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4.1-For the truss shown in figure 4.37(Q1),
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XXXXX</strong><XXXXXX class="XXXXXXXX-XXXXXX-converted">X</XXXXXX>: XX XXX immediately solve XXXXXXXXXX,<XXXXXX class="XXXXXXXX-inline-converted">0 = -50 + F(XX)</strong> <em>(Force in truss XX), </em>so <XXXXXX class="redactor-inline-converted"><u>F(BH) = XX</u></XXXXXX><br></p><p><XXXXXX class="XXXXXXXX-XXXXXX-XXXXXXXXX">XXXXX X:</strong>XXXXXXXXX,<XXXXXX XXXXX="XXXXXXXX-XXXXXX-XXXXXXXXX"><u>F(XX) = XX</u></XXXXXX></p><p><strong XXXXX="XXXXXXXX-inline-converted">Point X:</strong> XXX only vertical XXXXXXX to F(BH)XX X(XX).<strong class="XXXXXXXX-inline-converted">0 = -XX +X(XX)/XXX(45deg)</strong>, accounting XXX XXX XXXXX in XXX XXXXX. XXXX gives us<XXXXXX></XXXXXX><u><strong>F(XX)=XX root2.</XXXXXX></u> XXXXXXXXX,<XXXXXX><u>F(EF) = 50 XXXX 2</u></XXXXXX></p><p><XXXXXX class="redactor-XXXXXX-converted">Point</XXXXXX><XXXXXX class="XXXXXXXX-XXXXXX-converted">A:</XXXXXX> XX XXX XXXXXXXX XXXX<XXXXXX class="redactor-XXXXXX-XXXXXXXXX">XX=50,</strong> since it is XXX only XXXXXXXX XXXXX XXXXXXXXXX out X(XX). From that, we XXXXXXXX XXXX <XXXXXX class="redactor-XXXXXX-converted">XX=50</XXXXXX> and, accordingly,<XXXXXX XXXXX="XXXXXXXX-XXXXXX-XXXXXXXXX"><u>X(AB)= F(DE) = XX</u></strong></p><p>Then XX XXX XXXXXX a XXXXXXXX XXXX XXXX point X XX XXXXX X and XXX a quick horizontal XXXXXX of XXXXXXXXXXXXXX that:<XXXXXX class="XXXXXXXX-inline-XXXXXXXXX">X(AB) = X(XX)= X(CD) = F(DE) = XX</XXXXXX></p><p><strong class="XXXXXXXX-XXXXXX-XXXXXXXXX">Point X</XXXXXX>: XXXXXXXXXX out XXX horizontal XXXXXXXXX XX X(XX), we get<XXXXXX XXXXX="XXXXXXXX-XXXXXX-XXXXXXXXX">X(XX) = XX</strong>.</p><p><strong class="XXXXXXXX-XXXXXX-XXXXXXXXX">Point F:</strong>XXXXXXXXX,<strong><u>F(FG) = 50</u></XXXXXX></p><p><strong>Point X:</XXXXXX> All that's XXXXXXX XXX XX F(CG). XX XXXXXX no vertical XXXXXXXXXX XXX countering it, so we conclude that<strong><u>F(XX) = X</u></XXXXXX><br></p>">
