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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

$$Q1$$$$M=200\times 1+800\times \sin 70°\times (3+1)+800\times \cos 70°\times 2$$$$\approx 2798Nm$$$$$$$$Q2$$$$\tan \alpha =\frac{10}{5}=2\Rightarrow \sin \alpha =\frac{2}{\sqrt{5}},\cos \alpha =\frac{1}{\sqrt{5}}$$$$\tan \beta =\frac{4}{3}\Rightarrow \sin \beta =\frac{4}{5}$$$$F=200lb$$$$F_1=componentalongAB$$$$F_2=componentparalleltoCD$$$$\frac{F_1}{\sin (\alpha +\beta )}=\frac{F_2}{\sin \alpha }=\frac{F}{\sin \beta }$$$$F_1=\frac{\sin (\alpha +\beta )}{\sin \beta }\times F=(\frac{\sin \alpha }{\tan \beta }+\cos \alpha )\times F$$$$=(\frac{2\times 3}{\sqrt{5}\times 4}+\frac{1}{\sqrt{5}})\times 200=100\sqrt{5}=223.6lb$$$$F_2=\frac{\sin \alpha }{\sin \beta }\times F$$$$=\frac{2\times 5}{\sqrt{5}\times 4}\times 200=100\sqrt{5}=223.6lb$$

Commented by Eric002 last updated on 02/Mar/21

$$welldonesir$$

Commented by ajfour last updated on 02/Mar/21

$$Indeed.$$

Commented by mr W last updated on 05/Mar/21

$$canyouconsidersolvingQ134376sir?$$

Commented by ajfour last updated on 05/Mar/21

$$sureishalltry..$$

Commented by mr W last updated on 07/Mar/21

$$ihaveworkedoutasolution,but{i}^{\prime }m$$$$notquitesureif{it}^{\prime }sreallycorrect.$$$$pleasereview.$$

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