Zero Torque and Static Equilibrium pulley problem

[just.karl]

A string that passes over a pulley has a .301kg mass attached to one end and a 0.635 kg mass attached to the other end. The pulley, which is a disk of radius 9.50cm, 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?

I know I have to figure out the torque required by the axle to get both tensions equal. But I don't know what equations you use to figure out the torque with two masses on each side.


Thanks

 

[Chi Meson]

I know I have to figure out the torque required by the axle to get both tensions equal.

No, the tensions will not be equal. total torque will be zero. Two different tensions will be applied to the wheel at the given radius. The angles are assumed to be 90 degrees, since both masses will hang down (safe assumption).

torque = Fr sin(theta)

 

[Moetasim]

for your question. In order to solve this problem, we can use the equation for torque, which is T = rFsinθ, where T is torque, r is the radius of the pulley, F is the force applied, and θ is the angle between the force and the lever arm. In this case, the lever arm is the radius of the pulley and the force is the tension in the string.

To find the tension in the string, we can use the fact that the system is in static equilibrium, meaning that the net force and net torque must be equal to zero. This means that the tension on both sides of the pulley must be equal. We can set up an equation for the tension on the side with the 0.301kg mass: T1 = m1g. Similarly, for the other side with the 0.635kg mass, we have T2 = m2g.

Since the tensions must be equal, we can set T1 equal to T2 and solve for the tension, which is T = m1g = m2g. Plugging this into the torque equation, we get T = r(m1g)sinθ = r(m2g)sinθ. Since r and sinθ are the same for both sides, we can cancel them out, leaving us with m1g = m2g. This means that the frictional torque required by the axle must be equal to the difference in the masses multiplied by the gravitational acceleration, or Tfriction = (m2 - m1)g.

In this case, we have m1 = 0.301kg and m2 = 0.635kg, so the frictional torque required by the axle is (0.635kg - 0.301kg)(9.8m/s^2) = 3.34 Nm. This means that in order for the system to be in static equilibrium, the axle must exert a frictional torque of 3.34 Nm.