Question

Two masses
(m1
and
m2)
are attached to the ends of a cord that passes over a pulley as shown in the figure. If
m1
has a mass of 0.560 kg,
m2
has a mass of 0.379 kg, and the pulley has a radius of
R = 9.50 cm,
determine the magnitude of the frictional torque that must be exerted by the axle if the system is to be in static equilibrium.

Answer 1

For static equilibrium the tensions equal the weights: T1 = m1 g and T2 = m2 g. The frictional torque must balance the net rope torque on the pulley:
τ = (T1 − T2) R = (m1 − m2) g R.

Insert numbers (g = 9.80 m/s^2, R = 0.0950 m):
τ = (0.560 − 0.379)(9.80)(0.0950) = 0.169 N·m (approximately).

So the required frictional torque magnitude is about 1.69×10^−1 N·m.