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Gymnast force problem

  1. Jun 18, 2012 #1
    1. The problem statement, all variables and given/known data
    A gymnast with mass 46.0 kg stands on the end of a uniform balance beam. The beam is 5.00 m long and has a mass of 250 kg (excluding the mass of the two supports). Each support is 0.540 m from its end of the beam. In unit-vector notation, what are the forces on the beam due to (a) support 1 and (b) support 2?


    2. Relevant equations



    3. The attempt at a solution

    this is my teachers solution and i have a question about it..:

    For computing torques, we choose the axis to be at support 2 and consider torques that encourage counterclockwise rotation to be positive. Let m = mass of gymnast and M = mass of beam. Thus, equilibrium of torques leads to
    M g (1 . 9 6 m ) − m g ( 0 . 5 4 m ) − F ( 3 . 9 2 m ) = 0 .

    my question is where did 3.92 come from? i know .54 is given and 1.96 is 2.5-.54 but where does 3.92 come from? its driving me crazy!
     
  2. jcsd
  3. Jun 18, 2012 #2
    3.92 is the distance between the second support and the first support, solved by taking 5m and subtracting 0.54 twice.
     
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