# Why does mechanical equilibrium depend only on potential energy?

As far as I understand, for a system to be considered in equilibrium, the sum of the forces that is applied to it must be $0$:

$\vec F = 0$

which is

$\partial \frac{E_p}{\partial x}\bigg\rvert _{x=x_0} = 0$

But it only depends on the potential energy, what about kinetic energy?

• The derivative of kinetic energy with respect to position is zero anyway... After all, kinetic energy only depends on velocity. – Physicist137 Jun 7 '17 at 1:25

Equilibrium here just means that if a particle is placed at the location $x_0 = 0$ with $0$ kinetic energy, it will stay at that location. Imagine another location, say $x_1$, such that $\nabla V(x_1)\not = 0$, if a particle is placed there, it will start moving, if the kinetic energy is zero initially.
A simple example of this is a pendulum. Imagine a bar with ends points labeled $A$ and $B$ suspended from one of its end-points ($A$). If the bar is placed vertically with $B$ lower than $A$, the bar will stay at that configuration forever, unless you add kinetic energy. This is also the case if you place $B$ vertically above $A$. The difference between these two configurations is the type of equilibrium (stable vs unstable), but both of them are equilibrium points.