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="N106DF">
            <pb pagenum="32" xlink:href="037/01/079.jpg"/>
            <p id="id.2.1.265.0.0" type="main">
              <s id="id.2.1.265.1.0">
                <emph type="italics"/>
              Similmente dimoſtreraßi, che i peſi EF peſeranno tanto appiccati in qual ſi voglia al­
                <lb/>
              tro punto, quanto ſe l'vno, & l'altro foſſe pendente dal punto H della diuiſione.
                <lb/>
              </s>
              <s id="id.2.1.265.2.0">Percioche ſe, come di ſopra habbiamo inſegnato, ſi troueranno i peſi nella bilancia, à
                <lb/>
              i quali i peſi EF peſino egualmente; gli isteßi peſi EF pendenti da H peſeranno
                <lb/>
              egualmente co' medeſimi peſi trouati; per eſſere il punto P ſempre il centro della
                <lb/>
              grauezza loro; & la HP a piombo dell'orizonte.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="id.2.1.266.0.0" type="head">
              <s id="id.2.1.266.1.0">PROPOSITIONE VI. </s>
            </p>
            <p id="id.2.1.267.0.0" type="main">
              <s id="id.2.1.267.1.0">I peſi eguali nella bilancia appiccati hanno in grauezza quella pro­
                <lb/>
              portione, che hanno le diſtanze, dalle quali ſtanno pendenti. </s>
            </p>
            <figure id="id.037.01.079.1.jpg" xlink:href="037/01/079/1.jpg" number="69"/>
            <p id="id.2.1.269.0.0" type="main">
              <s id="id.2.1.269.1.0">
                <emph type="italics"/>
              Sia la bilancia BAC ſoſpeſa nel punto A; & ſia ſegata la AC, come pare in D. </s>
              <s id="N12F87">&
                <lb/>
              da i punti DC ſiano attaccati EF peſi eguali. </s>
              <s id="id.2.1.269.2.0">Dico, che il peſo F verſo il peſo E ba
                <lb/>
              quella proportione in grauezza, che hala diſtanza CA alla diſtanza AD. </s>
              <s id="id.2.1.269.3.0">Per­
                <lb/>
              cioche facciaſi come CA verſo AD, coſi il peſo F verſo vn'altro peſo, che ſia G.
                <lb/>
              </s>
              <s id="id.2.1.269.4.0">Dico prima i peſi GF pendenti dal punto C tanto peſare, quanto i peſi EF penden
                <lb/>
              ti da punti DC. </s>
              <s id="id.2.1.269.5.0">Tagliſi DC in due parti eguali in H, & da H ſiano fatti pendere
                <lb/>
              ambidue i peſi EF. </s>
              <s id="id.2.1.269.6.0">Peſeranno EF preſi inſieme in quel ſito tanto quanto peſano
                <emph.end type="italics"/>
                <arrow.to.target n="note87"/>
                <lb/>
                <emph type="italics"/>
              in DC. </s>
              <s id="id.2.1.269.7.0">Pongaſi BA eguale ad AH, & ſitagli BA in K, di modo, che KA
                <lb/>
              ſia eguale ad AD: dapoi dal punto B ſia ſatto pendente il peſo L, ilquale ſia il dop
                <lb/>
              pio del peſo F, cioè eguale a i due peſi EF, ilqual peſerà egualmente co'peſi EF ap
                <lb/>
              piccati in H, cioè appiccati in DC. </s>
              <s id="id.2.1.269.8.0">Percioche dunque, come CA verſo AD, così è
                <lb/>
              il peſo F verſo il peſo G, ſarà componendo come CA AD verſo AD, cioè come
                <lb/>
              CK verſo AD, così i peſi FG verſo il peſo G. </s>
              <s id="id.2.1.269.9.0">Ma per eſſer come CA verſo AD,
                <emph.end type="italics"/>
                <arrow.to.target n="note88"/>
                <lb/>
                <emph type="italics"/>
              così il peſo F al peſo G, ſarà anche conuertendo, come DA verſo AC, così il peſo
                <lb/>
              G verſo il peſo F; & i doppi dei conſeguenti, come DA alla doppia di eſſa AC,
                <lb/>
              così il peſo G al doppio del peſo F, cioè al peſo L. </s>
              <s id="id.2.1.269.10.0">Per laqual coſa come CK verſo
                <emph.end type="italics"/>
                <arrow.to.target n="note89"/>
                <lb/>
                <emph type="italics"/>
              DA, così i peſi FG al peſo G; & come AD alla doppia di AC, così il peſo G al
                <lb/>
              peſo L, adunque dalla egual proportione come CK alla doppia di AC, così i peſi FG
                <lb/>
              al peſo L. </s>
              <s id="id.2.1.269.11.0">Ma come CK alla doppia di AC, così la metà di CK, cioè AH, cioè
                <emph.end type="italics"/>
                <arrow.to.target n="note90"/>
                <lb/>
                <emph type="italics"/>
              BA verſo AC. </s>
              <s id="id.2.1.269.12.0">Adunque come BA verſo AC, così FG peſi al peſo L. </s>
              <s id="id.2.1.269.13.0">Per laqual
                <emph.end type="italics"/>
              </s>
            </p>
          </chap>
        </body>
      </text>
    </archimedes>
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