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Homework Help: Find the x coordinate of the center of mass of the system of three chocolate blocks

  1. Oct 21, 2007 #1
    1. The problem statement, all variables and given/known data

    Three odd-shaped blocks of chocolate have the following masses and center-of-mass coordinates:
    (1) 0.310 kg, ( 0.200 m, 0.310 m);
    (2) 0.410 kg, ( 0.110 m, -0.380 m);
    (3) 0.200 kg, ( -0.280 m, 0.610 m).

    Find the x coordinate of the center of mass of the system of three chocolate blocks.

    3. The attempt at a solution

    This is how I commenced,

    I took (mass1*x1+mass2*x2+mass3*x3)/(m1+m2+m3) and got .051

    this should be right but I got it wrong. I then redid the calculation and this time I got .05
    not that much difference, I only got one try left, and don't think this is right because they are so close together.

    I did the y the same way and got it right. :S

    Please and thank you.
    Last edited: Oct 21, 2007
  2. jcsd
  3. Oct 21, 2007 #2


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    Dearly Missed

    Well, then you have put in the wrong numbers somewhere (I assume that even though you have written m3*x2, that should read m3*x3)
    It is impossible for us to know how you clicked about on your calculator.

    You should, at least, calculate the following:
    which ought to be about what you've found already..
    Last edited: Oct 21, 2007
  4. Oct 21, 2007 #3
    for some reason, either my calculator is wack, for the numerator I keep on getting

    .062+.0451-.056 = .0511

    for the denominator I get .92

    .0511/.92 = .055543478


    edit: I tried it on windows, and I get .0511 too. :O
  5. Oct 21, 2007 #4
    for your x i got .055543. try that and see if it works. I did it how you had it up there and that is what I got. I did this on my homework yesterday and got it right so it should be right. :)
  6. Oct 21, 2007 #5
    yep, it's right thanks. I it just as it showed in the calculator this time without any rounding. :)

    thank you princessfrost and arildno for your help.
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