"> <head xml:id="echoid-head155" xml:space="preserve">AVERTISSEMENT.</head> <p style="it"> <s xml:id="echoid-s2128" xml:space="preserve">C’EST ici l’éxamen promis dans la réfléxion qui <lb/>ſuit la preuve de la premiére propoſition du Trojet <lb/>précédent. </s> <s xml:id="echoid-s2129" xml:space="preserve">On ne croyoit pas d’abord le pouvoir donner <lb/>ici; </s> <s xml:id="echoid-s2130" xml:space="preserve">mais s’étant trouvé fait plûtôt qu’on ne s’y étoit <lb/>attendu, on le joint à ce projet. </s> <s xml:id="echoid-s2131" xml:space="preserve">On a été naturelle-<lb/>ment conduit par les principes qu’on y ſuit, à une <lb/>propoſition ſur les proprietez des poids ſuſpendus par <lb/>des cordes, qui s’eſt trouvée la meſme que celle que <lb/>Monſieur Borelli avoit critiquée dans Stévin, <lb/>& </s> <s xml:id="echoid-s2132" xml:space="preserve">dans Hérigone; </s> <s xml:id="echoid-s2133" xml:space="preserve">& </s> <s xml:id="echoid-s2134" xml:space="preserve">ç’a été par la neceſſité de <lb/>la juſtiſier qu’on s’eſt trouvé engagé à l’examen de ſa <lb/>critique.</s> <s xml:id="echoid-s2135" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s2136" xml:space="preserve">On diviſe cet examen en deux Cbapitres: </s> <s xml:id="echoid-s2137" xml:space="preserve">dans le <lb/>premier on fait voir que le ſentiment que M. <lb/></s> <s xml:id="echoid-s2138" xml:space="preserve">Borelli reprend dans H’erigone, dans Stévin, & </s> <s xml:id="echoid-s2139" xml:space="preserve">dans <lb/>les autres; </s> <s xml:id="echoid-s2140" xml:space="preserve">bien loin d’ètre contraire, comme il l’a <lb/>crû, à la 68. </s> <s xml:id="echoid-s2141" xml:space="preserve">propoſition du Tome premier de ſon traité <lb/>du mouvement des Animaux, en eſt une ſuite ſi né-<lb/>ceſſaire, que s’il eut fait encore quelques pas, il y ſeroit <lb/>infailliblement entré. </s> <s xml:id="echoid-s2142" xml:space="preserve">On indique enſuite dans ce <lb/>meſme Chapitre quelques paralogiſmes que cet Au-<lb/>theur a commis, lors meſme qu’il croyoit en voir dans <lb/>les raiſonnemens qu’il a critiquez.</s> <s xml:id="echoid-s2143" xml:space="preserve"/> </p> <pb file="0114" n="114" rhead="AVERTISSEMENT."/> <p style="it"> <s xml:id="echoid-s2144" xml:space="preserve">Dans le ſecond Cbapitre, apres avoir encore donné <lb/>quelques démonſtrations du ſentiment d’Hèrigone & </s> <s xml:id="echoid-s2145" xml:space="preserve"><lb/>des autres, toutes différentes de celles que M. </s> <s xml:id="echoid-s2146" xml:space="preserve">Borelli <lb/>a critiqu@es; </s> <s xml:id="echoid-s2147" xml:space="preserve">on rend par la metbode du Projet <lb/>précedent les Lemmes ſur leſquels cet Au beur a fondé <lb/>tout ce qu’il a dit de la force des Muſcles, beaucoup <lb/>plus généraux qu’ils ne le peuvent eſtre par la ſienne.</s> <s xml:id="echoid-s2148" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s2149" xml:space="preserve">Au reſte ſi l’on attaque une erreur ou M. <lb/></s> <s xml:id="echoid-s2150" xml:space="preserve">Borelli eſt tombé, on n’en eſt pas moins perſuadé <lb/>du mérite ex@raordinaire de ce grand bomme, dont <lb/>les principaux Ouvrages doivent eſtre mis au nombre <lb/>des Livres les plus originaux qui ayent paru dans ce <lb/>ſiecle-ci; </s> <s xml:id="echoid-s2151" xml:space="preserve">mais il n’y a perſonne qui ne puiſſe faire un <lb/>fauxpas, ſur tout dans des matieres auſſi delicates que <lb/>celles-ci, & </s> <s xml:id="echoid-s2152" xml:space="preserve">où le paralogiſme ſe gliſſe auſſi facile-<lb/>ment.</s> <s xml:id="echoid-s2153" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s2154" xml:space="preserve">Tout ce qu’on citera de cet Auteur dans cet exa-<lb/>men, ſera pris du Tome premier de ſon Traité du mou-<lb/>vement des Animaux, de l’édition de Rome, faite en <lb/>1680. </s> <s xml:id="echoid-s2155" xml:space="preserve">On ſpécifie l’édition à cauſe des pages qu’on en <lb/>citera quelques fois.</s> <s xml:id="echoid-s2156" xml:space="preserve"/> </p> <pb o="89" file="0115" n="115"/> <figure> <image file="0115-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0115-01"/> </figure> </div> <div xml:id="echoid-div235" type="section" level="1" n="156"> <head xml:id="echoid-head156" xml:space="preserve">EXAMEN <lb/>DE L’OPINION <lb/>DE M. BORELLI <lb/>Sur les propriétez des Poids ſuſpendus <lb/>par des cordes.</head> <figure> <image file="0115-02" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0115-02"/> </figure> </div> <div xml:id="echoid-div236" type="section" level="1" n="157"> <head xml:id="echoid-head157" style="it" xml:space="preserve">ET AT DE LA QUESTION.</head> <p> <s xml:id="echoid-s2157" xml:space="preserve"><emph style="sc">MOnsieur</emph> <emph style="sc">Borelli</emph> dans ſon Traité du <lb/>mouvement des Animaux Tome 1. </s> <s xml:id="echoid-s2158" xml:space="preserve">Chapi-<lb/>tre 13. </s> <s xml:id="echoid-s2159" xml:space="preserve">a fait une fort longue digreſſion pour prou-<lb/>ver que Hérigone, Stévin, & </s> <s xml:id="echoid-s2160" xml:space="preserve">pluſieurs autres ſe <lb/>ſont trompez d’avoir avancé comme propoſition gé-<lb/>nérale que le poids T ſoutenu avec les cordes obliques AC <lb/> <anchor type="note" xlink:label="note-0115-01a" xlink:href="note-0115-01"/> & </s> <s xml:id="echoid-s2161" xml:space="preserve">BC par deux poids, ou deux puiſſances R & </s> <s xml:id="echoid-s2162" xml:space="preserve">S; </s> <s xml:id="echoid-s2163" xml:space="preserve">eſt à chacun <pb o="90" file="0116" n="116" rhead="EXAMEN DE L’OPINION"/> d’eux, ou d’elles, comme la partie HC de ſa ligne de di-<lb/> <anchor type="note" xlink:label="note-0116-01a" xlink:href="note-0116-01"/> rection à chacun des côtez CN & </s> <s xml:id="echoid-s2164" xml:space="preserve">MC du parallelogramme <lb/>MN, dont elle eſt diagonale. </s> <s xml:id="echoid-s2165" xml:space="preserve">Cet Autheur dit (pag. <lb/></s> <s xml:id="echoid-s2166" xml:space="preserve">” 137.) </s> <s xml:id="echoid-s2167" xml:space="preserve">que cette propoſition priſe dans toute ſon <lb/>étenduë & </s> <s xml:id="echoid-s2168" xml:space="preserve">ſans reſtriction, lui paroît ſuſpecte <lb/>pour bien des raiſons; </s> <s xml:id="echoid-s2169" xml:space="preserve">& </s> <s xml:id="echoid-s2170" xml:space="preserve">même qu’il la croit capa-<lb/>ble de jetter dans l’erreur.</s> <s xml:id="echoid-s2171" xml:space="preserve"/> </p> <div xml:id="echoid-div236" type="float" level="2" n="1"> <note position="right" xlink:label="note-0115-01" xlink:href="note-0115-01a" xml:space="preserve">ſig. @@</note> <note position="left" xlink:label="note-0116-01" xlink:href="note-0116-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> <p> <s xml:id="echoid-s2172" xml:space="preserve">Il réduit toutes ces prétenduës raiſons à trois: <lb/></s> <s xml:id="echoid-s2173" xml:space="preserve">1°. </s> <s xml:id="echoid-s2174" xml:space="preserve">il dit (pag. </s> <s xml:id="echoid-s2175" xml:space="preserve">138.) </s> <s xml:id="echoid-s2176" xml:space="preserve">avoir démontré dans le ſcholie <lb/>” de la 68. </s> <s xml:id="echoid-s2177" xml:space="preserve">propoſition du Tome 1. </s> <s xml:id="echoid-s2178" xml:space="preserve">de ce traité, que <lb/>les deux puiſſances R & </s> <s xml:id="echoid-s2179" xml:space="preserve">S appliquées au poids T <lb/>ſuivant des direction obliques, peuvent demeurer <lb/>en équilibre avec lui, non ſeulement quelque <lb/>raport qu’elles ayent entr’elles, fût-il plus grand, <lb/>ou moindre que celui de NC à CM, mais encore de <lb/>quelque maniére que le raport de la ſomme de ces <lb/>deux puiſſances à ce poids, fût différent de celui <lb/>de la ſomme de NC & </s> <s xml:id="echoid-s2180" xml:space="preserve">MC à CH. </s> <s xml:id="echoid-s2181" xml:space="preserve">2°. </s> <s xml:id="echoid-s2182" xml:space="preserve">il a fait, dit-<lb/>il, auſſi pluſieurs expériences qui lui paroiſſent <lb/>confirmer ce ſentiment. </s> <s xml:id="echoid-s2183" xml:space="preserve">3°. </s> <s xml:id="echoid-s2184" xml:space="preserve">Enfin il a crû voir du <lb/>paralogiſme dans deux démonſtrations qu’il a criti-<lb/>quées, dont la premiére paroît être du P.</s> <s xml:id="echoid-s2185" xml:space="preserve">Pardie, & </s> <s xml:id="echoid-s2186" xml:space="preserve">l’au-<lb/>tre commune au reſte des Autheurs qu’il attaque.</s> <s xml:id="echoid-s2187" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2188" xml:space="preserve">Il eſt conſtant que de toutes ces raiſons la premiére <lb/>eſt non ſeulement la principale, mais encore l’unique <lb/>qui puiſſe ſervir à la deciſion de ce différent: </s> <s xml:id="echoid-s2189" xml:space="preserve">Car 1°. </s> <s xml:id="echoid-s2190" xml:space="preserve">en <lb/>fait d’exactitude & </s> <s xml:id="echoid-s2191" xml:space="preserve">de préciſion, l’expérience ne <lb/>prouve rien; </s> <s xml:id="echoid-s2192" xml:space="preserve">ſur tout ici, ou la réſiſtance qui vient <lb/>du frotement des poulies avec leurs pivots, &</s> <s xml:id="echoid-s2193" xml:space="preserve">c. </s> <s xml:id="echoid-s2194" xml:space="preserve">rend <lb/>ces ſortes d’expériences poſſibles en tant de manieres <lb/>différentes, qu’il n’y a preſque point de ſentiment <lb/>pour, ou contre lequel on n’en puiſſe faire à ſon gré. <lb/></s> <s xml:id="echoid-s2195" xml:space="preserve">2°. </s> <s xml:id="echoid-s2196" xml:space="preserve">Qu’il y ait, ou qu’il n’y ait point de paralogiſme <pb o="91" file="0117" n="117" rhead="DE M. BORELLI."/> dans les raiſonnemens qué cet Autheur critique, on <lb/> <anchor type="note" xlink:label="note-0117-01a" xlink:href="note-0117-01"/> n’en peut rien conclure non plus contre le ſentiment <lb/>qu’il attaque; </s> <s xml:id="echoid-s2197" xml:space="preserve">puiſque la vérité ne dépend point du <lb/>tout de la maniere dont on la démontre.</s> <s xml:id="echoid-s2198" xml:space="preserve"/> </p> <div xml:id="echoid-div237" type="float" level="2" n="2"> <note position="right" xlink:label="note-0117-01" xlink:href="note-0117-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſe@-<lb/>lemen@.</note> </div> <p> <s xml:id="echoid-s2199" xml:space="preserve">Toute la queſtion préſente ſe réduit donc à ſça-<lb/>voir ſi en effet M. </s> <s xml:id="echoid-s2200" xml:space="preserve">Borelli a démontré dans le <lb/>ſcholie de ſa 6 8. </s> <s xml:id="echoid-s2201" xml:space="preserve">Propoſition Tome 1. </s> <s xml:id="echoid-s2202" xml:space="preserve">que les “ <lb/>deux puiſſances R & </s> <s xml:id="echoid-s2203" xml:space="preserve">S appliquées au poids T <lb/>ſuivant des directions obliques, peuvent demeu-<lb/>rer en équilibre avec lui, non ſeulement quelque <lb/>raport qu’elles ayent entr’elles, fût-il plus grand, <lb/>ou moindre que celui de NC à CM; </s> <s xml:id="echoid-s2204" xml:space="preserve">mais encore <lb/>de quelque maniére que le raport de la ſomme de <lb/>ces deux puiſſances à ce poids, fût différent de <lb/>celui de la ſomme de NC & </s> <s xml:id="echoid-s2205" xml:space="preserve">MC à CH. </s> <s xml:id="echoid-s2206" xml:space="preserve">Que dî-je? <lb/></s> <s xml:id="echoid-s2207" xml:space="preserve">Ce ſeroit aſſez pour détruire la propoſition qu’il <lb/>rejette, s’il avoit ſeulement démontré un cas ou ce <lb/>poids pût ainſi demeurer en équilibré avec ces puiſ-<lb/>ſances, ſans être à chacune d’elles, comme la partie <lb/>CH de ſa ligne de direction, qui fait la diagonale du <lb/>parallelogramme MN, à chacune des parties de leurs <lb/>cordes, qui lui ſervent de côtez. </s> <s xml:id="echoid-s2208" xml:space="preserve">Mais bien loin de <lb/>l’avoir fait, la propoſition d’où il tire le ſcholie en <lb/>queſtion, prouve tout le contraire, je veux dire, le <lb/>ſentiment d’où il a crû qu’elle le devoit cloigner.</s> <s xml:id="echoid-s2209" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2210" xml:space="preserve">C’eſt ce qu’on va faire voir dans le premier <lb/>Chapitre de cet Examen; </s> <s xml:id="echoid-s2211" xml:space="preserve">& </s> <s xml:id="echoid-s2212" xml:space="preserve">dans le ſecond, aprés <lb/>avoir encore donné quelques démonſtrations de ce <lb/>même ſentiment, toutes différentes de celles que <lb/>M. </s> <s xml:id="echoid-s2213" xml:space="preserve">Borelli a critiquées, on rendra par la méthode <lb/>du Projet précédent les Lemmes qu’il a déduits de ſa <lb/>68. </s> <s xml:id="echoid-s2214" xml:space="preserve">propoſition, beaucoup plus généraux qu’ils ne le <lb/>peuvent être par la ſienne.</s> <s xml:id="echoid-s2215" xml:space="preserve"/> </p> <pb o="92" file="0118" n="118" rhead="EXAMEN DE L’OPINION"/> <figure> <image file="0118-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0118-01"/> </figure> </div> <div xml:id="echoid-div239" type="section" level="1" n="158"> <head xml:id="echoid-head158" xml:space="preserve">CHAPITRE I. <lb/>SENTIMENT D’HERIGONE, <lb/>DE STEVIN, &c. <lb/>SUR LES PROPRIETEZ DES POIDS <lb/>ſuſpendus par des cordes, <lb/>Démontré par la propoſition même que M. BORELLI <lb/>avoit cru leur être contraire.</head> <p> <s xml:id="echoid-s2216" xml:space="preserve">LE ſcholie de la 68. </s> <s xml:id="echoid-s2217" xml:space="preserve">propoſition de M. </s> <s xml:id="echoid-s2218" xml:space="preserve">Borel-<lb/>li, dont il eſt ici queſtion, porte ” 1°. </s> <s xml:id="echoid-s2219" xml:space="preserve">Qu’a-<lb/>” vec les mêmes inclinaiſons de cordes, le poids qui <lb/>y eſt ſuſpendu, & </s> <s xml:id="echoid-s2220" xml:space="preserve">les forces qui le ſoutiennent, <lb/>peuvent varier en mille manieres différentes ſans <lb/> <anchor type="note" xlink:label="note-0118-01a" xlink:href="note-0118-01"/> que pour cela il ceſſe de faire équilibre avec elles; <lb/></s> <s xml:id="echoid-s2221" xml:space="preserve">pourvû que la puiſſance R ſoit à la partie X de ce poids, <lb/>comme GC à GF, ou comme AC à CH; </s> <s xml:id="echoid-s2222" xml:space="preserve">& </s> <s xml:id="echoid-s2223" xml:space="preserve">que <lb/>la puiſſance S ſoit à ſon autre partie Z, comme IC à <lb/>IK, ou comme BC à CH.</s> <s xml:id="echoid-s2224" xml:space="preserve">..</s> <s xml:id="echoid-s2225" xml:space="preserve">..</s> <s xml:id="echoid-s2226" xml:space="preserve">..</s> <s xml:id="echoid-s2227" xml:space="preserve">. 2° Ce même ſcholie <lb/>porte réciproquement qu’en retenant les mêmes <lb/>poids; </s> <s xml:id="echoid-s2228" xml:space="preserve">c’eſt - à - dire ici, les mêmes forces, (pourvû que <lb/>celui du milieu ſoit moindre que les deux extrê-<lb/>mes) on peut changer l’inclinaiſon de leurs cordes <lb/>ſans en rompre l’équilibre.</s> <s xml:id="echoid-s2229" xml:space="preserve"/> </p> <div xml:id="echoid-div239" type="float" level="2" n="1"> <note position="left" xlink:label="note-0118-01" xlink:href="note-0118-01a" xml:space="preserve">ſig. 1.</note> </div> <p> <s xml:id="echoid-s2230" xml:space="preserve">Il eſt clair que la premiére partie de ce ſcholie <lb/>peut avoir deux ſens bien différens. </s> <s xml:id="echoid-s2231" xml:space="preserve">1°. </s> <s xml:id="echoid-s2232" xml:space="preserve">Elle peut <lb/>ſignifier que dans cette variation de poids & </s> <s xml:id="echoid-s2233" xml:space="preserve">de <pb o="93" file="0119" n="119" rhead="DE M. BORELLI."/> forces, ou cet Autheur veut que l’équilibre ſe con-<lb/> <anchor type="note" xlink:label="note-0119-01a" xlink:href="note-0119-01"/> ſerve ſans changer l’inclinaiſon de leurs cordes, ce <lb/>poids demeure toujours à chacune de ces puiſſances <lb/>en même raiſon que la diagonale CH du parallelo-<lb/>gramme MN, à chaque partie de leurs cordes, qui <lb/>lui ſert de côté; </s> <s xml:id="echoid-s2234" xml:space="preserve">& </s> <s xml:id="echoid-s2235" xml:space="preserve">en cela on va voir que cette conſé-<lb/>quence eſt parfaitement juſte, mais auſſi (Cor. </s> <s xml:id="echoid-s2236" xml:space="preserve">12. <lb/></s> <s xml:id="echoid-s2237" xml:space="preserve">de la Prop. </s> <s xml:id="echoid-s2238" xml:space="preserve">fond. </s> <s xml:id="echoid-s2239" xml:space="preserve">des poids ſuſpendus par des cordes, du <lb/>Proiet précéd.) </s> <s xml:id="echoid-s2240" xml:space="preserve">parfaitement conforme au ſentiment <lb/>que cet Autheur attaque. </s> <s xml:id="echoid-s2241" xml:space="preserve">2°. </s> <s xml:id="echoid-s2242" xml:space="preserve">Au contraire, ſi on lui <lb/>fait ſigniſier que cet équilibre puiſſe ſubſiſter ſans un <lb/>tel raport; </s> <s xml:id="echoid-s2243" xml:space="preserve">alors on conclud tres-mal, & </s> <s xml:id="echoid-s2244" xml:space="preserve">même au-<lb/>tant contre cet Autheur que contre Hérigone, <lb/>&</s> <s xml:id="echoid-s2245" xml:space="preserve">c. </s> <s xml:id="echoid-s2246" xml:space="preserve">J’en dis tout autant de la ſeconde partie de <lb/>ce même ſcholie; </s> <s xml:id="echoid-s2247" xml:space="preserve">& </s> <s xml:id="echoid-s2248" xml:space="preserve">pour le démontrer, je vais <lb/>faire voir que la propoſition d’où M. </s> <s xml:id="echoid-s2249" xml:space="preserve">Borelli le <lb/>tire, prouve tout le contraire, & </s> <s xml:id="echoid-s2250" xml:space="preserve">que s’il y eût <lb/>fait un peu plus d’attention, elle l’auroit infaillible-<lb/>ment conduit au ſentiment d’où il a crû qu’elle le <lb/>devoit éloigner; </s> <s xml:id="echoid-s2251" xml:space="preserve">je veux dire, à croire (du moins <lb/>pour tous les cas qu’elle comprend) que le poids T <lb/>ſoutenu avec les cordes obliques AC & </s> <s xml:id="echoid-s2252" xml:space="preserve">BC par deux poids, <lb/>ou deux puiſſances R & </s> <s xml:id="echoid-s2253" xml:space="preserve">S, eſt toujours à chacun d’eux, ou <lb/>d’elles, comme la partie HC de ſa ligne de direction, à cbacun <lb/>descôtez CN & </s> <s xml:id="echoid-s2254" xml:space="preserve">MC du parallelogramme MN, dont elle <lb/>eſt diagonale. </s> <s xml:id="echoid-s2255" xml:space="preserve">Voici comment.</s> <s xml:id="echoid-s2256" xml:space="preserve"/> </p> <div xml:id="echoid-div240" type="float" level="2" n="2"> <note position="right" xlink:label="note-0119-01" xlink:href="note-0119-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avcc <lb/>des cordes ſeu-<lb/>lement.</note> </div> <p> <s xml:id="echoid-s2257" xml:space="preserve">Selon cet Autheur (Prop. </s> <s xml:id="echoid-s2258" xml:space="preserve">68) lorſque les puiſ-<lb/>ſances R & </s> <s xml:id="echoid-s2259" xml:space="preserve">S ſoutiennent enſemble le poids T, la <lb/>puiſſance R ſoutient pour ſa part une partie X de ce <lb/>poids, de même qu’elle feroit, ſi elle étoit appliquée <lb/>ſuivant ſa même direction AC avec cette partie X au <lb/>levier horizontal CG; </s> <s xml:id="echoid-s2260" xml:space="preserve">& </s> <s xml:id="echoid-s2261" xml:space="preserve">la puiſſance S ſoutient auſſi <lb/>pour la ſienne l’autre partie Z de ce même poids, <lb/>de même qu’elle feroit, ſi elle étoit auſſi appliquée <pb o="94" file="0120" n="120" rhead="EXAMEN DE L’OPINION"/> ſuivant ſa même direction CB avec cette partie Z au <lb/> <anchor type="note" xlink:label="note-0120-01a" xlink:href="note-0120-01"/> levier CI qu’on ſuppoſe encore horizontal & </s> <s xml:id="echoid-s2262" xml:space="preserve">égal <lb/>au premier: </s> <s xml:id="echoid-s2263" xml:space="preserve">& </s> <s xml:id="echoid-s2264" xml:space="preserve">par conſéquent ſi l’on regarde (Cor 2. <lb/></s> <s xml:id="echoid-s2265" xml:space="preserve">Lem. </s> <s xml:id="echoid-s2266" xml:space="preserve">3. </s> <s xml:id="echoid-s2267" xml:space="preserve">du Projet précéd.) </s> <s xml:id="echoid-s2268" xml:space="preserve">l’impreſſion que la puiſſance <lb/>S fait ſuivant CB ſur le noeud C qui retient enſemble <lb/>les cordes de ces puiſſances & </s> <s xml:id="echoid-s2269" xml:space="preserve">de ce poids, comme <lb/>compoſée de deux impreſſions particuliéres, dont <lb/>l’une eſt ſuivant l’horizontale CO, & </s> <s xml:id="echoid-s2270" xml:space="preserve">l’autre ſuivant <lb/>la perpendiculaire CH; </s> <s xml:id="echoid-s2271" xml:space="preserve">on trouvera que ce que cette <lb/>puiſſance lui en fait ſuivant CO, eſt égal à la réſiſ-<lb/>tance que feroit alors contre ce même point, & </s> <s xml:id="echoid-s2272" xml:space="preserve">ſui-<lb/>vant cette même ligne, le levier CG pour empêcher <lb/>la corde ACX de ſe redreſſer; </s> <s xml:id="echoid-s2273" xml:space="preserve">c’eſt-à-dire, égal à la <lb/>charge de l’apui G de ce même levier. </s> <s xml:id="echoid-s2274" xml:space="preserve">Or (Cor. </s> <s xml:id="echoid-s2275" xml:space="preserve">4. </s> <s xml:id="echoid-s2276" xml:space="preserve"><lb/>Prop. </s> <s xml:id="echoid-s2277" xml:space="preserve">fond. </s> <s xml:id="echoid-s2278" xml:space="preserve">des leviers du Proj. </s> <s xml:id="echoid-s2279" xml:space="preserve">précéd.) </s> <s xml:id="echoid-s2280" xml:space="preserve">la puiſſance R <lb/>eſt à la charge de cet apui, comme le côté AC du <lb/>parallelogramme AE, à ſa diagonale CG; </s> <s xml:id="echoid-s2281" xml:space="preserve">& </s> <s xml:id="echoid-s2282" xml:space="preserve">la force <lb/>de l’impreſſion que fait la puiſſance S ſur ce même <lb/>point C ſuivant CO, eſt (Cor. </s> <s xml:id="echoid-s2283" xml:space="preserve">3. </s> <s xml:id="echoid-s2284" xml:space="preserve">Lemm. </s> <s xml:id="echoid-s2285" xml:space="preserve">3. </s> <s xml:id="echoid-s2286" xml:space="preserve">du Projet <lb/>précéd.) </s> <s xml:id="echoid-s2287" xml:space="preserve">à celle de cette même puiſſance ſuivant CB, <lb/>c’eſt-à-dire, à cette puiſſance, elle-même, comme <lb/>le côté OC du parallelogramme OH, à ſa diagonale <lb/>BC; </s> <s xml:id="echoid-s2288" xml:space="preserve">c’eſt-à-dire, en faiſant IK perpendiculaire ſur <lb/>BC, comme CK à CI égale (Hyp.) </s> <s xml:id="echoid-s2289" xml:space="preserve">à CG: </s> <s xml:id="echoid-s2290" xml:space="preserve">Donc la <lb/>puiſſance R eſt à la puiſſance S, comme le produit <lb/>de AC par CK, au quarré de CG. </s> <s xml:id="echoid-s2291" xml:space="preserve">Or en faiſant GF <lb/>perpendiculaire ſur AC, les triangles AGC & </s> <s xml:id="echoid-s2292" xml:space="preserve">GFC <lb/>étant ſemblables, le produit de AC par CF eſt égal <lb/>au quarré de CG: </s> <s xml:id="echoid-s2293" xml:space="preserve">Donc la puiſſance R eſt à la puiſ-<lb/>ſance S, comme le produit de AC par CK au produit <lb/>de la même AC par CF, c’eſt-à-dire, comme CK à <lb/>CF; </s> <s xml:id="echoid-s2294" xml:space="preserve">ou bien, à cauſe des rayons IC & </s> <s xml:id="echoid-s2295" xml:space="preserve">CG (byp.) </s> <s xml:id="echoid-s2296" xml:space="preserve"><lb/>égaux, comme le ſinus de KIC égal à BCH, au ſinus <lb/>de FGC égal à ACH: </s> <s xml:id="echoid-s2297" xml:space="preserve">Donc la puiſſance R eſt à la <lb/>puiſſance S, comme le ſinus de l’angle BCH à celui <pb o="95" file="0121" n="121" rhead="DE M. BORELLI."/> de l’angle ACH qui, à cauſe du parallelogramme <lb/> <anchor type="note" xlink:label="note-0121-01a" xlink:href="note-0121-01"/> MN, eſt égal à CHM: </s> <s xml:id="echoid-s2298" xml:space="preserve">Donc (Lemm. </s> <s xml:id="echoid-s2299" xml:space="preserve">5. </s> <s xml:id="echoid-s2300" xml:space="preserve">du Projet <lb/>précéd) la puiſſance R eſt à la puiſſance S, comme <lb/>HM, ou comme ſon égale NC à CM.