Frictional Torque for Static Equilibrium of a Pulley

[BraedenP]

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

$$\tau=rFsin\theta$$

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:

$$\tau_{net} \approx 0.52 - 0.26 \approx 0.26$$

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

Thanks!

 

[Doc Al]

Your method sounds good to me. What's the radius of the pulley?

 

[BraedenP]

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.

 

[Doc Al]

Oh, sorry. The radius is 9.5cm, so I just converted to m and solved.

OK.

$$\tau_{net} \approx 0.52 - 0.26 \approx 0.26$$

Show how you made those calculations.

 

[BraedenP]

OK.


Show how you made those calculations.

$$\tau=rFsin\theta$$

$$\tau=(0.095m)((0.311 \cdot 9.81)sin(90))\approx 0.26 N \cdot m$$

$$\tau=(0.095m)((0.625 \cdot 9.81)sin(90))\approx 0.52 N \cdot m$$

 

[Doc Al]

Redo your calculations.

 

[BraedenP]

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!

 

[Doc Al]

Your formula is fine. But your arithmetic is a bit off. Plug those number into your calculator once more.

 

[BraedenP]

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?

 

[Doc Al]

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.

 

[BraedenP]

Great! Thanks for your help. :)