## Engineering Mechanics: Statics & Dynamics (14th Edition)

$F_{BC}=434lb$ $\phi=33.5^{\circ}$
We are asked to determine the force in member BC and the required angle $\phi$ Given that the force on member AB is 650lb and the force F is 850lb, we can apply the law of cosines to find force in member BC, $F_{BC}$ $c=\sqrt{a^2+b^2-2*a*b*\cos(C)}$ $c=\sqrt{850^2+650^2-2*850*650*\cos(30^{\circ})}$ $c=433.64$ $F_{BC}=434lb$ Using this result, we can apply the law of sines to find the$\phi$ $\sin(45^{\circ}+\phi) / 850 = \sin 30^{\circ} / 433.64$ Solving for $\phi$, we obtain: $\phi=33.5^{\circ}$