Answer
$k=2.81lb/ft$
Work Step by Step
We can determine the required stiffness of the springs as follows:
According to the potential energy equation
$V=V_g+V_e=Wy+\frac{1}{2}ks^2$
$\implies V=2(8)sin\theta +\frac{1}{2}k(4cos\theta-1)^2$
At equilibrium $\frac{dV}{d\theta}=0$
$\frac{d[2(8)sin\theta +\frac{1}{2}k(4cos\theta-1)^2)]}{d\theta}=16cos\theta-4ksin\theta(4cos\theta-2)=0$
We plug in $\theta=30^{\circ}$
$\implies 16cos30-4ksin30(4cos30-2)=0$
This simplifies to:
$k=2.81lb/ft$