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Proving what the reaction force equals with moments in mechanics.

  1. Feb 10, 2010 #1
    1. The problem statement, all variables and given/known data
    A horizontal piece of wood that has a weight 40N is held in equilibrium by two brackets, A and B. The wood is 1.6m long (CD). A is 0.4m from C and B is 0.2m from D. Meaning that there is 1.0m between A and B. At rest the reaction force at A is 24 N and at B 16N. When an object is put on the wood at xm from C, show that the vertical reaction force on the shelf at A is (24-W(x-1.4))N


    2. Relevant equations
    Moment=Force x Distance


    3. The attempt at a solution
    I understand that the W(x-1.4) is the moment of the object but I cant seem to get the 24.
    If taking moments about B:
    (x-1.4)W + (24 x 1) = 40 x 0.6
    which isn't right

    Any help at all would be greatly appreciated. many thanks.
     
  2. jcsd
  3. Feb 10, 2010 #2

    PhanthomJay

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    The 24 N force at A is the reaction force before the object W is placed on the plank. It is given to you (or it could be calculated). By the superposition principle, the additional reaction at A as caused by W acting alone , is added to the existing 24N reaction at A.

    EDIT:
    If you wish to start from scratch, it's
    (1.4 -x)W -A(1) + 40(.6) =0
    Solve for A. You have to watch your plus and minus signs (counterclockwise vs. clockwise moments) when summing moments about a point.
     
    Last edited: Feb 10, 2010
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