</s> <s xml:id="echoid-s2301" xml:space="preserve"/> </p> <div xml:id="echoid-div241" type="float" level="2" n="3"> <note position="left" xlink:label="note-0120-01" xlink:href="note-0120-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>l@ment.</note> <note position="right" xlink:label="note-0121-01" xlink:href="note-0121-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> <p> <s xml:id="echoid-s2302" xml:space="preserve">Voilà ce que M. </s> <s xml:id="echoid-s2303" xml:space="preserve">Borelli devoit premiérement <lb/>conclure de ſa 68. </s> <s xml:id="echoid-s2304" xml:space="preserve">propoſition, ſinon en géné-<lb/>ral, du moins pour tous les cas qu’elle com-<lb/>prend: </s> <s xml:id="echoid-s2305" xml:space="preserve">Sçavoir que lorſque deux puiſſances R & </s> <s xml:id="echoid-s2306" xml:space="preserve">S ſou-<lb/>tiennent enſemble quelque poids T avec des cordes ſeulement, <lb/>elles ſont toujours @ntr’elles en même raiſon que les parties <lb/>CN & </s> <s xml:id="echoid-s2307" xml:space="preserve">MC de leurs cordes, qui ſervent de côtez au pa-<lb/>rallelogramme MN, qui a pour diagonale une partie CH <lb/>de la li@ne de direction du poids qu’elles ſoutiennent. </s> <s xml:id="echoid-s2308" xml:space="preserve">De là <lb/>en faiſant MP & </s> <s xml:id="echoid-s2309" xml:space="preserve">NQ perpendiculaires ſur HC, ces <lb/>lignes marquant toujours CP égale à HQ, cet Au-<lb/>theur auroit trouvé, comme il a fait (pag. </s> <s xml:id="echoid-s2310" xml:space="preserve">137.) <lb/></s> <s xml:id="echoid-s2311" xml:space="preserve">que chacune des puiſſances R & </s> <s xml:id="echoid-s2312" xml:space="preserve">S, étant toujours (par <lb/>le Cor. </s> <s xml:id="echoid-s2313" xml:space="preserve">de ſa 69. </s> <s xml:id="echoid-s2314" xml:space="preserve">Prop.) </s> <s xml:id="echoid-s2315" xml:space="preserve">à tout ce poids, comme chacun <lb/>des côtez CN & </s> <s xml:id="echoid-s2316" xml:space="preserve">MC du parallelogramme MN, à la <lb/>ſomme de leurs ſublimitez CP & </s> <s xml:id="echoid-s2317" xml:space="preserve">CQ; </s> <s xml:id="echoid-s2318" xml:space="preserve">lui eſt auſſi <lb/>toujours comme cbacun de ces mêmes côtez à la diagonale <lb/>CH de ce même para@lelogramme. </s> <s xml:id="echoid-s2319" xml:space="preserve">Ce qu’il faloit démon-<lb/>trer.</s> <s xml:id="echoid-s2320" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2321" xml:space="preserve">Quoi que cette conſéquence ſuive néceſſairement de la <lb/>68. </s> <s xml:id="echoid-s2322" xml:space="preserve">Propoſition de M. </s> <s xml:id="echoid-s2323" xml:space="preserve">Borelli, cependant parce-que cette <lb/>Propoſition ne peut pas s’appliquer aux cas ou une de <lb/>ces pu@ſſances ſe trouve avoir ſa direction au deſſous de <lb/>l’borizontale qui paſſe par le point où leurs cordes ſe com-<lb/>muniquent, elle n’en eſt pas une ſuitte ſi générale que de la <lb/>Propoſition fondamentale des poids ſuſpendus par des cordes <lb/>du Projet précédent: </s> <s xml:id="echoid-s2324" xml:space="preserve">C’eſt pour cela qu’on ſe contente ici de <lb/>dire, que ſi cet Autbeur eût fait un peu plus d’attention à <lb/>ſa 68. </s> <s xml:id="echoid-s2325" xml:space="preserve">Prop. </s> <s xml:id="echoid-s2326" xml:space="preserve">il auroit aperçû que tout ce que nous venons <lb/>d’en conclure, eſt abſolument vrai, du moins pour tous <lb/>les cas qu’elle comprend.</s> <s xml:id="echoid-s2327" xml:space="preserve"/> </p> <pb o="96" file="0122" n="122" rhead="EXAMEN DE L’OPINION"/> <p> <s xml:id="echoid-s2328" xml:space="preserve">Telle eſt la conſéquence que Monſieur Borelli <lb/> <anchor type="note" xlink:label="note-0122-01a" xlink:href="note-0122-01"/> devoit tirer de ſa 68. </s> <s xml:id="echoid-s2329" xml:space="preserve">Propoſition; </s> <s xml:id="echoid-s2330" xml:space="preserve">s’il l’eût fait, il <lb/>auroit aperçû 1°. </s> <s xml:id="echoid-s2331" xml:space="preserve">que la premiére partie du ſcholie <lb/>qu’il en tire, n’eſt vraye qu’en cas que la variation <lb/>du poids T & </s> <s xml:id="echoid-s2332" xml:space="preserve">des forces ſ & </s> <s xml:id="echoid-s2333" xml:space="preserve">S, entre leſquels il dit <lb/>que l’équilibre ſe peut conſerver ſans changer l’in-<lb/>clinaiſon de leurs cordes, ſoit telle que ce poids de-<lb/>meure toujours à chacune de ces puiſſances en même <lb/>raiſon que la diagonale CH du parallelogramme <lb/>MN, à chaque partie de leurs cordes qui lui ſert de <lb/>côté. </s> <s xml:id="echoid-s2334" xml:space="preserve">2°. </s> <s xml:id="echoid-s2335" xml:space="preserve">Il auroit encore vû que la ſeconde partie <lb/>de ce même ſcholie eſt abſolument fauſſe; </s> <s xml:id="echoid-s2336" xml:space="preserve">puis qu’il <lb/>n’eſt pas poſſible de faire le moindre changement <lb/>auquel que ce ſoit des angles ACB, ACH, & </s> <s xml:id="echoid-s2337" xml:space="preserve">BCH, <lb/>ſans changer en même - tems le raport qui eſt, ou <lb/>entre les côtez du parallelogramme MN, ou entre <lb/>quelqu’un d’eux & </s> <s xml:id="echoid-s2338" xml:space="preserve">ſa diagonale; </s> <s xml:id="echoid-s2339" xml:space="preserve">c’eſt - à - dire, puis <lb/>que (byp.) </s> <s xml:id="echoid-s2340" xml:space="preserve">on ne change rien au raport qui eſt entre <lb/>ce poids & </s> <s xml:id="echoid-s2341" xml:space="preserve">chacune de ces puiſſances, ſans faire ceſſer <lb/>la reſſemblance de ces deux raports: </s> <s xml:id="echoid-s2342" xml:space="preserve">& </s> <s xml:id="echoid-s2343" xml:space="preserve">par conſe-<lb/>quent auſſi, ſuivant ce qui vient d’être conclu de la <lb/>68. </s> <s xml:id="echoid-s2344" xml:space="preserve">Propoſition de M. </s> <s xml:id="echoid-s2345" xml:space="preserve">Borelli, ſans rompre l’équili-<lb/>bre de ce poids avec ces puiſſances.</s> <s xml:id="echoid-s2346" xml:space="preserve"/> </p> <div xml:id="echoid-div242" type="float" level="2" n="4"> <note position="left" xlink:label="note-0122-01" xlink:href="note-0122-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des c<unsure/>otdes ſeu-<lb/>lement.</note> </div> </div> <div xml:id="echoid-div244" type="section" level="1" n="159"> <head xml:id="echoid-head159" xml:space="preserve"><emph style="sc">Remarque.</emph></head> <p> <s xml:id="echoid-s2347" xml:space="preserve">Ayant demontré, comme l’on vient de faire, <lb/>que le ſentiment, dont il eſt ici queſtion, bien loin <lb/>d’être contraire à la 68. </s> <s xml:id="echoid-s2348" xml:space="preserve">Propoſit. </s> <s xml:id="echoid-s2349" xml:space="preserve">de M. </s> <s xml:id="echoid-s2350" xml:space="preserve">Borelli, <lb/>comme cet Autheur l’a crû, en eſt une ſuitte ſi né-<lb/>ceſſaire que, s’il eût fait encore quelques pas, il <lb/>l’auroit linfailliblement trouvé: </s> <s xml:id="echoid-s2351" xml:space="preserve">c’eſt encore une nou-<lb/>velle raiſon de ne nous point arrêter aux expériences <lb/>qu’il objecte à Stévin, à Hérigone, & </s> <s xml:id="echoid-s2352" xml:space="preserve">aux autres, & </s> <s xml:id="echoid-s2353" xml:space="preserve"><lb/>de ne toucher à la critique qu’il a faite de leurs rai- <pb o="97" file="0123" n="123" rhead="DE M. BORELLI."/> ſonnemens, que pour indiquer les fauſſes ſuppo-<lb/> <anchor type="note" xlink:label="note-0123-01a" xlink:href="note-0123-01"/> ſitions ſur leſquelles il s’eſt appuyé pour y trouver <lb/>du paralogiſme: </s> <s xml:id="echoid-s2354" xml:space="preserve">Il y en a trois que voici.</s> <s xml:id="echoid-s2355" xml:space="preserve"/> </p> <div xml:id="echoid-div244" type="float" level="2" n="1"> <note position="right" xlink:label="note-0123-01" xlink:href="note-0123-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> <p> <s xml:id="echoid-s2356" xml:space="preserve">1°. </s> <s xml:id="echoid-s2357" xml:space="preserve">Dans la Critique qu’il a faite du premier de <lb/>ces raiſonnemens, qui paroît être du P. </s> <s xml:id="echoid-s2358" xml:space="preserve">Pardie, <lb/>aprés avoir dit que, ſi l’on regarde la corde AC com-<lb/>me une verge de fer mobile autour du point fixe A, <lb/> <anchor type="note" xlink:label="note-0123-02a" xlink:href="note-0123-02"/> à laquelle le poids T ſoit attaché, ce poids ſera ſou-<lb/>tenu avec cette verge par ce point fixe, de même que <lb/>ſur un plan CI perpendiculaire à AC; </s> <s xml:id="echoid-s2359" xml:space="preserve">il fait IL per-<lb/>pendiculaire à l’horizontale LC, & </s> <s xml:id="echoid-s2360" xml:space="preserve">(pag. </s> <s xml:id="echoid-s2361" xml:space="preserve">139.) </s> <s xml:id="echoid-s2362" xml:space="preserve">il <lb/>dit: </s> <s xml:id="echoid-s2363" xml:space="preserve">patet quod pondus T.</s> <s xml:id="echoid-s2364" xml:space="preserve">..</s> <s xml:id="echoid-s2365" xml:space="preserve">..</s> <s xml:id="echoid-s2366" xml:space="preserve">..</s> <s xml:id="echoid-s2367" xml:space="preserve">. ad vim quâ idem T inni-<lb/>titur, & </s> <s xml:id="echoid-s2368" xml:space="preserve">comprimit idem planum IC, eſt ut IC ad LC.</s> <s xml:id="echoid-s2369" xml:space="preserve"/> </p> <div xml:id="echoid-div245" type="float" level="2" n="2"> <note position="right" xlink:label="note-0123-02" xlink:href="note-0123-02a" xml:space="preserve">ſig. 2.</note> </div> <p> <s xml:id="echoid-s2370" xml:space="preserve">2°. </s> <s xml:id="echoid-s2371" xml:space="preserve">Dans la Critique qu’il fait enſuite du rai-<lb/>ſonnement de Hérigone, de Stévin, &</s> <s xml:id="echoid-s2372" xml:space="preserve">c. </s> <s xml:id="echoid-s2373" xml:space="preserve">aprés <lb/> <anchor type="note" xlink:label="note-0123-03a" xlink:href="note-0123-03"/> avoir regardé le poids T ſoutenu par les cordes AC <lb/>& </s> <s xml:id="echoid-s2374" xml:space="preserve">BC, comme s’il l’étoit ſur les plans CK perpendi-<lb/>culaire à AC, & </s> <s xml:id="echoid-s2375" xml:space="preserve">CG perpendiculaire à CB, inéga-<lb/>lement inclinez, il dit (pag. </s> <s xml:id="echoid-s2376" xml:space="preserve">141.) </s> <s xml:id="echoid-s2377" xml:space="preserve">Tunc pondus T <lb/>dum moveri niteretur per duas rectas inclinatas CK & </s> <s xml:id="echoid-s2378" xml:space="preserve">CG, <lb/>cogeretur moveri, aut niſum exercere per diagonalem CO <lb/>ſecantem angulum GCK bifariam.</s> <s xml:id="echoid-s2379" xml:space="preserve"/> </p> <div xml:id="echoid-div246" type="float" level="2" n="3"> <note position="right" xlink:label="note-0123-03" xlink:href="note-0123-03a" xml:space="preserve">ſig. 3.</note> </div> <p> <s xml:id="echoid-s2380" xml:space="preserve">Outre cette ſuppoſition, M. </s> <s xml:id="echoid-s2381" xml:space="preserve">Borelli ſe ſert en-<lb/>core ici de la premiére qu’il a déja faite contre le <lb/>P. </s> <s xml:id="echoid-s2382" xml:space="preserve">Pardie. </s> <s xml:id="echoid-s2383" xml:space="preserve">Il dit (pag. </s> <s xml:id="echoid-s2384" xml:space="preserve">141.) </s> <s xml:id="echoid-s2385" xml:space="preserve">aprés avoir fait CP <lb/>perpendiculaire à l’horizontale KG: </s> <s xml:id="echoid-s2386" xml:space="preserve">Idem pondus ab-<lb/>ſolutum T ad vim, quâ comprimit planum CO, eandem <lb/>rationem babebit quàm CO ad OP.</s> <s xml:id="echoid-s2387" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2388" xml:space="preserve">3°. </s> <s xml:id="echoid-s2389" xml:space="preserve">Enfin ces deux ſuppoſitions ne lui ſuffiſant pas <lb/>encore pour trouver à redire au raiſonnement d’Hé- <pb o="98" file="0124" n="124" rhead="EXAMEN DE L’OPINION"/> rigone, de Stévin, &</s> <s xml:id="echoid-s2390" xml:space="preserve">c. </s> <s xml:id="echoid-s2391" xml:space="preserve">il y en ajoûte une troi-<lb/> <anchor type="note" xlink:label="note-0124-01a" xlink:href="note-0124-01"/> ſiéme qui ne vaut pas mieux. </s> <s xml:id="echoid-s2392" xml:space="preserve">Vis, dit-il au même en-<lb/>droit, quam patitur planum CO à compreſſione ponderis T <lb/>aqualis eſt viribus ambarum potentiarum R & </s> <s xml:id="echoid-s2393" xml:space="preserve">S, quæ <lb/>ſuſtinendo idem pondus in tali ſitu plani CO inclinati vicem <lb/>ſupplent.</s> <s xml:id="echoid-s2394" xml:space="preserve"/> </p> <div xml:id="echoid-div247" type="float" level="2" n="4"> <note position="left" xlink:label="note-0124-01" xlink:href="note-0124-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> <p style="it"> <s xml:id="echoid-s2395" xml:space="preserve">On ne démontre point ici la fauſſeté de toutes ces ſup-<lb/>poſitions: </s> <s xml:id="echoid-s2396" xml:space="preserve">elle eſt trop évidente par la propoſition des Sur-<lb/>faces du Projet précédent, pour s’y arrêter davantage. <lb/></s> <s xml:id="echoid-s2397" xml:space="preserve">D’ailleurs c’eſt, ce me ſemble, avoir ſuffiſamment répondu <lb/>à M. </s> <s xml:id="echoid-s2398" xml:space="preserve">Borelli que d’avoir démontré, comme l’on vient de <lb/>faire, le ſentiment d’Hérigone, de Stévin, &</s> <s xml:id="echoid-s2399" xml:space="preserve">c. </s> <s xml:id="echoid-s2400" xml:space="preserve">par la <lb/>Propoſition même que cet Autbeur croyoit leur être contraire: </s> <s xml:id="echoid-s2401" xml:space="preserve"><lb/>c’eſt auſſi tout ce qu’on s’étoit propoſé dans ce premier Cba-<lb/>pitre; </s> <s xml:id="echoid-s2402" xml:space="preserve">paſſons au ſecond.</s> <s xml:id="echoid-s2403" xml:space="preserve"/> </p> <figure> <image file="0124-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0124-01"/> </figure> <pb o="99" file="0125" n="125" rhead="DE M. BORELLI"/> <figure> <image file="0125-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0125-01"/> </figure> </div> <div xml:id="echoid-div249" type="section" level="1" n="160"> <head xml:id="echoid-head160" xml:space="preserve">CHAPITRE II. <lb/>NOUVELLES DEMONSTRATIONS <lb/>du ſentiment d’Hérigone, de Stévin, &c.</head> <head xml:id="echoid-head161" xml:space="preserve">Sur les propriétez des poids ſuspendus par <lb/>des cordes.</head> <head xml:id="echoid-head162" xml:space="preserve">AVEC QUELQUES PROPOSITIONS <lb/>de M. Borelli renduës par la méthode du Projet <lb/>précédent beaucoup plus générales qu’elles <lb/>ne le peuvent être par la ſienne.</head> <figure> <image file="0125-02" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0125-02"/> </figure> </div> <div xml:id="echoid-div250" type="section" level="1" n="161"> <head xml:id="echoid-head163" xml:space="preserve">AVER TISSEMENT.</head> <p style="it"> <s xml:id="echoid-s2404" xml:space="preserve">LE Poids T étant ſoutenu par deux, ou pluſieurs puiſ-<lb/>ſances R, S, &</s> <s xml:id="echoid-s2405" xml:space="preserve">c. </s> <s xml:id="echoid-s2406" xml:space="preserve">Si des extrémitez G, H, &</s> <s xml:id="echoid-s2407" xml:space="preserve">c. </s> <s xml:id="echoid-s2408" xml:space="preserve">des <lb/> <anchor type="note" xlink:label="note-0125-01a" xlink:href="note-0125-01"/> parties CG, CH, &</s> <s xml:id="echoid-s2409" xml:space="preserve">c. </s> <s xml:id="echoid-s2410" xml:space="preserve">de leurs cordes qui leur ſoient pro-<lb/>portionnelles, on fait GP, HQ, &</s> <s xml:id="echoid-s2411" xml:space="preserve">c. </s> <s xml:id="echoid-s2412" xml:space="preserve">perpendiculaires ſur <lb/>ſa ligne de direction CD; </s> <s xml:id="echoid-s2413" xml:space="preserve">elles y déſigncront depuis leurs <lb/>points de rencontre P, Q, &</s> <s xml:id="echoid-s2414" xml:space="preserve">c. </s> <s xml:id="echoid-s2415" xml:space="preserve">juſqu’au point C ou cette <lb/>ligne concourt avec ces cordes, certaines parties CP, CQ <lb/>&</s> <s xml:id="echoid-s2416" xml:space="preserve">c. </s> <s xml:id="echoid-s2417" xml:space="preserve">d@nt nous parlerons ſouvent dans la ſuite; </s> <s xml:id="echoid-s2418" xml:space="preserve">C’eſt pour-<lb/>quoy nous leur allons donner des noms.</s> <s xml:id="echoid-s2419" xml:space="preserve"/> </p> <div xml:id="echoid-div250" type="float" level="2" n="1"> <note position="right" xlink:label="note-0125-01" xlink:href="note-0125-01a" xml:space="preserve">ſig. 4. <lb/>5.</note> </div> <pb o="100" file="0126" n="126" rhead="EXAMEN DE L’OPINION"/> </div> <div xml:id="echoid-div252" type="section" level="1" n="162"> <head xml:id="echoid-head164" xml:space="preserve"><emph style="sc">Definition</emph> I.</head> <note position="left" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> <p> <s xml:id="echoid-s2420" xml:space="preserve">Lorſque ces parties CP, CQ, &</s> <s xml:id="echoid-s2421" xml:space="preserve">c. </s> <s xml:id="echoid-s2422" xml:space="preserve">ſe trouveront <lb/>au-deſſus du point C, nous les appellerons les ſublimitez <lb/>des puiſſances qui les auront déterminées par leurs <lb/>proportionnelles.</s> <s xml:id="echoid-s2423" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div253" type="section" level="1" n="163"> <head xml:id="echoid-head165" xml:space="preserve"><emph style="sc">Definition</emph> II.</head> <p> <s xml:id="echoid-s2424" xml:space="preserve">Et celles de ces lignes qui ſe trouveront au-deſſous <lb/>de ce même point C, nous les appellerons les Pro-<lb/>fondeurs de ces mêmes puiſſances.</s> <s xml:id="echoid-s2425" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2426" xml:space="preserve">Selon ces définitions C P eſt la Sublimité de la puiſſance R <lb/>dans les fig. </s> <s xml:id="echoid-s2427" xml:space="preserve">4. </s> <s xml:id="echoid-s2428" xml:space="preserve">& </s> <s xml:id="echoid-s2429" xml:space="preserve">5. </s> <s xml:id="echoid-s2430" xml:space="preserve">CL eſt encore la Sublimité de la <lb/>puiſſance S dans la fig. </s> <s xml:id="echoid-s2431" xml:space="preserve">4. </s> <s xml:id="echoid-s2432" xml:space="preserve">Mais dans la fig. </s> <s xml:id="echoid-s2433" xml:space="preserve">5. </s> <s xml:id="echoid-s2434" xml:space="preserve">CL eſt la <lb/>Profondeur de cette mème puiſſance.</s> <s xml:id="echoid-s2435" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s2436" xml:space="preserve">On avertit encore que lorſqu’on comparera à la ſublimi-<lb/>té, ou à la profondeur de ces puiſſances, des parties de leurs <lb/>cordes qui leur ſoient proportionnelles, on ne l’entendra pas <lb/>indiffèremment de toutes les proportionnelles qu’on pouroit <lb/>leur aſſigner; </s> <s xml:id="echoid-s2437" xml:space="preserve">mais ſeulement de celles qui déterminent les ſu-<lb/>blimitez, ou les profondeurs en queſtion.</s> <s xml:id="echoid-s2438" xml:space="preserve"/> </p> <figure> <image file="0126-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0126-01"/> </figure> </div> <div xml:id="echoid-div254" type="section" level="1" n="164"> <head xml:id="echoid-head166" xml:space="preserve">PROPOSITION I.</head> <p style="it"> <s xml:id="echoid-s2439" xml:space="preserve">LE poids T ſoutenu avec les cordes AC & </s> <s xml:id="echoid-s2440" xml:space="preserve">BC par les <lb/> <anchor type="note" xlink:label="note-0126-02a" xlink:href="note-0126-02"/> puiſſances R & </s> <s xml:id="echoid-s2441" xml:space="preserve">S, & </s> <s xml:id="echoid-s2442" xml:space="preserve">en équilibre avec elles, eſt tou-<lb/>jours à chacune d’elles, comme la partie DC de ſa ligne de ái-<lb/>rection, à chacun des còtez GC & </s> <s xml:id="echoid-s2443" xml:space="preserve">HC du parallelogramme <lb/>GH, dont elle eſt diagonale.</s> <s xml:id="echoid-s2444" xml:space="preserve"/> </p> <div xml:id="echoid-div254" type="float" level="2" n="1"> <note position="left" xlink:label="note-0126-02" xlink:href="note-0126-02a" xml:space="preserve">ſig. 4. <lb/>5. <lb/>6. <lb/>7.</note> </div> <figure> <image file="0126-02" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0126-02"/> </figure> <pb o="101" file="0127" n="127" rhead="DE M. BORELLI."/> </div> <div xml:id="echoid-div256" type="section" level="1" n="165"> <head xml:id="echoid-head167" xml:space="preserve"><emph style="sc">Demonstrations.</emph></head> <note position="right" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> <p> <s xml:id="echoid-s2445" xml:space="preserve">1°. </s> <s xml:id="echoid-s2446" xml:space="preserve">Voyez celle qu’on a donnée de cette même <lb/>propoſition dans le Projet précédent.</s> <s xml:id="echoid-s2447" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2448" xml:space="preserve">2°. </s> <s xml:id="echoid-s2449" xml:space="preserve">Voyez ci-aprés la Remarque qui ſuitle Corol-<lb/>laire de la Prop. </s> <s xml:id="echoid-s2450" xml:space="preserve">3.</s> <s xml:id="echoid-s2451" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2452" xml:space="preserve">3°. </s> <s xml:id="echoid-s2453" xml:space="preserve">Soient conçûs les leviers MC & </s> <s xml:id="echoid-s2454" xml:space="preserve">NC placez <lb/> <anchor type="note" xlink:label="note-0127-02a" xlink:href="note-0127-02"/> chacun en ligne droite avec chacune des directions <lb/>AC & </s> <s xml:id="echoid-s2455" xml:space="preserve">BC des puiſſances R & </s> <s xml:id="echoid-s2456" xml:space="preserve">S. </s> <s xml:id="echoid-s2457" xml:space="preserve">De leurs points <lb/>d’appui M & </s> <s xml:id="echoid-s2458" xml:space="preserve">N pris à diſcretion, ſoient tirées MF <lb/>& </s> <s xml:id="echoid-s2459" xml:space="preserve">NK perpendiculairement à ces mêmes lignes ré-<lb/>ciproquement priſes, & </s> <s xml:id="echoid-s2460" xml:space="preserve">ML avec NO perpendicu-<lb/>laires auſſi à la ligne de direction DCE de ce même <lb/>poids. </s> <s xml:id="echoid-s2461" xml:space="preserve">Enfin de quelqu’un des points D de cette <lb/>même ligne faite DH & </s> <s xml:id="echoid-s2462" xml:space="preserve">DG paralleles à AC & </s> <s xml:id="echoid-s2463" xml:space="preserve">à <lb/>CB. </s> <s xml:id="echoid-s2464" xml:space="preserve">Cela fait, il eſt clair que le levier CN, étant <lb/>(hyp.) </s> <s xml:id="echoid-s2465" xml:space="preserve">en ligne droite avec la ligne de dìrection CB <lb/>de la puiſſance S, ſupplée néceſſairement tout l’effet <lb/>de cette puiſſance; </s> <s xml:id="echoid-s2466" xml:space="preserve">& </s> <s xml:id="echoid-s2467" xml:space="preserve">que par conſéquent la puiſ-<lb/>ſance R pourroit ſuivant ſa même direction AC ſou-<lb/>tenir ſeule le poids T avec ce levier ainſi placé, de <lb/>même qu’elle le ſoutient préſentement avec la puiſ-<lb/>ſance S. </s> <s xml:id="echoid-s2468" xml:space="preserve">Pour la même raiſon la puiſſance S pourroit <lb/>auſſi le ſoutenir ſeule avec le levier CM, de même <lb/>qu’elle le ſoutient préſentement avec la puiſſance <lb/>R: </s> <s xml:id="echoid-s2469" xml:space="preserve">lepoids T eſt donc ſoutenu par le concours d’ac-<lb/>tion des puiſſances R & </s> <s xml:id="echoid-s2470" xml:space="preserve">S, de même qu’il le ſeroit <lb/>par la ſeule puiſſance R appliquée avec lui au levier <lb/>NC, ou bien par la ſeule puiſſance S appliquée auſſi <lb/>avec lui au levier CM. </s> <s xml:id="echoid-s2471" xml:space="preserve">Or dans le premier cas la <lb/>puiſſance R ſeroit (Prop. </s> <s xml:id="echoid-s2472" xml:space="preserve">fond. </s> <s xml:id="echoid-s2473" xml:space="preserve">des leviers Cor. </s> <s xml:id="echoid-s2474" xml:space="preserve">13. </s> <s xml:id="echoid-s2475" xml:space="preserve">du <lb/>Projet précéd.) </s> <s xml:id="echoid-s2476" xml:space="preserve">au poids T, comme NO à NK; </s> <s xml:id="echoid-s2477" xml:space="preserve">c’eſt-<lb/>à-dire, comme le ſinus de l’angle NCO, ou de <pb o="102" file="0128" n="128" rhead="EXAMEN DE L’OPINION"/> DCH, à celui de l’angle NCK, ou de CHD. </s> <s xml:id="echoid-s2478" xml:space="preserve">Et <lb/> <anchor type="note" xlink:label="note-0128-01a" xlink:href="note-0128-01"/> dans le ſecond cas le poids T, pour la même raiſon, <lb/>ſeroit à la puiſſance S, comme MF à ML; </s> <s xml:id="echoid-s2479" xml:space="preserve">c’eſt-à-dire <lb/>encore, comme le ſinus de l’angle MCF, ou de CHD, <lb/>à celui de l’angle MCL, ou de HDC: </s> <s xml:id="echoid-s2480" xml:space="preserve">Donc la puiſ-<lb/>ſance R, le poids T, & </s> <s xml:id="echoid-s2481" xml:space="preserve">la puiſſance S, ſont entr’eux, <lb/>comme les ſinus des angles DCH, CHD, & </s> <s xml:id="echoid-s2482" xml:space="preserve">HDC; <lb/></s> <s xml:id="echoid-s2483" xml:space="preserve">c’eſt-à-dire, ( Demm. </s> <s xml:id="echoid-s2484" xml:space="preserve">5. </s> <s xml:id="echoid-s2485" xml:space="preserve">du Projet précéd.) </s> <s xml:id="echoid-s2486" xml:space="preserve">comme les <lb/>lignes DH, CD, & </s> <s xml:id="echoid-s2487" xml:space="preserve">CH: </s> <s xml:id="echoid-s2488" xml:space="preserve">Le poids T eſt done à la <lb/>puiſſance R, comme CD à DH, ou à CG; </s> <s xml:id="echoid-s2489" xml:space="preserve">& </s> <s xml:id="echoid-s2490" xml:space="preserve">à la <lb/>puiſſance S, comme la même CD à CH. </s> <s xml:id="echoid-s2491" xml:space="preserve">Ce qu’il faloit <lb/>démontrer.