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Finding shift in centre of gravuty

  1. Jul 19, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the minimum lateral shift in position of the centre of gravity of a standing man whose mass is 80 kg, when is carrying a 14 kg pail of water. Assume that he wishes to have an equal load on each foot. How is this shift in his centre of gravity accomplished?

    2. Relevant equations

    F=0

    [​IMG]

    3. The attempt at a solution

    I don't know where to start! With the pail, he has to have 58 kg on each foot. There's the weight of the bucket and the weight of the man, but since he's shifting, should I divide him in half? Like 40 kg on each foot?
     
  2. jcsd
  3. Jul 19, 2008 #2

    Doc Al

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    Staff: Mentor

    Start by finding the center of gravity of "man + pail". (It will be some distance to the left of the man's center of gravity.)
     
  4. Jul 19, 2008 #3
    How do I do that? :redface:
     
  5. Jul 19, 2008 #4

    Doc Al

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    Treat the pail and the man as two separate point masses and find the center of mass of both. (All you care about is the horizontal position of the center of mass of both compared to the original position of the man's center of mass.)

    If you don't know how to calculate the center of mass, read this: Center of Mass
     
  6. Jul 19, 2008 #5
    Aye, okay.

    x = 14(0) + 80 (36) / 14 + 80
    x = 31 cm

    Thanks Doc Al.
     
  7. Jul 19, 2008 #6

    Doc Al

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    Good. So how far must he shift to the side?
     
  8. Jul 19, 2008 #7
    I believe.. 5 cm to his left? Hopefully I'm understanding this now.
     
  9. Jul 20, 2008 #8

    Doc Al

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    Staff: Mentor

    Sounds good to me.
     
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