Answer
$\theta=90^{\circ}$, $\theta=36.1^{\circ}$
Work Step by Step
We can determine the required angle $\theta$ as follows:
$\Sigma M_B=0$
$\implies \implies 2R_A(150-5)(9.81)(\frac{150+250}{2})-1(9.81)(400)=0$
$\implies R_A=45.78N$
and $\Sigma Y_i=0$
$\implies R_C(150)+5(9.81)+1(9.81)-45.78=0$
$\implies R_C=16.35N$
Now, the virtual displacements are given as
$\delta_y=\frac{d(250 sin\theta)}{d\theta}=250 cos\theta$
and $\delta_x=\frac{d(250 cos\theta)}{d\theta}=-250 sin\theta$
According to the virtual-work equation
$\delta U=0$
$\implies R_A\delta_{yA}-R_C\delta_{yC}+F_s\delta x_{c}=0$
We plug in the known values to obtain:
$45(75)(250)cos\theta-16.35(250)cos\theta-50cos\theta(250) sin\theta=0$
This simplifies to:
$\theta=36.1^{\circ}$
