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4.1-For the truss shown in figure 4.37(Q1),
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Point</XXXXXX><strong class="XXXXXXXX-inline-converted">B</strong>: XX XXX XXXXXXXXXXX XXXXX XXXXXXXXXX,<strong class="XXXXXXXX-XXXXXX-converted">X = -XX + F(XX)</XXXXXX> <em>(Force in truss XX), </em>so <XXXXXX class="redactor-XXXXXX-XXXXXXXXX"><u>F(XX) = XX</u></XXXXXX><br></p><p><XXXXXX XXXXX="redactor-inline-converted">XXXXX D:</strong>Similarly,<strong class="redactor-inline-XXXXXXXXX"><u>X(XX) = 50</u></strong></p><p><XXXXXX class="XXXXXXXX-inline-XXXXXXXXX">Point X:</strong> XXX only XXXXXXXX counter to F(XX)is X(XX).<strong XXXXX="XXXXXXXX-XXXXXX-converted">0 = -XX +F(AH)/XXX(45deg)</strong>, accounting for XXX XXXXX in the XXXXX. That gives XX<strong></strong><u><strong>X(AH)=50 XXXXX.</XXXXXX></u> XXXXXXXXX,<XXXXXX><u>F(EF) = XX XXXX 2</u></XXXXXX></p><p><XXXXXX XXXXX="XXXXXXXX-inline-XXXXXXXXX">Point</strong><XXXXXX XXXXX="redactor-XXXXXX-XXXXXXXXX">A:</strong> XX can XXXXXXXX that<strong class="redactor-XXXXXX-converted">XX=XX,</XXXXXX> since it XX XXX XXXX vertical XXXXX cancelling out F(XX). From XXXX, XX conclude XXXX <strong XXXXX="redactor-inline-converted">Ax=XX</strong> and, XXXXXXXXXXX,<strong class="XXXXXXXX-XXXXXX-converted"><u>F(XX)= X(XX) = XX</u></XXXXXX></p><p>XXXX XX can XXXXXX a straight XXXX from XXXXX X XX XXXXX E XXX XXX a XXXXX XXXXXXXXXX XXXXXX of Pointsconclude XXXX:<strong XXXXX="redactor-inline-XXXXXXXXX">X(XX) = X(BC)= X(XX) = F(XX) = 50</XXXXXX></p><p><XXXXXX XXXXX="XXXXXXXX-inline-converted">XXXXX X</strong>: Cancelling out XXX horizontal component of F(XX), we XXX<XXXXXX class="redactor-inline-converted">F(XX) = 50</XXXXXX>.</p><p><strong XXXXX="XXXXXXXX-inline-converted">XXXXX X:</XXXXXX>XXXXXXXXX,<strong><u>F(FG) = XX</u></XXXXXX></p><p><XXXXXX>Point G:</strong> All XXXX's missing now XX F(XX). We XXXXXX no vertical components XXX countering it, so we conclude that<strong><u>X(XX) = X</u></XXXXXX><br></p>">
