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Center of mass on a beam

  1. Feb 26, 2014 #1
    1. The problem statement, all variables and given/known data
    A person with mass m1 = 61 kg stands at the left end of a uniform beam with mass m2 = 104 kg and a length L = 2.7 m. Another person with mass m3 = 68 kg stands on the far right end of the beam and holds a medicine ball with mass m4 = 8 kg (assume that the medicine ball is at the far right end of the beam as well). Let the origin of our coordinate system be the left end of the original position of the beam as shown in the drawing. Assume there is no friction between the beam and floor.

    3)After the ball is thrown to the person on the left, what is the new x-position of the person at the left end of the beam? (How far did the beam move when the ball was throw from person to person?)


    2. Relevant equations
    Center of mass equation (can't figure out how to type it on here)

    3. Attempt at a solution
    I kept the original center of mass of 1.53 since it doesn't move and created an equation to solve for the new position.

    1.53 = [itex]\frac{(61+8) (0-x) + 104 (1.35-x) + 68 (2.7-x)}{241}[/itex]

    And I get x = .186m, which is not correct.
     
  2. jcsd
  3. Feb 26, 2014 #2
    I would check your center of mass value to start -- I got 1.434.
     
  4. Feb 26, 2014 #3
    The system accepted 1.53 as the answer. I could try 1.434 though.
     
  5. Feb 26, 2014 #4
    Wow. I don't know why it accepted something so far off for the first question. Using 1.43 worked, thanks.
     
  6. Feb 26, 2014 #5
    No problem -- those darn online things.
     
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