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

$\theta=23.6^{\circ}$
We can determine the required angle $\theta$ as follows: $F_s=k(sin\theta-sin0)=6sin\theta$ The virtual displacements are given as $\delta_{yA}=\frac{d(3sin\theta)}{d\theta}=3cos\theta$ and $\delta_{yD}=\frac{d(1sin\theta)}{d\theta}=1cos\theta$ Now, the virtual-work equation is given as $\delta U=0$ $\implies P\delta_{yA}+F_x\delta_{yC}=0$ $\implies 0.8(3cos\theta)-6sin\theta cos\theta=0$ This simplifies to: $\theta=23.6^{\circ}$