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4.1-For the truss shown in figure 4.37(Q1),
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Point</XXXXXX><XXXXXX XXXXX="redactor-inline-converted">B</XXXXXX>: We can immediately XXXXX vertically,<strong class="redactor-inline-converted">X = -XX + F(BH)</strong> <em>(Force in truss XX), </em>so <XXXXXX XXXXX="redactor-inline-converted"><u>F(BH) = XX</u></strong><XX></p><p><strong class="redactor-inline-XXXXXXXXX">Point X:</XXXXXX>Similarly,<XXXXXX class="XXXXXXXX-XXXXXX-converted"><u>X(DF) = XX</u></XXXXXX></p><p><strong XXXXX="XXXXXXXX-inline-XXXXXXXXX">Point X:</XXXXXX> XXX only XXXXXXXX XXXXXXX XX X(BH)XX F(AH).<XXXXXX XXXXX="redactor-XXXXXX-converted">0 = -50 +F(XX)/XXX(XXXXX)</XXXXXX>, accounting XXX XXX slant in XXX XXXXX. That XXXXX us<strong></strong><u><XXXXXX>X(AH)=50 XXXXX.</strong></u> XXXXXXXXX,<XXXXXX><u>X(EF) = 50 XXXX X</u></XXXXXX></p><p><XXXXXX class="redactor-inline-converted">Point</strong><XXXXXX class="redactor-XXXXXX-XXXXXXXXX">X:</strong> XX can validate that<XXXXXX class="redactor-inline-converted">Ay=50,</strong> XXXXX it XX XXX XXXX XXXXXXXX force cancelling out X(AH). From that, we XXXXXXXX XXXX <strong class="XXXXXXXX-XXXXXX-converted">Ax=XX</strong> XXX, XXXXXXXXXXX,<XXXXXX XXXXX="redactor-inline-converted"><u>F(AB)= X(XX) = XX</u></strong></p><p>XXXX we XXX follow a straight XXXX XXXX XXXXX A to point X and use a quick horizontal XXXXXX XX XXXXXXXXXXXXXX that:<XXXXXX XXXXX="XXXXXXXX-inline-XXXXXXXXX">F(XX) = X(XX)= F(XX) = F(DE) = XX</XXXXXX></p><p><strong class="redactor-XXXXXX-converted">Point X</XXXXXX>: XXXXXXXXXX out the XXXXXXXXXX XXXXXXXXX XX X(AH), XX get<strong XXXXX="XXXXXXXX-inline-XXXXXXXXX">F(GH) = 50</XXXXXX>.</p><p><strong XXXXX="XXXXXXXX-inline-XXXXXXXXX">XXXXX X:</strong>Similarly,<XXXXXX><u>F(XX) = 50</u></XXXXXX></p><p><XXXXXX>Point X:</XXXXXX> All XXXX's missing XXX XX F(CG). We notice XX XXXXXXXX XXXXXXXXXX are XXXXXXXXXX it, so XX conclude that<strong><u>F(XX) = X</u></strong><br></p>">
