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Pulley and Platform

  1. Jun 22, 2013 #1
    1. The problem statement, all variables and given/known data

    Two construction workers each of mass m raise themselves on a hanging platform using pulleys as shown above. If the platform has a mass of 0.5m, the initial distance between the pulleys and the platform is d, and the workers each pull with a force f on the ropes, what is the acceleration a of the workers? Assume the pulleys and ropes are massless.

    2. Relevant equations



    3. The attempt at a solution

    Forces on one Man:
    FME (weight of the man),
    The normal force NMP (contact of the man with the platform),
    The tension force FMR (the response force that the rope applies to the man).
    The force applied by 2 men pointing down

    For the platform:
    FPE (weight of the platform),
    NPM (force on the platform due to the man, it is the pair of NMP) and remember you have two of this because you have 2 men,
    T and T: The two tensions acting on the platform due to the ropes
     

    Attached Files:

  2. jcsd
  3. Jun 22, 2013 #2

    mfb

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    I would consider men+platform as one system. This reduces the number of forces involved.

    What did you do after listing all those forces?
     
  4. Jun 22, 2013 #3
    Eliminated Normal force. What answer do you get? I tried doing it by considering them as a one system too. Still stuck.
     
  5. Jun 22, 2013 #4



    Considering them as one system:

    2N downwards
    2N upwards
    2mg + .5 mg downwards
    2f downwards (force applied by men)
    2f upwards. (The tension in the strings attached to the platform)
    2t upwards. (the tension in the strings held by men)

    Where am I going wrong?
     
  6. Jun 22, 2013 #5

    mfb

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    They do not apply a downwards force on the pulley/men system.
    Where is the difference between f and t?

    The other forces look good.
     
  7. Jun 22, 2013 #6
    But the question says they pull down with a force f. What are the correct forces?
    also, is there a difference between f and t in this case?
     
  8. Jun 22, 2013 #7

    mfb

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    They pull the rope down, and themself up.
    That was my question to you. Can the rope tension be different at both sides?
     
  9. Jun 22, 2013 #8
    Okay, so tension and force are the same in this case. Can you please list out all the forces?
     
  10. Jun 22, 2013 #9
    For the two men combined:
    [tex] 2N - 2F - 2Mg = 2Ma [/tex]
    For the platform
    [tex] 2f -.5Mg - 2N = .5Ma [/tex]

    Is that right?
     
  11. Jun 22, 2013 #10

    TSny

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    Why the minus sign in -2F here?
     
  12. Jun 22, 2013 #11
    Because they're pulling it down.
    Here's another way I did it in:
    Forces on 2 men
    tension upwards
    normal upwards
    weight downwards

    Forces on platform:
    normal downwards
    weight downwards

    I've ignored the forces applied by the two men to pull the string down, because they get balanced by the forces on platform pointing up
     
  13. Jun 22, 2013 #12

    mfb

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    F points upwards for the men, as I mentioned before. They are pulling the rope down, so they are pulling themself up.
    Apart from that, post 9 is right.

    That does not work.
     
  14. Jun 22, 2013 #13
    But aren't they trying to pull it down? How would it point upwards?
    And F gets balanced by the force on platform. Don't they?
     
  15. Jun 22, 2013 #14
    Why not? They are reaction forces after all? How does F point upwards for men when they're applying force downwards.
    Also, Isn't tension upwards?
     
  16. Jun 22, 2013 #15
    [tex] 2N + 2F - 2Mg = 2Ma [/tex]

    [tex] 2F -.5Mg - 2N = .5Ma [/tex]

    Solving them, this is what I get:

    [tex]4F - 25m = 2.5Ma [/tex]

    Is that correct?
     
  17. Jun 22, 2013 #16

    mfb

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    Did you ever try to push a rope? Did it work?
    You can only pull it, and get a positive tension in it.
    They are "trying" to pull at the anchor on the top of the building. As this anchor feels a force downwards, they feel a force upwards (action/reaction).

    Exactly, tension pulls them upwards.

    If you add the missing decimal point, yes.
     
  18. Jun 22, 2013 #17
    I'm confused. If I'm pulling the rope, am I not applying the force downwards? Of course, it's upwards on the platform. And isn't the equation incomplete without tension?
    Also, I did add the decimal points.
     
  19. Jun 22, 2013 #18
    I got the answer, but can someone explain to me the working of this problem? I haven't understood it.
    I just got the answer by blindly following the posts
     
  20. Jun 22, 2013 #19

    haruspex

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    I think judas_priest has substituted g = 10.
    Yes, downwards on the rope, so the rope is applying the same force upwards on you.
    The quick way to solve this problem is to consider the forces pulling upwards the pulleys. What do these add up to?
     
  21. Jun 22, 2013 #20
    Okay, got it. Thanks a lot!!
     
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