</s> <s xml:id="echoid-s2492" xml:space="preserve"/> </p> <div xml:id="echoid-div256" type="float" level="2" n="1"> <note position="right" xlink:label="note-0127-02" xlink:href="note-0127-02a" xml:space="preserve">fig. 4. <lb/>5.</note> <note position="left" xlink:label="note-0128-01" xlink:href="note-0128-01a" xml:space="preserve">DES POIDS <lb/>foutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> <p> <s xml:id="echoid-s2493" xml:space="preserve">4°. </s> <s xml:id="echoid-s2494" xml:space="preserve">Si au lieu des puiſſances R & </s> <s xml:id="echoid-s2495" xml:space="preserve">S, les cordes AC <lb/>& </s> <s xml:id="echoid-s2496" xml:space="preserve">BC étoient attachees aux extrémitez de quelque le-<lb/>vier AB, dont l’appui D fut dans la ligne direction de <lb/> <anchor type="note" xlink:label="note-0128-02a" xlink:href="note-0128-02"/> DCE du poids T. </s> <s xml:id="echoid-s2497" xml:space="preserve">Il eſt clair qu’en quelque fitua-<lb/>tion que ce levier ſe trouvât alors, (Lemm. </s> <s xml:id="echoid-s2498" xml:space="preserve">@. </s> <s xml:id="echoid-s2499" xml:space="preserve">du <lb/>Projet précéd.) </s> <s xml:id="echoid-s2500" xml:space="preserve">il y demeureroit, & </s> <s xml:id="echoid-s2501" xml:space="preserve">que la charge de <lb/>ſon point d’appui ſeroit alors égale au poids T. </s> <s xml:id="echoid-s2502" xml:space="preserve">Il <lb/>eſt encore clair que les extrémitez A & </s> <s xml:id="echoid-s2503" xml:space="preserve">B de ce mê-<lb/>me levier ſeroient auſſi tirées ſuivant AC & </s> <s xml:id="echoid-s2504" xml:space="preserve">BC, <lb/>chacune avec une force égale à celle de la puiſſance <lb/>R, ou S qu’elle ſupplée. </s> <s xml:id="echoid-s2505" xml:space="preserve">Or les forces avec leſquelles <lb/>les points A & </s> <s xml:id="echoid-s2506" xml:space="preserve">B de ce levier ſeroient ainſi tirez ſui-<lb/>vant AC & </s> <s xml:id="echoid-s2507" xml:space="preserve">BC, ſeroient entr’elles, ( Cor. </s> <s xml:id="echoid-s2508" xml:space="preserve">13. </s> <s xml:id="echoid-s2509" xml:space="preserve">prop. <lb/></s> <s xml:id="echoid-s2510" xml:space="preserve">fond. </s> <s xml:id="echoid-s2511" xml:space="preserve">des leviers du Projet précéd.) </s> <s xml:id="echoid-s2512" xml:space="preserve">comme DF & </s> <s xml:id="echoid-s2513" xml:space="preserve">DK <lb/>tirées du point D perpendiculairement ſur BC & </s> <s xml:id="echoid-s2514" xml:space="preserve"><lb/>AC; </s> <s xml:id="echoid-s2515" xml:space="preserve">c’eſt-à-dire, en faiſant le parallelogramme GH, <lb/>comme les ſinus des angles DCH & </s> <s xml:id="echoid-s2516" xml:space="preserve">CDH, ou <lb/>(Lemm. </s> <s xml:id="echoid-s2517" xml:space="preserve">5. </s> <s xml:id="echoid-s2518" xml:space="preserve">du Proiet précéd.) </s> <s xml:id="echoid-s2519" xml:space="preserve">comme les côtez DH & </s> <s xml:id="echoid-s2520" xml:space="preserve"><lb/>HC de ce parallelogramme. </s> <s xml:id="echoid-s2521" xml:space="preserve">Ces mêmes forces ſe-<lb/>roient auſſi ( Cor. </s> <s xml:id="echoid-s2522" xml:space="preserve">4. </s> <s xml:id="echoid-s2523" xml:space="preserve">Prop. </s> <s xml:id="echoid-s2524" xml:space="preserve">fond. </s> <s xml:id="echoid-s2525" xml:space="preserve">des leviers au Projet <lb/>précèd.) </s> <s xml:id="echoid-s2526" xml:space="preserve">chacune à la charge du point d’appui D de <lb/>ce levier, c’eſt-à-dire, au poids T, comme chacun <pb o="103" file="0129" n="129" rhead="DE M. BORELLI."/> ces de mêmes côtez à la diagonale DC: </s> <s xml:id="echoid-s2527" xml:space="preserve">les forces <lb/> <anchor type="note" xlink:label="note-0129-01a" xlink:href="note-0129-01"/> des puiſſances R & </s> <s xml:id="echoid-s2528" xml:space="preserve">S, c’eſt-à-dire, ces mêmes puiſ-<lb/>ſances, elles-mêmes, ſont donc entr’elles, comme <lb/>DH, ou GC & </s> <s xml:id="echoid-s2529" xml:space="preserve">HC; </s> <s xml:id="echoid-s2530" xml:space="preserve">& </s> <s xml:id="echoid-s2531" xml:space="preserve">au poids T, comme chacun <lb/>de ces mêmes côtez du parallelogramme GH, à ſa <lb/>diagonale DC. </s> <s xml:id="echoid-s2532" xml:space="preserve">Cequ’il faloit démontrer.</s> <s xml:id="echoid-s2533" xml:space="preserve"/> </p> <div xml:id="echoid-div257" type="float" level="2" n="2"> <note position="left" xlink:label="note-0128-02" xlink:href="note-0128-02a" xml:space="preserve">fig. 6. <lb/>7.</note> <note position="right" xlink:label="note-0129-01" xlink:href="note-0129-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement</note> </div> <p style="it"> <s xml:id="echoid-s2534" xml:space="preserve">On pourroit encore démontrer cette mème Propoſition en <lb/>ſe ſervant des plans inclinez, pourvû qu’on en prit un qui <lb/>fut perpendiculaire à la direction de quelqu’une des deux <lb/>puiſſances qui ſoutiennent ce poids: </s> <s xml:id="echoid-s2535" xml:space="preserve">Car cette puiſſance & </s> <s xml:id="echoid-s2536" xml:space="preserve"><lb/>la cbarge de ce plan alors égales, n’ayant qu’un mème <lb/>rapport avec ce poids, non plus qu’avec l’autre puiſſance <lb/>qu’on conſidére en ce cas comme le ſoutenant ſeule ſur ce plan; <lb/></s> <s xml:id="echoid-s2537" xml:space="preserve">on trouveroit par le Cor. </s> <s xml:id="echoid-s2538" xml:space="preserve">7. </s> <s xml:id="echoid-s2539" xml:space="preserve">de la Prop. </s> <s xml:id="echoid-s2540" xml:space="preserve">des Surfaces du <lb/>Projet précédent, que ce poids eſt toujours à chacune de ces <lb/>puiſſances, comme le ſinus de l’angle que leurs cordes font <lb/>entr’elles, à chacun des ſinus des angles que font avec la <lb/>ligne de direction de ce poids chacune de ces cordes récipro-<lb/>quement priſes. </s> <s xml:id="echoid-s2541" xml:space="preserve">Tout cela eſt préſentement trop clair pour <lb/>s’y arréter davantage.</s> <s xml:id="echoid-s2542" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div259" type="section" level="1" n="166"> <head xml:id="echoid-head168" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s2543" xml:space="preserve">On peut conclure généralement de ces démon-<lb/> <anchor type="note" xlink:label="note-0129-02a" xlink:href="note-0129-02"/> ſtrations, ce que nous n’avons conclu ( chap. </s> <s xml:id="echoid-s2544" xml:space="preserve">1.) </s> <s xml:id="echoid-s2545" xml:space="preserve">dela <lb/>68. </s> <s xml:id="echoid-s2546" xml:space="preserve">Propoſition de M. </s> <s xml:id="echoid-s2547" xml:space="preserve">Borelli, quepour les cas qu’elle <lb/>comprend: </s> <s xml:id="echoid-s2548" xml:space="preserve">Sçavoir qu’il n’y en à aucun de poſſible, <lb/>ou l’on puiſſe conſerver l’équilibre du poids T avec <lb/>les puiſſances R & </s> <s xml:id="echoid-s2549" xml:space="preserve">S, en changeant le rapport qu’elles <lb/>ont entr’elles, ou avec lui; </s> <s xml:id="echoid-s2550" xml:space="preserve">à moins qu’on ne change <lb/>en même-tems l’inclinaiſon de leurs cordes: </s> <s xml:id="echoid-s2551" xml:space="preserve">non plus <lb/>qu’en changeant l’inclinaiſon de ces cordes, ſans <lb/>changer auſſi le rapport de ces mêmes puiſſances, <lb/>ou entr’elles, ou avec ce poids: </s> <s xml:id="echoid-s2552" xml:space="preserve">parce que ſans cela <lb/>il n’eſt pas poſſible de faire que chacun des côtez CH <pb o="104" file="0130" n="130" rhead="EXAMEN DE L’OPINION"/> & </s> <s xml:id="echoid-s2553" xml:space="preserve">CG du parallelogramme GH, continuë d’être à ſa <lb/> <anchor type="note" xlink:label="note-0130-01a" xlink:href="note-0130-01"/> diagonale DC, comme chacune des puiſſances R & </s> <s xml:id="echoid-s2554" xml:space="preserve">S <lb/>au poids T; </s> <s xml:id="echoid-s2555" xml:space="preserve">ce qui doit cependant être, comme onle <lb/>vient de voir, pour qu’elles faſſent équilibre avec lui.</s> <s xml:id="echoid-s2556" xml:space="preserve"/> </p> <div xml:id="echoid-div259" type="float" level="2" n="1"> <note position="right" xlink:label="note-0129-02" xlink:href="note-0129-02a" xml:space="preserve">fig. 4. <lb/>5. <lb/>6. <lb/>7.</note> <note position="left" xlink:label="note-0130-01" xlink:href="note-0130-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> <p style="it"> <s xml:id="echoid-s2557" xml:space="preserve">On peut comparer ce Corollaire aux Scholies des Pro-<lb/>poſitions 68. </s> <s xml:id="echoid-s2558" xml:space="preserve">& </s> <s xml:id="echoid-s2559" xml:space="preserve">69. </s> <s xml:id="echoid-s2560" xml:space="preserve">M. </s> <s xml:id="echoid-s2561" xml:space="preserve">Borelli.</s> <s xml:id="echoid-s2562" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div261" type="section" level="1" n="167"> <head xml:id="echoid-head169" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s2563" xml:space="preserve">Il ſuit encore de ces démonſtrations que chacune <lb/>des puiſſances R & </s> <s xml:id="echoid-s2564" xml:space="preserve">S eſt au poids T, comme chacune <lb/> <anchor type="note" xlink:label="note-0130-02a" xlink:href="note-0130-02"/> des parties GC & </s> <s xml:id="echoid-s2565" xml:space="preserve">HC de leurs cordes, qui leurs ſont <lb/>proportionelles, à la ſomme ( fig. </s> <s xml:id="echoid-s2566" xml:space="preserve">4.) </s> <s xml:id="echoid-s2567" xml:space="preserve">de leurs ſublimi-<lb/>tez, ou à la difference (fig. </s> <s xml:id="echoid-s2568" xml:space="preserve">5.) </s> <s xml:id="echoid-s2569" xml:space="preserve">quieſt entre la ſublimité <lb/>de l’une & </s> <s xml:id="echoid-s2570" xml:space="preserve">la profondeur de l’autre: </s> <s xml:id="echoid-s2571" xml:space="preserve">parce que dans <lb/>le parallelogramme GH les angles GCD & </s> <s xml:id="echoid-s2572" xml:space="preserve">CDH <lb/>étant égaux, auſſi-bien que les lignes GC & </s> <s xml:id="echoid-s2573" xml:space="preserve">DH; <lb/></s> <s xml:id="echoid-s2574" xml:space="preserve">de plus les angles qui ſe font en P & </s> <s xml:id="echoid-s2575" xml:space="preserve">en Q, étant auſſi <lb/>(avert.) </s> <s xml:id="echoid-s2576" xml:space="preserve">égaux, les triangles GPC & </s> <s xml:id="echoid-s2577" xml:space="preserve">HQD ſont <lb/>non ſeulement ſemblables, mais encore leurs côtez <lb/>CP & </s> <s xml:id="echoid-s2578" xml:space="preserve">DQ ſont égaux: </s> <s xml:id="echoid-s2579" xml:space="preserve">Donc (fig. </s> <s xml:id="echoid-s2580" xml:space="preserve">4.) </s> <s xml:id="echoid-s2581" xml:space="preserve">CP plus CQ <lb/>eſt égal à DQ plus CQ, & </s> <s xml:id="echoid-s2582" xml:space="preserve">( fig. </s> <s xml:id="echoid-s2583" xml:space="preserve">5.) </s> <s xml:id="echoid-s2584" xml:space="preserve">CP moins CQ <lb/>ſera auſſi égal à DQ moins CQ. </s> <s xml:id="echoid-s2585" xml:space="preserve">Or ( fig. </s> <s xml:id="echoid-s2586" xml:space="preserve">4.) </s> <s xml:id="echoid-s2587" xml:space="preserve">DQ <lb/>plus CQ eſt égal à CD, de même (fig. </s> <s xml:id="echoid-s2588" xml:space="preserve">5.) </s> <s xml:id="echoid-s2589" xml:space="preserve">que <lb/>DQ moins CQ: </s> <s xml:id="echoid-s2590" xml:space="preserve">Donc ( fig. </s> <s xml:id="echoid-s2591" xml:space="preserve">4.) </s> <s xml:id="echoid-s2592" xml:space="preserve">CP plus CQ eſt égal <lb/>à CD, auſſi-bien (fig. </s> <s xml:id="echoid-s2593" xml:space="preserve">5.) </s> <s xml:id="echoid-s2594" xml:space="preserve">que CP moins CQ. </s> <s xml:id="echoid-s2595" xml:space="preserve">Or <lb/>ſelon les démonſtrations précédentes chacune des <lb/>puiſſances R & </s> <s xml:id="echoid-s2596" xml:space="preserve">S eſt au poids T, comme chacune de <lb/>leurs proportionelles CG & </s> <s xml:id="echoid-s2597" xml:space="preserve">HC à CD: </s> <s xml:id="echoid-s2598" xml:space="preserve">Donc cha-<lb/>cune de ces mêmes puiſſances eſt à ce poids, comme <lb/>chacune de ces mêmes proportionelles à CP ( fig. </s> <s xml:id="echoid-s2599" xml:space="preserve">4.) </s> <s xml:id="echoid-s2600" xml:space="preserve"><lb/>plus CQ, ou (fig. </s> <s xml:id="echoid-s2601" xml:space="preserve">5.) </s> <s xml:id="echoid-s2602" xml:space="preserve">à CP moins CQ; </s> <s xml:id="echoid-s2603" xml:space="preserve">c’eſt-à-dire, <lb/>(Def. </s> <s xml:id="echoid-s2604" xml:space="preserve">1. </s> <s xml:id="echoid-s2605" xml:space="preserve">& </s> <s xml:id="echoid-s2606" xml:space="preserve">2.) </s> <s xml:id="echoid-s2607" xml:space="preserve">à la ſomme ( fig. </s> <s xml:id="echoid-s2608" xml:space="preserve">4.) </s> <s xml:id="echoid-s2609" xml:space="preserve">de leurs ſublimitez, <lb/>ou bien ( fig. </s> <s xml:id="echoid-s2610" xml:space="preserve">5.) </s> <s xml:id="echoid-s2611" xml:space="preserve">à la différence qui eſt entre la ſublimi-<lb/>té de l’une & </s> <s xml:id="echoid-s2612" xml:space="preserve">la profondeur de l’autre.</s> <s xml:id="echoid-s2613" xml:space="preserve"/> </p> <div xml:id="echoid-div261" type="float" level="2" n="1"> <note position="left" xlink:label="note-0130-02" xlink:href="note-0130-02a" xml:space="preserve">fig. 4. <lb/>5.</note> </div> <pb o="105" file="0131" n="131" rhead="DE M. BORELLI."/> </div> <div xml:id="echoid-div263" type="section" level="1" n="168"> <head xml:id="echoid-head170" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <note position="right" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> <p> <s xml:id="echoid-s2614" xml:space="preserve">D’où l’on voit que la ſomme des deux puiſſances <lb/>qui ſoutiennent un poids avec des cordes, eſt tou-<lb/>jours à ce poids, comme la ſomme des longueurs de <lb/>leurs cordes, qui leur ſont proportionelles, à la ſomme <lb/>de leurs ſublimitez, ou à la différence qui eſt entre de <lb/>la ſublimité de l’une & </s> <s xml:id="echoid-s2615" xml:space="preserve">la profondeur de l’autre.</s> <s xml:id="echoid-s2616" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s2617" xml:space="preserve">On peut comparer encore ces deux derniers Corollaires à la <lb/>69. </s> <s xml:id="echoid-s2618" xml:space="preserve">Prop. </s> <s xml:id="echoid-s2619" xml:space="preserve">de M. </s> <s xml:id="echoid-s2620" xml:space="preserve">Borelli, & </s> <s xml:id="echoid-s2621" xml:space="preserve">au Corollaire qu’il entire.</s> <s xml:id="echoid-s2622" xml:space="preserve"/> </p> <figure> <image file="0131-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0131-01"/> </figure> </div> <div xml:id="echoid-div264" type="section" level="1" n="169"> <head xml:id="echoid-head171" xml:space="preserve">PROPOSITION II.</head> <p style="it"> <s xml:id="echoid-s2623" xml:space="preserve">DE quelque maniére qu’un poids T ſoit ſoutenu avee <lb/> <anchor type="note" xlink:label="note-0131-02a" xlink:href="note-0131-02"/> des cordes par quelque nombre de puiſſances A, B, D, <lb/>E, F, & </s> <s xml:id="echoid-s2624" xml:space="preserve">c. </s> <s xml:id="echoid-s2625" xml:space="preserve">que ce ſoit, appliquèes à un même nœud C; </s> <s xml:id="echoid-s2626" xml:space="preserve">ſi l’ on <lb/>prend ſur leurs cordes autant de parties CG, CR, CM, CN, <lb/>CP, &</s> <s xml:id="echoid-s2627" xml:space="preserve">c. </s> <s xml:id="echoid-s2628" xml:space="preserve">qui leurs ſoient proportionelles, & </s> <s xml:id="echoid-s2629" xml:space="preserve">que ſous deux <lb/>de ces parties, par exemple, ſous GC & </s> <s xml:id="echoid-s2630" xml:space="preserve">RC, l’on faſſe un <lb/>parallelogramme RG, dont la diagonale CH faſſe encore <lb/>avecune autre de ces parties CM le parallelogramme HM, <lb/>dont la diagonale CL faſſe encore avec une autre de ces <lb/>parties CN le parallelogramme LN, dont la diagonale <lb/>CQ faſſe encore avec une autre de ces parties CP le paral-<lb/>lelogramme PQ, & </s> <s xml:id="echoid-s2631" xml:space="preserve">ainſi juſqu’à la derniére de ces propor-<lb/>tionelles. </s> <s xml:id="echoid-s2632" xml:space="preserve">On verra 1°. </s> <s xml:id="echoid-s2633" xml:space="preserve">Que la diagonale du dernier de ces <lb/>parallelogrammes, qui eſt ici CK, ſera dans la ligne de di-<lb/>rection du poids T. </s> <s xml:id="echoid-s2634" xml:space="preserve">2°. </s> <s xml:id="echoid-s2635" xml:space="preserve">Et que chacune de ces puiſſances ſera <lb/>à ce poids, comme chacune de ces proportionelles, ſelon qu’el-<lb/>les leurs répoudent, eſt à cette même diagonale.</s> <s xml:id="echoid-s2636" xml:space="preserve"/> </p> <div xml:id="echoid-div264" type="float" level="2" n="1"> <note position="right" xlink:label="note-0131-02" xlink:href="note-0131-02a" xml:space="preserve">fig. 2.</note> </div> </div> <div xml:id="echoid-div266" type="section" level="1" n="170"> <head xml:id="echoid-head172" xml:space="preserve"><emph style="sc">Demonstration.</emph></head> <p> <s xml:id="echoid-s2637" xml:space="preserve">1°. </s> <s xml:id="echoid-s2638" xml:space="preserve">Puiſque ( hyp.) </s> <s xml:id="echoid-s2639" xml:space="preserve">la puiſſance A eſt à la puiſ- <pb o="106" file="0132" n="132" rhead="EXAMEN DE L’OPINION"/> ſance B, comme CG à CR, il réſultera (Lemm. </s> <s xml:id="echoid-s2640" xml:space="preserve">3. </s> <s xml:id="echoid-s2641" xml:space="preserve">du <lb/> <anchor type="note" xlink:label="note-0132-01a" xlink:href="note-0132-01"/> Projet précèd.) </s> <s xml:id="echoid-s2642" xml:space="preserve">de leur concours d’action ſur le point <lb/>C, un@ impreſſion compoſée ſuivant CH, d’une force <lb/>qui ſera (Cor. </s> <s xml:id="echoid-s2643" xml:space="preserve">3. </s> <s xml:id="echoid-s2644" xml:space="preserve">du même Lemm.) </s> <s xml:id="echoid-s2645" xml:space="preserve">à chacune de ces <lb/>puiſſances, comme CH à chacune des lignes CG & </s> <s xml:id="echoid-s2646" xml:space="preserve"><lb/>CR qui les répreſentent: </s> <s xml:id="echoid-s2647" xml:space="preserve">l’impreſſion, que font en-<lb/>ſemble ces deux puiſſances ſur le point C, eſt donc <lb/>la même que celle que feroit ſeule ſur ce même point <lb/>quelque nouvelle puiſſance qui, lui étant appliquée <lb/>ſuivant CH, au lieu d’elles, leur ſeroit à chacune, <lb/>comme CH à chacune des lignes CG & </s> <s xml:id="echoid-s2648" xml:space="preserve">CR: </s> <s xml:id="echoid-s2649" xml:space="preserve">& </s> <s xml:id="echoid-s2650" xml:space="preserve">par <lb/>conſèquent les 3. </s> <s xml:id="echoid-s2651" xml:space="preserve">puiſſances A, B, D, doivent faire <lb/>enſemble la même impreſſion ſur le point C que cette <lb/>nouvelle puiſſance ( je l’appelle H ) feroit alors avec <lb/>la puiſſance D. </s> <s xml:id="echoid-s2652" xml:space="preserve">Or (Lemm 3. </s> <s xml:id="echoid-s2653" xml:space="preserve">du Projet précéd.) </s> <s xml:id="echoid-s2654" xml:space="preserve">l’im-<lb/>preſſion qui réſulteroit alors du concours d’action <lb/>des puiſſances D & </s> <s xml:id="echoid-s2655" xml:space="preserve">H ſur le point C, ſe feroit ſui-<lb/>vant CL, d’une force qui ſeroit (Cor. </s> <s xml:id="echoid-s2656" xml:space="preserve">3. </s> <s xml:id="echoid-s2657" xml:space="preserve">du même <lb/>Lemm.) </s> <s xml:id="echoid-s2658" xml:space="preserve">à celle de la puiſſance D, comme CL à CM: <lb/></s> <s xml:id="echoid-s2659" xml:space="preserve">Donc l’impreſſion compoſée qui réſulte du concours <lb/>d’action des 3. </s> <s xml:id="echoid-s2660" xml:space="preserve">puiſſances A, B, D, ſur le point C, <lb/>ſe fait en effet ſuivant CL, d’une force qui eſt à <lb/>celle de la puiſſance D, comme CL à CM: </s> <s xml:id="echoid-s2661" xml:space="preserve">elles ne <lb/>font donc toutes trois enſemble ſur ce point que la <lb/>même impreſſion que feroit ſeule quelqu’autre puiſ-<lb/>ſance (je l’appelle L) qui appliquée ſuivant CL, au <lb/>lieu de ces trois, ſeroit à la puiſſance D, comme CL <lb/>à CM: </s> <s xml:id="echoid-s2662" xml:space="preserve">& </s> <s xml:id="echoid-s2663" xml:space="preserve">par conſéquent les 4. </s> <s xml:id="echoid-s2664" xml:space="preserve">puiſſances A, B, D, E, <lb/>ne doivent faire ſur le point C que la même impreſ-<lb/>ſion que feroit alors la puiſſance L avec la puiſſance <lb/>E. </s> <s xml:id="echoid-s2665" xml:space="preserve">Or (Lemm. </s> <s xml:id="echoid-s2666" xml:space="preserve">3. </s> <s xml:id="echoid-s2667" xml:space="preserve">du Projet Projet précéd.) </s> <s xml:id="echoid-s2668" xml:space="preserve">l’impreſſion qui <lb/>réſulteroit alors du concours d’action de ces deux <lb/>dernieres puiſſances ſur le point C, ſe feroit ſuivant <lb/>CQ, d’une force qui ſeroit ( Cor. </s> <s xml:id="echoid-s2669" xml:space="preserve">3. </s> <s xml:id="echoid-s2670" xml:space="preserve">du même Lemm.) </s> <s xml:id="echoid-s2671" xml:space="preserve"><lb/>à celle de la puiſſance E, comme CQ à CN: </s> <s xml:id="echoid-s2672" xml:space="preserve">Donc <pb o="107" file="0133" n="133" rhead="DE M. BORELLI."