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Bannana Monkey System

  1. Aug 27, 2008 #1
    1. The problem statement, all variables and given/known data
    The monkey wishes to get the bananas. It starts climbing up with an accleration 'a'.

    Consider the situation using Newton's Laws of motion:
    FBD for monkey:
    T-mg=ma
    For bananas:
    T-mg=ma'

    Its but obvious that a'=a. That is both the monkey and the bananas go up by the same acceleration.

    --------------------------------------------------------------------------------------
    Now consider this situation:

    The monkey pulley and bananas are in one system. The external forces acting on the system is just the gravitaional force by the earth which is 2mg(other components assumed to be massless) vertically downwards. And the tension in the string segment btween the rigid support and the pulley.

    How will we explain the movement of centre of mass?
     

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  3. Aug 27, 2008 #2

    Dick

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    If I'm guessing your picture correctly, then the support, pulley and the earth are in the same system as the monkey and bananas. The center of mass of the entire system doesn't move.
     
  4. Aug 27, 2008 #3

    tiny-tim

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    Hi Dick! :smile:

    But the centre of mass of the system minus the earth can move relative to the earth. :wink:

    (btw, I'm guessing that the picture has a lady monkey up another tree, and the first monkey is throwing the bananas to her as soon as he gets them! :biggrin:)
     
  5. Aug 27, 2008 #4
    No: The pulley, monkey string and bananas are assumed one system by me.
    [​IMG]
     
  6. Aug 27, 2008 #5
    I support Tim, its exactly what I presume, everything except the earth.




    But I wonder where's that lady monkey :biggrin:
     
  7. Aug 27, 2008 #6

    Dick

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    The pulley and support are free falling? Or is something holding them up?
     
  8. Aug 27, 2008 #7

    Doc Al

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    Don't forget the string that supports the pulley.
     
  9. Aug 27, 2008 #8
    They are not free falling.
     
  10. Aug 27, 2008 #9
    If you read my first post carefully: I have mentioned tension in this segment of string as the external force.
     
  11. Aug 27, 2008 #10

    Doc Al

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    You're right. So, what's the problem? If the monkey is to actually climb the rope, he must exert a tension greater than his weight. So there's a net upward force on the system.
     
  12. Aug 27, 2008 #11

    Dick

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    A system with an 'external force' operating on it is not isolated. It's center of mass can move. Why should that be a problem? Can you state exactly what principle you are questioning?
     
  13. Aug 27, 2008 #12
    I just wanted to get that Tension out without using Newton's Laws. Is it possible? Th centre of mass will have an accleration 'a' upwards. Right? But you know this only if you have the value of tension in the upper string.
     
  14. Aug 27, 2008 #13

    tiny-tim

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    Hi ritwik06! :smile:

    Consider the forces on the string that goes round the pulley …

    The mass of the monkey equals the mass of the bananas in this case … so why should the centre of mass move? :confused: :wink: :smile:
     
  15. Aug 27, 2008 #14

    Doc Al

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    You know that the tension in the upper string equals the sum of the tensions in the lower string. That explains where the external force comes form. But you already knew the acceleration of the masses from the tension in the lower string.

    What's your beef with Newton?
     
  16. Aug 27, 2008 #15

    Doc Al

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    You are one crazy monkey, tiny-tim! :wink:
     
  17. Aug 27, 2008 #16

    tiny-tim

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    more bananas, less beef!

    Why? Newton is your frend! :smile:
     
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