Answer
$\theta=12.1^{\circ}$, unstable equilibrium
Work Step by Step
We can determine the required angle $\theta$ as follows:
According to the potential energy equation
$V=V_{g,bar}-V_{g,block}=2W_{bar}y_1-W_{block}y_2$
$\implies V=2(3)(250)sin\theta-7(1500-1000cos\theta)$
At equilibrium $\frac{dV}{d\theta}=0$
$\implies \frac{d[2(3)(250)sin\theta-7(1500-1000cos\theta)]}{d\theta}=1500cos\theta-700sin\theta=0$
This simplifies to:
$\theta=12.095^{\circ}\approx12.1^{\circ}$
Now, $\frac{d^2V}{d\theta^2}=-1500sin\theta-700cos\theta$
For $\theta=12.095^{\circ}$
$\implies \frac{d^2V}{d\theta^2}=-1500sin(12.095^{\circ})-700cos(12.095^{\circ})=-7158.9\lt 0$
which implies unstable equilibrium.