/> l’impreſſion compoſée qui réſulte du concours d’ac-<lb/> <anchor type="note" xlink:label="note-0133-01a" xlink:href="note-0133-01"/> tion des 4. </s> <s xml:id="echoid-s2673" xml:space="preserve">puiſſances A, B, D, E, ſur le point C, <lb/>ſe fait en effet ſuivant la ligne CQ, d’une force qui <lb/>eſt à celle de la puiſſance E, comme CQ à CN. </s> <s xml:id="echoid-s2674" xml:space="preserve">On <lb/>prouvera de même que l’impreſſion compoſée qui <lb/>réſulte du concours d’action des 5. </s> <s xml:id="echoid-s2675" xml:space="preserve">puiſſances A, B, <lb/>D, E, F, ſe fait auſſi ſuivant CK, d’une force qui <lb/>eſt à la puiſſance F, comme CK à CP. </s> <s xml:id="echoid-s2676" xml:space="preserve">Et ainſi tou-<lb/>jours de même juſqu’à la derniére des puiſſances <lb/>appliquées à ce poids: </s> <s xml:id="echoid-s2677" xml:space="preserve">D’où il ſuit que l’impreſſion <lb/>compoſée qui réſulte du concours d’action de toutes <lb/>ces puiſſances ſur le point C, en quelque nombre <lb/>qu’elles ſoient, ſe fait toujours ſuivant la diagonale <lb/>du dernier des parallelogrammes faits comme l’on <lb/>vient de dire, c’eſt-à-dire ici, ſuivant CK: </s> <s xml:id="echoid-s2678" xml:space="preserve">& </s> <s xml:id="echoid-s2679" xml:space="preserve">par <lb/>conſéquent (n. </s> <s xml:id="echoid-s2680" xml:space="preserve">2. </s> <s xml:id="echoid-s2681" xml:space="preserve">Demonſt. </s> <s xml:id="echoid-s2682" xml:space="preserve">Prop. </s> <s xml:id="echoid-s2683" xml:space="preserve">fond. </s> <s xml:id="echoid-s2684" xml:space="preserve">des poids ſoutenus <lb/>par des cordes du Projet précéd.) </s> <s xml:id="echoid-s2685" xml:space="preserve">cette diagonale eſt <lb/>toujours en ligne droite avec CT, c’eſt-à-dire, dans <lb/>la ligne de direction du poids T. </s> <s xml:id="echoid-s2686" xml:space="preserve">Ce qu’il faloit dé-<lb/>montrer.</s> <s xml:id="echoid-s2687" xml:space="preserve"/> </p> <div xml:id="echoid-div266" type="float" level="2" n="1"> <note position="left" xlink:label="note-0132-01" xlink:href="note-0132-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſcu-<lb/>lement.</note> <note position="right" xlink:label="note-0133-01" xlink:href="note-0133-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> <p> <s xml:id="echoid-s2688" xml:space="preserve">2°. </s> <s xml:id="echoid-s2689" xml:space="preserve">On vient de voir que le poids T eſt ſoutenu <lb/>par les puiſſances A, B, D, E, F, &</s> <s xml:id="echoid-s2690" xml:space="preserve">c. </s> <s xml:id="echoid-s2691" xml:space="preserve">De même qu’il <lb/>le ſeroit, par exemple ici, par la puiſſance F aidée <lb/>d’une autre appliquée ſuivant CQ, à qui elle ſeroit <lb/>comme CP à CQ ( les directions de toutes demeu-<lb/>rant toujours les mêmes): </s> <s xml:id="echoid-s2692" xml:space="preserve">Donc (prop. </s> <s xml:id="echoid-s2693" xml:space="preserve">1.) </s> <s xml:id="echoid-s2694" xml:space="preserve">la puiſ-<lb/>ſance F eſt au poids T, comme CP à CK. </s> <s xml:id="echoid-s2695" xml:space="preserve">Or ( hyp.) <lb/></s> <s xml:id="echoid-s2696" xml:space="preserve">la puiſſance F eſt à chacune des puiſſances E, D, B, A, <lb/>&</s> <s xml:id="echoid-s2697" xml:space="preserve">c. </s> <s xml:id="echoid-s2698" xml:space="preserve">comme CP à chacune des parties de leurs cordes, <lb/>CN, CM, CR, CG, &</s> <s xml:id="echoid-s2699" xml:space="preserve">c. </s> <s xml:id="echoid-s2700" xml:space="preserve">Donc chacune de ces <lb/>puiſſances eſt au poids T, comme chacune de ces <lb/>proportionelles, ſelon qu’elles leurs répondent, eſt à la <lb/>diagonale CK. </s> <s xml:id="echoid-s2701" xml:space="preserve">Ce qu’on vient de dire de la puiſſance <lb/>F, ſe prouvera de même de toute autre dont la pro- <pb o="108" file="0134" n="134" rhead="EXAMEN DE L’OPINION"/> portionelle feroit un des côtez du parallelogramme <lb/> <anchor type="note" xlink:label="note-0134-01a" xlink:href="note-0134-01"/> qu’on vient de démontrer avoir toujours ſa diago-<lb/>nale, comme ici CK, dans la ligne de direction du <lb/>poids T: </s> <s xml:id="echoid-s2702" xml:space="preserve">Ainſi en général de quelque maniére qu’un <lb/>poids ſoit ſoutenu avec des cordes par quelque nom-<lb/>bre de puiſſances que ce ſoit, appliquées à un même <lb/>nœud, chacune de ces puiſſances eſt toujours à ce <lb/>poids, comme chacune de leurs proportionelles qui <lb/>ſervent de côtez aux parallelogrammes dont il eſt ici <lb/>queſtion, eſt à la diagonale du dernier, qu’on vient <lb/>de voir ſe trouver toujours dans ſa ligne de direction. <lb/></s> <s xml:id="echoid-s2703" xml:space="preserve">Ce qu’il faloit démontrer.</s> <s xml:id="echoid-s2704" xml:space="preserve"/> </p> <div xml:id="echoid-div267" type="float" level="2" n="2"> <note position="left" xlink:label="note-0134-01" xlink:href="note-0134-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cord@s ſeu-<lb/>lement.</note> </div> </div> <div xml:id="echoid-div269" type="section" level="1" n="171"> <head xml:id="echoid-head173" xml:space="preserve"><emph style="sc">Corollaire.</emph></head> <p> <s xml:id="echoid-s2705" xml:space="preserve">D’où l’on voit que toutes ces puiſſances priſes <lb/>enſemble ſont toujours au poids T qu’elles ſoutien-<lb/>nent, comme la ſomme de leurs proportionelles CG, <lb/>CR, CM, CN, CP, &</s> <s xml:id="echoid-s2706" xml:space="preserve">c. </s> <s xml:id="echoid-s2707" xml:space="preserve">à la diagonale du paralle-<lb/>logramme qu’on vient de démontrer (n. </s> <s xml:id="echoid-s2708" xml:space="preserve">1.) </s> <s xml:id="echoid-s2709" xml:space="preserve">ſe trou-<lb/>ver toujours dans la ligne de direction de ce poids: <lb/></s> <s xml:id="echoid-s2710" xml:space="preserve">De ſorte que, lorſque toutes ces puiſſances ſont éga-<lb/>les entr’elles, ces mêmes proportionelles l’étant auſſi, <lb/>la ſomme de toutes ces puiſſances eſt à ce poids, com-<lb/>me une de ces proportionelles à une partie de cette <lb/>diagonale diviſée en autant d’égales qu’il y a de telles <lb/>puiſſances; </s> <s xml:id="echoid-s2711" xml:space="preserve">c’eſt-à-dire ici, comme laquelle que ce <lb/>ſoit, des lignes CG, CR, CM, CN, CP, à {1/5} de <lb/>CK.</s> <s xml:id="echoid-s2712" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s2713" xml:space="preserve">Pour exprimer le raport que nous venons de trouver <lb/>entre ce poids & </s> <s xml:id="echoid-s2714" xml:space="preserve">les puiſſances qui le ſoutiennent, d’une <lb/>maniére qui en rende le calcul plus facile, ſoit le Lemme <lb/>ſuivant.</s> <s xml:id="echoid-s2715" xml:space="preserve"/> </p> <figure> <image file="0134-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0134-01"/> </figure> <pb o="109" file="0135" n="135" rhead="DE M. BORELLI."/> <note position="right" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſcu-<lb/>lement.</note> <figure> <image file="0135-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0135-01"/> </figure> </div> <div xml:id="echoid-div270" type="section" level="1" n="172"> <head xml:id="echoid-head174" xml:space="preserve">LEMME.</head> <p style="it"> <s xml:id="echoid-s2716" xml:space="preserve">DE quelque maniére que la ligne droite CP paſſe par <lb/> <anchor type="note" xlink:label="note-0135-02a" xlink:href="note-0135-02"/> une des pointes C du parallelogramme IE, ſi des trois <lb/>autres pointes G, I, E, on tire ſur la même CP les trois <lb/>perpendiculaires GL, IP, VE: </s> <s xml:id="echoid-s2717" xml:space="preserve">ſa partie CL compriſe <lb/>entre le point C, & </s> <s xml:id="echoid-s2718" xml:space="preserve">la perpendiculaire GL qui part de la <lb/>pointe G qui lui eſt oppoſée, eſt toujours égale à la ſomme de <lb/>ſes deux autres parties CP & </s> <s xml:id="echoid-s2719" xml:space="preserve">CV compriſes entre ce même <lb/>point C & </s> <s xml:id="echoid-s2720" xml:space="preserve">les perpendiculaires IP & </s> <s xml:id="echoid-s2721" xml:space="preserve">EV, lors que ces <lb/>deux perpendiculaires tombent du même côté de C; </s> <s xml:id="echoid-s2722" xml:space="preserve">ou à la <lb/>diffèrence de ces deux parties, lors que ces deux perpendiculaires <lb/>tombent de différens côtez.</s> <s xml:id="echoid-s2723" xml:space="preserve"/> </p> <div xml:id="echoid-div270" type="float" level="2" n="1"> <note position="right" xlink:label="note-0135-02" xlink:href="note-0135-02a" xml:space="preserve">fig. 9. <lb/>10. <lb/>11. <lb/>12. <lb/>13. <lb/>14. <lb/>15. <lb/>16.</note> </div> </div> <div xml:id="echoid-div272" type="section" level="1" n="173"> <head xml:id="echoid-head175" xml:space="preserve"><emph style="sc">Demonstration.</emph></head> <p> <s xml:id="echoid-s2724" xml:space="preserve">Joignez IE & </s> <s xml:id="echoid-s2725" xml:space="preserve">GC qui ſe coupent par la moitié <lb/>l’une & </s> <s xml:id="echoid-s2726" xml:space="preserve">l’autre en K, & </s> <s xml:id="echoid-s2727" xml:space="preserve">apres avoir fait QK perpen-<lb/>diculaire à CP, concevez un plan qui paſſe par QK, <lb/>à qui CP ſoit perpendiculaire, & </s> <s xml:id="echoid-s2728" xml:space="preserve">ſur lequel des <lb/>points I & </s> <s xml:id="echoid-s2729" xml:space="preserve">E, tombent auſſi perpendiculairement <lb/>IM, & </s> <s xml:id="echoid-s2730" xml:space="preserve">EN; </s> <s xml:id="echoid-s2731" xml:space="preserve">Enfin joignez QM & </s> <s xml:id="echoid-s2732" xml:space="preserve">QN. </s> <s xml:id="echoid-s2733" xml:space="preserve">Cela fait, <lb/>ſoit que QK, QM, & </s> <s xml:id="echoid-s2734" xml:space="preserve">QN, ſe confondent en une <lb/>ſeule ligne, ſoit qu’elles en faſſent trois différentes, <lb/>il eſt clair que puis que les lignes IM, PQ, NE, & </s> <s xml:id="echoid-s2735" xml:space="preserve"><lb/>VQ, ſont toutes (Hyp.) </s> <s xml:id="echoid-s2736" xml:space="preserve">perpendiculaires à ce plan, <lb/>elles ſont auſſi toutes paralleles entr’elles; </s> <s xml:id="echoid-s2737" xml:space="preserve">& </s> <s xml:id="echoid-s2738" xml:space="preserve">par <lb/>conſéquent 1°. </s> <s xml:id="echoid-s2739" xml:space="preserve">IM & </s> <s xml:id="echoid-s2740" xml:space="preserve">PQ ſont dans un même plan <lb/>avec PI & </s> <s xml:id="echoid-s2741" xml:space="preserve">QM: </s> <s xml:id="echoid-s2742" xml:space="preserve">Ainſi les angles en M, Q, & </s> <s xml:id="echoid-s2743" xml:space="preserve">P, <lb/>étant (hyp) droits, MP ſera un parallelogramme. <lb/></s> <s xml:id="echoid-s2744" xml:space="preserve">On prouvera de même que VN eſt auſſi un paralle-<lb/>logramme: </s> <s xml:id="echoid-s2745" xml:space="preserve">Donc IM eſt égale à PQ, & </s> <s xml:id="echoid-s2746" xml:space="preserve">EN égale <lb/>à VQ. </s> <s xml:id="echoid-s2747" xml:space="preserve">2°. </s> <s xml:id="echoid-s2748" xml:space="preserve">De ce que IM & </s> <s xml:id="echoid-s2749" xml:space="preserve">EN ſont paralleles en- <pb o="110" file="0136" n="136" rhead="EXAMEN DE L’OPINION"/> tr’elles, il ſuit auſſi que les angles MIK & </s> <s xml:id="echoid-s2750" xml:space="preserve">NEK <lb/> <anchor type="note" xlink:label="note-0136-01a" xlink:href="note-0136-01"/> ſont égaux; </s> <s xml:id="echoid-s2751" xml:space="preserve">& </s> <s xml:id="echoid-s2752" xml:space="preserve">par conſéquent, ſi l’on joint KM & </s> <s xml:id="echoid-s2753" xml:space="preserve"><lb/>KN, les angles en M & </s> <s xml:id="echoid-s2754" xml:space="preserve">en N étant (hyp.) </s> <s xml:id="echoid-s2755" xml:space="preserve">égaux, <lb/>auſſi-bien que les lignes IK & </s> <s xml:id="echoid-s2756" xml:space="preserve">KE, les triangles IMK <lb/>& </s> <s xml:id="echoid-s2757" xml:space="preserve">ENK ſeront non ſeulement ſemblables, mais en-<lb/>core IM ſera égale à EN. </s> <s xml:id="echoid-s2758" xml:space="preserve">Or on vient de voir (n. </s> <s xml:id="echoid-s2759" xml:space="preserve">1.) <lb/></s> <s xml:id="echoid-s2760" xml:space="preserve">que IM eſt égale à PQ, & </s> <s xml:id="echoid-s2761" xml:space="preserve">EN égale à VQ: </s> <s xml:id="echoid-s2762" xml:space="preserve"><lb/>Donc PQ eſt égale à VQ: </s> <s xml:id="echoid-s2763" xml:space="preserve">Donc (fig. </s> <s xml:id="echoid-s2764" xml:space="preserve">9. </s> <s xml:id="echoid-s2765" xml:space="preserve">11. </s> <s xml:id="echoid-s2766" xml:space="preserve">12. </s> <s xml:id="echoid-s2767" xml:space="preserve">13. </s> <s xml:id="echoid-s2768" xml:space="preserve"><lb/>& </s> <s xml:id="echoid-s2769" xml:space="preserve">14.) </s> <s xml:id="echoid-s2770" xml:space="preserve">CP plus CV, ou ( fig. </s> <s xml:id="echoid-s2771" xml:space="preserve">10. </s> <s xml:id="echoid-s2772" xml:space="preserve">14. </s> <s xml:id="echoid-s2773" xml:space="preserve">15. </s> <s xml:id="echoid-s2774" xml:space="preserve">& </s> <s xml:id="echoid-s2775" xml:space="preserve">16.) </s> <s xml:id="echoid-s2776" xml:space="preserve"><lb/>CP moins CV, eſt égal à deux fois CQ. </s> <s xml:id="echoid-s2777" xml:space="preserve">Or à cauſe <lb/>que les triangles CGL, & </s> <s xml:id="echoid-s2778" xml:space="preserve">CKQ ſont ſemblables, & </s> <s xml:id="echoid-s2779" xml:space="preserve"><lb/>que CG eſt double de CK; </s> <s xml:id="echoid-s2780" xml:space="preserve">CL ſera auſſi double de <lb/>CQ: </s> <s xml:id="echoid-s2781" xml:space="preserve">Donc (fig. </s> <s xml:id="echoid-s2782" xml:space="preserve">9. </s> <s xml:id="echoid-s2783" xml:space="preserve">11. </s> <s xml:id="echoid-s2784" xml:space="preserve">12. </s> <s xml:id="echoid-s2785" xml:space="preserve">13. </s> <s xml:id="echoid-s2786" xml:space="preserve">& </s> <s xml:id="echoid-s2787" xml:space="preserve">14.) </s> <s xml:id="echoid-s2788" xml:space="preserve">CP plus CV, <lb/>ou (fig. </s> <s xml:id="echoid-s2789" xml:space="preserve">10. </s> <s xml:id="echoid-s2790" xml:space="preserve">14. </s> <s xml:id="echoid-s2791" xml:space="preserve">16. </s> <s xml:id="echoid-s2792" xml:space="preserve">& </s> <s xml:id="echoid-s2793" xml:space="preserve">16.) </s> <s xml:id="echoid-s2794" xml:space="preserve">CP moins CV, eſt égale <lb/>à CL. </s> <s xml:id="echoid-s2795" xml:space="preserve">Ce qu’il faloit dèmontrer.</s> <s xml:id="echoid-s2796" xml:space="preserve"/> </p> <div xml:id="echoid-div272" type="float" level="2" n="1"> <note position="left" xlink:label="note-0136-01" xlink:href="note-0136-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> <figure> <image file="0136-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0136-01"/> </figure> </div> <div xml:id="echoid-div274" type="section" level="1" n="174"> <head xml:id="echoid-head176" xml:space="preserve">PROPOSITION III.</head> <p style="it"> <s xml:id="echoid-s2797" xml:space="preserve">T Outes choſes étant les mèmes que dans la propoſition <lb/> <anchor type="note" xlink:label="note-0136-02a" xlink:href="note-0136-02"/> précédente, on trouvera préſentement que chacune des <lb/>puiſſances A, B, D, E, F, &</s> <s xml:id="echoid-s2798" xml:space="preserve">c. </s> <s xml:id="echoid-s2799" xml:space="preserve">eſt au poids T qu’elles <lb/>ſoutiennent, comme chacune de leurs proportionelles CG, CR, <lb/>CM, CN, CP, &</s> <s xml:id="echoid-s2800" xml:space="preserve">c. </s> <s xml:id="echoid-s2801" xml:space="preserve">à la ſomme de leurs ſublimitez moins <lb/>celle de leurs profondeurs.</s> <s xml:id="echoid-s2802" xml:space="preserve"/> </p> <div xml:id="echoid-div274" type="float" level="2" n="1"> <note position="left" xlink:label="note-0136-02" xlink:href="note-0136-02a" xml:space="preserve">fig. 8. <lb/>17.</note> </div> </div> <div xml:id="echoid-div276" type="section" level="1" n="175"> <head xml:id="echoid-head177" xml:space="preserve"><emph style="sc">Demonstration.</emph></head> <p> <s xml:id="echoid-s2803" xml:space="preserve">De toutes les pointes des parallelogrammes GR, <lb/> <anchor type="note" xlink:label="note-0136-03a" xlink:href="note-0136-03"/> HM, LN, QP, &</s> <s xml:id="echoid-s2804" xml:space="preserve">c. </s> <s xml:id="echoid-s2805" xml:space="preserve">tirez Gg, Hh, Rr, Ll, Mm, Qq, <lb/>Nn, Pp, &</s> <s xml:id="echoid-s2806" xml:space="preserve">c. </s> <s xml:id="echoid-s2807" xml:space="preserve">perpendiculairement ſur la ligne de di-<lb/>rection du poids T, prolongée indéfiniment de part <lb/>& </s> <s xml:id="echoid-s2808" xml:space="preserve">d’autre. </s> <s xml:id="echoid-s2809" xml:space="preserve">Cela fait, vous trouverez par le Lemme <lb/>précédent. </s> <s xml:id="echoid-s2810" xml:space="preserve">1°. </s> <s xml:id="echoid-s2811" xml:space="preserve">Ch = Cg Cr. </s> <s xml:id="echoid-s2812" xml:space="preserve">2°. </s> <s xml:id="echoid-s2813" xml:space="preserve">Cl = Cm -<lb/>Ch: </s> <s xml:id="echoid-s2814" xml:space="preserve">Donc Cl = Cm - Cg + Cr. </s> <s xml:id="echoid-s2815" xml:space="preserve">3°. </s> <s xml:id="echoid-s2816" xml:space="preserve">Cq = <pb o="111" file="0137" n="137" rhead="DE M. BORELLI."/> Cl + Cn: </s> <s xml:id="echoid-s2817" xml:space="preserve">Donc Cq = Cm - Cg + Cr + Cn. <lb/></s> <s xml:id="echoid-s2818" xml:space="preserve"> <anchor type="note" xlink:label="note-0137-01a" xlink:href="note-0137-01"/> 4°. </s> <s xml:id="echoid-s2819" xml:space="preserve">Ck = Cq - Cp: </s> <s xml:id="echoid-s2820" xml:space="preserve">Donc Ck = Cm - Cg + <lb/>Cr + Cn - Cp. </s> <s xml:id="echoid-s2821" xml:space="preserve">Enfin continuant toujours ainſi <lb/>juſqu’à la diagonale qui ſetrouve toujours ( Prop. </s> <s xml:id="echoid-s2822" xml:space="preserve">2.) <lb/></s> <s xml:id="echoid-s2823" xml:space="preserve">dans la ligne de ditection du poids T, on trouvera <lb/>de même que cette diagonale eſt toujours égale à <lb/>Cm - Cg + Cr + Cn - Cp ± &</s> <s xml:id="echoid-s2824" xml:space="preserve">c. </s> <s xml:id="echoid-s2825" xml:space="preserve">Or on vient <lb/>de voir ( Prop. </s> <s xml:id="echoid-s2826" xml:space="preserve">2.) </s> <s xml:id="echoid-s2827" xml:space="preserve">que chacune des puiſſances A, B, <lb/>D, E, F, &</s> <s xml:id="echoid-s2828" xml:space="preserve">c. </s> <s xml:id="echoid-s2829" xml:space="preserve">eſt auſſi toujours au poids T qu’elles <lb/>ſoutiennent, comme chacune de leurs proportio-<lb/>nelles CG, CR, CM, CN, CP, &</s> <s xml:id="echoid-s2830" xml:space="preserve">c. </s> <s xml:id="echoid-s2831" xml:space="preserve">à cette même <lb/>diagonale: </s> <s xml:id="echoid-s2832" xml:space="preserve">Donc chacune de ces puiſſances eſt à ce <lb/>poids, comme chacune de ces proportionelles à Cm + <lb/>Cr + Cn - Cg - Cp ± &</s> <s xml:id="echoid-s2833" xml:space="preserve">c. </s> <s xml:id="echoid-s2834" xml:space="preserve">C’eſt-à-dire, ( Def. </s> <s xml:id="echoid-s2835" xml:space="preserve"><lb/>1. </s> <s xml:id="echoid-s2836" xml:space="preserve">& </s> <s xml:id="echoid-s2837" xml:space="preserve">2.) </s> <s xml:id="echoid-s2838" xml:space="preserve">à la ſomme de leurs ſublimitez Cm, Cr, <lb/>Cn, &</s> <s xml:id="echoid-s2839" xml:space="preserve">c. </s> <s xml:id="echoid-s2840" xml:space="preserve">moins la ſomme de leurs profondeurs Cg, <lb/>Cp, &</s> <s xml:id="echoid-s2841" xml:space="preserve">c. </s> <s xml:id="echoid-s2842" xml:space="preserve">D’où l’on voit en général, que de qu@lque <lb/>maniére qu’un poids ſoit ſoutenu avec des cordes <lb/>par quelque nombre de puiſſances que ce ſoit, appli-<lb/>quées à un même nœud, chacune de ces puiſſances <lb/>eſt toujours à ce poids, comme chacune de leurs pro-<lb/>portionelles, à la ſomme de leurs ſublimitez moins <lb/>celle de leurs profondeurs. </s> <s xml:id="echoid-s2843" xml:space="preserve">Ce qu’il faloit démon-<lb/>trer.</s> <s xml:id="echoid-s2844" xml:space="preserve"/> </p> <div xml:id="echoid-div276" type="float" level="2" n="1"> <note position="left" xlink:label="note-0136-03" xlink:href="note-0136-03a" xml:space="preserve">fig. 8.</note> <note position="right" xlink:label="note-0137-01" xlink:href="note-0137-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> </div> <div xml:id="echoid-div278" type="section" level="1" n="176"> <head xml:id="echoid-head178" xml:space="preserve"><emph style="sc">Autre</emph> <emph style="sc">Demonstration.</emph></head> <p> <s xml:id="echoid-s2845" xml:space="preserve">Soient encore les lignes CG, CR, CM, CN, CP, <lb/>&</s> <s xml:id="echoid-s2846" xml:space="preserve">c. </s> <s xml:id="echoid-s2847" xml:space="preserve">proportionelles aux puiſſances A, B, D, E, F, <lb/> <anchor type="note" xlink:label="note-0137-02a" xlink:href="note-0137-02"/> &</s> <s xml:id="echoid-s2848" xml:space="preserve">c. </s> <s xml:id="echoid-s2849" xml:space="preserve">proportionelles aux puiſſances A, B, D, E, F, <lb/>&</s> <s xml:id="echoid-s2850" xml:space="preserve">c. </s> <s xml:id="echoid-s2851" xml:space="preserve">concevez par le point C, où elles ſe communi-<lb/>quent, un plan horizontal OH, c’eſt-à-dire, per-<lb/>pendiculaire à la ligne de direction du poids T; </s> <s xml:id="echoid-s2852" xml:space="preserve">tirez <lb/>enſuite des extrémitez de ces proportionelles G, R, <lb/>M, N, P, &</s> <s xml:id="echoid-s2853" xml:space="preserve">c. </s> <s xml:id="echoid-s2854" xml:space="preserve">autant de perpendiculaires ſur le <lb/>plan OH, & </s> <s xml:id="echoid-s2855" xml:space="preserve">ſur la ligne de direction du poids T <lb/>indéfiniment prolongée de part & </s> <s xml:id="echoid-s2856" xml:space="preserve">d’™autre: </s> <s xml:id="echoid-s2857" xml:space="preserve">en faiſant <pb o="112" file="0138" n="138" rhead="EXAMEN DE L’OPINION"/> depuis C ſur le plan OH autant de lignes CH, CQ, <lb/> <anchor type="note" xlink:label="note-0138-01a" xlink:href="note-0138-01"/> CL, CO, CK, &</s> <s xml:id="echoid-s2858" xml:space="preserve">c. </s> <s xml:id="echoid-s2859" xml:space="preserve">qui joignent ce point avec les <lb/>perpendiculaires qui tombent ſur ce plan, on aura <lb/>autant de parallelogrammes rectangles Hg, Qr, Lm, <lb/>On, Kp, &</s> <s xml:id="echoid-s2860" xml:space="preserve">c. </s> <s xml:id="echoid-s2861" xml:space="preserve">qui exprimeront ( Lemm 3. </s> <s xml:id="echoid-s2862" xml:space="preserve">Cor. </s> <s xml:id="echoid-s2863" xml:space="preserve">2. </s> <s xml:id="echoid-s2864" xml:space="preserve">du <lb/>Projet précédent) que chacune de ces puiſſances, par <lb/>exemple la puiſſance A, fait la même impreſſion ſur le <lb/>point C, que feroient deux autres puiſſances appliquées <lb/>à ce point, l’une ſuivant CH, & </s> <s xml:id="echoid-s2865" xml:space="preserve">l’autre ſuivant Cg, à <lb/>chacune deſquelles celle-ci ſeroit comme @G à cha-<lb/>cune de ces mêmes lignes: </s> <s xml:id="echoid-s2866" xml:space="preserve">Le point C eſt donc tiré <lb/>vers bas ſuivant la ligne de direction du poids T par <lb/>la puiſſance A, d’une force (Cor. </s> <s xml:id="echoid-s2867" xml:space="preserve">3. </s> <s xml:id="echoid-s2868" xml:space="preserve">du même Lemm) <lb/>à qui cette puiſſance eſt comme CG à C@. </s> <s xml:id="echoid-s2869" xml:space="preserve">Pour la <lb/>même raiſon, il eſt encore tiré ſuivant la ligne de di-<lb/>rection de même poids, 1°. </s> <s xml:id="echoid-s2870" xml:space="preserve">Vers bas, par la puiſſance <lb/>F, d’une force à qui elle eſt comme CP à Cp; </s> <s xml:id="echoid-s2871" xml:space="preserve">&</s> <s xml:id="echoid-s2872" xml:space="preserve">c. <lb/></s> <s xml:id="echoid-s2873" xml:space="preserve">2°. </s> <s xml:id="echoid-s2874" xml:space="preserve">Vers haut, par la puiſſance B, d’une force à qui <lb/>elle eſt comme CR à Cr; </s> <s xml:id="echoid-s2875" xml:space="preserve">par la puiſſance D, d’une <lb/>force à qui elle eſt comme CM à Cm; </s> <s xml:id="echoid-s2876" xml:space="preserve">par la puiſſance <lb/>E, d’une force à qui elle eſt, comme CN à Cn; </s> <s xml:id="echoid-s2877" xml:space="preserve">&</s> <s xml:id="echoid-s2878" xml:space="preserve">c. </s> <s xml:id="echoid-s2879" xml:space="preserve"><lb/>Or (hyp.) </s> <s xml:id="echoid-s2880" xml:space="preserve">la puiſſance A eſt à chacune des puiſſances <lb/>B, D, E, F, &</s> <s xml:id="echoid-s2881" xml:space="preserve">c. </s> <s xml:id="echoid-s2882" xml:space="preserve">comme ſa proportionelle CG à cha-<lb/>cune des leurs CR, CM, CN, CP, &</s> <s xml:id="echoid-s2883" xml:space="preserve">c. </s> <s xml:id="echoid-s2884" xml:space="preserve">Donc la <lb/>puiſſance A eſt à chacune des forces avec leſquelles <lb/>le point C eſt tiré ſuivant la ligne de direction du <lb/>poids T, 1°. </s> <s xml:id="echoid-s2885" xml:space="preserve">Vers bas par les puiſſances A, F, <lb/>&</s> <s xml:id="echoid-s2886" xml:space="preserve">c. </s> <s xml:id="echoid-s2887" xml:space="preserve">comme CG à chacune de leurs profondeurs <lb/>Cg, Cp, &</s> <s xml:id="echoid-s2888" xml:space="preserve">c. </s> <s xml:id="echoid-s2889" xml:space="preserve">2°. </s> <s xml:id="echoid-s2890" xml:space="preserve">Vers haut par les puiſſances <lb/>B, D, E, &</s> <s xml:id="echoid-s2891" xml:space="preserve">c. </s> <s xml:id="echoid-s2892" xml:space="preserve">comme la même CG à chacune de <lb/>leurs ſublimitez Cr, Cm, Cn, &</s> <s xml:id="echoid-s2893" xml:space="preserve">c. </s> <s xml:id="echoid-s2894" xml:space="preserve">Donc cette <lb/>même puiſſance A eſt à la ſomme de toutes les forces <lb/>avec leſquelles le point C eſt tiré ſuivant la ligne de <lb/>direction du poids T, 1°. </s> <s xml:id="echoid-s2895" xml:space="preserve">Vers bas par les puiſſances <lb/>A, F, &</s> <s xml:id="echoid-s2896" xml:space="preserve">c. </s> <s xml:id="echoid-s2897" xml:space="preserve">comme ſa proportionelle CG à la ſomme <pb o="113" file="0139" n="139" rhead="DE M. BORELLI."