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Equilibrium of two rods

  1. Mar 24, 2012 #1
    1. The problem statement, all variables and given/known data
    Weight of AB = CD = 200 N
    Length of AB = 3 m
    Length of CD = 2 m
    A boy moves from point A toward B. At what distance from point A the rod will topple?
    a. 0.4 m b. 0.5 m c. 0.6 m d. 1 m e. 2 m
    untitled-6.jpg


    2. Relevant equations
    newton's law
    torque


    3. The attempt at a solution
    When rod AB topples, the normal force at B = normal force at C = 0 right?

    Will there be normal force at point A? Where should I take the pivot?

    Thanks
     
  2. jcsd
  3. Mar 24, 2012 #2

    ehild

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    What is the boy's weight?
    The rod AB topples only when DC does it. It happens when the torque of the weight of DC around E can not balance the torque arising from the other rod. In that case the support C does not exert force. Put the pivot of rod DC at E.

    ehild
     
  4. Mar 24, 2012 #3
    Oh I'm sorry. The boy's weight is 600 N. Is 0.5 m the correct answer? Thanks
     
  5. Mar 24, 2012 #4

    ehild

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    I got the same. :smile:

    ehild
     
  6. Mar 24, 2012 #5
    Thanks for the help :)
     
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