A string that passes over a pulley has a 0.321 kg mass attached to one end and a 0.655 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?
To find the magnitude of the frictional torque exerted by the axle, we need to consider the torques acting on the pulley due to the two masses.
1. Calculate the gravitational forces acting on the two masses:
- Gravitational force on the first mass (m1 = 0.321 kg): F1 = m1 * g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
- Gravitational force on the second mass (m2 = 0.655 kg): F2 = m2 * g.
2. Determine the tensions in the string:
- Tension on the side of mass m1: T1 = F1.
- Tension on the side of mass m2: T2 = F2.
3. Calculate the torques acting on the pulley:
- Torque due to mass m1: τ1 = T1 * r, where r is the radius of the pulley (9.50 cm = 0.095 m).
- Torque due to mass m2: τ2 = T2 * r.
4. The system is in static equilibrium, so the sum of the torques must be zero:
τ1 + τ2 = 0.
5. Solve for the frictional torque exerted by the axle:
τ_friction = - (τ1 + τ2).
Let's now calculate the magnitude of the frictional torque step-by-step:
Step 1: Calculate the gravitational forces:
F1 = 0.321 kg * 9.8 m/s^2 = 3.1488 N
F2 = 0.655 kg * 9.8 m/s^2 = 6.419 N
Step 2: Determine the tensions in the string:
T1 = F1 = 3.1488 N
T2 = F2 = 6.419 N
Step 3: Calculate the torques:
τ1 = T1 * r = 3.1488 N * 0.095 m = 0.2990416 N·m
τ2 = T2 * r = 6.419 N * 0.095 m = 0.609905 N·m
Step 4: Sum of torques for static equilibrium:
τ1 + τ2 = 0.2990416 N·m + 0.609905 N·m = 0.9089466 N·m
Step 5: Calculate the magnitude of the frictional torque:
τ_friction = - (τ1 + τ2) = -0.9089466 N·m
Therefore, the magnitude of the frictional torque exerted by the axle is approximately 0.9089 N·m.
To find the magnitude of the frictional torque exerted by the axle, we need to analyze the forces and torques acting on the system.
First, let's consider the forces acting on each mass:
For the 0.321 kg mass:
- The weight (mg) acts downwards with a magnitude of (0.321 kg)(9.8 m/s^2) = 3.1482 N.
For the 0.655 kg mass:
- The weight (mg) acts downwards with a magnitude of (0.655 kg)(9.8 m/s^2) = 6.419 N.
Next, let's consider the forces and torques acting on the pulley:
- The tension in the string is the same on both sides of the pulley, as the string does not stretch. Let's call this tension T.
- The tension T exerts a radial force on the pulley which causes a torque. Since the pulley has friction in its axle, the frictional torque must counteract this torque to keep the system in static equilibrium. Let's call the magnitude of the frictional torque Tf.
- The torque due to the weight of the masses can be calculated using the formula:
Torque = force × lever arm
The lever arm is the distance from the center of the pulley to where the force is applied, which is the radius of the pulley (9.50 cm = 0.095 m).
The torque due to the weight on the 0.321 kg mass is:
T1 = (3.1482 N) × (0.095 m) = 0.299073 N·m
The torque due to the weight on the 0.655 kg mass is:
T2 = (6.419 N) × (0.095 m) = 0.609405 N·m
For the system to be in static equilibrium, the sum of the torques acting on the pulley must be zero:
Tf - T1 - T2 = 0
Rearranging the equation:
Tf = T1 + T2 = 0.299073 N·m + 0.609405 N·m = 0.908478 N·m
Therefore, the magnitude of the frictional torque that must be exerted by the axle is approximately 0.908478 N·m.