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How far does the canoe move during this process?

  1. Oct 31, 2007 #1
    [SOLVED] Center of Mass

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

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

    [​IMG]

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


    2. The attempt at a solution

    So she is moving to the other side of the canoe.

    What do they mean by "If you ignore resistance to motion of the canoe in the water, how far does the canoe move during this process?"

    That the water does not move the canoe?

    Ma = 45kg
    Mb = 60kg

    There is no applied external force, so velocity is constant..(is this right)

    How do I proceed...

    (MaX + MbX) / (Ma+Mb) for finding center of mass, so

    45X + 60X / 105 ....
     
  2. jcsd
  3. Oct 31, 2007 #2
    Right, you've answered one of your own questions. There's no applied external force, so we can forget about any outside forces on the system. You know the equation for the center of mass, so you want to find the initial center of mass of the system, the system being the lady and the canoe. Next you want to find the center of mass of the system after the lady has walked one meter from the end of the canoe. The difference in the two cm's is the distance the conoe moves, right? Don't forget that we assume the canoe is of uniform mass, in other words, we take its center of mass, not to be confused with the center of mass of the system, to be right in the middle of the canoe. Does this help?
     
    Last edited: Oct 31, 2007
  4. Oct 31, 2007 #3

    Dick

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    Better to write (MaXa+MbXb)/(Ma+Mb) as center of mass, where a is woman and b is canoe. When the woman (and canoe) move to new positions Y, the new center of mass is (MaYa+MbYb)/(Ma+Mb). But since there is no force from the water, the center of mass doesn't move. Why? So those two centers of mass are equal. Subtract the two and get zero. What does this tell you about the relation between (Xa-Ya) and (Xb-Yb)?
     
  5. Nov 1, 2007 #4
    After simplifying I get that

    MaXa+MbXb = MaYa+MbYb

    this should mean that there was no movement in the center of mass.

    I see, the rest was plug in the equation above. And solve for Yb, and since it's the center of mass, Ya (Position of person) must be added 1.5 for her to reach the other side of the canoe. :)
     
    Last edited: Nov 1, 2007
  6. Dec 7, 2010 #5
    Re: [SOLVED] Center of Mass

    Hey is the answer to this problem 1.29 m? Sorry for the huge bump, I didn't want to post something that was already posted. I just need someone to confirm this please?
     
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