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Statics -- Belt Friction

  1. Apr 1, 2015 #1
    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. jcsd
  3. Apr 1, 2015 #2
    You're on the right track.

    Chet
     
  4. Apr 1, 2015 #3
    Nice avatar lol.

    I'm not quite sure where to go from here.
     
  5. Apr 1, 2015 #4
    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
     
  6. Apr 1, 2015 #5
    T = T2/(5.466)

    So if I get what you mean moment about O:

    75 * T1 + 450 * P = 150
     
  7. Apr 1, 2015 #6
    No. Please first show how you get T.
     
  8. Apr 1, 2015 #7
    T1 = T

    T2/T = 5.466
    T = T2/5.466
    I'm lost after this.
     
  9. Apr 1, 2015 #8
    $$(T_2-T_1)0.15=150$$

    Chet
     
  10. Apr 1, 2015 #9
    Where did .15 come from?
     
  11. Apr 1, 2015 #10
    The radius of the wheel. We're taking moments around the axis.

    Chet
     
  12. Apr 1, 2015 #11
    I'm not sure why it's the axis and not O. Is it because we're trying to stop the rotation?
     
  13. Apr 1, 2015 #12
    We're going to use both. First this.
     
  14. Apr 1, 2015 #13
    (T2 - T2/5.466)*.15 = 150
    T2 = 1223.914 N

    If I'm understnading correctly.
     
  15. Apr 2, 2015 #14
    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
     
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