# Question-134341

Question Number 134341 by Eric002 last updated on 02/Mar/21
Answered by mr W last updated on 02/Mar/21
Commented by mr W last updated on 02/Mar/21
$${Q}\mathrm{1} \\$$$${M}=\mathrm{200}×\mathrm{1}+\mathrm{800}×\mathrm{sin}\:\mathrm{70}°×\left(\mathrm{3}+\mathrm{1}\right)+\mathrm{800}×\mathrm{cos}\:\mathrm{70}°×\mathrm{2} \\$$$$\approx\mathrm{2798}\:{Nm} \\$$$$\\$$$${Q}\mathrm{2} \\$$$$\mathrm{tan}\:\alpha=\frac{\mathrm{10}}{\mathrm{5}}=\mathrm{2}\:\Rightarrow\mathrm{sin}\:\alpha=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}},\:\mathrm{cos}\:\alpha=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\$$$$\mathrm{tan}\:\beta=\frac{\mathrm{4}}{\mathrm{3}}\:\Rightarrow\mathrm{sin}\:\beta=\frac{\mathrm{4}}{\mathrm{5}} \\$$$${F}=\mathrm{200}\:{lb} \\$$$${F}_{\mathrm{1}} ={component}\:{along}\:{AB} \\$$$${F}_{\mathrm{2}} ={component}\:{parallel}\:{to}\:{CD} \\$$$$\frac{{F}_{\mathrm{1}} }{\mathrm{sin}\:\left(\alpha+\beta\right)}=\frac{{F}_{\mathrm{2}} }{\mathrm{sin}\:\alpha}=\frac{{F}}{\mathrm{sin}\:\beta} \\$$$${F}_{\mathrm{1}} =\frac{\mathrm{sin}\:\left(\alpha+\beta\right)}{\mathrm{sin}\:\beta}×{F}=\left(\frac{\mathrm{sin}\:\alpha}{\mathrm{tan}\:\beta}+\mathrm{cos}\:\alpha\right)×{F} \\$$$$=\left(\frac{\mathrm{2}×\mathrm{3}}{\:\sqrt{\mathrm{5}}×\mathrm{4}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\right)×\mathrm{200}=\mathrm{100}\sqrt{\mathrm{5}}=\mathrm{223}.\mathrm{6}\:{lb} \\$$$${F}_{\mathrm{2}} =\frac{\mathrm{sin}\:\alpha}{\mathrm{sin}\:\beta}×{F} \\$$$$=\frac{\mathrm{2}×\mathrm{5}}{\:\sqrt{\mathrm{5}}×\mathrm{4}}×\mathrm{200}=\mathrm{100}\sqrt{\mathrm{5}}=\mathrm{223}.\mathrm{6}\:{lb} \\$$
Commented by Eric002 last updated on 02/Mar/21
$${well}\:{done}\:{sir} \\$$
Commented by ajfour last updated on 02/Mar/21
$${Indeed}. \\$$
Commented by mr W last updated on 05/Mar/21
$${can}\:{you}\:{consider}\:{solving}\:{Q}\mathrm{134376}\:{sir}? \\$$
Commented by ajfour last updated on 05/Mar/21
$${sure}\:{i}\:{shall}\:{try}.. \\$$
Commented by mr W last updated on 07/Mar/21
$${i}\:{have}\:{worked}\:{out}\:{a}\:{solution},\:{but}\:{i}'{m} \\$$$${not}\:{quite}\:{sure}\:{if}\:{it}'{s}\:{really}\:{correct}. \\$$$${please}\:{review}. \\$$