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4.1-For the truss shown in figure 4.37(Q1),
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Point</XXXXXX><strong XXXXX="redactor-XXXXXX-XXXXXXXXX">B</strong>: We can immediately XXXXX XXXXXXXXXX,<strong class="redactor-inline-XXXXXXXXX">0 = -XX + F(XX)</XXXXXX> <em>(XXXXX in truss XX), </em>so <XXXXXX class="XXXXXXXX-XXXXXX-converted"><u>X(XX) = XX</u></strong><br></p><p><XXXXXX class="XXXXXXXX-inline-XXXXXXXXX">XXXXX D:</XXXXXX>XXXXXXXXX,<XXXXXX XXXXX="XXXXXXXX-inline-converted"><u>F(XX) = XX</u></strong></p><p><XXXXXX XXXXX="redactor-inline-converted">Point X:</strong> The XXXX XXXXXXXX XXXXXXX to X(BH)is F(XX).<XXXXXX XXXXX="XXXXXXXX-XXXXXX-XXXXXXXXX">X = -50 +F(XX)/cos(XXXXX)</XXXXXX>, accounting XXX XXX slant in the XXXXX. That XXXXX XX<XXXXXX></XXXXXX><u><strong>F(AH)=XX XXXXX.</strong></u> Similarly,<strong><u>X(EF) = XX root X</u></XXXXXX></p><p><strong class="XXXXXXXX-XXXXXX-XXXXXXXXX">XXXXX</XXXXXX><strong class="XXXXXXXX-XXXXXX-converted">A:</strong> XX XXX validate XXXX<XXXXXX XXXXX="redactor-inline-XXXXXXXXX">XX=50,</XXXXXX> XXXXX it XX XXX XXXX XXXXXXXX force XXXXXXXXXX out F(AH). XXXX that, XX conclude XXXX <XXXXXX class="redactor-inline-XXXXXXXXX">Ax=XX</XXXXXX> and, accordingly,<strong XXXXX="XXXXXXXX-XXXXXX-XXXXXXXXX"><u>F(XX)= X(DE) = 50</u></strong></p><p>Then XX can follow a straight XXXX XXXX XXXXX X to point E XXX XXX a XXXXX XXXXXXXXXX Method of Pointsconclude XXXX:<strong XXXXX="XXXXXXXX-XXXXXX-converted">X(XX) = F(XX)= F(CD) = F(XX) = 50</XXXXXX></p><p><XXXXXX class="XXXXXXXX-XXXXXX-converted">XXXXX X</XXXXXX>: XXXXXXXXXX out XXX XXXXXXXXXX component XX X(AH), we get<XXXXXX class="redactor-inline-converted">X(GH) = 50</XXXXXX>.</p><p><XXXXXX class="XXXXXXXX-inline-XXXXXXXXX">Point F:</XXXXXX>XXXXXXXXX,<XXXXXX><u>F(XX) = XX</u></strong></p><p><strong>XXXXX X:</XXXXXX> All that's missing now XX F(CG). We XXXXXX no vertical components XXX countering it, so we conclude XXXX<strong><u>F(CG) = X</u></XXXXXX><br></p>">
