Rigid Objects in Equilibrium & Center of Gravity

[tem_osu]

[SOLVED] Rigid Objects in Equilibrium & Center of Gravity

Homework Statement

http://img149.imageshack.us/img149/8568/0968lr6sc4.gif

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)

Homework Equations

Torque = force * distance

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.

 

[Doc Al]

Mug 0.350*9.8* 0.300= 1.029 Nm

Check that distance.

And I have no idea how to solve for F.

You can chose another pivot point to calculate torques or you can consider the sum of vertical forces.

 

[tem_osu]

Check that distance.

Do I use the distance between the thumb & the mug? = 0.320 m

 

[Doc Al]

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.

 

[tem_osu]

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?

 

[Doc Al]

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?

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.

 

[tem_osu]

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