Both forces act perpendicular to the tray, which is being held parallel to the ground.?

Question

A lunch tray is being held in one hand, as the drawing illustrates. The mass of the tray itself is 0.200 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 0.165 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)


http://www.webassign.net/CJ/09_68.gif

The first answer is correct (50.5) but the second is not and i can not figure it out... help anyone?

Answer 1

this is like a complicated see-saw. the fingers on the underside are the pivot point. the forces downward, times the distance from the pivot point, must sum to zero.

The three loads of/ on the tray are
tray = .200 x 9.8 x .100m = .196Nm
food = 1kg x 9.8 x .140m = 1.372Nm
cup = .165 x 9.8 x .280m = .453Nm
total = 2.02Nm

this is opposed by the thumb
TN x .040m = 2.02Nm
T = 50.5

The fingers support the entire downward loads
F = 50.5 + 2.02 = 52.52N