A fixed pulley carries a weightless thread with masses `m_(2) and m_(1)` at its ends `(m_(2) gt m_(1))`. There is friction between the thread and the pulley. If it is such that the thread starts slipping when the ratio of `m_(2)` to `m_(1)` is `n_(0)`. Find :
(a) the friction coefficient
(b) the acceleration of the masses when `(m_(2))/(m_(1))=n gt n_(0)`
[Hint : Consider an element of the thread of width `d theta` at angular distance `theta `from the horzontal diameter of the pulley. The forces at the ends are T and T = dT. If dN is the normal reaction of the pulley on the element than df (frictional force on the element) `= mu dN` Resolving the tensions at the ends along and perpendicular to the radius vector of the element, we have for equilibrium of the element `dT = df and dN = T d theta` Integrate from `T_(1)` to `T_(2).T_(1)=m_(1)g and T_(2)=m_(2)g`. These are obtained by considering equilibrium of `m_(1) and m_(2)`]

Answer

`mu-(ln n_(0))/(pi)a=(g(n-n_(0)))/((n+n_(0)))`