A 1 kg mass and a 0.7 kg mass are attached to the two ends of a string that goes over a pulley with a radius of 0.1 m. Because of friction, the pulley does not begin to rotate. What is the magnitude of the frictional torque on the bearing of the pulley if the system is in static equilibrium?
A)0.03 N m
B)0.29 N m
C)0.58 N m
D)2.90 N m

Question

Quizplus · Accepted Answer

The system is in static equilibrium, meaning the net force and net torque on the system are zero. The torque due to friction must balance the torque due to the weight of the masses. The torque ($τ$) is given by $τ = rF$, where $r$ is the radius of the pulley and $F$ is the force applied. The force due to each mass is its weight ($mg$), where $m$ is the mass and $g$ is the acceleration due to gravity (approximately $9.8 m/s^2$). The difference in weight between the two masses provides the net force causing the torque. For the 1 kg mass, the force is $1 kg \times 9.8 m/s^2 = 9.8 N$, and for the 0.7 kg mass, it's $0.7 kg \times 9.8 m/s^2 = 6.86 N$. The difference is $9.8 N - 6.86 N = 2.94 N$. The torque is then $0.1 m \times 2.94 N = 0.294 Nm$, which rounds to $0.29 Nm$, making option B correct.

A 1 Kg Mass and a 0