# Questions about potential energy and its corresponding force

The question arises when working at the following problem:

A light wire is bent into a right angle and a heavy ball is attached to the bend. The wire is placed onto supports with height difference $$h$$ and horizontal distance $$a$$. Find the position of the wire in its equilibrium. Express the position as the angle between the bisector of the right angle and the vertical. Neglect any friction between the wire and the supports;the supports have little grooves keeping all motion in the plane of the wire and the figure.

So it is not a hard problem. The solutions is as follows:

Since the wire always forms a right angle the ball is constrained to move in a circle whose diameter is the line that connects the points where the wire is contact with the support.

Then if it is in equilibrium its potential energy is minimum. That happens in the lowest point of the circle and it is solved.

But there is something I do not understand about that last idea. I thought that you could apply that only if all forces acting on the body were conservative, if that is not the case the potential would be minimum when the conservative forces are $$0$$ and since non-conservative forces are acting the body would not be in equilibrium. And here you have the forces exerted by the parts of the wire so you wouldn't be able to apply it.

I know that solution can't be wrong but I don't know what's wrong in what I think, so can you tell what is confusing me?

• But then if have $- \Delta U = \int F \cdot dr$, that $F$ in this case is equal to $m \vec g$ and at the point where the potential is minimum that $F$ would be $0$, but since there are other forces acting it would not be in equilibrium. That is what I mean. – Vmimi Feb 17 at 18:52
• If there are other forces for which a potential function exists, such as elastic force in a spring, then the position of equilibrium could be different. The total force acting on the heavy ball will not be $mg$ in this case. – sammy gerbil Feb 17 at 18:59
• That $F$ is not the force acting on the heavyball. $F$ would be the forces for which a potential function is known which in this case is just the weight, otherwise you would not be able to write that equality. – Vmimi Feb 17 at 19:01