DelMonte, Guidubaldo, Le mechaniche

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      <text id="id.0.0.0.0.3">
        <body id="id.2.0.0.0.0">
          <chap id="N13354">
            <pb xlink:href="037/01/088.jpg"/>
            <p id="id.2.1.346.0.0" type="head">
              <s id="id.2.1.346.1.0">Altramente. </s>
            </p>
            <p id="id.2.1.347.0.0" type="main">
              <s id="id.2.1.347.1.0">
                <emph type="italics"/>
              Sia la leua AB, il cui ſoſtegno ſia B, & il peſo E ſia pendente dal punto C, &
                <lb/>
              ſia in A la forza, che ſostiene l peſo E. </s>
              <s id="id.2.1.347.2.0">Dico, che ſi come BC à BA, coſi è
                <emph.end type="italics"/>
                <lb/>
                <figure id="id.037.01.088.1.jpg" xlink:href="037/01/088/1.jpg" number="82"/>
                <lb/>
                <emph type="italics"/>
              anco la poſſanza di A verſo il peſo E. </s>
              <s id="id.2.1.347.3.0">Allunghiſi AB in D, & facciaſi
                <lb/>
              BD eguale à BC; & appicchiſi il peſo F al punto D, che ſia eguale al peſo E;
                <lb/>
              & parimente dal punto A ſi faccia pendere il punto G in modo, che il peſo F hab
                <lb/>
              bia la proportione iſteſſa verſo il peſo G, che ha AB à BD. </s>
              <s id="id.2.1.347.4.0">i peſi FG verranno
                <lb/>
              à peſar egualmente: & concioſia che CB ſia eguale à BD, anco i peſi FE egua
                <lb/>
              li peſeranno egualmente. </s>
              <s id="id.2.1.347.5.0">Ma i peſi FEG nella bilancia, ouero nella leua DBA
                <lb/>
              appiccati, il cui ſoſtegno è B, non peſeranno egualmente, ma inchineranno à baſſo
                <lb/>
              dalla parte di A. </s>
              <s id="id.2.1.347.6.0">Per laqual coſa pongaſi in A tanta forza, che i peſi FEG peſi­
                <lb/>
              no egualmente, ſarà la poſſanza in A eguale al peſo G; peroche i peſi FE peſa­
                <lb/>
              no egualmente, & la forza in A niente altro deue fare, che ſoſtenere il peſo G, ac­
                <lb/>
              cioche non deſcenda. </s>
              <s id="id.2.1.347.7.0">Et percio che i peſi FEG, & la poſſanza in A peſano egual
                <lb/>
              mente, leuati dunque via i peſi FG, i quali peſano egualmente, i reſtanti peſeran­
                <lb/>
              no pur egualmente, cioè la poſſanza in A co'l peſo E, cioè la poſſanza in A ſo­
                <lb/>
              sterra il peſo E, ſi che la leua AB rimanga, come era prima. </s>
              <s id="id.2.1.347.8.0">Et per eſſere la
                <lb/>
              poſſanza in A eguale al peſo G, & il peſo E eguale al peſo F, haurà la poſſanza
                <lb/>
              in A la proportione isteſſa al peſo E, che hà BD, cioè BC à BA, che biſogna
                <lb/>
              ua moſtrare.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="id.2.1.349.0.0" type="head">
              <s id="id.2.1.349.1.0">COROLLARIO I. </s>
            </p>
            <p id="id.2.1.350.0.0" type="main">
              <s id="id.2.1.350.1.0">Da queſto etiandio, come prima, puote eſſere manifeſto, che ſe il peſo
                <lb/>
              E ſarà poſto piu vicino al ſoſtegno B, come in H, minore
                <lb/>
              poſſanza poſta in A puote ſoſtener il detto peſo. </s>
            </p>
            <p id="id.2.1.351.0.0" type="main">
              <s id="id.2.1.351.1.0">
                <emph type="italics"/>
              Percioche minor proportione ha HB à BA, che CB à BA. </s>
              <s id="N1365A">& quanto piu da
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="note110"/>
                <emph type="italics"/>
              vicino il peſo ſarà al ſoſtegno, ſempre anco ſi moſtrerà ſimilmente minor poſſanza
                <lb/>
              poter ſoſtener il peſo E.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="id.2.1.352.0.0" type="margin">
              <s id="id.2.1.352.1.0">
                <margin.target id="note110"/>
                <emph type="italics"/>
              Per la
                <emph.end type="italics"/>
              8.
                <emph type="italics"/>
              del quinto.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="id.2.1.353.0.0" type="head">
              <s id="id.2.1.353.1.0">COROLLARIO II. </s>
            </p>
            <p id="id.2.1.354.0.0" type="main">
              <s id="id.2.1.354.1.0">Segue etiandio, che la poſſanza in A ſempre è minore del peſo E: </s>
            </p>
          </chap>
        </body>
      </text>
    </archimedes>
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