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Equilibrium problem and a beam of mass

  1. May 17, 2008 #1
    [SOLVED] Equilibrium problem

    1. The problem statement, all variables and given/known data

    A uniform beam of mass 76 kg and length 3 m rests on two pivots, one at the left edge and one 2.4 m from the left edge. How far to the right of the right pivot can a mass of 150 kg be placed without the beam tipping?


    2. Relevant equations



    3. The attempt at a solution

    net forces in y direction, Fy : F1(pivot on egde) - Fg(weight of board) +F1(2nd pivot) - mg

    Sum of torque(im torquing around the rigt piovt): F1(2.4m) - Fg(.9m) -mg(x)

    the problem im having is that i have two unknown forces from the pivots, and when i solve for one of the forces, the results cancel out and x = 0, i dont what i am supposed to assume?
     
  2. jcsd
  3. May 17, 2008 #2

    Chi Meson

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    F1 and mg would both cause clockwise rotation, so they should have the same sign.
     
  4. May 17, 2008 #3
    which F1 are you talking about? and which mg?
     
  5. May 17, 2008 #4
    Try three equations - three unknowns system (x, normal force one, normal force two)

    You can pick first torque center at N1/N2, second at 150g mass .. and this give 2 eqns ..

    and third is simply adding all forces ...

    F1(pivot on egde) - Fg(weight of board) +F1(2nd pivot) "+" mg .. i think as said by chi meson
     
  6. May 17, 2008 #5

    Redbelly98

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    He meant the ones in your torque equation.
     
  7. May 17, 2008 #6

    Redbelly98

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    What is F1 (on the left edge pivot) when the 150 kg is placed right at the tipping point?
     
  8. May 18, 2008 #7

    Chi Meson

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    One step further, is F1 needed at all?
     
  9. May 18, 2008 #8
    i got it, thanks for the help
     
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