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Equilibrium under parallel forces

  1. Feb 18, 2012 #1
    Three identical bricks A , B and C are placed on the edge of a table as shown. Each brick is of length 24 cm and C overhangs the table by x cm. What is the maximum value of x so that the system can still be in equilibrium?

    Y-1.jpg


    Any idea?
     
  2. jcsd
  3. Feb 18, 2012 #2

    Delphi51

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    Homework Helper

    Welcome to PF!
    We aren't supposed to offer help until you post an attempt, but can make an exception on your first post. When it just begins to fall, C will turn ever so slightly with the edge of the table as fulcum. In turn, B will tip ever so slightly with its left end as fulcrum. You need to figure out the torques on B and C, which are related by the force B exerts on C.
    Setting the torque for C equal to zero (just at the point of rotating), solve for x.
     
  4. Feb 18, 2012 #3
    Oops thanks.. Sorry for not following the template. :redface:

    So, first take moment at the bottom-left corner of B.
    CFB(18) - mgB(12) = 0
    CFB = BFC = (2/3)(mgB) (Action-reaction pair)

    Then take moment at the edge of the table.
    BFC(24-x) - mgC(x-12) = 0
    (2/3)(mgB)(24-x) = mgC(x-12)
    ∵mgB = mgC
    ∴xmax = 16.8

    But is it necessary to consider the horizontal force acting on A by C? As C slightly overturns, it will push A slightly towards the left.
     
  5. Feb 19, 2012 #4

    Delphi51

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    Looks good! The question doesn't mention a coefficient of friction, so I doubt if you are expected to include the forces of the brick pushing sideways.
     
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