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 0.150 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.

Answer 1

As, the diagram is not given in the question, so let the distance are L1, L2, L3, L4, and L5.

This is a torque question. The sum of the counterclockwise(CCW) torques must be equal to the sum of the clockwise torques(CW). You can set the point of rotation anywhere you want.

Let's solve for F first. Set the point of rotation at L1 m from the edge of the tray, where T acts.

As, Torque = fdsinx, but x is 90 degrees for all of them, so sin90=1 . You have to subtract L1 m from every distance shown (because you want the distance from the T force, not the end of the tray).

The torque by F is the only CW torque. The rest are CCW, from the tray, food, and coffee.

F* d = W_tray*d_tray + W_food*d_food + W_coffee*d_coffee

F* (L2-L1) = (0.200kg)(9.8m/s^2)(L3-L1) + (1.00kg)(9.8m/s^2)(L4-L1) + (0.150kg)(9.8m/s^2)(L5-L1)

This will give value of F (up), the net torque about an axis through the contact point between the tray and the thumb.

Now, you could use torques to solve for T, but it's easier to use forces.
Sum of forces up = sum of forces down
F = T + W_tray + W_food + W_coffee

T = F - W_tray - W_food - W_coffee
T = F - (0.200kg)(9.8m/s^2) - (1.00kg)(9.8m/s^2) - (0.150kg)(9.8m/s^2)

And, this will give net torque about an axis through the point of contact between the tray and the finger.