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Static equilibrum

  1. Oct 17, 2009 #1
    1. The problem statement, all variables and given/known data
    "A uniform plank, length of 4M,has a mass of 120N and rests on top of two stools. One Stool, A, is 0.5M into the plank on the left side. The other stool, B, is 1.0M in on the right side.

    An 80N weight is placed between the two stools. Find the Force upon stools A and B.
    Thus meaning that the 80N weight is 1.75M in on the left hand side, and 2.25 M in on the right hand side.

    2. Relevant equations
    Moment of a force = Force * distance.


    3. The attempt at a solution
    First I find the moment of the force of both A and B.
    A = (1.5M * 120N) + ( 1.25M * 80N) = 280 N m.
    B = (1m * 120) + (1.25M * 80) = 220N

    So, we now have a ratio of 280/500:220/500 = 0.56:0.44
    0.56 * 200 = 112, and 0.44 * 200 = 88.
    Therefore, A = 112N and B = 88N.

    This seems all well and good, except according to my answer book it is the other way round. A should be 88 and B 112. I can't see where I've gone wrong.

    I'm thankful for any help given, I get so annoyed when I can't figure out a question =/.
     
  2. jcsd
  3. Oct 18, 2009 #2

    rl.bhat

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    Homework Helper

    Hi Doomoo, welcome to PF.
    Your approach to the problem is not correct.
    Take a reference point, say the stool A.
    First condition is ΣFY = O. Weights act in the downward direction and reaction on A and B act in the upward direction.
    Then take the sum of the moments of the forces about A and equate it to zero.
    The solve for A and B.
     
    Last edited: Oct 18, 2009
  4. Oct 18, 2009 #3
    Ahh, I see.
    Thanks a lot!
     
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