Answer
$\theta=41.2^{\circ}$
Work Step by Step
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:
$\theta=41.2^{\circ}$
