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Homework Help: Monkey climbing up a rope.

  1. Dec 17, 2013 #1
    1. The problem statement, all variables and given/known data
    I'd like to find P=F dot v that a monkey of mass m must exert upon climbing up a closed loop of rope of mass M, wrapped around a massless pulley, so that it (viz. the monkey) at any given moment remains in the same place. The rope is constantly tight and at t=0 the rope and the monkey are at rest. No friction is involved! And the monkey has no volume.

    2. Relevant equations

    3. The attempt at a solution
    I am not sure whether "monkey stays in the same place" entails dy/dt=0 for the monkey or, rather, that its acceleration is basically that of the rope but in the opposite direction.
    Further, I wrote the following equation for the forces which "feels" the rope:
    -(M+m)g(y^) + F(y^) = My''
    where F is the force with which the monkey pulls the rope.
    I am also not sure whether this is correct.
    I'd appreciate some insight on all of the above.
  2. jcsd
  3. Dec 17, 2013 #2
    "Monkey stays in the same place" means that its velocity is always zero (and therefore its acceleration is zero too). What does that tell you about the net force acting on the monkey?
  4. Dec 17, 2013 #3
    Well, in that case, obviously the net force acting on the monkey must be zero as well (Newton). However, if its velocity is always zero, does that mean that the power the monkey needs to exert in order to remain in the same place is F dot v hence zero?
  5. Dec 17, 2013 #4


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    yep. but if the pulley is completely frictionless, and the rope is a loop, what must the monkey do to keep tension in the rope? what place (or places) must he hold the rope, to keep tension in the rope, so that he doesn't fall?
  6. Dec 17, 2013 #5
    Hmm... Won't the monkey need to pull the loop with a force equal to (M+m)g/2?
  7. Dec 17, 2013 #6
    So if gravity exerts a force on the monkey of mg downwards and the net force on the monkey is zero, what other force is there on the monkey?

    The work done on the monkey is zero, but there is another part of this system whose energy is increasing. But you don't need to think about this to answer the question.
  8. Dec 17, 2013 #7
    I think the problem assumes that the monkey can only hold the descending side of the rope, otherwise there is no unique solution.
  9. Dec 17, 2013 #8
    The forces acting on the monkey are -F(y^) and -mg(y^), where F is the force with which the monkey pulls the rope, right? The forces acting on the rope are F(y^) and -Mg(y^), right? And yet, if the monkey has no volume (and it is merely a point of mass m on the rope), won't that entail that the mass of the rope would actually be (M+m), hence yielding -(M+m)g(y^) acting on the rope?
  10. Dec 17, 2013 #9
    Right but for a sign - the monkey pulls down on the rope so the reaction force of the rope on the monkey is F, not -F.

    Again the monkey pulls down on (one side of) the rope so this is -F not F. And the rope is supported by a pulley so the -Mg/2 of gravity on the side of the rope held by the monkey is opposed by Mg/2 from the pulley, but the centre of gravity of the rope does not change so you can ignore the effects of gravity on the rope.

    No, you have already accounted for the force of the monkey on the rope (it is -F) so it is wrong to also add the weight of the monkey.
  11. Dec 17, 2013 #10
    So the monkey pulls the rope with a force equal to its weight, right (yielding zero net force on monkey)? But the answer to the question would still be zero, as the velocity of the monkey remains zero, right?
  12. Dec 17, 2013 #11
    Yes! So there is no net force on the monkey, and the force on the rope is equal to the monkey's weight.

    Which question - what is ## \vec{F} \cdot \vec{v} ##? Yes this is zero - the monkey does not move so the rate of work done on the monkey is zero. But something moves in response to the force the monkey exerts - what is it? It will help you if you stop using vector equations of motion and think in terms of the scalar quantities of energy, work and power.
  13. Dec 17, 2013 #12
    Well, the rope moves, despite the monkey's staying put. Net force on the rope is not zero.
  14. Dec 17, 2013 #13
    So the rope accelerates, you can work out its speed after time t and therefore the power = force x speed.
  15. Dec 17, 2013 #14
    But that of course would not change the fact that the power needed to be exerted by the monkey in order to remain in the same place at any given moment will remain zero, true?
  16. Dec 17, 2013 #15
    Not true, the monkey is doing work on the rope all the time - at an increasing rate - just to stay still.
  17. Dec 17, 2013 #16
    ... and if you substitute some numbers in you can see that even with a very light and powerful monkey and a very heavy rope, he is not going to be able to keep this up for very long!
  18. Dec 17, 2013 #17
    Okay, so the monkey must move upwards at v=(m/M)gt(y^) in order to remain in the same place. But, if the net force on the monkey is zero, how come F dot v would not still yield zero, where F is the net force on the monkey?
    NB I realise the answer is probably (m^2/M)(g^2)*t for the power to be exerted by the monkey, as the force acted on the rope is (-mg)(y^) and its velocity is (-m^2/M)(gt)(y^). However, considering my argument above, I am not quite sure why that is indeed the case.
    Last edited: Dec 17, 2013
  19. Dec 17, 2013 #18
    Let's be quite specific:

    ... in the reference frame of the rope ...

    ... in the static reference frame.

    In the static reference frame ## \vec{F}_\textrm{monkey} ## and ## \vec{v}_\textrm{monkey} ## are both zero vectors so their dot product is of course zero.


    Because your argument above shows that there is no work done on the monkey - the power of the monkey is converted only into kinetic energy of the rope.
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