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Moveable pulleys you multiply the mechanical advantage

  1. Dec 22, 2005 #1
    Question: http://img486.imageshack.us/img486/5593/pulley9dm.jpg

    Ok so i understand that for every moveable pulleys you multiply the mechanical advantage by 2. In this scenario there are 3 moveable pulleys, and they all look like they are supported by the ceiling so I need to divide the load by a half (i think...) so does that mean that:

    (6000N + 300N)/(2*2*2(MA)*2(ceiling) = 196,9N?
  2. jcsd
  3. Dec 22, 2005 #2
    Is this right?
    cmon, someone has to know this
  4. Dec 22, 2005 #3


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    whether the uppermost pulley is moveble?
  5. Dec 22, 2005 #4
    No, whether my answer is correct. I'm really not sure if i worked the problem correctly.
  6. Dec 22, 2005 #5


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    only want to find correct answer or want to learn?

    If the uppermost pulley is not movable then how many movable pullies are there and what will be MA?
  7. Dec 22, 2005 #6
    Of course i want to learn, if i have the answer right it means that the calculation is right.

    I think that the highest pulley is fixed and the 3 others are moveable. So the mechanical advantage is 2*2*2 = 8. But since the 3 moveable pulleys are attached to the ceiling I think it means that the work force is divised by 2, only once or three times? no clue, i'll say 3 times. So then MA = 8*2*2*2=64
    Seems a little high. Is this right?
  8. Dec 22, 2005 #7


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    no, the mechenical adwantage is 2*2*2 = 8 only
    means you have to apply 1/8 of the load.

    for a single movable pulley one end of string is attached to the ceiling and the other end is held so that you have to apply a force equal to the tension in the string which is half in case of frictionless system. hence MA is 2.
  9. Dec 22, 2005 #8
    How do you work the math dealing with the ceiling in all this?

    Edit: ok i think i understand what you wrote.
  10. Dec 22, 2005 #9


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    it is 2 for a single pulley
    in our system we have three pulleies and hence it is 2*2*2 = 8

    now what is your answer for the effort, to lift the load?
  11. Dec 22, 2005 #10
    (6000N + 300N)/8MA = 787.5N

    So the only thing you have to do is basicly ignore the fixed pully and multiply the MA by 2 for each moveable pulleys in any possible system?

    This is like childs play... i was complicating things for nothing
  12. Dec 23, 2005 #11


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    As nothing is said about where and how the friction is acting you may add it with load, otherwise the things will be different and a bit complicated.

  13. Dec 23, 2005 #12
    When I read the problem it sounds like frictional force is a net force. Also note the subscript is 't' as in 'total'. I am wondering if the Ft should be devided by 8 too. If the given 300 N is a net fric. force that the man has to work against, then he really have to apply 6000/8 + 300 = 1050 N.
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