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Homework Help: Center of Mass Canoe Problem

  1. Nov 14, 2006 #1
    Hi there,

    I have a question about what I'm doing wrong with the following problem:

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

    What I did was found the x coordinate of the center of mass by doing the following math:

    m_w= 45.0kg, x_w1 = 1m
    m_c= 60.0kg, x_c1= 0m

    x_cm = [(45.0kg*1m) + 0] / (45.0kg + 60.0kg)
    x_cm = 0.429m

    Then I used the same x-coordinate of the center of mass and used the second coordinate for the woman to find the second coordinate of the canoe:

    x_w2 = 4m

    0.429m = [(45.0kg*4m) + (60.0kg*x_c2)] / (45.0kg + 60.0kg)
    x_c2 = -2.25m

    Now that seems right since the ratio between the masses is 45/60 and the distance the woman traveled was 3m so 3m*0.75 = 2.25, but the answer listed is 1.29m... I would like to know if someone can see anything that I'm missing or have done wrong, it is really frustrating me. :(
     
  2. jcsd
  3. Nov 14, 2006 #2

    radou

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    Your center of mass calculation seems to be wrong, since the canoe is not included.
     
  4. Nov 14, 2006 #3
    Well I included the mass of the canoe when I divided. I have the initial position of the canoe equal to 0 is there a different number I should use for that?
     
  5. Nov 14, 2006 #4

    radou

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    How you set the coordinate system does not matter. Let's say you set it on the beginning of the canoe. You would have: Xc = (Mw Xw + Mc Xc) / (Mw + Mc) = (45*1 + 60*2.5)/(60 + 45) = 1.857. If you set the coordinate system in the middle of the canou, you would have Xc = (45*(-1.5) + 60*0)/(60 + 45) = -0.643. You can convince yourself easily that this is the same point.
     
  6. Nov 14, 2006 #5
    awww I see, so if I want the initial position of the canoe to be 0 I would have to set the coordinate in the center of the canoe making the woman's initial position -1.5. Ok, well that makes sense, thanks Radou.
     
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