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Torque Equalibrium

  1. May 7, 2007 #1
    Problem: A 40kg, 5.0m long beam is supported by, but not attached to, the two posts in the figure. A 20kg boy starts walking along the beam. How close can he get to the right end of the beam without it falling over?
    [​IMG]

    My answer:
    \tau_net = (F_1*x_F_1)-(F_Beam*X_Beam)-(F_Boy*X_Boy)
    Since, \tau_net = 0

    0 = (F_1*x_F_1)-(F_Beam*X_Beam)-(F_Boy*X_Boy)
    0 = (40*3)-(40*2)-(20*X_boy)

    Comes out to X_Boy = 2?
    So the boy has to step at the very edge of the beam for it to start falling over?
     
  2. jcsd
  3. May 7, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Looks like you are treating the beam as have two parts, which is OK. But:
    (1) 40 kg is the mass of the entire beam, not each piece;
    (2) The weight of an object acts through its center of mass.
    You need to redo those torque calculations.

    Even better is to treat the beam as one piece--that way you only have two torques to worry about. Where is its center of mass with respect to the pivot?
     
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