/> de leurs profondeurs Cg, Cp, &</s> <s xml:id="echoid-s2898" xml:space="preserve">c. </s> <s xml:id="echoid-s2899" xml:space="preserve">2°. </s> <s xml:id="echoid-s2900" xml:space="preserve">Vers haut <lb/> <anchor type="note" xlink:label="note-0139-01a" xlink:href="note-0139-01"/> par les puiſſances B, D, E, &</s> <s xml:id="echoid-s2901" xml:space="preserve">c. </s> <s xml:id="echoid-s2902" xml:space="preserve">comme la même CG <lb/>à la ſomme de leurs ſublimitez Cr, Cm, Cn, &</s> <s xml:id="echoid-s2903" xml:space="preserve">c. <lb/></s> <s xml:id="echoid-s2904" xml:space="preserve">Or la ſomme faite de la peſanteur de ce poids, & </s> <s xml:id="echoid-s2905" xml:space="preserve"><lb/>des forces avec leſquelles le point C eſt tiré vers <lb/>bas ſuivant la ligne de direction de ce même poids <lb/>par les puiſſances A, F, &</s> <s xml:id="echoid-s2906" xml:space="preserve">c. </s> <s xml:id="echoid-s2907" xml:space="preserve">étant diamétralement <lb/>oppoſée à la ſomme de celles avec leſquelles ce même <lb/>point eſt tiré en même-tems vers haut ſuivant cette <lb/>même ligne par les puiſſances B, D, E, &</s> <s xml:id="echoid-s2908" xml:space="preserve">c. </s> <s xml:id="echoid-s2909" xml:space="preserve">& </s> <s xml:id="echoid-s2910" xml:space="preserve">au-<lb/>cune de ces deux ſommes de forces ne l’emportant <lb/>ſur l’autre; </s> <s xml:id="echoid-s2911" xml:space="preserve">puiſque (hyp.) </s> <s xml:id="echoid-s2912" xml:space="preserve">le poids T ne monte n’y <lb/>deſcend: </s> <s xml:id="echoid-s2913" xml:space="preserve">c’eſt une conſéquence néceſſaire qu’elles <lb/>ſoient égales: </s> <s xml:id="echoid-s2914" xml:space="preserve">Donc la puiſſance A eſt non ſeulement <lb/>à la ſomme des forces avec leſquelles le point C eſt <lb/>tiré vers bas, ſuivant la ligne de direction du poids <lb/>T par les puiſſances A, F, &</s> <s xml:id="echoid-s2915" xml:space="preserve">c. </s> <s xml:id="echoid-s2916" xml:space="preserve">comme ſa propor-<lb/>tionelle CG à la ſomme de leurs profondeurs Cg, <lb/>Cp, &</s> <s xml:id="echoid-s2917" xml:space="preserve">c. </s> <s xml:id="echoid-s2918" xml:space="preserve">Mais auſſi à la ſomme faite de cette pre-<lb/>miére & </s> <s xml:id="echoid-s2919" xml:space="preserve">de la peſanteur de ce même poids, comme <lb/>la même CG à la ſomme des ſublimitez Cr, Cm, <lb/>Cn, &</s> <s xml:id="echoid-s2920" xml:space="preserve">c. </s> <s xml:id="echoid-s2921" xml:space="preserve">des puiſſances B, D, E, &</s> <s xml:id="echoid-s2922" xml:space="preserve">c. </s> <s xml:id="echoid-s2923" xml:space="preserve">Donc la puiſ-<lb/>ſance A eſt à cette derniére ſomme moins la pre-<lb/>miére; </s> <s xml:id="echoid-s2924" xml:space="preserve">c’eſt-à-dire, à la peſanteur ſeule du poids <lb/>T, où à ce poids lui - même, comme ſa proportio-<lb/>nelle CG à la ſomme des ſublimitez Cr, Cm, Cn, <lb/>&</s> <s xml:id="echoid-s2925" xml:space="preserve">c. </s> <s xml:id="echoid-s2926" xml:space="preserve">moins la ſomme des profondeurs Cg, Cp, &</s> <s xml:id="echoid-s2927" xml:space="preserve">c. </s> <s xml:id="echoid-s2928" xml:space="preserve"><lb/>Or ( Hyp.) </s> <s xml:id="echoid-s2929" xml:space="preserve">chacune des puiſſances B, D, E, F, <lb/>&</s> <s xml:id="echoid-s2930" xml:space="preserve">c. </s> <s xml:id="echoid-s2931" xml:space="preserve">eſt à la puiſſance A, comme chacune de leurs <lb/>proportionelles CR, CM, CN, CP, &</s> <s xml:id="echoid-s2932" xml:space="preserve">c. </s> <s xml:id="echoid-s2933" xml:space="preserve">à ſa pro-<lb/>portionelle CG: </s> <s xml:id="echoid-s2934" xml:space="preserve">Donc chacune des puiſſances A, <lb/>B, D, E, F, &</s> <s xml:id="echoid-s2935" xml:space="preserve">c. </s> <s xml:id="echoid-s2936" xml:space="preserve">eſt au poids T qu’elles ſoutien-<lb/>nent, comme chacune de leurs proportionelles à la <lb/>ſomme de leurs ſublimitez moins celle de leurs pro-<lb/>fondeurs. </s> <s xml:id="echoid-s2937" xml:space="preserve">Ce qu’il faloit démontrer.</s> <s xml:id="echoid-s2938" xml:space="preserve"/> </p> <div xml:id="echoid-div278" type="float" level="2" n="1"> <note position="right" xlink:label="note-0137-02" xlink:href="note-0137-02a" xml:space="preserve">fig. 17.</note> <note position="left" xlink:label="note-0138-01" xlink:href="note-0138-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> <note position="right" xlink:label="note-0139-01" xlink:href="note-0139-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> <pb o="114" file="0140" n="140" rhead="EXAMEN DE L’OPINION"/> </div> <div xml:id="echoid-div280" type="section" level="1" n="177"> <head xml:id="echoid-head179" xml:space="preserve"><emph style="sc">Corollaire.</emph></head> <note position="left" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> <p> <s xml:id="echoid-s2939" xml:space="preserve">On voit préſentement en général que la ſomme <lb/>de toutes les puiſſances qui ſoutiennent un poids <lb/>avec des cordes qui ſe tiennent par un même nœud, <lb/>en quelque nombre qu’elles ſoient, quelque pro-<lb/>portion qu’elles ayent entr’elles, & </s> <s xml:id="echoid-s2940" xml:space="preserve">de quelque ma-<lb/>niére qu’elles lui ſoient appliquées; </s> <s xml:id="echoid-s2941" xml:space="preserve">eſt toujours à <lb/>ce poids, comme la ſomme des parties de leurs cor-<lb/>des qui leurs ſont (Chap. </s> <s xml:id="echoid-s2942" xml:space="preserve">2. </s> <s xml:id="echoid-s2943" xml:space="preserve">Avert ) proportionelles, <lb/>à la ſomme de leurs ſublimitez moins celle de leurs <lb/>profondeurs.</s> <s xml:id="echoid-s2944" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s2945" xml:space="preserve">On peut comparer tout ceci avec les Propoſitions 70. </s> <s xml:id="echoid-s2946" xml:space="preserve">73. <lb/></s> <s xml:id="echoid-s2947" xml:space="preserve">74. </s> <s xml:id="echoid-s2948" xml:space="preserve">de M. </s> <s xml:id="echoid-s2949" xml:space="preserve">Borelli, & </s> <s xml:id="echoid-s2950" xml:space="preserve">on verra non ſeulement qu’elles ſont <lb/>tres-limitées; </s> <s xml:id="echoid-s2951" xml:space="preserve">mais encore qu’avec ſa méthode on ne peut pas <lb/>aller ſi loin.</s> <s xml:id="echoid-s2952" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div281" type="section" level="1" n="178"> <head xml:id="echoid-head180" xml:space="preserve"><emph style="sc">Remarque.</emph></head> <p> <s xml:id="echoid-s2953" xml:space="preserve">En faiſant la ſeconde des deux démonſtrations <lb/>précédentes, il m’en eſt encore venu une de la pre-<lb/>miére Propoſition: </s> <s xml:id="echoid-s2954" xml:space="preserve">la voici.</s> <s xml:id="echoid-s2955" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s2956" xml:space="preserve">Le poids T étant donc ſoutenu avec des cordes par deux <lb/>puiſſances R & </s> <s xml:id="echoid-s2957" xml:space="preserve">S; </s> <s xml:id="echoid-s2958" xml:space="preserve">des angles G & </s> <s xml:id="echoid-s2959" xml:space="preserve">H du parallelogramme <lb/> <anchor type="note" xlink:label="note-0140-02a" xlink:href="note-0140-02"/> GH, dont la diagonale CD fait partie de la ligne de di-<lb/>rection de ce poids, ſoient faites GM & </s> <s xml:id="echoid-s2960" xml:space="preserve">HN paralleles à <lb/>cette diagonale, & </s> <s xml:id="echoid-s2961" xml:space="preserve">perpendiculaires à MCN; </s> <s xml:id="echoid-s2962" xml:space="preserve">achevez <lb/>les parallelogrammes MP & </s> <s xml:id="echoid-s2963" xml:space="preserve">NQ. </s> <s xml:id="echoid-s2964" xml:space="preserve">Cela fait, vous trou-<lb/>verez encore de la maniére que nous avons fait la ſeconde <lb/>des deux démonſtrations précédentes, que le poids T eſt aux <lb/>puiſſances R & </s> <s xml:id="echoid-s2965" xml:space="preserve">S, comme la partie CD de ſa ligne de di-<lb/>rection aux parties CG & </s> <s xml:id="echoid-s2966" xml:space="preserve">CH de leurs cordes, qui font <lb/>les côtez du parallelogramme GH, dont elle eſt diagonale.</s> <s xml:id="echoid-s2967" xml:space="preserve"/> </p> <div xml:id="echoid-div281" type="float" level="2" n="1"> <note position="left" xlink:label="note-0140-02" xlink:href="note-0140-02a" xml:space="preserve">fig. 18. <lb/>19.</note> </div> <pb o="115" file="0141" n="141" rhead="DE M. BORELLI."/> <p> <s xml:id="echoid-s2968" xml:space="preserve">Car (Cor. </s> <s xml:id="echoid-s2969" xml:space="preserve">2. </s> <s xml:id="echoid-s2970" xml:space="preserve">Lem. </s> <s xml:id="echoid-s2971" xml:space="preserve">3. </s> <s xml:id="echoid-s2972" xml:space="preserve">du Projet précéd.) </s> <s xml:id="echoid-s2973" xml:space="preserve">la puiſſance <lb/> <anchor type="note" xlink:label="note-0141-01a" xlink:href="note-0141-01"/> R fait ſur le point C la même impreſſion que fe-<lb/>roient deux autres puiſſances appliquées à ce point, <lb/>l’une ſuivant CP, & </s> <s xml:id="echoid-s2974" xml:space="preserve">l’autre ſuivant CM, à chacune <lb/>deſquelles celle-ci ſeroit, comme CG à chacune de <lb/>ces lignes: </s> <s xml:id="echoid-s2975" xml:space="preserve">Le point C reçoit donc en même - tems <lb/>deux impreſſions différentes de la puiſſance R, l’une <lb/>ſuivant CP, d’une force qui eſt à celle de cette puiſ-<lb/>ſance, (Cor. </s> <s xml:id="echoid-s2976" xml:space="preserve">3. </s> <s xml:id="echoid-s2977" xml:space="preserve">du même Lemm.) </s> <s xml:id="echoid-s2978" xml:space="preserve">comme CP à CG, & </s> <s xml:id="echoid-s2979" xml:space="preserve"><lb/>l’autre ſuivant CM, d’une force qui eſt auſſi (par le <lb/>mème Cor.) </s> <s xml:id="echoid-s2980" xml:space="preserve">à celle de cette même puiſſance, comme CM <lb/>à CG. </s> <s xml:id="echoid-s2981" xml:space="preserve">Pour la même raiſon ce même point C reçoit <lb/>encore en même - tems deux impreſſions différentes <lb/>de la puiſſance S, l’une ſuivant CQ, d’une force <lb/>qui eſt à celle de cette puiſſance, comme CQ à CH; </s> <s xml:id="echoid-s2982" xml:space="preserve">& </s> <s xml:id="echoid-s2983" xml:space="preserve"><lb/>l’autre ſuivant CN, d’une force qui eſt auſſi à celle de <lb/>cette même puiſſance, comme CN à CH. </s> <s xml:id="echoid-s2984" xml:space="preserve">Or 1°. </s> <s xml:id="echoid-s2985" xml:space="preserve">La <lb/>force de l’impreſſion que reçoit le point C de la puiſ-<lb/>ſance R ſuivant CM, eſt égale à celle qu’il reçoit <lb/>en même - tems de la puiſſance S ſuivant CN; </s> <s xml:id="echoid-s2986" xml:space="preserve">puis <lb/>qu’elles ſont diamétralement oppoſées, & </s> <s xml:id="echoid-s2987" xml:space="preserve">qu’aucu-<lb/>ne des deux (byp.) </s> <s xml:id="echoid-s2988" xml:space="preserve">ne ſurmonte l’autre: </s> <s xml:id="echoid-s2989" xml:space="preserve">La force <lb/>de la puiſſance R eſt donc à celle de l’impreſſion que <lb/>reçoit le point C de la puiſſance S ſuivant CN, com-<lb/>me CG à CM. </s> <s xml:id="echoid-s2990" xml:space="preserve">Or CM eſt égale à CN; </s> <s xml:id="echoid-s2991" xml:space="preserve">puis que <lb/>les triangles GPD & </s> <s xml:id="echoid-s2992" xml:space="preserve">HQC ſemblables, & </s> <s xml:id="echoid-s2993" xml:space="preserve">GD <lb/>égale à CH rendent GP égale à HQ; </s> <s xml:id="echoid-s2994" xml:space="preserve">& </s> <s xml:id="echoid-s2995" xml:space="preserve">que les <lb/>parallelogrammes MP & </s> <s xml:id="echoid-s2996" xml:space="preserve">NQ rendent auſſi GP <lb/>égale à CM, & </s> <s xml:id="echoid-s2997" xml:space="preserve">HQ égale à CN: </s> <s xml:id="echoid-s2998" xml:space="preserve">Donc la puiſſance <lb/>R eſt à la force de l’impreſſion que le point C reçoit <lb/>de la puiſſance S ſuivant CN, comme CG à CN. <lb/></s> <s xml:id="echoid-s2999" xml:space="preserve">Or on vient de voir que la force de cette même im-<lb/>preſſion eſt à la puiſſance S, comme CN à CH: </s> <s xml:id="echoid-s3000" xml:space="preserve"><lb/>Donc la puiſſance R eſt à la puiſſance S, comme <lb/>CG à CH: </s> <s xml:id="echoid-s3001" xml:space="preserve">2°. </s> <s xml:id="echoid-s3002" xml:space="preserve">On vient de voir auſſi que la puiſſan- <pb o="116" file="0142" n="142" rhead="EXAMEN DE L’OPINION"/> ce S eſt à la force de l’impreſſion qu’elle fait ſur le <lb/> <anchor type="note" xlink:label="note-0142-01a" xlink:href="note-0142-01"/> point C ſuivant CQ, comme CH à CQ: </s> <s xml:id="echoid-s3003" xml:space="preserve">Donc la <lb/>puiſſance R eſt auſſi à la force de cette même im-<lb/>preſſion, comme CG à CQ; </s> <s xml:id="echoid-s3004" xml:space="preserve">c’eſt - à - dire, comme <lb/>CG à DP; </s> <s xml:id="echoid-s3005" xml:space="preserve">puis que les triangles GPD & </s> <s xml:id="echoid-s3006" xml:space="preserve">HQC <lb/>ſemblables, & </s> <s xml:id="echoid-s3007" xml:space="preserve">GD égale à CH, rendent DP égale <lb/>à CQ. </s> <s xml:id="echoid-s3008" xml:space="preserve">On vient de voir encore que cette même <lb/>puiſſance R eſt à la force de l’impreſſion qu’elle fait <lb/>ſur ce même point C ſuivant CP, comme CG à CP: <lb/></s> <s xml:id="echoid-s3009" xml:space="preserve">Donc la puiſſance R eſt à la ſomme, où à la diffé-<lb/>rence des forces de ces deux impreſſions faites ſur <lb/>le point C ſuivant CP & </s> <s xml:id="echoid-s3010" xml:space="preserve">CQ, par elle & </s> <s xml:id="echoid-s3011" xml:space="preserve">par la puiſ-<lb/>ſance S, comme CG à la ſomme, où à la différence <lb/>de ces deux lignes. </s> <s xml:id="echoid-s3012" xml:space="preserve">Or (fig. </s> <s xml:id="echoid-s3013" xml:space="preserve">18.) </s> <s xml:id="echoid-s3014" xml:space="preserve">la ſomme de ces <lb/>deux lignes, où (fig. </s> <s xml:id="echoid-s3015" xml:space="preserve">19.) </s> <s xml:id="echoid-s3016" xml:space="preserve">leur différence, eſt égale <lb/>à la diagonale CD du parallelogramme GH; </s> <s xml:id="echoid-s3017" xml:space="preserve">& </s> <s xml:id="echoid-s3018" xml:space="preserve"><lb/>(fig. </s> <s xml:id="echoid-s3019" xml:space="preserve">18.) </s> <s xml:id="echoid-s3020" xml:space="preserve">la ſomme, où (fig. </s> <s xml:id="echoid-s3021" xml:space="preserve">19.) </s> <s xml:id="echoid-s3022" xml:space="preserve">la différence des <lb/>forces de ces deux impreſſions, eſt auſſi égale au <lb/>poids T: </s> <s xml:id="echoid-s3023" xml:space="preserve">Donc la puiſſance R eſt au poids T, com-<lb/>me CG à CD: </s> <s xml:id="echoid-s3024" xml:space="preserve">On vient de démontrer (n. </s> <s xml:id="echoid-s3025" xml:space="preserve">1.) </s> <s xml:id="echoid-s3026" xml:space="preserve">que <lb/>cettemême puiſſance R eſt auſſi à la puiſſance S, com-<lb/>me CG à CH: </s> <s xml:id="echoid-s3027" xml:space="preserve">Donc les puiſſances R & </s> <s xml:id="echoid-s3028" xml:space="preserve">S, & </s> <s xml:id="echoid-s3029" xml:space="preserve">le <lb/>poids T ſont entr’eux , comme les lignes CG, CH, <lb/>& </s> <s xml:id="echoid-s3030" xml:space="preserve">CD: </s> <s xml:id="echoid-s3031" xml:space="preserve">& </s> <s xml:id="echoid-s3032" xml:space="preserve">par conſéquent ce poids eſt à chacune <lb/>d’elles, comme la partie CD de ſa ligne de direction <lb/>à chacune des parties de leurs cordes, qui font les <lb/>côtez du parallelogramme GH, dont elle eſt diago-<lb/>nale. </s> <s xml:id="echoid-s3033" xml:space="preserve">Ce qu’il faloit démontrer.</s> <s xml:id="echoid-s3034" xml:space="preserve"/> </p> <div xml:id="echoid-div282" type="float" level="2" n="2"> <note position="right" xlink:label="note-0141-01" xlink:href="note-0141-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement</note> <note position="left" xlink:label="note-0142-01" xlink:href="note-0142-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> <p> <s xml:id="echoid-s3035" xml:space="preserve">On voit de-là, que ſi par le point C où ſe com-<lb/>muniquent les deux cordes qui ſoutiennent quel-<lb/>que poids que ce ſoit, on fait MN perpendiculaire à <lb/>la ligne de direction de ce poids, & </s> <s xml:id="echoid-s3036" xml:space="preserve">qu’aprés avoir pris <lb/>de part & </s> <s xml:id="echoid-s3037" xml:space="preserve">d’autre ſur cette ligne CM & </s> <s xml:id="echoid-s3038" xml:space="preserve">CN ègales en-<lb/>tr’elles, on faſſe aux points M & </s> <s xml:id="echoid-s3039" xml:space="preserve">N les perpendiculaires <pb o="117" file="0143" n="143" rhead="DE M. BORELLI."/> MG & </s> <s xml:id="echoid-s3040" xml:space="preserve">NH qui rencontrent auxpoints G & </s> <s xml:id="echoid-s3041" xml:space="preserve">H les <lb/> <anchor type="note" xlink:label="note-0143-01a" xlink:href="note-0143-01"/> cordes des puiſſances qui ſoutiennent ce poids: </s> <s xml:id="echoid-s3042" xml:space="preserve">elles <lb/>en détermineront des parties CG, CH, qui ſeront <lb/>toujours proportionelles à ces mêmes puiſſances.</s> <s xml:id="echoid-s3043" xml:space="preserve"/> </p> <div xml:id="echoid-div283" type="float" level="2" n="3"> <note position="right" xlink:label="note-0143-01" xlink:href="note-0143-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> <p style="it"> <s xml:id="echoid-s3044" xml:space="preserve">Si M. </s> <s xml:id="echoid-s3045" xml:space="preserve">Borelli eut fait rèflèxion que les puiſſances R & </s> <s xml:id="echoid-s3046" xml:space="preserve"><lb/>S n’agiſſent pas ſeulement contre le poids T, mais auſſi <lb/>l’une contre l’autre ; </s> <s xml:id="echoid-s3047" xml:space="preserve">& </s> <s xml:id="echoid-s3048" xml:space="preserve">que de même qu’elles concourent en-<lb/>ſemble pour empêcber que ce poids n’attire à lui (fig. </s> <s xml:id="echoid-s3049" xml:space="preserve">18.) <lb/></s> <s xml:id="echoid-s3050" xml:space="preserve">le nœud C, de même auſſi cbacune d’elles concourt avec <lb/>lui pour empêcber que l’autre ne l’emporte. </s> <s xml:id="echoid-s3051" xml:space="preserve">Si, dî - je, il avoit <lb/>fait cette rèfléxion, il auroit vû ſans doute que chacune. </s> <s xml:id="echoid-s3052" xml:space="preserve">de <lb/>ces puiſſances fait impreſſion ſur ce nœud, non ſeulement <lb/>ſuivant la direction du poids qu’elles ſoutiennent, pour le <lb/>tenir toujours à même hauteur; </s> <s xml:id="echoid-s3053" xml:space="preserve">mais auſſi ſuivant l’hori-<lb/>zontale MCN, pour empêcher qu’aucune d’elles ne l’attire <lb/>n’y à droit, n’y à gaucbe: </s> <s xml:id="echoid-s3054" xml:space="preserve">D'où il auroit infailliblement conclu <lb/>que ces impreſſions horizontales, ètant diamétralement <lb/>oppoſées, doivent toujours être égales. </s> <s xml:id="echoid-s3055" xml:space="preserve">De-là voyant qu’el-<lb/>les augmentent, où diminuent néceſſairement à meſure que <lb/>les angles que font les cordes de ces puiſſances avec la ligne <lb/>de direction du poids qu’elles ſoutiennent, s’aprochent, où <lb/>s’éloignent de l’angle droit, il auroit enfin aperçû l’impoſſibi-<lb/>litè de faire, finon aucun, du moins un tel changement à leurs <lb/>d. </s> <s xml:id="echoid-s3056" xml:space="preserve">rections ſans en rompre l’équilibre.</s> <s xml:id="echoid-s3057" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3058" xml:space="preserve">@e dis ſinon aucun changement, parce qu’il a été <lb/>démontré (Cor. </s> <s xml:id="echoid-s3059" xml:space="preserve">1. </s> <s xml:id="echoid-s3060" xml:space="preserve">Prop. </s> <s xml:id="echoid-s3061" xml:space="preserve">1.) </s> <s xml:id="echoid-s3062" xml:space="preserve">qu’il n’eſt pas poſſible d’y <lb/>en faire aucun ſans rompre l’équilibre qui eſt (hyp.) </s> <s xml:id="echoid-s3063" xml:space="preserve">entre <lb/>ces puiſſances, & </s> <s xml:id="echoid-s3064" xml:space="preserve">le poids qu’elles ſoutiennent. </s> <s xml:id="echoid-s3065" xml:space="preserve">Nous <lb/>l’avons même conclu (Chap. </s> <s xml:id="echoid-s3066" xml:space="preserve">1.) </s> <s xml:id="echoid-s3067" xml:space="preserve">de la 68. </s> <s xml:id="echoid-s3068" xml:space="preserve">Prop. </s> <s xml:id="echoid-s3069" xml:space="preserve">d’où cet <lb/>Autheur tire un ſcbolie tout contraire par un raiſonnement dont <lb/>le défaut eſt préſentement aiſé à découvrir; </s> <s xml:id="echoid-s3070" xml:space="preserve">Voyez-le.</s> <s xml:id="echoid-s3071" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s3072" xml:space="preserve">Sur ce qu’on vient de dire de l’uſage des impreſſions hori- <pb o="118" file="0144" n="144" rhead="EXAMEN DE L’OPINION"/> zontales que font ſur le nœud C (fig. </s> <s xml:id="echoid-s3073" xml:space="preserve">18. </s> <s xml:id="echoid-s3074" xml:space="preserve">& </s> <s xml:id="echoid-s3075" xml:space="preserve">19.) </s> <s xml:id="echoid-s3076" xml:space="preserve">les puiſſan-<lb/> <anchor type="note" xlink:label="note-0144-01a" xlink:href="note-0144-01"/> ces R & </s> <s xml:id="echoid-s3077" xml:space="preserve">S, il eſt aiſé de juger de celui des impreſſions ſem-<lb/>blables que font auſſi ſur le nœud C de leurs cordes ſuivant <lb/>le plan OH (fig. </s> <s xml:id="echoid-s3078" xml:space="preserve">17.) </s> <s xml:id="echoid-s3079" xml:space="preserve">les puiſſances A, B, D, E, F, <lb/>& </s> <s xml:id="echoid-s3080" xml:space="preserve">C. </s> <s xml:id="echoid-s3081" xml:space="preserve">auſſi ne s’y arrêtera-t-on pas davantage.