# Thread: Coefficient of friction problem

1. ## Coefficient of friction problem

Another problem from an old UK A-Level textbook that I'm stuck with. Any advice gratefully received.

In the diagram BAC is a rigid fixed rough wire and angle BAC is 60°. P and Q are two identical rings of mass m connected by a light elastic string of natural length 2a and modulus of elasticity mg. If P and Q are in equilibrium when PA = AQ = 3a, find the least coefficient of friction between the rings and the wire.

I have made a crude attempt at a diagram of what I think is going on and have marked the forces I believe are acting on the two rings.

Each ring is of mass m, so there is a force mg acting due to gravity vertically downward at each of P and Q. The tension in the string is the same at both P and Q and acts upwards parallel to the wire. There is a frictional force inhibiting motion acting in the opposite direction to the tension at both P and Q, with magnitude μ times the normal reaction at each point, where μ is the coefficient of friction. The normal reactions at P and Q, acting perpendicular to the wire, are S and R, respectively.

Considering the mass at P and resolving forces parallel and perpendicular to the wire, I get

T = μS + mg sin 60° i.e. T = μS + ½mg√3
and S = mg cos 60° i.e. S = ½mg

Then T = ½mg(√3 + μ)

Using Hooke's Law, T = mg(4a/2a) = 2mg, so

2mg = ½mg(√3 + μ)
so μ = 4 - √3

I'm sure I'm missing something here, as I have not considered the other ring at Q, and, moreover, the book gives the answer as μ = 5√3 - 8, which is obviously completely different to my answer.