Statics -- Belt Friction

1. Apr 1, 2015

bnosam

1. The problem statement, all variables and given/known data
http://oi59.tinypic.com/2lacy0o.jpg

2. Relevant equations
P = ?
T2/T1 = e^(u)(θ)
M = 150 N*m

3. The attempt at a solution

T2/T1 = e(.4)(7π/6)

T2/T1 = 5.466

I don't even think I'm on the right track.

2. Apr 1, 2015

Staff: Mentor

You're on the right track.

Chet

3. Apr 1, 2015

bnosam

Nice avatar lol.

I'm not quite sure where to go from here.

4. Apr 1, 2015

Staff: Mentor

Thanks. The avatar is a Mentors' April Fools joke. Hopefully, tomorrow it changes back.

Call T1 = T, and you know T2 in terms of T. So you can now do a moment balance on the flywheel to find T. Then you can do a moment balance of the lever arm to get P.

Chet

5. Apr 1, 2015

bnosam

T = T2/(5.466)

So if I get what you mean moment about O:

75 * T1 + 450 * P = 150

6. Apr 1, 2015

Staff: Mentor

No. Please first show how you get T.

7. Apr 1, 2015

bnosam

T1 = T

T2/T = 5.466
T = T2/5.466
I'm lost after this.

8. Apr 1, 2015

Staff: Mentor

$$(T_2-T_1)0.15=150$$

Chet

9. Apr 1, 2015

bnosam

Where did .15 come from?

10. Apr 1, 2015

Staff: Mentor

The radius of the wheel. We're taking moments around the axis.

Chet

11. Apr 1, 2015

bnosam

I'm not sure why it's the axis and not O. Is it because we're trying to stop the rotation?

12. Apr 1, 2015

Staff: Mentor

We're going to use both. First this.

13. Apr 1, 2015

bnosam

(T2 - T2/5.466)*.15 = 150
T2 = 1223.914 N

14. Apr 2, 2015

Staff: Mentor

Yes. Now, what is T1?

After that, you can do a moment balance around O to get the value of P. Don't forget to include both T1 and T2, and don't forget to use the correct moment arm on T1.

Chet