91Ó°ÊÓ

Force \(\mathbf{F}\) acts on the frame such that its component acting along member \(A B\) is \(650 \mathrm{lb}\), directed from \(B\) towards \(A\), and the component acting along member \(B C\) is 500 lb, directed from \(B\) towards \(C\).Determine the magnitude of \(\mathbf{F}\) and its direction \(\theta .\) Set \(\phi=60^{\circ}\).

Force \(\mathbf{F}\) acts on the frame such that its component acting along member \(A B\) is \(650 \mathrm{lb}\), directed from \(B\) towards \(A\). Determine the required angle \(\phi\left(0^{\circ} \leq \phi \leq 45^{\circ}\right)\) and the component acting along member \(B C\). Set \(F=850 \mathrm{lb}\) and \(\theta=30^{\circ}\).

Two forces act on the screw eye. If \(F_{1}=400 \mathrm{~N}\) and \(F_{2}=600 \mathrm{~N}\), determine the angle \(\theta\left(0^{\circ} \leq \theta \leq 180^{\circ}\right)\) between them, so that the resultant force has a magnitude of \(F_{R}=800 \mathrm{~N}\)

Two forces \(\mathbf{F}_{1}\) and \(\mathbf{F}_{2}\) act on the screw eye. If their lines of action are at an angle \(\theta\) apart and the magnitude of each force is \(F_{1}=\mathrm{F}_{2}=F\), determine the magnitude of the resultant force \(\mathbf{F}_{R}\) and the angle between \(\mathbf{F}_{R}\) and \(\mathbf{F}_{1}\).

Given the three vectors \(\mathbf{A}, \mathbf{B}\), and \(\mathbf{D}\), show that \(\mathbf{A} \cdot(\mathbf{B}+\mathbf{D})=(\mathbf{A} \cdot \mathbf{B})+(\mathbf{A} \cdot \mathbf{D})\)