A string that passes over a pulley has a 0.375 kg mass attached to one end and a 0.630 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?
How do you set up friction in the equation. Also, which direction would friction go?
To set up the equation for frictional torque, you need to understand a couple of concepts. Frictional torque is the torque exerted by a rotating object due to the friction between its surfaces. In this case, the frictional torque is exerted by the axle of the pulley.
To find the frictional torque, you first need to determine the force of static friction between the axle and the pulley. The force of static friction acts tangentially along the surface of the axle in the direction that opposes the rotation. In this case, we can assume that the rotation will be counterclockwise.
The maximum force of static friction (F_friction) can be calculated using the equation:
F_friction = μ * F_normal
Where:
- μ is the coefficient of static friction
- F_normal is the normal force between the axle and the pulley
Since the system is in static equilibrium, the sum of all torques must be zero. The torque due to friction (τ_friction) is given by:
τ_friction = F_friction * r
Where:
- r is the radius of the pulley (in this case, 9.50 cm or 0.095 m)
Thus, the magnitude of the frictional torque exerted by the axle can be found by substituting the earlier equation for F_friction:
τ_friction = (μ * F_normal) * r
Now, you need to determine the normal force, F_normal. In this case, it equals the weight of the object attached to the string, since the system is in equilibrium.
F_normal = m * g
Where:
- m is the mass of the object
- g is the acceleration due to gravity (approximately 9.8 m/s²)
Substituting this into the equation for τ_friction, we get:
τ_friction = (μ * m * g) * r
Remember that the direction of friction is opposite to the rotation, so the frictional torque will be exerted clockwise in this case.