Rigid Bodies and Center of Gravity?

Question

A lunch tray is being held in one hand, as the drawing illustrates. The mass of the tray itself is 0.160 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 0.180 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.

http://i21.photobucket.com/albums/b272/Cajunboiler/09_68.gif

Find F in Newtons, and T in Newtons.

Okay, Whenever I worked this problem out, I kept getting the wrong answer (homework site is telling me it's the wrong answer). I know this has to do with External Torques equally 0, but I don't know if I'm plugging the numbers in correctly. Torque = r (distance) x F (Force) x Sin(theta). It would be greatly appreciated if anyone could tell me what are the correct numbers to plug into the equation.

Answer 1

Any point can be chosen as the pivot. I choose the fingers and I choose CCW torque to be negative.

Torque of thumb + tray + food + coffee = 0
-T*.04 + .16*g*.1 + 1.0*g*.14 + .18*g*.28 = 0

T = g*(.016 + .14 + .05)/.04 = 50.5 Nt

The fingers have to be providing the upward force at the chosen pivot to keep the whole thing up. So
F = (.16 + 1.0 + .18)*g = 13.1 Nt

Answer 2

This is a question about torques, as you have said, but you need some relative positions to solve this problem. You need to know where the coffee and the food are, and also where the fingers are.

Answer 3

To keep equilibrium around 4 fingers there must be:
(g=9.8 m/s^2)
T*(0.1-0.06)+(0.16/0.4)*0.1*g*(0.1/2)=
=1*g*0.24+0.18*g*0.38+(0.16/0.4)*(0.4-0.1)*g*(0.4-0.1)/2.
Thus T=79.48 N
Now F=2*T; awful!! athlete of a waiter. Check it.