A string that passes over a pulley has a 0.352 kg mass attached to one end and a 0.683 kg mass attached to the other end. The pulley, which is a disk of radius 9.17 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?

To solve this problem, we need to understand that in static equilibrium, the sum of all forces and the sum of all torques must be zero.

1. First, we need to calculate the forces acting on the system. The force due to gravity on each mass is given by F = mg, where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s^2).

   For the 0.352 kg mass: F1 = 0.352 kg * 9.8 m/s^2 = 3.45 N For the 0.683 kg mass: F2 = 0.683 kg * 9.8 m/s^2 = 6.69 N

2. Next, we need to calculate the torques acting on the system. Torque is given by τ = rFsinθ, where r is the radius, F is the force, and θ is the angle between the force and the radius. In this case, the string is pulling directly downward, so θ = 90 degrees and sinθ = 1.

   The radius needs to be in meters, so convert 9.17 cm to meters: r = 9.17 cm * (1 m/100 cm) = 0.0917 m

   For the 0.352 kg mass: τ1 = 0.0917 m * 3.45 N * 1 = 0.316 Nm For the 0.683 kg mass: τ2 = 0.0917 m * 6.69 N * 1 = 0.613 Nm

3. In static equilibrium, the sum of all torques must be zero, so the frictional torque must balance out the torques due to the two masses.

   The frictional torque is therefore: τ_friction = τ2 - τ1 = 0.613 Nm - 0.316 Nm = 0.297 Nm

So, the magnitude of the frictional torque that must be exerted by the axle for the system to be in static equilibrium is 0.297 Nm.