Varignon, Pierre, Projet d' une nouvelle mechanique : avec Un examen de l' opinion de M. Borelli sur les propriétez des poids suspendus par des cordes

Table of Notes

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            puiſſances, celle-ci ſoutiendra ce poids en cet état. </s>
            <s xml:id="echoid-s1021" xml:space="preserve">Ce
              <lb/>
              <note position="left" xlink:label="note-0064-01" xlink:href="note-0064-01a" xml:space="preserve">DES
                <lb/>
              POULIES.</note>
            qu’il faloit faire.</s>
            <s xml:id="echoid-s1022" xml:space="preserve"/>
          </p>
          <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/>
              <note position="left" xlink:label="note-0064-02" xlink:href="note-0064-02a" xml:space="preserve">fig. 29.</note>
            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>
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        <div xml:id="echoid-div113" type="section" level="1" n="77">
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            <emph style="sc">Demonstration.</emph>
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          <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>
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          </p>
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        <div xml:id="echoid-div114" type="section" level="1" n="78">
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            <emph style="sc">Corollaire.</emph>
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            <s xml:id="echoid-s1052" xml:space="preserve">On voit que l’on peut toujours faire ſoutenir </s>
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