Answer
$k=166\frac{N}{m}$
Work Step by Step
We can determine the required stiffness of the spring as follows:
$\Sigma M_B=0$
$\implies 2R_A(150)-5(9.81)\frac{150+250}{2}-1(9.81)(400)=0$
$\implies R_A=45.78N$
and $\Sigma Y_i=0$
$\implies R_c(150)+5(9.81)+1(9.81)-45.78=0$
$\implies R_C=16.35N$
Now $F_s=k(250cos 45^{\circ}-250cos 90^{\circ})=176.77k$
The virtual displacements are given as
$\delta_y=\frac{d(250sin\theta)}{d\theta}=250cos\theta$
$\delta_{xC}=\frac{d(250cos\theta)}{d\theta}=-250sin\theta$
The virtual-work equation is
$\delta_U=0$
$\implies R_A\delta_{yA}-R_c\delta_{yC}+F_s\delta_{xC}=0$
We plug in the known values to obtain:
$45(75)(250) cos 45-16.35(250)cos45-176.77k(250) sin 45=0$
This simplifies to:
$k=166\frac{N}{m}$
