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4.1-For the truss shown in figure 4.37(Q1),
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Point</XXXXXX><XXXXXX XXXXX="XXXXXXXX-XXXXXX-XXXXXXXXX">X</strong>: We XXX XXXXXXXXXXX solve XXXXXXXXXX,<strong class="redactor-XXXXXX-XXXXXXXXX">0 = -XX + F(BH)</XXXXXX> <em>(Force in XXXXX BH), </em>so <strong XXXXX="redactor-XXXXXX-XXXXXXXXX"><u>X(BH) = 50</u></strong><XX></p><p><strong XXXXX="XXXXXXXX-XXXXXX-XXXXXXXXX">Point X:</XXXXXX>XXXXXXXXX,<XXXXXX XXXXX="XXXXXXXX-inline-XXXXXXXXX"><u>F(XX) = XX</u></XXXXXX></p><p><strong XXXXX="XXXXXXXX-inline-converted">Point X:</XXXXXX> XXX XXXX XXXXXXXX counter XX F(BH)is X(AH).<XXXXXX class="XXXXXXXX-inline-XXXXXXXXX">0 = -XX +F(XX)/XXX(XXXXX)</XXXXXX>, XXXXXXXXXX for the XXXXX in XXX truss. XXXX XXXXX us<strong></strong><u><XXXXXX>F(AH)=50 XXXXX.</strong></u> Similarly,<strong><u>F(EF) = XX root 2</u></strong></p><p><XXXXXX XXXXX="XXXXXXXX-inline-XXXXXXXXX">XXXXX</XXXXXX><strong class="redactor-inline-XXXXXXXXX">A:</strong> XX can XXXXXXXX XXXX<strong XXXXX="redactor-inline-XXXXXXXXX">Ay=XX,</strong> XXXXX it XX XXX XXXX XXXXXXXX force cancelling out F(XX). XXXX that, we XXXXXXXX XXXX <strong XXXXX="redactor-inline-converted">XX=50</strong> and, accordingly,<strong XXXXX="redactor-inline-converted"><u>F(AB)= X(DE) = XX</u></XXXXXX></p><p>Then we can XXXXXX a straight XXXX from XXXXX X XX XXXXX E XXX XXX a quick XXXXXXXXXX Method XX XXXXXXXXXXXXXX XXXX:<strong class="XXXXXXXX-inline-converted">X(AB) = F(BC)= F(CD) = F(XX) = XX</strong></p><p><strong class="XXXXXXXX-XXXXXX-XXXXXXXXX">XXXXX X</strong>: XXXXXXXXXX out the XXXXXXXXXX XXXXXXXXX XX F(AH), we get<XXXXXX class="redactor-XXXXXX-XXXXXXXXX">X(XX) = XX</XXXXXX>.</p><p><XXXXXX XXXXX="XXXXXXXX-inline-XXXXXXXXX">XXXXX X:</strong>XXXXXXXXX,<XXXXXX><u>X(XX) = XX</u></XXXXXX></p><p><strong>XXXXX X:</strong> XXX XXXX's XXXXXXX now is F(XX). XX notice XX XXXXXXXX components are countering it, so we conclude that<strong><u>X(CG) = 0</u></strong><XX></p>">
