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Two forces act on the hook. Determine the magnitude of the resultant force.

 

 

\( F_{R}=\sqrt{(200)^{2}+(500)^{2}-2(200)(50) \cos 140} \)           

\( =666 N \)


If \(\theta=60^{\circ}\)  and \( T=5 \mathrm{kN}\), determine the magnitude of the resultant force acting on the eyebolt and its direction measured clockmise from the positive  \(x\) axis

 

\( F_{R}=\sqrt{(5)^{2}+(8)^{2}-2(5)(8)\left(05 | 05^{\circ}\right.} \)

\( =10.47 \mathrm{kN} \)

 

 

\( \frac{\sin \phi}{5}=\frac{\sin 105}{10.47} \)

\( \rightarrow \phi=27.5^{\circ} \)

\( ∴ \alpha=45^{0}-27 \cdot 5=17.5^{\circ} \)


Determine the angle of \(\theta\) for connecting member \(A\) to the plate so that the resultant force of \(\mathrm{F}_{A}\) and \(\mathrm{F}_{B}\) is directed horizontally to the right. Also, what is the magnitude of the resultant force?

 

\( \frac{6}{\sin \phi}=\frac{8}{\sin 50} \)

\( ∴ \phi=35.1^{\circ} \)

\( ∴ \theta=90-35.1=54.9^{\circ} \)

\( \frac{F_{R}}{\sin 94.9}=\frac{8}{\sin 50^{\circ}} \rightarrow F_{R}=10.4 k N \)

 


 

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