</s> <s xml:id="echoid-s3082" xml:space="preserve"/> </p> <div xml:id="echoid-div284" type="float" level="2" n="4"> <note position="left" xlink:label="note-0144-01" xlink:href="note-0144-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> <figure> <image file="0144-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0144-01"/> </figure> </div> <div xml:id="echoid-div286" type="section" level="1" n="179"> <head xml:id="echoid-head181" xml:space="preserve">PROPOSITION IV.</head> <p style="it"> <s xml:id="echoid-s3083" xml:space="preserve">DE quelque maniére qu’un poids ſoit ſoutenu avec des <lb/>cordes par quelque nombre de puiſſances que ce ſoit, <lb/>appliquées à tant de nœuds qu’on voudra; </s> <s xml:id="echoid-s3084" xml:space="preserve">chacune d’elles eſt <lb/>toujours à ce poids en raiſon compoſée d’autant d’autres raiſons <lb/>qu’il y a de nœuds entre cette puiſſance & </s> <s xml:id="echoid-s3085" xml:space="preserve">ce poids: </s> <s xml:id="echoid-s3086" xml:space="preserve">Sçavoir <lb/>par chaque nœud, de la raiſon qui eſt entre la proportionelle <lb/>à la force dont ce nœud eſt tiré ſuivant la corde qui lui donne <lb/>communication avec cette puiſſance, & </s> <s xml:id="echoid-s3087" xml:space="preserve">la ſomme des ſu-<lb/>blimitez moins celle des profondeurs de toutes les forces dont <lb/>les branches dans leſquelles ce même nœud ſe diviſe, ſont <lb/>tirèes chacune ſuivant ſa direction contre la réſiſtance qui leur <lb/>vient par la corde de communication de lui à ce poids.</s> <s xml:id="echoid-s3088" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div287" type="section" level="1" n="180"> <head xml:id="echoid-head182" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s3089" xml:space="preserve">Soit le poids T dont la corde C p ſe diviſe en tant <lb/> <anchor type="note" xlink:label="note-0144-02a" xlink:href="note-0144-02"/> de branches CZ, CX, CY, Cφ, &</s> <s xml:id="echoid-s3090" xml:space="preserve">c. </s> <s xml:id="echoid-s3091" xml:space="preserve">qu’on vou-<lb/>dra, dont celles qu’on voudra, ſe diviſent encore <lb/>en pluſieurs branches, & </s> <s xml:id="echoid-s3092" xml:space="preserve">celles qu’on voudra encore <lb/>de celles-ci en pluſieurs autres de la maniére qu’on <lb/>voit ici; </s> <s xml:id="echoid-s3093" xml:space="preserve">& </s> <s xml:id="echoid-s3094" xml:space="preserve">toujours de même juſqu’auſſi loin qu’on <lb/>voudra. </s> <s xml:id="echoid-s3095" xml:space="preserve">Commencez au premier nœud C à marquer <lb/>ſur les branches CZ, XC, YC, Cφ, &</s> <s xml:id="echoid-s3096" xml:space="preserve">c. </s> <s xml:id="echoid-s3097" xml:space="preserve">des parties <lb/>CM, CN, CP, Cθ, &</s> <s xml:id="echoid-s3098" xml:space="preserve">c. </s> <s xml:id="echoid-s3099" xml:space="preserve">qui ſoient entr’elles com-<lb/>me les forces avec leſquelles ces cordes ſont tirées <lb/>chacune ſuivant ſa direction. </s> <s xml:id="echoid-s3100" xml:space="preserve">Faites - en autant ſur <pb o="119" file="0145" n="145" rhead="DE M. BORELLI"/> les branches dans leſquelles celles-ci ſe ſubdiviſent; <lb/></s> <s xml:id="echoid-s3101" xml:space="preserve"> <anchor type="note" xlink:label="note-0145-01a" xlink:href="note-0145-01"/> & </s> <s xml:id="echoid-s3102" xml:space="preserve">toujours de même juſqu’aux derniéres auſquelles <lb/>les puiſſances A, E, D, B, F, G, H, I, K, φ, &</s> <s xml:id="echoid-s3103" xml:space="preserve">c. <lb/></s> <s xml:id="echoid-s3104" xml:space="preserve">ſont appliquées. </s> <s xml:id="echoid-s3105" xml:space="preserve">Cela fait, aprés avoir marqué des <lb/>extrémitez de toutes ces proportionelles (Avertiſſ. </s> <s xml:id="echoid-s3106" xml:space="preserve"><lb/>Chap. </s> <s xml:id="echoid-s3107" xml:space="preserve">2.) </s> <s xml:id="echoid-s3108" xml:space="preserve">les ſublimitez & </s> <s xml:id="echoid-s3109" xml:space="preserve">les profondeurs de toutes <lb/>ces forces, on trouvera que chacune de ces puiſ-<lb/>ſances, par exemple, la puiſſance D eſt toujours à <lb/>ce poids en raiſon compoſée d’autant d’autres rai-<lb/>ſons telles que cette propoſition porte, qu’il y a de <lb/>nœuds entre cette puiſſance & </s> <s xml:id="echoid-s3110" xml:space="preserve">lui: </s> <s xml:id="echoid-s3111" xml:space="preserve">Car 1°. </s> <s xml:id="echoid-s3112" xml:space="preserve">La puiſ-<lb/>ſance D étant (hyp) à la puiſſance E, comme O S <lb/>à OV.</s> <s xml:id="echoid-s3113" xml:space="preserve">, elle eſt auſſi (Prop. </s> <s xml:id="echoid-s3114" xml:space="preserve">3.) </s> <s xml:id="echoid-s3115" xml:space="preserve">à la force dont<unsure/> le <lb/>nœud O leur réſiſte ſuivant OZ, comme OS à la <lb/>ſomme de leurs ſublimitez O ſ & </s> <s xml:id="echoid-s3116" xml:space="preserve">O u. </s> <s xml:id="echoid-s3117" xml:space="preserve">2°. </s> <s xml:id="echoid-s3118" xml:space="preserve">Cette <lb/>même force étant auſſi (hyp.) </s> <s xml:id="echoid-s3119" xml:space="preserve">aux puiſſances A & </s> <s xml:id="echoid-s3120" xml:space="preserve">B, <lb/>comme ZR à ZL & </s> <s xml:id="echoid-s3121" xml:space="preserve">ZQ, elle eſt de méme (Prop. </s> <s xml:id="echoid-s3122" xml:space="preserve"><lb/>3.) </s> <s xml:id="echoid-s3123" xml:space="preserve">à la réſiſtance que leur fait le nœud Z ſuivant <lb/>Z C, comme Z R à la ſomme des ſublimitez Z r & </s> <s xml:id="echoid-s3124" xml:space="preserve"><lb/>Z q moins la profondeur Z l. </s> <s xml:id="echoid-s3125" xml:space="preserve">3°. </s> <s xml:id="echoid-s3126" xml:space="preserve">Enfin la valeur <lb/>de cette réſiſtance étant encore (hyp.) </s> <s xml:id="echoid-s3127" xml:space="preserve">aux forces <lb/>dont le nœud C eſt tiré ſuivant C X, C Y, C φ. </s> <s xml:id="echoid-s3128" xml:space="preserve">&</s> <s xml:id="echoid-s3129" xml:space="preserve">c. </s> <s xml:id="echoid-s3130" xml:space="preserve"><lb/>comme CM à CN, CP, C θ, &</s> <s xml:id="echoid-s3131" xml:space="preserve">c. </s> <s xml:id="echoid-s3132" xml:space="preserve">elle eſt auſſi <lb/>(Prop. </s> <s xml:id="echoid-s3133" xml:space="preserve">3.) </s> <s xml:id="echoid-s3134" xml:space="preserve">au poids T, comme CM à la ſomme des <lb/>ſublimitez C m, C n, &</s> <s xml:id="echoid-s3135" xml:space="preserve">c. </s> <s xml:id="echoid-s3136" xml:space="preserve">moins celle des profondeurs <lb/>Cλ, C p, &</s> <s xml:id="echoid-s3137" xml:space="preserve">c. </s> <s xml:id="echoid-s3138" xml:space="preserve">Donc en multipliant par ordre ces <lb/>trois rangées de proportionelles, la puiſſance D ſe <lb/>trouvera au poids T, comme le produit fait des <lb/>trois antécédens OS, ZR, & </s> <s xml:id="echoid-s3139" xml:space="preserve">CM, au produit fait <lb/>de leur trois conſéquens Oſ + Ou, Zr + Zq -<lb/>Zl, & </s> <s xml:id="echoid-s3140" xml:space="preserve">Cm + Cn - Cp - Cλ. </s> <s xml:id="echoid-s3141" xml:space="preserve">C’eſt - à - dire, en <lb/>raiſon compoſée des trois raiſons de OS à Oſ + <lb/>Ou, de Z R à Zr + Zq - Zl, & </s> <s xml:id="echoid-s3142" xml:space="preserve">de CM à Cm + <lb/>Cn - Cp - Cλ, qu’on voit telles que cette pro-<lb/>poſition porte. </s> <s xml:id="echoid-s3143" xml:space="preserve">Or il n’y a en eſfet que trois nœuds <pb o="120" file="0146" n="146" rhead="EXAMEN DE L’OPINION"/> O, Z, & </s> <s xml:id="echoid-s3144" xml:space="preserve">C, entre cette puiſſance & </s> <s xml:id="echoid-s3145" xml:space="preserve">ce poids: </s> <s xml:id="echoid-s3146" xml:space="preserve">Donc <lb/> <anchor type="note" xlink:label="note-0146-01a" xlink:href="note-0146-01"/> la puiſſance D eſt au poids T en raiſon compoſée <lb/>d’autant d’autres raiſons telles que cette propoſition <lb/>porte, qu’il y a de nœuds entre cette puiſſance & </s> <s xml:id="echoid-s3147" xml:space="preserve"><lb/>ce poids. </s> <s xml:id="echoid-s3148" xml:space="preserve">On prouvera de même que la puiſſance A <lb/>eſt à ce poids en raiſon compoſée de ZL à Z r + <lb/>Z q - Zl, & </s> <s xml:id="echoid-s3149" xml:space="preserve">de CM à C m + C n - C p - C λ. <lb/></s> <s xml:id="echoid-s3150" xml:space="preserve">On trouvera encore de même que la puiſſance F <lb/>eſt à ce même poids en raiſon compoſée de X β à <lb/>X b + X f, & </s> <s xml:id="echoid-s3151" xml:space="preserve">de CN à C m + C n - C p - C λ; </s> <s xml:id="echoid-s3152" xml:space="preserve"><lb/>& </s> <s xml:id="echoid-s3153" xml:space="preserve">ainſi de toutes les autres puiſſances, en quelque <lb/>nombre qu’elles ſoient, de quelque maniére, & </s> <s xml:id="echoid-s3154" xml:space="preserve">à <lb/>quelque nombre de nœuds qu’elles ſoient appli-<lb/>quées. </s> <s xml:id="echoid-s3155" xml:space="preserve">D’où l’on voit en général, que de quelque <lb/>maniére qu’un poids ſoit ſoutenu avec des cordes <lb/>par quelque nombre de puiſſances que ce ſoit, ap-<lb/>pliquées à tant de nœuds qu’on voudra, chacune <lb/>d’elles eſt toujours à ce poids en raiſon compoſée <lb/>d’autant d’autres telles que cette propoſition porte, <lb/>qu’il y a de nœuds entre cette puiſſance & </s> <s xml:id="echoid-s3156" xml:space="preserve">ce poids. </s> <s xml:id="echoid-s3157" xml:space="preserve"><lb/>Ce qu’il faloit démontrer.</s> <s xml:id="echoid-s3158" xml:space="preserve"/> </p> <div xml:id="echoid-div287" type="float" level="2" n="1"> <note position="left" xlink:label="note-0144-02" xlink:href="note-0144-02a" xml:space="preserve">fig. 20.</note> <note position="right" xlink:label="note-0145-01" xlink:href="note-0145-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> <note position="left" xlink:label="note-0146-01" xlink:href="note-0146-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> </div> <div xml:id="echoid-div289" type="section" level="1" n="181"> <head xml:id="echoid-head183" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s3159" xml:space="preserve">On voit qu’en prenant ZR égale à O ſ + O u, <lb/>avec ZL & </s> <s xml:id="echoid-s3160" xml:space="preserve">ZQ à ZR en même proportion qu’elles <lb/>ſont ici; </s> <s xml:id="echoid-s3161" xml:space="preserve">de plus CM égale à Zq + Zr - Zl, <lb/>avec CN, CP, Cθ, &</s> <s xml:id="echoid-s3162" xml:space="preserve">c. </s> <s xml:id="echoid-s3163" xml:space="preserve">auſſi à CM en même <lb/>proportion qu’elles ſont ici; </s> <s xml:id="echoid-s3164" xml:space="preserve">la puiſſance D ſera au <lb/>poids T, comme OS à Cm + Cn - Cλ - Cp ±<unsure/> <lb/>&</s> <s xml:id="echoid-s3165" xml:space="preserve">c. </s> <s xml:id="echoid-s3166" xml:space="preserve">c’eſt-à-dire, comme ſa proportionelle à la ſomme <lb/>des ſublimitez moins celle des profondeurs des for-<lb/>ces avec leſquelles les branches du premier nœud <lb/>C ſont tirées chacune ſuivant ſa direction. </s> <s xml:id="echoid-s3167" xml:space="preserve">Il en faut <lb/>penſer autant de toutes les autres puiſſances appli-<lb/>quées au poids T, ſoit de prés, ſoit de loin.</s> <s xml:id="echoid-s3168" xml:space="preserve"/> </p> <pb o="121" file="0147" n="147" rhead="DE M. BORELLI."/> <p style="it"> <s xml:id="echoid-s3169" xml:space="preserve">On peut comparer cette propoſition avec ce Corollaire à la <lb/> <anchor type="note" xlink:label="note-0147-01a" xlink:href="note-0147-01"/> propoſition 78. </s> <s xml:id="echoid-s3170" xml:space="preserve">de M. </s> <s xml:id="echoid-s3171" xml:space="preserve">Borelli.</s> <s xml:id="echoid-s3172" xml:space="preserve"/> </p> <div xml:id="echoid-div289" type="float" level="2" n="1"> <note position="right" xlink:label="note-0147-01" xlink:href="note-0147-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> </div> <div xml:id="echoid-div291" type="section" level="1" n="182"> <head xml:id="echoid-head184" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s3173" xml:space="preserve">Lors qu’un poids attaché à une corde qui a plu-<lb/> <anchor type="note" xlink:label="note-0147-02a" xlink:href="note-0147-02"/> ſieurs nœuds par chacun deſquels, entre toutes <lb/>les branches qui en naiſſent, il n’y en a qu’une qui <lb/>ſe ſubdiviſe en d’autres branches: </s> <s xml:id="echoid-s3174" xml:space="preserve">lors dî-je que le <lb/>poids T attaché à une telle corde, eſt ſoutenu par <lb/>pluſieurs puiſſances Y, X, S, R, V, Z, &</s> <s xml:id="echoid-s3175" xml:space="preserve">c. </s> <s xml:id="echoid-s3176" xml:space="preserve">telle-<lb/>ment appliquées aux derniéres de ces branches que <lb/>tous les nœuds F, E, C, &</s> <s xml:id="echoid-s3177" xml:space="preserve">c. </s> <s xml:id="echoid-s3178" xml:space="preserve">d’où elles naiſſent, ſe <lb/>trouvent dans la ligne de direction de ce poids; <lb/></s> <s xml:id="echoid-s3179" xml:space="preserve">chacune de ces puiſſances, en quelque nombre qu’el-<lb/>les ſoient, eſt toujours à ce poids, comme la pro-<lb/>portionelle de cette même puiſſance à la ſomme des <lb/>ſublimitez moins celle des profondeurs de tout ce <lb/>qu’il y en a d’appliquées à ce même poids: </s> <s xml:id="echoid-s3180" xml:space="preserve">car ſi l’on <lb/>prend ſur les branches de chaque nœud des parties <lb/>OF, EI, CB, CA, EH, FK, FN, EM, &</s> <s xml:id="echoid-s3181" xml:space="preserve">c. </s> <s xml:id="echoid-s3182" xml:space="preserve"><lb/>proportionelles aux forces avec leſquelles chacune <lb/>d’elles eſt tirée ſuivant ſa direction, & </s> <s xml:id="echoid-s3183" xml:space="preserve">que des ex-<lb/>trémitez de ces mêmes parties on marque (avert. </s> <s xml:id="echoid-s3184" xml:space="preserve"><lb/>Chap. </s> <s xml:id="echoid-s3185" xml:space="preserve">2.) </s> <s xml:id="echoid-s3186" xml:space="preserve">leurs ſublimitez avec leurs profondeurs; </s> <s xml:id="echoid-s3187" xml:space="preserve"><lb/>on trouvera 1°. </s> <s xml:id="echoid-s3188" xml:space="preserve">que les proportionelles FN, EM, <lb/>&</s> <s xml:id="echoid-s3189" xml:space="preserve">c. </s> <s xml:id="echoid-s3190" xml:space="preserve">qui ſe trouvent dans la ligne de direction de ce <lb/>poids, ſont égales aux ſublimitez FN, FM, &</s> <s xml:id="echoid-s3191" xml:space="preserve">c. </s> <s xml:id="echoid-s3192" xml:space="preserve">des <lb/>forces avec leſquelles elles ſont tirées ſuivant leur <lb/>direction, c’eſt-à-dire, ſuivant cette même ligne. </s> <s xml:id="echoid-s3193" xml:space="preserve"><lb/>2°. </s> <s xml:id="echoid-s3194" xml:space="preserve">On trouvera encore que chacune de ces mêmes <lb/>proportionelles, par exemple FN eſt auſſi toujours <lb/>égale à la ſomme des ſublimitez moins celle des pro-<lb/>fondeurs des forces, ou des puiſſances appliquées au <pb o="122" file="0148" n="148" rhead="EXAMEN DE L’OPINION"/> nœud E qui eſt immédiatement au-deſſus du nœud <lb/> <anchor type="note" xlink:label="note-0148-01a" xlink:href="note-0148-01"/> F depuis lequel cette proportionelle a été priſe: <lb/></s> <s xml:id="echoid-s3195" xml:space="preserve">puis que (hyp) cette même proportionelle, & </s> <s xml:id="echoid-s3196" xml:space="preserve">(Prop. </s> <s xml:id="echoid-s3197" xml:space="preserve"><lb/>3.) </s> <s xml:id="echoid-s3198" xml:space="preserve">cette même ſomme ſont à la proportionelle EI <lb/>de la puiſſance X, comme la force dont le nœud E <lb/>eſt tire ſuivant la corde EF, eſt à cette même puiſſan-<lb/>ce. </s> <s xml:id="echoid-s3199" xml:space="preserve">Pour la même raiſon EM eſt égale à la ſomme <lb/>des ſublimitez moins celle des profondeurs des forces <lb/>où des puiſſances appliquées au nœud C qui eſt im-<lb/>médiatement au-deſſus de E; </s> <s xml:id="echoid-s3200" xml:space="preserve">& </s> <s xml:id="echoid-s3201" xml:space="preserve">ainſi des autres <lb/>proportionelles qui ſe trouvent dans la ligne de di-<lb/>rection du poids T. </s> <s xml:id="echoid-s3202" xml:space="preserve">De-là on verra que chacune des <lb/>ſublimitez FN, EM, &</s> <s xml:id="echoid-s3203" xml:space="preserve">c. </s> <s xml:id="echoid-s3204" xml:space="preserve">des forces qui ſuivent la direc-<lb/>tion de ce poids, eſt toujours égale à la ſomme des ſu-<lb/>blimitez moins celle des profondeurs des forces, ou <lb/>des puiſſances appliquées au nœud qui eſt immédia-<lb/>tement au-deſſus de celui depuis lequel elle ſe prend: </s> <s xml:id="echoid-s3205" xml:space="preserve"><lb/>D’où il ſuit que la ſublimité FN qui ſe prend depuis <lb/>le plus bas de tous ces nœuds, eſt égale à la ſomme <lb/>des ſublimitez moins celle des profondeurs de toutes <lb/>les puiſſances X, V, S, R, &</s> <s xml:id="echoid-s3206" xml:space="preserve">c. </s> <s xml:id="echoid-s3207" xml:space="preserve">appliquées à tous <lb/>les autres nœuds E, C, &</s> <s xml:id="echoid-s3208" xml:space="preserve">c. </s> <s xml:id="echoid-s3209" xml:space="preserve">Or on vient de voir <lb/>dans le Corollaire précédent que chacune de toutes <lb/>les puiſſances qui ſoutiennent le poids T, par exem-<lb/>ple S, ou Y, eſt à ce poids, comme ſa proportio-<lb/>nelle CB, ou OF à la ſomme des ſublimitez moins <lb/>celle des profondeurs des forces avec leſquelles <lb/>toutes les branches du plus bas nœud F ſont tirées <lb/>chacune ſuivant ſa direction contre le poids T; </s> <s xml:id="echoid-s3210" xml:space="preserve"><lb/>c’eſt-à-dire, à la ſomme faite de la ſublimité FN, <lb/>& </s> <s xml:id="echoid-s3211" xml:space="preserve">de la ſomme des ſublimitez moins les profondeurs <lb/>des puiſſances Y, Z, &</s> <s xml:id="echoid-s3212" xml:space="preserve">c. </s> <s xml:id="echoid-s3213" xml:space="preserve">immédiatement appliquées <lb/>au nœud F: </s> <s xml:id="echoid-s3214" xml:space="preserve">Donc chacune des puiſſances Y, X, S, <lb/>R, V, Z, &</s> <s xml:id="echoid-s3215" xml:space="preserve">c. </s> <s xml:id="echoid-s3216" xml:space="preserve">eſt en ce cas au poids T, comme ſa <pb o="123" file="0149" n="149" rhead="DE M. BORELLI."/> proportionelle à la ſomme des ſublimitez, moins celle <lb/> <anchor type="note" xlink:label="note-0149-01a" xlink:href="note-0149-01"/> des profondeurs de tout ce qu’il y en a d’appliquées <lb/>à ce même poids.</s> <s xml:id="echoid-s3217" xml:space="preserve"/> </p> <div xml:id="echoid-div291" type="float" level="2" n="1"> <note position="right" xlink:label="note-0147-02" xlink:href="note-0147-02a" xml:space="preserve">fig. 21. <lb/>22.</note> <note position="left" xlink:label="note-0148-01" xlink:href="note-0148-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> <note position="right" xlink:label="note-0149-01" xlink:href="note-0149-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> </div> <div xml:id="echoid-div293" type="section" level="1" n="183"> <head xml:id="echoid-head185" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s3218" xml:space="preserve">D’où l’on voit que la ſomme de toutes ces puiſ-<lb/>ſances eſt à ce poids, comme la ſomme de leurs pro-<lb/>portionelles à la ſomme de leurs ſublimitez moins <lb/>celle de leurs profondeurs: </s> <s xml:id="echoid-s3219" xml:space="preserve">De ſorte que s’il n’y <lb/>en avoit que deux d’appliquées à chaque nœud <lb/>dont l’une tirât à droit & </s> <s xml:id="echoid-s3220" xml:space="preserve">l’autre à gauche, & </s> <s xml:id="echoid-s3221" xml:space="preserve">que <lb/>toutes celles de chaque côté fuſſent égales entr’elles, <lb/>& </s> <s xml:id="echoid-s3222" xml:space="preserve">avec des directions paralleles entr’elles; </s> <s xml:id="echoid-s3223" xml:space="preserve">la ſomme <lb/>(fig. </s> <s xml:id="echoid-s3224" xml:space="preserve">21.) </s> <s xml:id="echoid-s3225" xml:space="preserve">des ſublimitez, par exemple F o, + F k, où <lb/>Ei + Eh, où Cb + Ca, &</s> <s xml:id="echoid-s3226" xml:space="preserve">c. </s> <s xml:id="echoid-s3227" xml:space="preserve">des deux puiſſances ap-<lb/>pliquées au quel que ce ſoit des nœuds F, E, C, &</s> <s xml:id="echoid-s3228" xml:space="preserve">c. <lb/></s> <s xml:id="echoid-s3229" xml:space="preserve">ou bien la différence (fig. </s> <s xml:id="echoid-s3230" xml:space="preserve">22.) </s> <s xml:id="echoid-s3231" xml:space="preserve">de la ſublimité de l’une <lb/>à la profondeur de l’autre, par exemple Fk,- Fo, <lb/>ou Eh - Ei, au Ca - Cb, &</s> <s xml:id="echoid-s3232" xml:space="preserve">c. </s> <s xml:id="echoid-s3233" xml:space="preserve">étant alors la même <lb/>pour tous ces nœuds, auſſi-bien que les proportio-<lb/>nelles de ces puiſſances; </s> <s xml:id="echoid-s3234" xml:space="preserve">la ſomme de ces mêmes <lb/>puiſſances ſeroit alors au poids T, comme la ſomme <lb/>des proportionelles de deux d’entr’elles appliquées à un <lb/>même nœud, quel qu’il ſoit, eſt à la ſomme (fig. </s> <s xml:id="echoid-s3235" xml:space="preserve">21.) </s> <s xml:id="echoid-s3236" xml:space="preserve"><lb/>des ſublimitez de ces deux puiſſances, où (fig. </s> <s xml:id="echoid-s3237" xml:space="preserve">22.) </s> <s xml:id="echoid-s3238" xml:space="preserve">à la <lb/>différence qui eſt entre la ſublimité de l’une, & </s> <s xml:id="echoid-s3239" xml:space="preserve">la <lb/>profondeur del ’autre.</s> <s xml:id="echoid-s3240" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div294" type="section" level="1" n="184"> <head xml:id="echoid-head186" xml:space="preserve"><emph style="sc">Corollaire</emph> IV.</head> <p> <s xml:id="echoid-s3241" xml:space="preserve">Ce qui fait enfin voir que ſi toutes les puiſſances <lb/>Y, X, S, R, V, Z, &</s> <s xml:id="echoid-s3242" xml:space="preserve">c. </s> <s xml:id="echoid-s3243" xml:space="preserve">étoient égales entr’elles, & </s> <s xml:id="echoid-s3244" xml:space="preserve"><lb/>que toutes leurs directions fiſſent avec celle de ce <lb/>poids des angles égaux auſſi entr’eux, leur ſomme <lb/>ſeroit alors à ce même poids, (fig. </s> <s xml:id="echoid-s3245" xml:space="preserve">21.) </s> <s xml:id="echoid-s3246" xml:space="preserve">comme une <lb/>de leurs proportionelles à une de leurs ſublimitez, <pb o="124" file="0150" n="150" rhead="EXAMEN DE L’OPINION"/> l’une & </s> <s xml:id="echoid-s3247" xml:space="preserve">l’autre choiſie à diſcretion; </s> <s xml:id="echoid-s3248" xml:space="preserve">c’eſt - à - dire, <lb/> <anchor type="note" xlink:label="note-0150-01a" xlink:href="note-0150-01"/> comme le ſinus total au ſinus du complement de <lb/>celui qu’on voudra de ces mêmes angles.</s> <s xml:id="echoid-s3249" xml:space="preserve"/> </p> <div xml:id="echoid-div294" type="float" level="2" n="1"> <note position="left" xlink:label="note-0150-01" xlink:href="note-0150-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> <p style="it"> <s xml:id="echoid-s3250" xml:space="preserve">On peut encore comparer ces trois Corollaires à la Propo-<lb/>ſition 71. </s> <s xml:id="echoid-s3251" xml:space="preserve">de M. </s> <s xml:id="echoid-s3252" xml:space="preserve">Borelli & </s> <s xml:id="echoid-s3253" xml:space="preserve">au Corollaire qu’il en tire.</s> <s xml:id="echoid-s3254" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div296" type="section" level="1" n="185"> <head xml:id="echoid-head187" xml:space="preserve"><emph style="sc">Corollaire</emph> V.