Engineering Mechanics: Statics & Dynamics (14th Edition)

$\theta=72.9^{\circ}$ stable equilibrium, $\theta=0^{\circ}$ unstable equilibrium
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.