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

$P=\frac{W}{2}cot\theta$
We can determine the magnitude of the required force $P$ as follows: The virtual displacements are given as $\delta_y=\frac{d(0.5lsin\theta)}{d\theta}=0.5lsin\theta$ and $\delta_{xC}=\frac{d(lsin\theta)}{d\theta}=-lsin\theta$ Now, according to the virtual-work equation $\delta U=0$ $\implies W\cdot\delta_y+P\cdot \delta x=0$ $\implies W(0.5)lcos\theta+P(-lsin\theta)=0$ This simplifies to: $P=\frac{W}{2}cot\theta$