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Homework Help: Monkey Climbing a Rope

  1. Oct 24, 2007 #1
    1. The problem statement, all variables and given/known data
    A bunch of bananas hangs from the end of a rope that passes over a light, frictionless pulley. A monkey of mass equal to the mass of bananas hangs from the other end of the rope. The monkey and the bananas are initially balanced and at rest. Now the monkey starts to climb up the rope, moving away from the ground with speed v. What happens to the bananas?
    A) They move up at speed 2v
    B) They move downward at speed v
    C) They move up at speed 1/2 v.
    D) They remain stationary
    E) They move up at speed v.

    3. The attempt at a solution
    I chose D. Since the monkey moves up the rope, his mass is still equal to that on the other side. The rope I took to be negligible in mass, as every rope we've used in physics so far has been. The mass of the bananas = the mass of the monkey.
    [tex]a = \frac{g(m2 - m1)}{(m1 + m2)}[/tex]
    Since the masses do not change, the acceleration is zero, and the ropes themselves do not move.

    Is this right? My professor said E is the right answer, but I was wondering if there is an explanation for this.
  2. jcsd
  3. Oct 24, 2007 #2

    Doc Al

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    Staff: Mentor

    Your professor is correct. In order for the monkey to accelerate (which he must do if he starts from rest) he must give an additional pull on the rope. In turn, the increased tension in the rope exerts a similar force on the bananas.
  4. Oct 24, 2007 #3


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    Science Advisor
    Homework Helper

    The way the monkey climbs and increases his velocity to v is by increasing the tension T in the rope so T>mg for long enough to reach velocity v. The equal massed bunch of bananas will experience the same increased tension for the same time. So it is also moving up at velocity v.
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