What is the angle of the string with the vertical?

A heavy uniform sphere of radius $a$ has a light inextensible string attached to a point on its surface. The other end of the string is fixed to a point on a rough vertical wall. The sphere rests in equilibrium touching the wall at a point distant $h$ below the fixed point.

If the point of the sphere in contact with the wall is about to slip downwards and the coefficient of friction between the sphere and the wall is $\mu$ find the inclination of the string to the vertical.

If $\mu=h/(2a)$ and the weight of the sphere is $W$, show that the tension in the string is $$\frac{(1+\mu^2)^{1/2}}{2\mu}W$$

I am trying to solve the problem and I have no idea how to get to the answer that's at the back of the book. Apparently the angle is $\arctan{(a/(h-a\mu))}$ and I cannot see why the friction coefficient is involved at all. As far as I can see, the centre of the sphere, and the ends of the string should be collinear, which would give the tangent of the angle $a/h$. Could someone please help me?

• If you imagine a tennis ball attached to a wall by a string. Pull the ball away from the wall, keeping the string and the centre of the ball collinear. Now swing the ball to the wall and let it rest, keeping colinearity. That's what you're describing, but that's not what the question is asking. The ball in the question is "about to slip", so take your tennis ball and rotate it up a bit on the wall, keeping the string taut. It will stay. Rotate it up a bit more. Keep doing this until the ball won't stay there - it will slip down. This is the point they're asking about. – twistor59 Dec 17 '12 at 21:04
• I'm guessing that $a$ is the radius of the spehere, and $h$ the length of the string? It would be nice if you edited your question with that information... – Jaime Dec 17 '12 at 21:32
• Always start with a Free Body Diagram. Once you have this, the solution is at hand. – ja72 Dec 18 '12 at 10:49
• Thank you so much for your help! Twistor59, the way you explained the problem was really helpful! Hwlau, thank you for editing my question! – user16989 Dec 19 '12 at 12:52