Frictional Torque for Static Equilibrium of a Pulley

In summary, the homework statement states that a mass of 0.311kg is attached to one end of a string, and a mass of 0.625kg is handing off the other end. The string goes over a pulley which has friction in its axle. To maintain static equilibrium, the torque required is 0.52 - 0.26 Nm.
  • #1
BraedenP
96
0

Homework Statement


A mass of 0.311kg is attached to one end of a string, and a mass of 0.625kg is handing off the other end. The string goes over a pulley which has friction in its axle.

What is the magnitude of frictional torque required to maintain static equilibrium?


Homework Equations



[tex]\tau=rFsin\theta[/tex]

The Attempt at a Solution



I figured that I'd just have to find the forces acted on the pulley by both masses, calculate the resulting torques created by both of them, and subtract them to find the torque required to "cancel" the net torque in the system.

Thus:

[tex]\tau_{net} \approx 0.52 - 0.26 \approx 0.26[/tex]

My answer wasn't right.. Am I getting the concept right and just making a calculation mistake? Or is my method completely wrong?

Thanks!
 
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  • #2
Your method sounds good to me. What's the radius of the pulley?
 
  • #3
Doc Al said:
Your method sounds good to me. What's the radius of the pulley?

Oh, sorry. The radius is 9.5cm, so I just converted to m and solved.
 
  • #4
BraedenP said:
Oh, sorry. The radius is 9.5cm, so I just converted to m and solved.
OK.

BraedenP said:
[tex]\tau_{net} \approx 0.52 - 0.26 \approx 0.26[/tex]
Show how you made those calculations.
 
  • #5
Doc Al said:
OK.


Show how you made those calculations.

[tex]\tau=rFsin\theta[/tex]

[tex]\tau=(0.095m)((0.311 \cdot 9.81)sin(90))\approx 0.26 N \cdot m[/tex]

[tex]\tau=(0.095m)((0.625 \cdot 9.81)sin(90))\approx 0.52 N \cdot m[/tex]
 
  • #6
Redo your calculations.
 
  • #7
Hmm, I'm not getting it. I'm using the right formula, aren't I? And I'm pretty sure the angle is 90, since the force is directed straight down.

And I really hope I didn't make a mistake in the force/radius, because those are just simple!
 
  • #8
Your formula is fine. But your arithmetic is a bit off. Plug those number into your calculator once more.
 
  • #9
Oh.. Ugh! I had it in Radian mode the whole time from my Calculus homework..

So those should be 0.29 and 0.58 respectively, right?
 
  • #10
BraedenP said:
Oh.. Ugh! I had it in Radian mode the whole time from my Calculus homework..
D'oh! But you should know that sin90° = 1 without using a calculator!

So those should be 0.29 and 0.58 respectively, right?
Much better.
 
  • #11
Great! Thanks for your help. :)
 

FAQ: Frictional Torque for Static Equilibrium of a Pulley

What is frictional torque?

Frictional torque is the rotational force that arises due to the friction between two surfaces in contact. It is caused by the resistance between the surfaces and is always in the opposite direction to the motion of the surfaces.

How is frictional torque calculated?

To calculate frictional torque, you need to know the coefficient of friction between the two surfaces, the normal force acting on the surfaces, and the radius of the pulley. The equation for frictional torque is T = μF*r, where μ is the coefficient of friction, F is the normal force, and r is the radius of the pulley.

What is static equilibrium?

Static equilibrium is a state in which an object is at rest and not moving. In this state, the forces acting on the object are balanced, meaning that the net force and net torque on the object are both equal to zero.

Why is frictional torque important in the static equilibrium of a pulley?

Frictional torque is important in the static equilibrium of a pulley because it is one of the forces that needs to be balanced for the pulley to remain in a state of rest. If the frictional torque is greater than the applied torque, the pulley will not be in equilibrium and will start to rotate.

How does the direction of frictional torque affect the static equilibrium of a pulley?

The direction of frictional torque is important in the static equilibrium of a pulley because it determines the direction in which the pulley will rotate. If the frictional torque is in the opposite direction to the applied torque, the pulley will experience a net torque of zero and remain in static equilibrium. However, if the frictional torque is in the same direction as the applied torque, the pulley will start to rotate and will not be in equilibrium.

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