## Engineering Mechanics: Statics & Dynamics (14th Edition)

Stable equilibrium at $x=\frac{1}{6}$.
We can determine the required equilibrium positions along with stability at these positions as follows: As given that, $V=8x^3-2x^2-10$ At equilibrium, we have $\frac{dV}{dx}=0$ $\implies \frac{d(8x^3-2x^2-10)}{dx}=24x^2-4x=0$ eq(1) This simplifies to: $x_1=0$, $x_2=\frac{1}{6}$ We take the second derivative of eq(1) $\frac{d^2V}{dx^2}=48x-4$ for $x=\frac{1}{6}$, the above equation becomes $\frac{d^2V}{dx^2}=48(\frac{1}{6})-4=4\gt 0$ which means stable equilibrium. for $x=0$, we get: $\frac{d^2V}{dx^2}=48(0)-4=-4\lt 0$, unstable