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Rigid Objects in Equilibrium & Center of Gravity

  1. Feb 29, 2008 #1
    [SOLVED] Rigid Objects in Equilibrium & Center of Gravity

    1. The problem statement, all variables and given/known data

    [​IMG]

    A lunch tray is being held in one hand, as the drawing illustrates. The mass of the tray itself is 0.220 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 0.350 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
    Torque = force * distance

    3. The attempt at a solution
    Mug 0.350*9.8* 0.300= 1.029 Nm
    Plate 1.00*9.8*0.140= 1.372 Nm
    Tray 0.220*9.8*0.100= 0.2156 Nm
    Total Clockwise torque = 2.6166 Nm

    2.6166 = 0.04T
    T = 65.415 N - wrong answer

    And I have no idea how to solve for F. Please help me.
     
    Last edited: Feb 29, 2008
  2. jcsd
  3. Feb 29, 2008 #2

    Doc Al

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    Staff: Mentor

    Check that distance.

    You can chose another pivot point to calculate torques or you can consider the sum of vertical forces.
     
  4. Feb 29, 2008 #3
    Do I use the distance between the thumb & the mug? = 0.320 m
     
  5. Feb 29, 2008 #4

    Doc Al

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    Staff: Mentor

    You must measure all distances from the same point when calculating torques. Based on your other calculations, I thought you were using the point of application of force F as your pivot point.
     
  6. Feb 29, 2008 #5
    Using distance measure from the same spot:

    Mug: 0.350 x 9.8 x 0.380 = 1.3034
    Plate: 1.00 x 9.8 x 0.240 = 2.352
    Tray: 0.220 x 9.8 x 0.400 = 0.8624

    Total = 4.5178 Nm
    4.5178Nm / 0.06 m = 75.297 N for T?
     
  7. Feb 29, 2008 #6

    Doc Al

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    Staff: Mentor

    For some reason, you changed your pivot point to be the left side of the tray. In that case: (1) T also provides a clockwise torque; (2) You need to include the torque from F (which would be counterclockwise).

    Combine this torque equation (once you correct it) with an equation for the total force on the tray. Solve them together to find T and F.

    Your original choice of using the point of application of force F as your pivot point was a good one, since it eliminates force F and allows you to immediately calculate T. (Your only error was using an incorrect distance for one of the forces.) You'd still need another equation to find F.
     
  8. Feb 29, 2008 #7
    Thank you for your explanation. I understood it now.

    Solve for T:
    Mug 0.350*9.8* 0.280= 0.9604 Nm
    Plate 1.00*9.8*0.140= 1.372 Nm
    Tray 0.220*9.8*0.100= 0.2156 Nm
    Total Clockwise torque = 2.548 Nm

    2.548 = 0.04T
    T = 63.7 N

    Solve for F:
    F is the upward force so it must be equal to the sum of the down forces
    Mug 0.35*9.8 = 3.43 N
    Plate 1.00*9.8 = 9.8 N
    Tray 0.220*9.8 = 2.156 N
    Upward force = 63.7 N

    F = 3.43 + 9.8 + 2.156 + 63.7
    F = 79.086 N
     
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