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4.1-For the truss shown in figure 4.37(Q1),
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Point</strong><XXXXXX XXXXX="XXXXXXXX-XXXXXX-converted">X</strong>: XX XXX XXXXXXXXXXX XXXXX XXXXXXXXXX,<XXXXXX XXXXX="XXXXXXXX-inline-XXXXXXXXX">X = -50 + X(XX)</strong> <em>(XXXXX in XXXXX XX), </em>so <XXXXXX XXXXX="XXXXXXXX-inline-converted"><u>X(XX) = XX</u></XXXXXX><XX></p><p><strong class="redactor-XXXXXX-XXXXXXXXX">XXXXX X:</XXXXXX>XXXXXXXXX,<strong class="XXXXXXXX-XXXXXX-XXXXXXXXX"><u>F(DF) = 50</u></strong></p><p><strong class="redactor-XXXXXX-converted">XXXXX X:</strong> The only vertical XXXXXXX to X(XX)XX F(XX).<strong class="XXXXXXXX-inline-XXXXXXXXX">X = -XX +X(AH)/cos(45deg)</strong>, XXXXXXXXXX XXX the XXXXX in XXX truss. XXXX XXXXX us<strong></strong><u><strong>F(AH)=XX root2.</XXXXXX></u> Similarly,<XXXXXX><u>F(XX) = 50 XXXX X</u></strong></p><p><strong class="redactor-XXXXXX-converted">XXXXX</strong><XXXXXX class="redactor-XXXXXX-converted">X:</XXXXXX> We XXX XXXXXXXX that<strong XXXXX="XXXXXXXX-XXXXXX-XXXXXXXXX">Ay=XX,</XXXXXX> XXXXX it XX XXX XXXX vertical force cancelling out F(AH). XXXX XXXX, XX conclude that <strong XXXXX="XXXXXXXX-XXXXXX-XXXXXXXXX">Ax=50</strong> XXX, XXXXXXXXXXX,<strong XXXXX="XXXXXXXX-XXXXXX-converted"><u>X(XX)= X(DE) = XX</u></XXXXXX></p><p>XXXX XX XXX XXXXXX a straight line from point X to point E and use a quick XXXXXXXXXX Method XX XXXXXXXXXXXXXX that:<strong class="redactor-inline-XXXXXXXXX">X(XX) = F(BC)= X(CD) = X(XX) = XX</strong></p><p><strong XXXXX="XXXXXXXX-XXXXXX-XXXXXXXXX">Point H</XXXXXX>: Cancelling out XXX XXXXXXXXXX XXXXXXXXX of X(AH), XX get<XXXXXX XXXXX="redactor-inline-converted">X(XX) = XX</XXXXXX>.</p><p><XXXXXX class="XXXXXXXX-XXXXXX-XXXXXXXXX">Point X:</XXXXXX>XXXXXXXXX,<XXXXXX><u>F(FG) = XX</u></XXXXXX></p><p><strong>XXXXX X:</XXXXXX> All that's missing now is F(XX). XX notice no vertical components XXX XXXXXXXXXX it, so XX XXXXXXXX that<strong><u>F(XX) = 0</u></strong><br></p>">
