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Equilibrium and Torque of a table

  1. Dec 1, 2013 #1
    1. The problem statement, all variables and given/known data
    The top view of a table, with weight Wt, is shown in the figure. (Figure 1) The table has lost the leg at (Lx, Ly), in the upper right corner of the diagram, and is in danger of tipping over. Company is about to arrive, so the host tries to stabilize the table by placing a heavy vase (represented by the green circle) of weight Wv at ( X, Y). Denote the magnitudes of the upward forces on the table due to the legs at (0, 0), (Lx, 0), and (0, Ly) as F0, Fx, and Fy, respectively.

    Find Fx, the magnitude of the upward force on the table due to the leg at (Lx, 0).
    Express the force in terms of Wv, Wt, X, Y, Lx, and/or Ly. Note that not all of these quantities may appear in the answer.


    http://session.masteringphysics.com/problemAsset/1010953/19/MFS_st_0_a.jpg


    2. Relevant equations
    ƩTorque = 0


    3. The attempt at a solution
    I have no idea where to begin. The hint says to "find the y-component of the torque", but I can't find torque cause I don't know where the pivot point (axis of rotation) is.
    Please do not answer the question for me, but rather help me understand what to do.
     
  2. jcsd
  3. Dec 1, 2013 #2

    Dick

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    If the table is stablized then it's not rotating around any axis. So the torques around any axis must balance. Picking the y-axis as the axis of rotation is handy for your problem. Do you see why?
     
  4. Dec 1, 2013 #3
    Still not seeing it. Don't I need to choose a pivot point to find Torque? If I choose this pivot point to be at (0,0) then I don't need to worry about the force of the leg at (0,0)? Is that what you want me to see?
     
  5. Dec 1, 2013 #4

    Dick

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    Close. But, no, you don't have to pick a pivot point. Pick a rotation axis. If the table is stablized then it's not rotating around the y-axis. In that case the forces at (0,0) AND (0,Ly) exert no torque around the y-axis. Do you see what the hint is getting at now?
     
    Last edited: Dec 1, 2013
  6. Dec 1, 2013 #5
    So then the only forces that exert torque are W[table], W[vase], and F[x].

    I've always used a pivot point for torque, but what I think you are getting at is that you can use a "pivot axis" in about the same way, so...

    (F[x] * Lx) - (W[table] * Lx/2) - (W[vase] * X) = 0

    F[x] = (W[table]/2) + ((W[vase] * X) / Lx)

    Thank you!
     
  7. Dec 1, 2013 #6

    Dick

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    Right! If the table isn't rotating then the sum of torques around any pivot point must be zero. But so must the torques around any pivot axis. The y-axis is just a useful one to chose.
     
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