</head> <p> <s xml:id="echoid-s3255" xml:space="preserve">On voit encore de cette propoſition que dans l’hy-<lb/> <anchor type="note" xlink:label="note-0150-02a" xlink:href="note-0150-02"/> pothéſe ou les lignes de direction de tous les points <lb/>du corps A D concourent au centre E de la terre, <lb/>de quelque maniére que ce poids ſoit ſoutenu par <lb/>tant de puiſſances F, G, H, I, K, L, M, N, &</s> <s xml:id="echoid-s3256" xml:space="preserve">c. <lb/></s> <s xml:id="echoid-s3257" xml:space="preserve">qu’on voudra avec des cordes qui lui ſoient appli-<lb/>quées en tant de points A, B, C, D, &</s> <s xml:id="echoid-s3258" xml:space="preserve">c. </s> <s xml:id="echoid-s3259" xml:space="preserve">qu’on vou-<lb/>dra; </s> <s xml:id="echoid-s3260" xml:space="preserve">chacune de ces puiſſances ſera toujours à ce <lb/>poids, comme chacune de leurs proportionelles à la <lb/>ſomme des ſublimitez des forces avec leſquelles ces <lb/>points A, B, C, D, &</s> <s xml:id="echoid-s3261" xml:space="preserve">c. </s> <s xml:id="echoid-s3262" xml:space="preserve">ſont tirez ſuivant les lignes <lb/>AE, BE, CE, DE, &</s> <s xml:id="echoid-s3263" xml:space="preserve">c. </s> <s xml:id="echoid-s3264" xml:space="preserve">par le concours d’action <lb/>des puiſſances qui y ſont appliquées: </s> <s xml:id="echoid-s3265" xml:space="preserve">des ſublimitez, <lb/>dî-je, déterminées comme dans le Corollaire 1. </s> <s xml:id="echoid-s3266" xml:space="preserve">Car <lb/>il eſt clair que ce poids agit contre toutes ces puiſſan-<lb/>ces de même que feroit une force qui lui ſeroit égale, <lb/>ſi AE, BE, CE, DE, &</s> <s xml:id="echoid-s3267" xml:space="preserve">c. </s> <s xml:id="echoid-s3268" xml:space="preserve">étoient autant de cor-<lb/>des attachées enſemble au point E par un nœud <lb/>commun auquel cette force fut appliquée ſuivant la <lb/>direction ZE du centre de gravité de ce poids. </s> <s xml:id="echoid-s3269" xml:space="preserve">Or <lb/>en ce cas les points A, B, C, D, &</s> <s xml:id="echoid-s3270" xml:space="preserve">c. </s> <s xml:id="echoid-s3271" xml:space="preserve">étant comme <lb/>autant de nœuds auſquels ſont appliquées, chacune <lb/>ſuivant ſa direction, les puiſſances F, G, H, I, K, <lb/>L, M, N, &</s> <s xml:id="echoid-s3272" xml:space="preserve">c. </s> <s xml:id="echoid-s3273" xml:space="preserve">ſi l’on prend depuis E ſur chacune <lb/>des lignes AE, BE, CE, DE, &</s> <s xml:id="echoid-s3274" xml:space="preserve">c. </s> <s xml:id="echoid-s3275" xml:space="preserve">une partie Eg, <lb/>Ef, Ec, Eb, &</s> <s xml:id="echoid-s3276" xml:space="preserve">c. </s> <s xml:id="echoid-s3277" xml:space="preserve">égale à la ſomme des ſublimitez <pb o="125" file="0151" n="151" rhead="DE M. BORELLI."/> moins celle des profondeurs des puiſſances appli-<lb/> <anchor type="note" xlink:label="note-0151-01a" xlink:href="note-0151-01"/> quées à chacun des points A, B, C, D, &</s> <s xml:id="echoid-s3278" xml:space="preserve">c. </s> <s xml:id="echoid-s3279" xml:space="preserve">On <lb/>trouvera (Cor 1.) </s> <s xml:id="echoid-s3280" xml:space="preserve">que chacune de toutes ces puiſ-<lb/>ſances F, G, H, I, K, L, M, N, &</s> <s xml:id="echoid-s3281" xml:space="preserve">c. </s> <s xml:id="echoid-s3282" xml:space="preserve">ſeroit alors à <lb/>la force qu’on ſuppoſe en E, comme chacune de leurs <lb/>proportionelles DO, CP, BQ, DX, AR, CV, BT, <lb/>AS, &</s> <s xml:id="echoid-s3283" xml:space="preserve">c. </s> <s xml:id="echoid-s3284" xml:space="preserve">à la ſomme des ſublimitez, El, Ee, Ed, <lb/>Ea, &</s> <s xml:id="echoid-s3285" xml:space="preserve">c. </s> <s xml:id="echoid-s3286" xml:space="preserve">des forces dont les nœuds A, B, C, D, &</s> <s xml:id="echoid-s3287" xml:space="preserve">c. <lb/></s> <s xml:id="echoid-s3288" xml:space="preserve">ſeroient alors tirez, chacun ſuivant la ligne qui le <lb/>joint avec le point E: </s> <s xml:id="echoid-s3289" xml:space="preserve">Donc chacune de ces mêmes <lb/>puiſſances eſt en effet au poids AD, comme chacune <lb/>de leurs proportionelles à la ſomme de telles ſubli-<lb/>mitez des forces avec leſquelles les points A, B, C, D, <lb/>&</s> <s xml:id="echoid-s3290" xml:space="preserve">c. </s> <s xml:id="echoid-s3291" xml:space="preserve">ſont tirez ſuivant les lignes AE, BE, CE, DE, <lb/>&</s> <s xml:id="echoid-s3292" xml:space="preserve">c. </s> <s xml:id="echoid-s3293" xml:space="preserve">par le concours d’action de celles qui y ſont ap-<lb/>pliquées.</s> <s xml:id="echoid-s3294" xml:space="preserve"/> </p> <div xml:id="echoid-div296" type="float" level="2" n="1"> <note position="left" xlink:label="note-0150-02" xlink:href="note-0150-02a" xml:space="preserve">fig. 23. <lb/>24.</note> <note position="right" xlink:label="note-0151-01" xlink:href="note-0151-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> <p style="it"> <s xml:id="echoid-s3295" xml:space="preserve">Si les forces, avec leſquelles les différens points A, B, C, <lb/>D, &</s> <s xml:id="echoid-s3296" xml:space="preserve">c. </s> <s xml:id="echoid-s3297" xml:space="preserve">du corps AD, ſont tirez ſuivant des lignes qui con-<lb/>courent au centre E de la terre, par le concours d’action des <lb/>puiſſances qui y ſont appliquées, avoient quelque profondeur, <lb/>on trouveroit de même que chacune de toutes les puiſſances qui <lb/>ſoutiennent ce poids, lui ſeroit en même raiſon que chacune de <lb/>leurs proportionnelles à la ſomme de telles ſublimitez moins <lb/>celle des profondeurs de ces mêmes forces: </s> <s xml:id="echoid-s3298" xml:space="preserve">mais ce cas étant <lb/>naturellement impoſſible; </s> <s xml:id="echoid-s3299" xml:space="preserve">puiſqu’il faudroit pour cela que ce <lb/>poids comprît pour le moins plus du quart de la circonférence <lb/>de la terre, on n’a pas crû qu’il fût néceſſaire de l’exprimer.</s> <s xml:id="echoid-s3300" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div298" type="section" level="1" n="186"> <head xml:id="echoid-head188" xml:space="preserve"><emph style="sc">Corollaire</emph> VI.</head> <p> <s xml:id="echoid-s3301" xml:space="preserve">On voit préſentement que dans l’hypotheſe ordi-<lb/>naire, ou l’on regarde les directions AE, BE, CE, <lb/>DE, &</s> <s xml:id="echoid-s3302" xml:space="preserve">c. </s> <s xml:id="echoid-s3303" xml:space="preserve">comme paralleles entr’elles, chacune des <lb/>ſublimitez El, Ee, Ed, Ea, &</s> <s xml:id="echoid-s3304" xml:space="preserve">c. </s> <s xml:id="echoid-s3305" xml:space="preserve">déterminées ſur <pb o="126" file="0152" n="152" rhead="EXAMEN DE L’OPINION"/> ZE par chacune des proportionelles Eg, Ef, Ec, <lb/> <anchor type="note" xlink:label="note-0152-01a" xlink:href="note-0152-01"/> Eb, &</s> <s xml:id="echoid-s3306" xml:space="preserve">c. </s> <s xml:id="echoid-s3307" xml:space="preserve">qu’on vient de prendre egales à la ſomme <lb/>des ſublimitez moins celle des profondeurs des puiſ-<lb/>ſances appliquées à chacun des points A, B, C, D, <lb/>&</s> <s xml:id="echoid-s3308" xml:space="preserve">c. </s> <s xml:id="echoid-s3309" xml:space="preserve">étant alors égales à ces mêmes proportionelles; <lb/></s> <s xml:id="echoid-s3310" xml:space="preserve">chacune de ce qu’il y a de puiſſances ainſi appliquées <lb/>à ce poids, ſçavoir F, G, H, I, K, L, M, N, &</s> <s xml:id="echoid-s3311" xml:space="preserve">c. </s> <s xml:id="echoid-s3312" xml:space="preserve">eſt <lb/>toujours en ce cas à ce même poids, comme chacune <lb/>de leurs proportionelles à la ſomme de toutes leurs <lb/>ſublimitez moins celle de toutes leurs profondeurs.</s> <s xml:id="echoid-s3313" xml:space="preserve"/> </p> <div xml:id="echoid-div298" type="float" level="2" n="1"> <note position="left" xlink:label="note-0152-01" xlink:href="note-0152-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> </div> <div xml:id="echoid-div300" type="section" level="1" n="187"> <head xml:id="echoid-head189" xml:space="preserve"><emph style="sc">Corollaire</emph> VII.</head> <p> <s xml:id="echoid-s3314" xml:space="preserve">D’où il ſuit que dans la même hypothêſe la ſom-<lb/>me de toutes ces puiſſances eſt à ce poids, comme <lb/>la ſomme de leurs proportionelles à la ſomme de leurs <lb/>ſublimitez moins celle de leurs profondeurs: </s> <s xml:id="echoid-s3315" xml:space="preserve">De <lb/>ſorte que s’il n’y en avoit que deux d’appliquées à <lb/>chaque point, dont l’une tirât à droit, & </s> <s xml:id="echoid-s3316" xml:space="preserve">l’autre à <lb/>gauche; </s> <s xml:id="echoid-s3317" xml:space="preserve">& </s> <s xml:id="echoid-s3318" xml:space="preserve">que toutes celles de chaque côté fuſſent <lb/>égales entr’elles, & </s> <s xml:id="echoid-s3319" xml:space="preserve">avec des directions qui fiſſent <lb/>avec celle du point où elles ſont appliquées, des angles <lb/>de@chaque côté égaux entr’eux: </s> <s xml:id="echoid-s3320" xml:space="preserve">la ſomme (fig. </s> <s xml:id="echoid-s3321" xml:space="preserve">23.) </s> <s xml:id="echoid-s3322" xml:space="preserve">des <lb/>ſublimitez, par exemple Ar + Aſ, ou Bq + Bt, <lb/>ou Cp + Cu, ou Do + Dx, &</s> <s xml:id="echoid-s3323" xml:space="preserve">c. </s> <s xml:id="echoid-s3324" xml:space="preserve">des deux puiſſances <lb/>appliquées au quel que ce ſoit des nœuds A, B, <lb/>C, D, &</s> <s xml:id="echoid-s3325" xml:space="preserve">c. </s> <s xml:id="echoid-s3326" xml:space="preserve">ou bien (fig. </s> <s xml:id="echoid-s3327" xml:space="preserve">24.) </s> <s xml:id="echoid-s3328" xml:space="preserve">la différence de la ſu-<lb/>blimité de l’une à la profondeur de l’autre, par <lb/>exemple Ar - Aſ, ou Bq - Bt, ou Cp - Cu, ou <lb/>Do - Dx, &</s> <s xml:id="echoid-s3329" xml:space="preserve">c. </s> <s xml:id="echoid-s3330" xml:space="preserve">étant alors la même pour tous ces <lb/>points, auſſi-bien que les proportionelles de ces puiſ-<lb/>ſances; </s> <s xml:id="echoid-s3331" xml:space="preserve">la ſomme de toutes ces puiſſances ſeroit <lb/>alors au poids AD, comme la ſomme des proportio-<lb/>nelles de deux d’entr’elles appliquées à un même <lb/>point, quel qu’il ſoit, eſt à la ſomme (fig. </s> <s xml:id="echoid-s3332" xml:space="preserve">23.) </s> <s xml:id="echoid-s3333" xml:space="preserve">de leurs <pb o="127" file="0153" n="153" rhead="DE M. BORELLI."/> ſublimitez, ou (fig. </s> <s xml:id="echoid-s3334" xml:space="preserve">24.) </s> <s xml:id="echoid-s3335" xml:space="preserve">à la différence qui eſt entre <lb/> <anchor type="note" xlink:label="note-0153-01a" xlink:href="note-0153-01"/> la ſublimité de l’une & </s> <s xml:id="echoid-s3336" xml:space="preserve">la profondeur de l’autre.</s> <s xml:id="echoid-s3337" xml:space="preserve"/> </p> <div xml:id="echoid-div300" type="float" level="2" n="1"> <note position="right" xlink:label="note-0153-01" xlink:href="note-0153-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> </div> <div xml:id="echoid-div302" type="section" level="1" n="188"> <head xml:id="echoid-head190" xml:space="preserve"><emph style="sc">Corollaire</emph> VIII.</head> <p> <s xml:id="echoid-s3338" xml:space="preserve">Ce qui fait enfin voir que ſi toutes les puiſſances <lb/>F, G, H, I, K, L, M, N, &</s> <s xml:id="echoid-s3339" xml:space="preserve">c. </s> <s xml:id="echoid-s3340" xml:space="preserve">étoient égales en-<lb/>tr’elles, & </s> <s xml:id="echoid-s3341" xml:space="preserve">que toutes leurs directions fiſſent avec <lb/>celles des points où elles ſont appliquées, des angles <lb/>égaux entr’eux; </s> <s xml:id="echoid-s3342" xml:space="preserve">leur ſomme ſeroit alors au poids <lb/>CD (fig. </s> <s xml:id="echoid-s3343" xml:space="preserve">23.) </s> <s xml:id="echoid-s3344" xml:space="preserve">comme une de leurs proportionelles <lb/>à une de leurs ſublimitez, de quelque maniére qu’on <lb/>les prenne; </s> <s xml:id="echoid-s3345" xml:space="preserve">c’eſt-à-dire, comme le ſinus total au ſinus <lb/>du complement de celui qu’on voudra de ces mêmes <lb/>angles.</s> <s xml:id="echoid-s3346" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s3347" xml:space="preserve">On peut enfin comparer ces quatre derniers corollaires à <lb/>la Propoſition 72. </s> <s xml:id="echoid-s3348" xml:space="preserve">de M. </s> <s xml:id="echoid-s3349" xml:space="preserve">Borelli, & </s> <s xml:id="echoid-s3350" xml:space="preserve">au corollaire qu’il <lb/>en tire.</s> <s xml:id="echoid-s3351" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s3352" xml:space="preserve">Tels ſont les principes généraux de tout ce que cet Au-<lb/>theur a dit des poids ſuſpendus par des cordes, & </s> <s xml:id="echoid-s3353" xml:space="preserve">de l’uſage <lb/>qu’il en a fait pour exprimer la force des muſcles; </s> <s xml:id="echoid-s3354" xml:space="preserve">c’eſt ce <lb/>qu’on s’étoit propoſé d’établir dans ce Chapitre par la mé-<lb/>@hode du Projet précédent: </s> <s xml:id="echoid-s3355" xml:space="preserve">qu’on voye préſentement ſi la <lb/>ſienne peut aller juſques-là, & </s> <s xml:id="echoid-s3356" xml:space="preserve">ſi elle peut conduire à la So-<lb/>lution du Problême ſuivant.</s> <s xml:id="echoid-s3357" xml:space="preserve"/> </p> <figure> <image file="0153-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0153-01"/> </figure> </div> <div xml:id="echoid-div303" type="section" level="1" n="189"> <head xml:id="echoid-head191" xml:space="preserve">PROBLEME.</head> <p style="it"> <s xml:id="echoid-s3358" xml:space="preserve">DIx puiſſances φ, A, E, D, B, F, G, H, I, K, <lb/> <anchor type="note" xlink:label="note-0153-02a" xlink:href="note-0153-02"/> appliquées à pluſieurs nœuds de cordes, étant données <lb/>avec les angles que toutes ces cordes font entr’elles; </s> <s xml:id="echoid-s3359" xml:space="preserve">trouver <pb o="128" file="0154" n="154" rhead="EXAMEN DE L’OPINION"/> la valeur du poids T que toutes ces puiſſances ainſi appli-<lb/> <anchor type="note" xlink:label="note-0154-01a" xlink:href="note-0154-01"/> quées ſoutiennent enſemble.</s> <s xml:id="echoid-s3360" xml:space="preserve"/> </p> <div xml:id="echoid-div303" type="float" level="2" n="1"> <note position="right" xlink:label="note-0153-02" xlink:href="note-0153-02a" xml:space="preserve">fig. 20.<unsure/></note> <note position="left" xlink:label="note-0154-01" xlink:href="note-0154-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> </div> <div xml:id="echoid-div305" type="section" level="1" n="190"> <head xml:id="echoid-head192" xml:space="preserve"><emph style="sc">Solution</emph>.</head> <p> <s xml:id="echoid-s3361" xml:space="preserve">Soit la valeur de chaque puiſſance, & </s> <s xml:id="echoid-s3362" xml:space="preserve">de chaque <lb/>angle donné dans la Table ſuivante.</s> <s xml:id="echoid-s3363" xml:space="preserve"/> </p> <note position="right" xml:space="preserve"> <lb/>Puiſſance. # Livres. # # Angle. # Deg. # M. <lb/>φ # 5. # # θCT # 45. # 30. <lb/>A # 4. # {1/4} # MCT # 150. # 20. <lb/>E # 7. # {1/4} # LZC # 58. # 30. <lb/>D # 12. # {1/2} # RZC # 112. # 15. <lb/>B # 14. # # VOZ # 151. <lb/>F # 11. # {1/2} # SOZ # 110. <lb/>G # 17. # # QZC # 143. <lb/>H # 7. # {1/2} # NCT # 145. <lb/>I # 16. # # βXC # 131. # 30. <lb/>K # 13. # {1/2} # hXC # 123. # 30. <lb/># PCT # 64. # 40. <lb/># δYC # 62. <lb/># zYC # 107. # 20. <lb/># eYC # 151. # 40. <lb/></note> <pb file="0155" n="155"/> <figure> <image file="0155-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0155-01"/> </figure> <pb file="0155a" n="156"/> <figure> <image file="0155a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0155a-01"/> </figure> <pb file="0156" n="157"/> <pb file="0157" n="158"/> <pb file="0157a" n="159"/> <figure> <image file="0157a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0157a-01"/> </figure> <pb file="0158" n="160"/> <pb file="0159" n="161"/> <pb file="0159a" n="162"/> <figure> <image file="0159a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0159a-01"/> </figure> <pb file="0160" n="163"/> <pb file="0161" n="164"/> <pb file="0161a" n="165"/> <figure> <image file="0161a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0161a-01"/> </figure> <pb file="0162" n="166"/> <pb file="0163" n="167"/> <pb file="0163a" n="168"/> <figure> <image file="0163a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0163a-01"/> </figure> <pb file="0164" n="169"/> <pb file="0165" n="170"/> <pb file="0165a" n="171"/> <figure> <image file="0165a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0165a-01"/> </figure> <pb file="0166" n="172"/> <pb file="0167" n="173"/> <pb file="0167a" n="174"/> <figure> <image file="0167a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0167a-01"/> </figure> <pb file="0168" n="175"/> <pb file="0169" n="176"/> <pb file="0169a" n="177"/> <figure> <image file="0169a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0169a-01"/> </figure> <pb file="0170" n="178"/> <pb file="0171" n="179"/> <pb file="0171a" n="180"/> <figure> <image file="0171a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0171a-01"/> </figure> <pb file="0172" n="181"/> <pb file="0173" n="182"/> <pb file="0173a" n="183"/> <figure> <image file="0173a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0173a-01"/> </figure> <pb file="0174" n="184"/> <pb file="0175" n="185"/> <pb file="0175a" n="186"/> <figure> <image file="0175a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0175a-01"/> </figure> <pb file="0176" n="187"/> <pb file="0177" n="188"/> <pb file="0177a" n="189"/> <figure> <image file="0177a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0177a-01"/> </figure> <pb file="0178" n="190"/> <pb file="0179" n="191"/> <pb file="0179a" n="192"/> <figure> <image file="0179a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0179a-01"/> </figure> <pb file="0180" n="193"/> <pb o="129" file="0181" n="194" rhead="DE M. BORELLI."/> <p> <s xml:id="echoid-s3364" xml:space="preserve">Cela ſuppoſé, ſur les branches des cordes auſquel-<lb/> <anchor type="note" xlink:label="note-0181-01a" xlink:href="note-0181-01"/> les les puiſſances φ, A, E, D, B, F, G, H, I, K, <lb/>ſont immédiatement appliquées, ſoient priſes depuis <lb/>leurs nœuds des parties θC, LZ, VO, SO, QZ, βX, <lb/>hX, δY, zY, eY, qui ſoient entr’elles, comme les <lb/>forces de ces mêmes puiſſances, c’eſt-à-dire, comme <lb/>les chifres qui leur répondent dans la Table pré-<lb/>cédente.</s> <s xml:id="echoid-s3365" xml:space="preserve"/> </p> <div xml:id="echoid-div305" type="float" level="2" n="1"> <note position="right" xlink:label="note-0181-01" xlink:href="note-0181-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> <p> <s xml:id="echoid-s3366" xml:space="preserve">Préſentement ſi l’on regarde chacune de ces pro-<lb/>portionelles comme un ſinus total, le ſinus de la diffé-<lb/>rence d’un angle droit à l’angle d’application de la <lb/>puiſſance qui rèpond à cette proportionelle, ſera la <lb/>ſublimité, ou la profondeur de cette même puiſſance: <lb/></s> <s xml:id="echoid-s3367" xml:space="preserve">par exemple, ſi l’on prend la proportionelle C θ <lb/>de la puiſſance φ, pour un ſinus total, ſa profondeur <lb/>Cλ ſera le ſinus de l’angle Cθλ, qui eſt la différen-<lb/>ce de θCλ angle d’application de cette puiſſance, à <lb/>un angle droit. </s> <s xml:id="echoid-s3368" xml:space="preserve">De même en prenant la proportio-<lb/>nelle V O de la puiſſance E, pour un ſinus total, ſa <lb/>ſublimité Ou ſera le ſinus de l'angle OVu, qui eſt la <lb/>diffèrence d’un angle droit à ſon angle d’application <lb/>V O Z; </s> <s xml:id="echoid-s3369" xml:space="preserve">de cette facon nous aurons les ſublimi-<lb/>tez & </s> <s xml:id="echoid-s3370" xml:space="preserve">les profondeurs de toutes ces puiſſances par les <lb/>analogies ſuivantes.</s> <s xml:id="echoid-s3371" xml:space="preserve"/> </p> <pb o="130" file="0182" n="195" rhead="EXAMEN DE L’OPINION"/> <note position="right" xml:space="preserve"> <lb/>Comme # au # # # # Ainſi # # à <lb/>Le ſinus total # Sinus # de \\ l’angle # de \\ Deg. M. \\ différence \\ de 90. deg. \\ à l’angle \\ d’applica- \\ tion # de la \\ puiſ- \\ ſan- \\ ce. # La propor- \\ tionelle \\ de cette \\ méme puiſ- \\ ſance # de # Saſubli- \\ mité, ou \\ à ſa pro- \\ fondeur # de <lb/># 7009093. # Cθλ # 44. 30. # φ # Cθ # 5. # Cλ # 3. {1013093./2000000.} <lb/># 5224986. # ZLl # 31. 30. # A # ZL # 4. {1/4} # Z l # 2. {4412381./20000000.} <lb/># 8746197. # OVu # 61. # E # OV # 7. {1/4} # Ou # 6. {13639713./40000000.} <lb/>10000000.{ # 3420202. # OSſ # 20. # D # OS # 12. {1/2} # Oſ # 4. {110101./400000.} <lb/># 7986355. # ZQq # 53. # B # ZQ # 14. # Zq # 11. {180897./1000000.} <lb/># 6626201. # Xβf # 41. 30. # F # X β # 11. {1/2} # Xf # 7. {6439043./20000000.} <lb/># 5519370. # Xhb # 33. 30. # G # Xh # 17. # Xh # 9. {382929./1000000.} <lb/># 4694716. # Yδd # 28. # H # Yδ # 7. {1/2} # Yd # 3. {521037./1000000.} <lb/># 2979303. # Yzx # 17. 20. # I # Yz # 16. # Yx # 4. {479303./625000.} <lb/># 8802014. # Yeg # 61. 40. # K # Ye # 13. {1/2} # Yg # 11. {8827189./10000000.} <lb/></note> <p> <s xml:id="echoid-s3372" xml:space="preserve">Aprés avoir ainſi trouvé la valeur de chacune des ſu-<lb/>blimitez & </s> <s xml:id="echoid-s3373" xml:space="preserve">des profondeurs de toutes les puiſſances qui <lb/>ſoutiennent le poids T; </s> <s xml:id="echoid-s3374" xml:space="preserve">ſoit priſe Z R égale à Ou <lb/>plus O ſ; </s> <s xml:id="echoid-s3375" xml:space="preserve">c’eſt-à-dire, ſuivant les analogies précéden-<lb/>tes, égale à 6. </s> <s xml:id="echoid-s3376" xml:space="preserve">{13639713.</s> <s xml:id="echoid-s3377" xml:space="preserve">/40000000.</s> <s xml:id="echoid-s3378" xml:space="preserve">} plus 4. </s> <s xml:id="echoid-s3379" xml:space="preserve">{110101.</s> <s xml:id="echoid-s3380" xml:space="preserve">/400000.</s> <s xml:id="echoid-s3381" xml:space="preserve">}; </s> <s xml:id="echoid-s3382" xml:space="preserve">ou bien en ré-<lb/>duiſant ces deux fractions à une même dénomina- <pb o="131" file="0183" n="196" rhead="DEM. BORELLI."/> tion, égale à 10. </s> <s xml:id="echoid-s3383" xml:space="preserve">{24649813.</s> <s xml:id="echoid-s3384" xml:space="preserve">/40000000.</s> <s xml:id="echoid-s3385" xml:space="preserve">}. </s> <s xml:id="echoid-s3386" xml:space="preserve">Aprés cela O V étant à Z R, <lb/> <anchor type="note" xlink:label="note-0183-01a" xlink:href="note-0183-01"/> comme la puiſſance E à la force dont le point Z eſt <lb/>tiré ſuivant ZO par le concours d’action des puiſ-<lb/>ſances D & </s> <s xml:id="echoid-s3387" xml:space="preserve">E; </s> <s xml:id="echoid-s3388" xml:space="preserve">Z R ſera la proportionelle de cette <lb/>force, & </s> <s xml:id="echoid-s3389" xml:space="preserve">l’angle R Z C étant (hyp.) </s> <s xml:id="echoid-s3390" xml:space="preserve">de 112. </s> <s xml:id="echoid-s3391" xml:space="preserve">deg. </s> <s xml:id="echoid-s3392" xml:space="preserve">15. <lb/></s> <s xml:id="echoid-s3393" xml:space="preserve">min. </s> <s xml:id="echoid-s3394" xml:space="preserve">ſa diffèrence à un angle droit; </s> <s xml:id="echoid-s3395" xml:space="preserve">c’eſt-à-dire, <lb/>l’angle ZRr ſera de 22. </s> <s xml:id="echoid-s3396" xml:space="preserve">deg. </s> <s xml:id="echoid-s3397" xml:space="preserve">15. </s> <s xml:id="echoid-s3398" xml:space="preserve">min. </s> <s xml:id="echoid-s3399" xml:space="preserve">Ce qui don-<lb/>nera par une analogie ſemblable aux précédentes, <lb/>3. </s> <s xml:id="echoid-s3400" xml:space="preserve">{74265285916559.