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Homework Help: Accelerating pulley

  1. Oct 6, 2007 #1
    Hello, this is my first post here. I hope you can answer my current question, and that i will be able to answer other's questions later on!

    I have all the footwork (equations) under control, but I can't figure out how the forces are divided in this situation:


    There is a force applied upwards.

    When the force F isn't great enough to move any of the blocks, the force on A and on B both equal ½F (since the pulley's strings are parallel to the direction of F)

    Now, we increase the force F so that ½F > Weight of B, meaning B accelerates upwards, while A remains at rest.

    Now my question is:

    how does F split up? Does F exert the same amount of force on B as on A?

    And also, if F is great enough to move both blocks off the ground, is F still exerting the same amount of force on A as on B? (excluding the amount of force that A will exert on B and opposite)

    I thought about this the whole day, and couldn't make up my mind between two possible ways that the force could be split in the pulley.
    The first idea was that the force, no matter what, is split 50/50.
    The other idea was that if there is any acceleration, the force coming from F will split as follows: (mA and mB are masses of the blocks)
    F(on A) = mB/(mA+mB)*F
    F(on B) = mA/(mA+mB)*F
    This means that the force on the heavy block is smaller than the force on the lighter block. And together, they make up F
    This makes most sense to me, but my physics book doesn't say anything about it :rolleyes:
    Last edited: Oct 6, 2007
  2. jcsd
  3. Oct 6, 2007 #2


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    Hi Laban. I'm wondering, is this a question you had yourself, or is it a textbook question?

    I would not think of F splitting up into A and B... I think that's a problematic way of thinking about it...

    F exerts a force on the pulley... the cords exert a force on the masses... they are different forces...

    Is the pulley given as massless?
  4. Oct 6, 2007 #3
    The pulley is massless and so are the cords.

    Basically, I have a force F = 424 N and I need to find the accelerations of each block.
    To solve it, I gave block A an unknown force upwards + tension in the rope - weight. Block B I gave a different unkown force upwards + the same tension on the rope - weight.
    So, to solve it, i need to know how F splits into the two unknown forces. That's all i need.

    I also tried thinking about the problem in an entirely different way. I calculated the accelerations of the blocks, supposing the pulley was fastened to the ceiling and not accelerating.
    Then I calculated the acceleration of the entire system: [tex]\frac{424 N - g\cdot m_{A} - g\cdot m_{B}}{m_{A}+m_{B}}[/tex] and then just added that acceleration to each of the accelerations for the blocks.
    BUT i discarded that idea since I believe it is faulty.
  5. Oct 6, 2007 #4


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    The only forces on the blocks are the tension the weight... there is no other unknown force... can you explain why you think there's another force involved?

    The tension in both cords is the same. First get the tension using the freebody diagram of the pulley...

    What is the net force on the pulley? Solve for tension...

    Then using the tension you can get the acceleration of each mass.
  6. Oct 6, 2007 #5
    You say "both cords", but we all know there is only one cord... which is why i'm getting confused.
    If there were two cords, and they were simply fastened to the upper rope without a pulley, then the downwards force in the free-body diagram for that junction of ropes would contain the sum of the tensions from each lower rope.
    But since it's a pulley, i'm confused how the tension works... is the downward force 2T or just T ?

    I can explain why i thought there was two unknown forces, but I want to get clear on this first ;)
    Thanks for your help so far, it's really amazing :)
  7. Oct 6, 2007 #6


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    It is one cord, but there are two forces involved for each part of the cord on each side of the pulley...

    In this particular case those two forces are the same... ie the tension on either side of the one cord is the same... generally this is not true... it is only true here because the pulley massless.

    The downward force is 2T.
  8. Oct 6, 2007 #7
    Hmm.. I tried this and ended up with negative acceleration for mass A. That should be impossible.

    The equation for the pulley looks like this
    [tex]\sum F_{y} = F - 2T = \left( m_{A} + m_{B} \right)\cdot a_{sys}[/tex]
    where a_sys is the acceleration of the entire system.

    The equation for mass A looks like this
    [tex]\sum F_{y} = T - m_{A}\cdot g = m_{A} \cdot a_{A}[/tex]

    The equation for mass B looks like this
    [tex]\sum F_{y} = T - m_{B}\cdot g = m_{B} \cdot a_{B}[/tex]

    AND then I use the fact that the acceleration relative to the pulley is equal of magnitude for both block A and B, but with opposite direction.
    [tex]a_{A} = a_{sys} - a_{r}[/tex]
    [tex]a_{B} = a_{sys} + a_{r}[/tex]
    where "a_r" is the relative acceleration.

    What's wrong?
  9. Oct 6, 2007 #8


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    No. this isn't right. we are looking at the freebody diagram of the pulley. you shouldn't have mA or mB here...

    F - 2T = mpulley*apulley.
  10. Oct 6, 2007 #9
    So since the pulley is massless, the net force will be zero. Meaning T = ½F.
    That's what I thought from the very beginning, but my father talked me out of it (without knowing why) ^^ hehe
  11. Oct 6, 2007 #10


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    lol! :)
  12. Oct 6, 2007 #11
    I just want to say a formal thank you... Thank you! :wink:
    You really got me out of a dire strait here.

    Looks tricky at first,
    but turns out to be really simple because it goes against the spontaneous common sense (of course i can only speak for myself, but i know my classmates had great difficulty with this one)
    Last edited: Oct 6, 2007
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