A string that passes over a pulley has a 0.373 kg mass attached to one end and a 0.620 kg mass attached to the other end. The pulley, which is a disk of radius 9.50 cm, has friction in its axle. What is the magnitude of the frictional torque that must be exerted by the axle if the system is to be in static equilibrium?

Answer 1

To prevent the pulley from turning, the frictional torque must equal the difference of the weight-applied torques.

Tf = (0.620 - 0.373)kg*9.8 m/s^2*0.095 m
= 0.230 kg*m^2/s^2 (Newton-meter)