</s> <s xml:id="echoid-s3401" xml:space="preserve">/200000000000000.</s> <s xml:id="echoid-s3402" xml:space="preserve">} pour la valeur de Zr ſublimité de cette <lb/>force: </s> <s xml:id="echoid-s3403" xml:space="preserve">puiſque ZR de 10. </s> <s xml:id="echoid-s3404" xml:space="preserve">{24642813.</s> <s xml:id="echoid-s3405" xml:space="preserve">/40000000.</s> <s xml:id="echoid-s3406" xml:space="preserve">} eſt à 3. </s> <s xml:id="echoid-s3407" xml:space="preserve">{17426@285916559.</s> <s xml:id="echoid-s3408" xml:space="preserve">/200000000000000.</s> <s xml:id="echoid-s3409" xml:space="preserve">} <lb/>comme le ſinus total 10000000. </s> <s xml:id="echoid-s3410" xml:space="preserve">à 3786486. </s> <s xml:id="echoid-s3411" xml:space="preserve">ſinus <lb/>de l’angle ZRr de 22. </s> <s xml:id="echoid-s3412" xml:space="preserve">deg. </s> <s xml:id="echoid-s3413" xml:space="preserve">15. </s> <s xml:id="echoid-s3414" xml:space="preserve">min.</s> <s xml:id="echoid-s3415" xml:space="preserve"/> </p> <div xml:id="echoid-div306" type="float" level="2" n="2"> <note position="right" xlink:label="note-0183-01" xlink:href="note-0183-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> <p> <s xml:id="echoid-s3416" xml:space="preserve">Soit enſuitte 1°. </s> <s xml:id="echoid-s3417" xml:space="preserve">C M égale à Zq plus Zr moins <lb/>Zl; </s> <s xml:id="echoid-s3418" xml:space="preserve">c’eſt-à-dire, ſuivant les analogies que nous ve-<lb/>nons de trouver, égale à 11. </s> <s xml:id="echoid-s3419" xml:space="preserve">{180897.</s> <s xml:id="echoid-s3420" xml:space="preserve">/1000000} plus 3. </s> <s xml:id="echoid-s3421" xml:space="preserve">{174265285916559.</s> <s xml:id="echoid-s3422" xml:space="preserve">/200000000@00000.</s> <s xml:id="echoid-s3423" xml:space="preserve">} <lb/>moins 2. </s> <s xml:id="echoid-s3424" xml:space="preserve">{4412381.</s> <s xml:id="echoid-s3425" xml:space="preserve">/20000000.</s> <s xml:id="echoid-s3426" xml:space="preserve">}; </s> <s xml:id="echoid-s3427" xml:space="preserve">ou bien en réduiſant ces trois frac-<lb/>tions à une même dénomination, égale à 12. </s> <s xml:id="echoid-s3428" xml:space="preserve">{1663208759165@9.</s> <s xml:id="echoid-s3429" xml:space="preserve">/200000000000000}. <lb/></s> <s xml:id="echoid-s3430" xml:space="preserve">Ce qui donnera par une analogie ſemblable aux pré-<lb/>cédentes, 11. </s> <s xml:id="echoid-s3431" xml:space="preserve">{74566272432665199141.</s> <s xml:id="echoid-s3432" xml:space="preserve">/500000000000000000000.</s> <s xml:id="echoid-s3433" xml:space="preserve">} pour la valeur de la <lb/>ſublimité Cm: </s> <s xml:id="echoid-s3434" xml:space="preserve">puis que 12. </s> <s xml:id="echoid-s3435" xml:space="preserve">{166320875916559.</s> <s xml:id="echoid-s3436" xml:space="preserve">/200000000000000.</s> <s xml:id="echoid-s3437" xml:space="preserve">} eſt à 11. </s> <s xml:id="echoid-s3438" xml:space="preserve"><lb/>{745662724;</s> <s xml:id="echoid-s3439" xml:space="preserve">2665199141.</s> <s xml:id="echoid-s3440" xml:space="preserve">/500000000000000000000.</s> <s xml:id="echoid-s3441" xml:space="preserve">}, comme le ſinus total 10000000. </s> <s xml:id="echoid-s3442" xml:space="preserve">à <lb/>8689196. </s> <s xml:id="echoid-s3443" xml:space="preserve">ſinus de l’angle CMm de 60. </s> <s xml:id="echoid-s3444" xml:space="preserve">deg. </s> <s xml:id="echoid-s3445" xml:space="preserve">20. </s> <s xml:id="echoid-s3446" xml:space="preserve"><lb/>min. </s> <s xml:id="echoid-s3447" xml:space="preserve">qui eſt la différence d’un angle droit à l’angle <lb/>M C T de (hyp.) </s> <s xml:id="echoid-s3448" xml:space="preserve">150. </s> <s xml:id="echoid-s3449" xml:space="preserve">deg. </s> <s xml:id="echoid-s3450" xml:space="preserve">20. </s> <s xml:id="echoid-s3451" xml:space="preserve">min.</s> <s xml:id="echoid-s3452" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3453" xml:space="preserve">2°. </s> <s xml:id="echoid-s3454" xml:space="preserve">Faite de même C N égale à Xb plus Xf; </s> <s xml:id="echoid-s3455" xml:space="preserve">c’eſt-<lb/>à-dire, ſuivant les analogies de la table précédente, <lb/>égale à 9. </s> <s xml:id="echoid-s3456" xml:space="preserve">{382929.</s> <s xml:id="echoid-s3457" xml:space="preserve">/1000000.</s> <s xml:id="echoid-s3458" xml:space="preserve">} plus 7. </s> <s xml:id="echoid-s3459" xml:space="preserve">{64,9043.</s> <s xml:id="echoid-s3460" xml:space="preserve">/20000000.</s> <s xml:id="echoid-s3461" xml:space="preserve">}; </s> <s xml:id="echoid-s3462" xml:space="preserve">ou bien en réduiſant <lb/>ces deux fractions à une même dénomination, égale <lb/>à 16. </s> <s xml:id="echoid-s3463" xml:space="preserve">{14097623.</s> <s xml:id="echoid-s3464" xml:space="preserve">/20000000}. </s> <s xml:id="echoid-s3465" xml:space="preserve">Ce qui donnera par une analogie ſem-<lb/>blable aux précédentes, 13. </s> <s xml:id="echoid-s3466" xml:space="preserve">{136767694854583/200000000000000.</s> <s xml:id="echoid-s3467" xml:space="preserve">} pour la va-<lb/>leur de la ſublimité Cn: </s> <s xml:id="echoid-s3468" xml:space="preserve">puis que 16. </s> <s xml:id="echoid-s3469" xml:space="preserve">{14097623.</s> <s xml:id="echoid-s3470" xml:space="preserve">/20000000.</s> <s xml:id="echoid-s3471" xml:space="preserve">} eſt à <lb/>13. </s> <s xml:id="echoid-s3472" xml:space="preserve">{136767694854583.</s> <s xml:id="echoid-s3473" xml:space="preserve">/200000000000000.</s> <s xml:id="echoid-s3474" xml:space="preserve">}, comme le ſinus total 10000000. </s> <s xml:id="echoid-s3475" xml:space="preserve">à <pb o="132" file="0184" n="197" rhead="EXAMEN DE L’OPINION"/> 8191521. </s> <s xml:id="echoid-s3476" xml:space="preserve">ſinus de l’angle CNn de 55. </s> <s xml:id="echoid-s3477" xml:space="preserve">deg. </s> <s xml:id="echoid-s3478" xml:space="preserve">qui eſt <lb/> <anchor type="note" xlink:label="note-0184-01a" xlink:href="note-0184-01"/> la différence d’un angle droit à l’angle NCT de <lb/>(hyp.) </s> <s xml:id="echoid-s3479" xml:space="preserve">145. </s> <s xml:id="echoid-s3480" xml:space="preserve">deg.</s> <s xml:id="echoid-s3481" xml:space="preserve"/> </p> <div xml:id="echoid-div307" type="float" level="2" n="3"> <note position="left" xlink:label="note-0184-01" xlink:href="note-0184-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>de@cordes ſeu-<lb/>lement.</note> </div> <p> <s xml:id="echoid-s3482" xml:space="preserve">3°. </s> <s xml:id="echoid-s3483" xml:space="preserve">Enſin ſoit encore C P égale à Yg plus Yx <lb/>moins Yd; </s> <s xml:id="echoid-s3484" xml:space="preserve">c’eſt-à-dire, ſuivant les analogies de la <lb/>table précédente, égale à 11. </s> <s xml:id="echoid-s3485" xml:space="preserve">{8827189.</s> <s xml:id="echoid-s3486" xml:space="preserve">/10000000.</s> <s xml:id="echoid-s3487" xml:space="preserve">} plus 4. </s> <s xml:id="echoid-s3488" xml:space="preserve">{479303.</s> <s xml:id="echoid-s3489" xml:space="preserve">/625000.</s> <s xml:id="echoid-s3490" xml:space="preserve">} moins <lb/>3. </s> <s xml:id="echoid-s3491" xml:space="preserve">{521037.</s> <s xml:id="echoid-s3492" xml:space="preserve">/1000000.</s> <s xml:id="echoid-s3493" xml:space="preserve">}; </s> <s xml:id="echoid-s3494" xml:space="preserve">ou bien en réduiſant ces trois fractions à <lb/>une même dénomination, égale à 12. </s> <s xml:id="echoid-s3495" xml:space="preserve">{232141683.</s> <s xml:id="echoid-s3496" xml:space="preserve">/282000000.</s> <s xml:id="echoid-s3497" xml:space="preserve">}. </s> <s xml:id="echoid-s3498" xml:space="preserve">Ce qui <lb/>donnera par une analogie encore ſemblable aux <lb/>précédentes, 5. </s> <s xml:id="echoid-s3499" xml:space="preserve">{686442223302177.</s> <s xml:id="echoid-s3500" xml:space="preserve">/1410000000000000.</s> <s xml:id="echoid-s3501" xml:space="preserve">} pour la valeur de la pro-<lb/>fondeur C p: </s> <s xml:id="echoid-s3502" xml:space="preserve">puis que 12. </s> <s xml:id="echoid-s3503" xml:space="preserve">{232141683.</s> <s xml:id="echoid-s3504" xml:space="preserve">/282000000.</s> <s xml:id="echoid-s3505" xml:space="preserve">} eſt à 5. </s> <s xml:id="echoid-s3506" xml:space="preserve">{686442223302177.</s> <s xml:id="echoid-s3507" xml:space="preserve">/1410000000000000.</s> <s xml:id="echoid-s3508" xml:space="preserve">}, <lb/>comme le ſinus total 10000000. </s> <s xml:id="echoid-s3509" xml:space="preserve">à 4278838. </s> <s xml:id="echoid-s3510" xml:space="preserve">ſinus <lb/>de l’angle CPp de 25. </s> <s xml:id="echoid-s3511" xml:space="preserve">deg. </s> <s xml:id="echoid-s3512" xml:space="preserve">20. </s> <s xml:id="echoid-s3513" xml:space="preserve">min. </s> <s xml:id="echoid-s3514" xml:space="preserve">qui eſt la diffè-<lb/>rence d’un angle droit à l’angle P C T de (hyp.) </s> <s xml:id="echoid-s3515" xml:space="preserve">64. <lb/></s> <s xml:id="echoid-s3516" xml:space="preserve">deg. </s> <s xml:id="echoid-s3517" xml:space="preserve">40. </s> <s xml:id="echoid-s3518" xml:space="preserve">min.</s> <s xml:id="echoid-s3519" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3520" xml:space="preserve">De tout cela on voit preſentement que <lb/>la {Subl. </s> <s xml:id="echoid-s3521" xml:space="preserve">Cm \\ Subl. </s> <s xml:id="echoid-s3522" xml:space="preserve">Cn \\ Prof. </s> <s xml:id="echoid-s3523" xml:space="preserve">Cλ \\ Prof. </s> <s xml:id="echoid-s3524" xml:space="preserve">Cp} eſt ègale à {11. </s> <s xml:id="echoid-s3525" xml:space="preserve">{74566272432665199141.</s> <s xml:id="echoid-s3526" xml:space="preserve">/500000000000000000000.</s> <s xml:id="echoid-s3527" xml:space="preserve">}. </s> <s xml:id="echoid-s3528" xml:space="preserve">\\ 13. </s> <s xml:id="echoid-s3529" xml:space="preserve">{136767694854583.</s> <s xml:id="echoid-s3530" xml:space="preserve">/200000000000000.</s> <s xml:id="echoid-s3531" xml:space="preserve">} \\ 3. </s> <s xml:id="echoid-s3532" xml:space="preserve">{1013093.</s> <s xml:id="echoid-s3533" xml:space="preserve">/2000000.</s> <s xml:id="echoid-s3534" xml:space="preserve">} \\ 5. </s> <s xml:id="echoid-s3535" xml:space="preserve">{686442223302377.</s> <s xml:id="echoid-s3536" xml:space="preserve">/1410000000000000.</s> <s xml:id="echoid-s3537" xml:space="preserve">}</s> </p> <p> <s xml:id="echoid-s3538" xml:space="preserve">De ſorte qu’en réduiſant toutes ces fractions à une <lb/>même dénomination, on aura Cm + Cn - Cλ -<lb/>Cp = 15. </s> <s xml:id="echoid-s3539" xml:space="preserve">{5919081693@137450578881.</s> <s xml:id="echoid-s3540" xml:space="preserve">/70500000000000000000000}. </s> <s xml:id="echoid-s3541" xml:space="preserve">Or ayant pris, comme <lb/>nous venons de faire, 1° CR = Oſ + Ou. </s> <s xml:id="echoid-s3542" xml:space="preserve">2°. </s> <s xml:id="echoid-s3543" xml:space="preserve">CM <lb/>= Zq + Zr - Zl. </s> <s xml:id="echoid-s3544" xml:space="preserve">3° CN = Xf + Xb. </s> <s xml:id="echoid-s3545" xml:space="preserve">4°. </s> <s xml:id="echoid-s3546" xml:space="preserve">CP <lb/>= Yg + Yx - Yd; </s> <s xml:id="echoid-s3547" xml:space="preserve">chacune des puiſſances qui <lb/>ſoutiennent ainſi le poids T; </s> <s xml:id="echoid-s3548" xml:space="preserve">par exemple, la puiſ-<lb/>ſance E eſt (Prop. </s> <s xml:id="echoid-s3549" xml:space="preserve">4. </s> <s xml:id="echoid-s3550" xml:space="preserve">Cor. </s> <s xml:id="echoid-s3551" xml:space="preserve">1.) </s> <s xml:id="echoid-s3552" xml:space="preserve">à ce poids, comme ſa <lb/>proportionnelle OV de (hyp.) </s> <s xml:id="echoid-s3553" xml:space="preserve">7 {1/4} à Cm + Cn - <pb o="133" file="0185" n="198" rhead="DE M. BORELLI."/> Cλ - Cp: </s> <s xml:id="echoid-s3554" xml:space="preserve">Donc cette même puiſſance E eſt à ce <lb/> <anchor type="note" xlink:label="note-0185-01a" xlink:href="note-0185-01"/> poids, comme 7. </s> <s xml:id="echoid-s3555" xml:space="preserve">{1/4} à 15. </s> <s xml:id="echoid-s3556" xml:space="preserve">{1919081693413745<unsure/>0578881.</s> <s xml:id="echoid-s3557" xml:space="preserve">/705<unsure/>00000000000000000000.</s> <s xml:id="echoid-s3558" xml:space="preserve">}; </s> <s xml:id="echoid-s3559" xml:space="preserve">& </s> <s xml:id="echoid-s3560" xml:space="preserve">par con-<lb/>ſèquent la valeur de cette puiſſance étant (hyp.) <lb/></s> <s xml:id="echoid-s3561" xml:space="preserve">de 7. </s> <s xml:id="echoid-s3562" xml:space="preserve">{1/4} liv. </s> <s xml:id="echoid-s3563" xml:space="preserve">ce même poids eſt auſſi juſtement de <lb/>15. </s> <s xml:id="echoid-s3564" xml:space="preserve">{5919081692@137450@78881.</s> <s xml:id="echoid-s3565" xml:space="preserve">/70500000000000000000000.</s> <s xml:id="echoid-s3566" xml:space="preserve">} liv. </s> <s xml:id="echoid-s3567" xml:space="preserve">c’eſt-à-dire, de 15. </s> <s xml:id="echoid-s3568" xml:space="preserve">livres, & </s> <s xml:id="echoid-s3569" xml:space="preserve"><lb/>un peu plus de cinq ſeptiémes de livre. </s> <s xml:id="echoid-s3570" xml:space="preserve">Ce qu’il faloit <lb/>trouver.</s> <s xml:id="echoid-s3571" xml:space="preserve"/> </p> <div xml:id="echoid-div308" type="float" level="2" n="4"> <note position="right" xlink:label="note-0185-01" xlink:href="note-0185-01a" xml:space="preserve">DES POIDS <lb/>ſoutenus avec <lb/>des cordes ſeu-<lb/>lement.</note> </div> </div> <div xml:id="echoid-div310" type="section" level="1" n="191"> <head xml:id="echoid-head193" xml:space="preserve">FIN.</head> <head xml:id="echoid-head194" xml:space="preserve">AU RELIEUR.</head> <p style="it"> <s xml:id="echoid-s3572" xml:space="preserve">LE S neuf planches au pied deſquelles on voit Projet, <lb/>doivent ètre reliées ſelon l’ordre de leurs chifres immé-<lb/>diatement à la fin de la Méchanique; </s> <s xml:id="echoid-s3573" xml:space="preserve">en ſorte que le quart <lb/>du papier où il n’y a rien, ſoit enfermé dans le livre, & </s> <s xml:id="echoid-s3574" xml:space="preserve">que <lb/>l’autre quart où ſont les figures, puiſſe, en les dépliant, en <lb/>ſortir tout entier, pour ſervir à tous les endroits de cette Mé-<lb/>chanique où l’on en aura beſoin. </s> <s xml:id="echoid-s3575" xml:space="preserve">Pour les quatre autres plan-<lb/>ches, au pied deſquelles on voit Examen, elles doivent auſſi <lb/>être reliées de la même maniére à la fin de l’Examen que voicy <lb/>de M. </s> <s xml:id="echoid-s3576" xml:space="preserve">Borelli.</s> <s xml:id="echoid-s3577" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div311" type="section" level="1" n="192"> <head xml:id="echoid-head195" xml:space="preserve">A PARIS,</head> <p> <s xml:id="echoid-s3578" xml:space="preserve">De l’Imprimerie dela Veuve Clement Gaſſe, 1687.</s> <s xml:id="echoid-s3579" xml:space="preserve"/> </p> <pb file="0186" n="199"/> </div> <div xml:id="echoid-div312" type="section" level="1" n="193"> <head xml:id="echoid-head196" xml:space="preserve">EXTRAIT DV PRIVILE’GE</head> <head xml:id="echoid-head197" style="it" xml:space="preserve">du Roy.</head> <p> <s xml:id="echoid-s3580" xml:space="preserve"><emph style="sc">PAr</emph> Lettres Patentes du Roy, données à Ver-<lb/>ſailles 19. </s> <s xml:id="echoid-s3581" xml:space="preserve">jourde Juin 1687. </s> <s xml:id="echoid-s3582" xml:space="preserve">ſignées par le Roy <lb/>en ſon Conſeil, <emph style="sc">Galloys</emph> & </s> <s xml:id="echoid-s3583" xml:space="preserve">ſcellées. </s> <s xml:id="echoid-s3584" xml:space="preserve">Il eſt permis <lb/>à <emph style="sc">Jean</emph> <emph style="sc">Boudot</emph> Libraire à Paris, d’imprimer un <lb/>livre intitulé Projet d’une nouvelle Méchanique, avec un <lb/>Examen de l’Opinion de M. </s> <s xml:id="echoid-s3585" xml:space="preserve">Borelli, ſur les propriétez des <lb/>Poids ſuſpendus par des Cordes, par M. </s> <s xml:id="echoid-s3586" xml:space="preserve">V * * * * * * * en <lb/>tel volume, marge, & </s> <s xml:id="echoid-s3587" xml:space="preserve">caractére, en autant de volumes, <lb/>& </s> <s xml:id="echoid-s3588" xml:space="preserve">tout autant de fois que bon luy ſemblera, pendant <lb/>le tems de huit années conſécutives, à commencer <lb/>du jour qu’il aura été achevé d’imprimer pour la <lb/>premiére fois. </s> <s xml:id="echoid-s3589" xml:space="preserve">Et défenſes ſont faites à toutes ſortes <lb/>de perſonnes d’imprimer ledit livre, d’en vendre n’y <lb/>diſtribuër d’autre impreſſion que de celle dudit <lb/><emph style="sc">Boudot</emph>, ou de ceux qui auront ſon droit, à peine <lb/>de mil livres d’amende, & </s> <s xml:id="echoid-s3590" xml:space="preserve">de tous dèpens, dom-<lb/>mages & </s> <s xml:id="echoid-s3591" xml:space="preserve">intérêts, & </s> <s xml:id="echoid-s3592" xml:space="preserve">de confiſcation des Exemplaires <lb/>contrefaits, ainſi qu’il eſt porté plus au long dans ledit <lb/>Privilége.</s> <s xml:id="echoid-s3593" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3594" xml:space="preserve">Regiſtré ſur le Livre de la Communauté des Li-<lb/>braires & </s> <s xml:id="echoid-s3595" xml:space="preserve">Imprimeurs de Paris, le 30. </s> <s xml:id="echoid-s3596" xml:space="preserve">Juillet 1687.</s> <s xml:id="echoid-s3597" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3598" xml:space="preserve">Signé, <emph style="sc">Coignard</emph>, Syndic.</s> <s xml:id="echoid-s3599" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s3600" xml:space="preserve">Achevè d’imprimer pour la premiére fois le 30. </s> <s xml:id="echoid-s3601" xml:space="preserve">Aouſt 1687.</s> <s xml:id="echoid-s3602" xml:space="preserve"/> </p> <pb file="0187" n="200"/> </div> <div xml:id="echoid-div313" type="section" level="1" n="194"> <head xml:id="echoid-head198" style="it" xml:space="preserve">FAVTES A CORRIGER.</head> <p style="it"> <s xml:id="echoid-s3603" xml:space="preserve">PAg. </s> <s xml:id="echoid-s3604" xml:space="preserve">4. </s> <s xml:id="echoid-s3605" xml:space="preserve">lig. </s> <s xml:id="echoid-s3606" xml:space="preserve">20. </s> <s xml:id="echoid-s3607" xml:space="preserve">AH liſez AK</s> </p> <p style="it"> <s xml:id="echoid-s3608" xml:space="preserve">pag. </s> <s xml:id="echoid-s3609" xml:space="preserve">9. </s> <s xml:id="echoid-s3610" xml:space="preserve">ligne 14. </s> <s xml:id="echoid-s3611" xml:space="preserve">Lemme 3.</s> <s xml:id="echoid-s3612" xml:space="preserve">, liſez Lemme 3.</s> <s xml:id="echoid-s3613" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s3614" xml:space="preserve">pag. </s> <s xml:id="echoid-s3615" xml:space="preserve">14. </s> <s xml:id="echoid-s3616" xml:space="preserve">ligne 11. </s> <s xml:id="echoid-s3617" xml:space="preserve">poid liſez poids</s> </p> <p style="it"> <s xml:id="echoid-s3618" xml:space="preserve">pag. </s> <s xml:id="echoid-s3619" xml:space="preserve">17. </s> <s xml:id="echoid-s3620" xml:space="preserve">ligne 4. </s> <s xml:id="echoid-s3621" xml:space="preserve">elle ſe liſez elle le</s> </p> <p style="it"> <s xml:id="echoid-s3622" xml:space="preserve">pag. </s> <s xml:id="echoid-s3623" xml:space="preserve">23. </s> <s xml:id="echoid-s3624" xml:space="preserve">à la marge Vis-à-vis du problême marquez fig. </s> <s xml:id="echoid-s3625" xml:space="preserve">19.</s> <s xml:id="echoid-s3626" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s3627" xml:space="preserve">pag. </s> <s xml:id="echoid-s3628" xml:space="preserve">29. </s> <s xml:id="echoid-s3629" xml:space="preserve">ligne 22. </s> <s xml:id="echoid-s3630" xml:space="preserve">AH liſez AD</s> </p> <p style="it"> <s xml:id="echoid-s3631" xml:space="preserve">pag. </s> <s xml:id="echoid-s3632" xml:space="preserve">35. </s> <s xml:id="echoid-s3633" xml:space="preserve">à la marge fig. </s> <s xml:id="echoid-s3634" xml:space="preserve">26. </s> <s xml:id="echoid-s3635" xml:space="preserve">liſez fig. </s> <s xml:id="echoid-s3636" xml:space="preserve">26. <lb/></s> <s xml:id="echoid-s3637" xml:space="preserve">27.</s> <s xml:id="echoid-s3638" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s3639" xml:space="preserve">pag. </s> <s xml:id="echoid-s3640" xml:space="preserve">38. </s> <s xml:id="echoid-s3641" xml:space="preserve">ligne 25. </s> <s xml:id="echoid-s3642" xml:space="preserve">à EF; </s> <s xml:id="echoid-s3643" xml:space="preserve">liſez HEF; </s> <s xml:id="echoid-s3644" xml:space="preserve">& </s> <s xml:id="echoid-s3645" xml:space="preserve">ligne 26. </s> <s xml:id="echoid-s3646" xml:space="preserve">effacéz ou EHF,</s> </p> <p style="it"> <s xml:id="echoid-s3647" xml:space="preserve">pag. </s> <s xml:id="echoid-s3648" xml:space="preserve">48. </s> <s xml:id="echoid-s3649" xml:space="preserve">ligne 31. </s> <s xml:id="echoid-s3650" xml:space="preserve">points liſez point</s> </p> <p style="it"> <s xml:id="echoid-s3651" xml:space="preserve">pag. </s> <s xml:id="echoid-s3652" xml:space="preserve">74. </s> <s xml:id="echoid-s3653" xml:space="preserve">ligne 1. </s> <s xml:id="echoid-s3654" xml:space="preserve">puiſſnces, liſez puiſſances,</s> </p> <p style="it"> <s xml:id="echoid-s3655" xml:space="preserve">pag. </s> <s xml:id="echoid-s3656" xml:space="preserve">99. </s> <s xml:id="echoid-s3657" xml:space="preserve">ligne 14. </s> <s xml:id="echoid-s3658" xml:space="preserve">& </s> <s xml:id="echoid-s3659" xml:space="preserve">pag. </s> <s xml:id="echoid-s3660" xml:space="preserve">100. </s> <s xml:id="echoid-s3661" xml:space="preserve">lignes 5. </s> <s xml:id="echoid-s3662" xml:space="preserve">16. </s> <s xml:id="echoid-s3663" xml:space="preserve">& </s> <s xml:id="echoid-s3664" xml:space="preserve">17. </s> <s xml:id="echoid-s3665" xml:space="preserve">propotionnelles liſez <lb/>proportionelles</s> </p> <p style="it"> <s xml:id="echoid-s3666" xml:space="preserve">pag. </s> <s xml:id="echoid-s3667" xml:space="preserve">103. </s> <s xml:id="echoid-s3668" xml:space="preserve">ligne 1. </s> <s xml:id="echoid-s3669" xml:space="preserve">ces de liſez de ces</s> </p> <p style="it"> <s xml:id="echoid-s3670" xml:space="preserve">pag. </s> <s xml:id="echoid-s3671" xml:space="preserve">104. </s> <s xml:id="echoid-s3672" xml:space="preserve">ligne 6. </s> <s xml:id="echoid-s3673" xml:space="preserve">69. </s> <s xml:id="echoid-s3674" xml:space="preserve">M. </s> <s xml:id="echoid-s3675" xml:space="preserve">liſez 69. </s> <s xml:id="echoid-s3676" xml:space="preserve">de M.</s> <s xml:id="echoid-s3677" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s3678" xml:space="preserve">pag. </s> <s xml:id="echoid-s3679" xml:space="preserve">105. </s> <s xml:id="echoid-s3680" xml:space="preserve">ligne 27. </s> <s xml:id="echoid-s3681" xml:space="preserve">répoudent, liſez répondent,</s> </p> <pb file="0188" n="201"/> <pb file="0189" n="202"/> <pb file="0190" n="203"/> <pb file="0191" n="204"/> <pb file="0192" n="205"/> <pb file="0193" n="206"/> <pb file="0194" n="207"/> <pb file="0195" n="208"/> <pb file="0196" n="209"/> </div></text> </echo>