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Statics Mechanics: Looking for the flaw in my logic

  1. Mar 25, 2012 #1

    Femme_physics

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    Yep...back to old statics. Not because I have a test (already passed it in flying colours) just for fun seeing what I can solve and can't...now I ran into this:


    1. The problem statement, all variables and given/known data
    http://img705.imageshack.us/img705/1533/pulleyf.jpg [Broken]

    Statics:

    The weight is 50kg. Find force P to hold it in equilibrium.

    3. The attempt at a solution



    http://img18.imageshack.us/img18/5787/pulleya.jpg [Broken]


    Somehow I got the result in minus, trying to find where is the logic in this. Or, otherwise, where is the flaw in my logic.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Mar 25, 2012 #2

    I like Serena

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    Hey Fp! :smile:

    Only saw this thread just now.

    Weren't you giving lessons in mechanics nowadays?

    Anyway, you are assuming that all cables give the same tensional force.
    They don't.
    It's not all the same cable.
    There are 3 cables, each with its own tensional force.
     
  4. Mar 27, 2012 #3

    Femme_physics

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    I am giving lessons in mechanics, but with a very focused set of materials that I'm well-practiced with. In our course we never gone through several cables connected to system of pulleys. In fact, we hardly gave pulleys much time. At most we had one cable if I recall correctly.

    Where do you see 3 cables?
     
  5. Mar 27, 2012 #4

    Femme_physics

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    Last edited by a moderator: May 5, 2017
  6. Mar 27, 2012 #5

    Femme_physics

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    And I just solved 2 more such exercises correctly. Don't make me prove myself :) Just needed to understand one "Graphical" principle!
     
  7. Mar 27, 2012 #6

    ehild

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    Very good! You are :cool:

    ehild
     
  8. Mar 27, 2012 #7

    I like Serena

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    Good!

    If you feel up to the challenge... what would happen if there was indeed 1 cable and holes in the pulleys?
     
  9. Apr 4, 2012 #8

    Femme_physics

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    Well then P would equal 1/6 of the weight. :)
     
  10. Apr 4, 2012 #9

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    Suppose P would equal 1/6 of the weight, what would be the resultant force on the leftmost pulley?
     
  11. Apr 4, 2012 #10

    Femme_physics

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    2/6 of the weight
     
  12. Apr 5, 2012 #11

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    That would mean that ƩFy is not zero... what would that mean?
     
  13. Apr 5, 2012 #12

    Femme_physics

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    Of course ƩFy is zero. The resultant force at Y of the rightest pulley is 2/6.... Of the middle pulley is also 2/6....and the right 2/6.... these are the up vectors in the Y direction...the down vector is just the weight....so of course it ends up being ƩFy=0
     
  14. Apr 5, 2012 #13

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    Which up vectors do you mean?

    I see only 2 downward tensional forces on each pulley and the pulleys are not attached to anything else (since we assumed that the cables were not attached to the pulleys).
     
  15. Apr 5, 2012 #14

    Femme_physics

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    Last edited by a moderator: May 5, 2017
  16. Apr 5, 2012 #15

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    Ah, I meant that I didn't see any up vectors on for instance pulley A.
     
  17. Apr 5, 2012 #16

    Femme_physics

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    Wait a second....That was a trick question! If there are no pins attaching the pulleys to the rope they're gonna slide down!
     
  18. Apr 5, 2012 #17

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    Yep. The whole thing would collapse! :eek:
     
  19. Apr 5, 2012 #18

    Femme_physics

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    :) Thanks for the stimulation!
     
  20. Apr 5, 2012 #19

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    I'm just thinking of those poor cardboard men that were standing below that weight!! :bugeye: :wink:
     
  21. Apr 5, 2012 #20

    Femme_physics

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    LOL! Remeber them eh? :)
     
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