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Friction of a rope wound around a stationary rod

  1. Oct 4, 2013 #1
    1. The problem statement, all variables and given/known data

    A 100 kg man holds up a 1000 kg weight by holding on to a rope that is wrapped around a stationary horizontal rod whose other end is tied to the suspended weight. If the coefficient of static friction between the rope and rod is 0.5, what is the minimum number of times that the rope must be wrapped around the rod?
    Edit: Assume the rope is massless.
    Edit 2: The rope that is not in contact with the stationary rod is vertical.

    2. Relevant equations

    F(friction) = μ*m(weight)*g*θ/2∏, m(man)*g+F(friction)=m(weight)*g

    3. The attempt at a solution

    If the equations I wrote are correct, the problem is easy to solve, but I think they might be wrong. Could someone please explain why they are or aren't correct? Thanks.
     
    Last edited: Oct 5, 2013
  2. jcsd
  3. Oct 5, 2013 #2

    haruspex

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    The normal force between the rope and the rod is not the weight of the suspended mass. It will not even be constant along the section of rope touching the rod.
    You need to use calculus. Consider a short section of rope around the circumference of the rod making angle dθ at the centre. The tension is T(θ) at one end and T(θ+dθ) at the other. Deduce the normal force and hence the max frictional force. Get an equation relating that to T(θ) etc.
     
  4. Oct 5, 2013 #3

    Doc Al

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    When a rope is wrapped around a pole, things get a bit more interesting. The governing equation that you want is called the Capstan Equation. (Which can be derived with a bit of calculus.)
     
  5. Oct 5, 2013 #4
    I forgot to mention that the rope is massless in the problem. I believe that in this case the normal force per unit angle will be constant along the section of rope touching the rod, right?
     
  6. Oct 5, 2013 #5

    ehild

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    The rod is stationary, so it does not matter if massless or not. The rope can slide on it, friction between the rope and rod opposes sliding.


    ehild
     
  7. Oct 5, 2013 #6

    haruspex

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    ... so if you consider a short section of the rope, the friction is acting along the rope. It follows that the tension at one end is different from that at the other.
     
  8. Oct 5, 2013 #7
    Thank you very much haruspex and Doc Al. I just finished deriving the correct formula and answering the question. The minimum angle is 4.6 radians. Since this is greater than pi radians, the rope has to loop an extra time around the rod so the angle will be 3 pi radians (since the rope is vertical at both ends).

    ehild, if you want to understand the problem, this youtube video might help: I found it much clearer than the derivation on wikipedia.
     
    Last edited by a moderator: Sep 25, 2014
  9. Oct 5, 2013 #8

    haruspex

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    I think ehild understands it already :smile:.
     
    Last edited by a moderator: Sep 25, 2014
  10. Oct 5, 2013 #9

    ehild

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    Well, I derived the formula before anybody answered, but was not sure it was correct. It was too simple. :biggrin:

    ehild
     
  11. Oct 5, 2013 #10

    Doc Al

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    :tongue:
     
  12. Oct 5, 2013 #11

    ehild

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    Thanks for the link, Doc. I never heard Capstan equation before. :smile:

    ehild
     
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