Center of gravity

1. Nov 20, 2009

lilaznjewel

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
couples?

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)
1600(x-8)+150(x-0)+150(x-10)+150(x-19)=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. Nov 20, 2009

PhanthomJay

see correction above. Your method is OK, although it is a bit cumbersome.

3. Nov 21, 2009

lilaznjewel

omg I am so stupid. Thank you for pointing out my mistake!