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?
