Answer
$\theta=72.9^{\circ}$ stable equilibrium, $\theta=0^{\circ}$ unstable equilibrium
Work Step by Step
The required angle $\theta$ can be determined as follows:
According to the potential energy equation
$V=V_g+V_e=Wy+\frac{1}{2}ks^2$
We plug in the known values to obtain:
$V=20(9.81)(0.75)cos\theta+\frac{1}{2}(2000)(0.5cos\theta)^2$
At equilibrium $\frac{dV}{d\theta}=0$
$\implies \frac{d[20(9.81)(0.75)cos\theta+\frac{1}{2}2000(0.5cos\theta)^2]}{d\theta}=-147.15sin\theta-250sin(2\theta)=0$
This simplifies to:
$\theta_1=72.9^{\circ}$
and $\theta_2=0^{\circ}$
Now $\frac{d^2V}{d\theta^2}=\frac{d[20(9.81)(0.75)cos\theta+\frac{1}{2}(2000)(0.5cos\theta)^2]}{d\theta^2}$
$\implies \frac{d^2V}{d\theta^2}=-147.15cos\theta-500 cos(2\theta)$
For $\theta_1=-147.15cos(72.9)-500cos(2\times 72.9)=456.67\gt 0$
which implies stable equilibrium.
For $\theta_2=0^{\circ}$ $\frac{d^2V}{d\theta^2}=-147.15(cos 0)-500cos(2\cdot 0)=-647.15\lt 0$
which implies unstable equilibrium.
