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Center of Mass problem

  1. Mar 7, 2008 #1
    [SOLVED] Center of Mass problem

    1. The problem statement, all variables and given/known data
    A 3.12 kg particle has the xy coordinates (-1.76 m, 0.652 m), and a 2.89 kg particle has the xy coordinates (0.380 m, -0.636 m). Both lie on a horizontal plane. At what (a)x and (b)y coordinates must you place a 3.74 kg particle such that the center of mass of the three-particle system has the coordinates (-0.565 m, -0.601 m)?


    2. Relevant equations



    3. The attempt at a solution
    I know this problem is extremely easy but I'm just having trouble getting it. I solve for x(com) and y(com) =

    -.565 = (3.12(-1.76)+2.89(.380)+3.74(x))/3.12+2.89+3.74
    -.601 = (3.12(.652)+2.89(-.636)+3.74(y))/3.12+2.89+3.74

    This is correct right? So how do I pull out to solve for x and y?
     
  2. jcsd
  3. Mar 7, 2008 #2
    Try simplifying your expressions and cross-multiplying?
     
  4. Mar 7, 2008 #3
    I've tried that and I'm just doing something wrong I think
     
  5. Mar 7, 2008 #4

    Dick

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    There is nothing wrong with your approach. Maybe if you could post the steps to get the values of x and y someone could figure out where you are going wrong.
     
  6. Mar 7, 2008 #5
    M total = 9.75

    So -.565 = (-5.4912+1.0982+3.74(x)) / 9.75 -> (-.653(x))/9.75 -> -.066974359(x) = -.565

    So x = 8.436064315 but I did it wrong apparently.
     
  7. Mar 7, 2008 #6
    In the numerator you have (-5.4912 + 1.0982 + 3.74x).
    You can't add the numbers together to get (-.653x).

    Instead you have (-4.393 + 3.74x).
    Only the last number is multiplied by x.
     
  8. Mar 8, 2008 #7
    Can I divide the 3.74x by 9 though?
     
  9. Mar 8, 2008 #8

    Dick

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    Why would you want to? As mace2 pointed out the equation you want to solve is (3.74*x-4.393)/9.75=-.565.
     
  10. Mar 8, 2008 #9
    So what should my next step be? Sorry my algebra is horrible but I have no problem in calculus. Go figure.
     
  11. Mar 8, 2008 #10

    Dick

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    I would multiply both sides of the equation by 9.75. Shouldn't you? Clear the fraction first. BTW problems in algebra should be causing you HUGE problems in calculus. I've never seen the reverse. Go figure.
     
  12. Mar 8, 2008 #11
    Yeah I was just about to ask that. Thanks for the help got it figured out :P
     
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