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Center of Mass

  1. Nov 1, 2009 #1
    1. The problem statement, all variables and given/known data

    A 45.0-kg woman stands up in a 60.0-kg canoe 5.00 m long. She walks from a point 1.00 m from one end to a point 1.00 m from the other end.

    If you ignore resistance to motion of the canoe in the water, how far does the canoe move during this process?

    2. Relevant equations

    xcm = (mAxA + mBxB) / (mA + mB)

    3. The attempt at a solution

    I know that I should find the position of the center of mass when the woman is just standing on the canoe, and then plug that into a second center of mass equation to find the new x-value for the canoe, but I can't seem to get it. I am not sure what to plug into the first equation for the x value of the canoe, when the woman isn't walking. Would it be 0, 5, or 2.5 (half of 5, and thus the center of its mass?) or neither of these? Thanks.
  2. jcsd
  3. Nov 1, 2009 #2


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    Hi am13! :smile:

    Hint: Whereabout in the canoe is the joint centre of mass before and after?
  4. Nov 1, 2009 #3
    Would the joint center of mass be where the person is walking?
  5. Nov 2, 2009 #4


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    (just got up :zzz: …)
    uhh? :confused:

    The joint centre of mass is given by the equation in your first post.
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