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Torque has to equal zero, but introduce F

  1. Mar 9, 2009 #1
    1. The problem statement, all variables and given/known data
    A lunch tray is being held in one hand, as the drawing illustrates. The mass of the tray itself is 0.180 kg, and its center of gravity is located at its geometrical center. On the tray is a 1.00 kg plate of food and a 0.235 kg cup of coffee. Obtain the force T exerted by the thumb and the force F exerted by the four fingers. Both forces act perpendicular to the tray, which is being held parallel to the ground.

    T = ? N (downward)
    F = ? N (upward)


    2. Relevant equations

    [tex]\Sigma[/tex]Fx = 0
    [tex]\Sigma[/tex]Fy = 0
    [tex]\Sigma[/tex]T = 0

    3. The attempt at a solution
    [tex]\Sigma[/tex]T = 0
    [tex]\Sigma[/tex]T = -Tthumb + Tfingers - Tplate - Tcup
    0 = (-Tthumb*lthumb) + (Ffingers*lfingers) - (Wplate*lplate) - (Wcup*lcup)

    I'm not sure where to go from here since I don't have Tthumb or Ffingers.
    Also I'm unsure how to find the lever arms because if the pivot point is where I think it is (at the end of the tray in the hand) would they just be the distances given?

    If so, it would look like this

    0 = (-Tthumb*.0600) + (Ffingers*.100) - (.240) - (.0893)

    Attached Files:

  2. jcsd
  3. Mar 9, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Don't forget the weight of the tray itself.

    You'll need two equations to solve for those two unknowns. The net torque = 0 is one of them. What's another condition for equilibrium?
    Since the tray is in equilibrium, the net torque about any point must be zero. So it doesn't matter which point you choose as the reference for computing torques.
  4. Mar 9, 2009 #3
    Thanks for the help, I figured it out. After I factored in the weight of the tray and found T in terms of F it was easy.
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