Physics help-Rotational Dynamics?

Question

Question 2:
A lunch tray is being held in one hand, as the drawing illustrates. The mass of the tray itself is 0.260 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 0.305 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. The length of the tray is .4m. The cup is .38m from the end. Plate is .24m from end. