Zero Torque and Static Equilibrium pulley problem

So, in summary, the question is asking for the magnitude of the frictional torque required by the axle to maintain static equilibrium in a system where a 0.301kg mass and a 0.635kg mass are attached to opposite ends of a pulley with friction in its axle. The torque equation to use is Fr sin(theta), with the assumption that both masses hang down at a 90 degree angle.
  • #1
just.karl
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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
 
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  • #2
just.karl said:
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)
 
  • #3
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.
 

1. What is zero torque in a pulley system?

Zero torque in a pulley system means that the forces acting on the pulley are balanced, resulting in no rotational movement. This is achieved when the net torque, which is the product of the force and the distance from the axis of rotation, is equal to zero.

2. How is static equilibrium achieved in a pulley problem?

Static equilibrium is achieved in a pulley problem when the system is at rest and all forces acting on the pulley are balanced. This means that the sum of all the forces in the system is equal to zero and the net torque is also equal to zero.

3. What is the role of the pulley in a zero torque and static equilibrium problem?

The pulley serves as a mechanical advantage in a zero torque and static equilibrium problem. It helps to distribute the force required to lift an object, making it easier to lift heavy objects with less effort.

4. How does the number of pulleys affect the zero torque and static equilibrium problem?

The more pulleys there are in a system, the easier it is to achieve zero torque and static equilibrium. This is because each additional pulley reduces the force needed to lift an object, resulting in a more balanced system.

5. What are some real-life applications of zero torque and static equilibrium pulley problems?

Zero torque and static equilibrium pulley problems are used in various real-life applications, such as cranes and elevators. These systems use pulleys to distribute the force required to lift heavy objects, making it easier and more efficient to perform tasks.

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