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Energy required to lift a heavy box (pulley problem)

  1. Oct 24, 2009 #1
    1.

    As you are trying to move a heavy box of mass m, you realize that it is too heavy for you to lift by yourself. There is no one around to help, so you attach an ideal pulley to the box and a massless rope to the ceiling, which you wrap around the pulley. You pull up on the rope to lift the box.

    Use g for the magnitude of the acceleration due to gravity and neglect friction forces.

    What is the magnitude F of the upward force you must apply to the rope to start raising the box with constant velocity?
    Express the magnitude of the force in terms of m, the mass of the box.




    2. F = ma



    3. I have never done an ideal pulley problem before, so I think if someone just gets me started with a general strategy I'll be able to figure it out. Obviously there's a tension force T pointing upward, opposing the gravity force mg. Is there anything else I need to consider with an ideal pulley that I wouldn't normally? I know that if you're pulling up with constant velocity, then acceleration is zero... but then force is zero. Doesn't make sense to me.
     
  2. jcsd
  3. Oct 24, 2009 #2

    cepheid

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    Hello Linus Pauling,

    That's right!

    Not really. EDIT: All a pulley really does is change the direction of a force. You can check your intuition about the forces acting on the box by drawing a free body diagram for it and making sure that you haven't missed anything. It helps you to be methodical and is a consistent method for solving these problems.

    The box will not accelerate unless if there is a NET force acting on it. The net force is the "F" in Newton's second law. If it is moving at a constant velocity, then this means that all of the forces acting on it balance each other out, so that there is no net force in any direction. It does not mean that the applied (pulling) force is zero.
     
  4. Oct 24, 2009 #3
    So, there is a tension force of equal magnitude and opposite to the gravity force.

    Fnet = T-mg
    T=mg

    This is incorrect. It says the pulley reduces the force required to life the box, and my answer should be in terms of the mass only. I do not understand.

    EDIT: Does the pulley have two tension forces? So it should be 2T=mg ---> T=0.5mg?
     
  5. Oct 24, 2009 #4
    Yep, I got it. nevermind.
     
  6. Oct 24, 2009 #5
    New question:



    Consider lifting a box of mass m to a height h using two different methods: lifting the box directly or lifting the box using a pulley (as in the previous part).

    What is {W_{\rm d}}/{W_{\rm p}}, the ratio of the work done lifting the box directly to the work done lifting the box with a pulley?


    W_d = mg*h
    W_p = 0.5mg*h

    W_d/W_p = mgh/0.5mgh = 2

    This is incorrect. Why?
     
  7. Oct 24, 2009 #6

    cepheid

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    Where do you get the factor of 0.5 from?

    In general, using a pulley system doesn't change the amount of work done.
     
  8. Oct 24, 2009 #7
    the 0.5 is because there are two upward tension forces. Could explain (conceptually) why having more upward tension force does not change the amount of work done? Is it simply because of the fact that work depends on displacement, and either way the displacement is the same?
     
  9. Oct 24, 2009 #8

    cepheid

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    The weight of the box is shared amongst two upward tension forces, so you're only pulling with a force equal to half the weight of the box.

    So, why isn't the work done using the pulley system equal to half as much? Because the distance over which the force is applied is equal to the length of rope pulled, which is always twice the height gained by the box. Wikipedia seems to have a good explanation with useful diagrams:

    http://en.wikipedia.org/wiki/Pulley

    especially in section 2.2, "How it Works." Quoting from the article:

    "It is important to notice that a system of pulleys does not change the amount of work done. The work is given by the force times the distance moved. The pulley simply allows trading force for distance: you pull with less force, but over a longer distance."

    EDIT: By the way, what I said in my first response about a pulley only changing the direction of a force was incorrect. I was thinking of a fixed pulley, which doesn't provide any mechanical advantage. The problem is talking about a moveable pulley, which in this case provides a mechanical advantage of 2.
     
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