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Homework Help: Pulley with Chair Question

  1. Oct 24, 2012 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    F = ma

    3. The attempt at a solution

    I know that the rope is pulling the man up twice as much because there are two points of contact between him and the rope. I'm really just looking for somebody to explain why that is?

    It seems to me that when he pulls down on the rope, the chair will move with a force upwards, but the other side will move with a force downwards and he will basically be in the same position unless he lets go of the rope with his hands.

    What I'm trying to say is, the tension on each side of the rope is going in opposite directions. Why would it count as if the man is being pulled up twice when, the way I see it, the two forces should cancel each other out?
  2. jcsd
  3. Oct 24, 2012 #2


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    Homework Helper

    The rope is like a spring, it stretches a bit when pulled from both sides and a tension evolves all along its length to withstand that stretching external force. According to Newton's Third Law, the rope also exerts force on the objects pulling it, at both ends. These forces are of equal magnitude (the tension T) and point inward, to the bulk of the rope. So the rope exerts an upward force T on the chair, and also an upward force T on the hand of the man. When the man is in rest, the sum of these upward forces balance the weight of the chair/man.
    The man can rise by moving his hand down, pulling the rope, and then grabbing the rope with the other hand higher. So the rope gets shorter and shorter between the chair and his hand. That can be if the chair rises.

    The chair/man moves upward either with uniform velocity or with a given acceleration under the resultant force F= 2T-mg.


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    Last edited: Oct 24, 2012
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