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

$\theta=41.2^{\circ}$
We can determine the required angle $\theta$ as follows: $F_s=k(600sin\theta-600sin15)-(200sin\theta-200sin15)$ $\implies F_s=6000(sin\theta-sin15)$ The virtual displacements are given as $\delta_{yA}=\frac{d(200sin\theta)}{d\theta}=200cos\theta$ $\delta_{yB}=\frac{d(600sin\theta)}{d\theta}=200cos\theta$ and $\delta_{yC}=\frac{d(1600sin\theta)}{d\theta}=1600cos\theta$ Now, according to the virtual-work equation $\delta U=0$ $\implies P\delta_{yC}+F_s(\delta_{yB}-\delta_{yB})=0$ We plug in the known values to obtain: $600(1600cos\theta)-6000(sin\theta-sin15)(600cos\theta-200cos\theta)=0$ This simplifies to: $\theta=41.2^{\circ}$