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

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?

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

\( ∴ \phi=35.1^{\circ} \)

\( ∴ \theta=90-35.1=54.9^{\circ} \)

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

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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