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Center of gravity of a dancer

  1. Nov 20, 2009 #1
    1. The problem statement, all variables and given/known data
    Given: A flat dance floor of dimensions ℓx =
    19 m by ℓy = 23 m and has a mass of M =
    1600 kg. Use the bottom left corner of the
    dance floor as the origin. Three dance couples,
    each of mass m = 150 kg start in the top left,
    top right, and bottom left corners.
    What is the initial y coordinate of the cen-
    ter of gravity of the dance floor and three
    Answer in units of m.

    The couple in the bottom left corner moves
    ℓx = 10 m to the right. What is the new x
    coordinate of the center of gravity?

    2. Relevant equations
    xcm =
    m1x1 + m2x2/
    m1 + m2

    3. The attempt at a solution

    for the first part I did 1600(y-11.5)+150(y-23)+150(y-0)+150(y-23)=0
    and solved for y and I got 12.34
    and I got it right

    my problem is I did the exact same thing for the next part just you know used the X coordinates and moved (0,0) to (10,0)
    I came out with 8.366
    but my online hw says that answer is incorrect I don't know why isn't it basically the same question but with a different number
  2. jcsd
  3. Nov 20, 2009 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    see correction above. Your method is OK, although it is a bit cumbersome.
  4. Nov 21, 2009 #3
    omg I am so stupid. Thank you for pointing out my mistake!
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