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Torque: Table tipping

  1. Mar 1, 2015 #1
    1. The problem statement, all variables and given/known data
    http://imgur.com/dMw79w1 [Broken]

    The image contains the problem as well as the answers to the previous problems.

    2. Relevant equations
    Net torque = 0

    3. The attempt at a solution
    First, I set the axis of rotation to be the bottom of the table (where it goes through Fx and F0)
    τdue to downward forces = τdue to upward forces
    τvase + τtable + τcat = τdue to Fy (Fy (the solution to Fy is in the image)
    WvY + WtLy/2 + WmaxLy = Wt/2 + WvY/Ly

    Rearrange for Wmax:
    Wmax = Wt/2Ly + WvY/Ly2 - WvY/Ly - Wt/2

    The answer to the question is:
    Wmax = Wv(1 - X/Lx - Y/Ly)

    I don't know how to obtain this answer from my answer above.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Mar 1, 2015 #2
    It seems that I forgot to include the distance the cat is from the axis of rotation (Ly). After including it, I got Wmax = 0.
     
  4. Mar 1, 2015 #3
    So here it the attempt once again:
    WvY + WtLy/2 + WmaxLy = (Wt/2 + WvY/Ly)(Ly)

    Rearranging for Wmax:
    WmaxLy = (Wt/2 + WvY/Ly)(Ly) - WvY - WtLy/2
    Wmax = Wt/2 + WvY/Ly - WvY/Ly - Wt/2
    Wmax = 0 N
     
  5. Mar 1, 2015 #4

    haruspex

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    The upward forces were calculated in the absence of a cat.
     
  6. Mar 1, 2015 #5
    The cat exerts a downward force on the table though (the weight), which is shonw by WmaxLy
     
  7. Mar 1, 2015 #6

    haruspex

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    Yes, but you used the expressions obtained for Fx etc. in the first part. There was no cat then. The forces will be different now.
     
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