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Tug Between Two People On Friction-less Ice.

  1. Dec 16, 2012 #1
    Okay, suppose we have two people, one being 50 kg and the other being 100 kg; they are separated by 600 m, with each person holding on the end of a rope of negligible mass. Why is it, that when one of the two people creates a tension in the string, they meet at their center of mass?

    Also, how my friend solved a problem similar to this one, was that he said since the one person is half as heavy, that person will travel twice as far; so, 2x + x = 600 m.

    Why do both of these methods work? Could someone fill in the reasoning behind each method, please? I'd like to understand why both work.
     
  2. jcsd
  3. Dec 16, 2012 #2

    Doc Al

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    Ask yourself: How can you calculate the location of the center of mass? Can the location of the center of mass change?
     
  4. Dec 16, 2012 #3
    It would seem like if the two people approached it other, then the center of mass would change.
     
  5. Dec 16, 2012 #4

    Doc Al

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    Consider it more generally. Imagine a system at rest. For the center of mass to start moving, what is required?
     
  6. Dec 16, 2012 #5
    Oooh, would it be that one of the people has to start moving, and only one?
     
  7. Dec 16, 2012 #6

    Doc Al

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    Well, that would do it. But it's not what I was going for. What is required for a system to accelerate?
     
  8. Dec 16, 2012 #7
    It would be a force.
     
  9. Dec 16, 2012 #8

    Doc Al

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    Right. In order for the center of mass of a system to accelerate, there needs to be a net force on the system. An external force. But here there are no such forces available, since the ice is frictionless.
     
  10. Dec 16, 2012 #9
    Hmm, okay, I see. So, how can I apply that knowledge to the two different methods of solving a problem like this, to understand the reasoning behind them?
     
  11. Dec 16, 2012 #10

    Doc Al

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    Well, the first method is already done. Since there are no external forces acting, the only place they can meet is at the center of mass.

    As far as the second method goes: If the 100 kg person moved 1 m to the left, how far must the 50 kg person move to keep the center of mass unchanged?
     
  12. Dec 16, 2012 #11
    I would guess two meters. But how can I justify that?
     
  13. Dec 16, 2012 #12

    Doc Al

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    What's the definition of the center of mass? How would you calculate it?
     
  14. Dec 16, 2012 #13
    For one dimensional motion: [itex]x_{cm}= \Large\frac{m_1x_1+m_2x_2+m_3x_3+ \cdots m_nx_n}{m_1+m_2+m_3+ \cdots m_n}[/itex]. And the center of mass would be constant?
     
    Last edited: Dec 16, 2012
  15. Dec 16, 2012 #14

    Doc Al

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    Good. Now write it in terms of Δx for two masses. Then you can relate Δx1 to Δx2.
     
  16. Dec 16, 2012 #15
    Oh, so if I know the center of mass, and I know the change in position of one of the people, then I can find the corresponding change in position of the other person, so that the center of mass remains the same, is that right?
     
  17. Dec 16, 2012 #16

    Doc Al

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    Exactly.
     
  18. Dec 16, 2012 #17
    You can also think about it in the form of Newton's laws of motion:

    [tex]a_1m_1=F=-a_2m_2[/tex]

    If [itex]m_1[/itex] is twice the mass of [itex]m_2[/itex], then [itex]a_2[/itex] must be twice the acceleration of [itex]a_1[/itex] (but in the opposite direction).
     
  19. Dec 16, 2012 #18
    Okay, I believe I understand. Thank you both for your insight.
     
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