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Center of mass with 2 people, a medicine ball, and a beam

  1. Oct 22, 2013 #1
    1. The problem statement, all variables and given/known data
    A person with mass m1 = 50 kg stands at the left end of a uniform beam with mass m2 = 95 kg and a length L = 2.4 m. Another person with mass m3 = 60 kg stands on the far right end of the beam and holds a medicine ball with mass m4 = 14 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.

    1) What is the location of the center of mass of the system?

    2) The medicine ball is throw to the left end of the beam (and caught). What is the location of the center of mass now?


    2. Relevant equations

    1/Mtotal ƩMiRcm,i

    3. The attempt at a solution

    So for number 1 I got the center of mass pretty easily by doing this:
    ((50 x 0) + (95 x 1.2) + ((60+14) x 2.4)) / (50 + 95 + 60 + 14) ≈ 1.332 meters
    That I completely understand.

    Now for number 2 I thought it'd have to be this

    (((50+14) x 0) + (95 x 1.2) + (60 x 2.4)) / (50 + 94 + 60 + 14) ≈ 1.178 meters

    However the correct answer for 2 is still the 1.332 meters. How can that be? What is wrong with my above equation for 2?
     
  2. jcsd
  3. Oct 23, 2013 #2
    Why do you think it is mentioned that there is no friction between the beam and the floor?
     
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