Answer
$k=157N/m$ stable at $\theta=60^{\circ}$
Work Step by Step
We can determine the required stiffness of the spring as follows:
According to the potential energy equation
$V=V_g+V_e=W_1y_1+W_2y_2+\frac{1}{2}ks^2$
$\implies V=8(9.81)(0.75)cos\theta+8(9.81)(2.25)cos\theta+\frac{1}{2}k(3cos\theta-1)^2$
$\implies V=235.44cos\theta+\frac{1}{2}k(3cos\theta-1)^2$
At equilibrium $\frac{dV}{d\theta}=0$
$\implies \frac{d(235.44cos\theta+\frac{1}{2}k(3cos\theta-1)^2)}{d\theta}=-235.44sin\theta-3ksin\theta(3cos\theta-1)=0$
It is given that $\theta=60^{\circ}$
$\implies -235.44sin60-3ksin60(3cos60-1)=0$
This simplifies to:
$k=157N/m$
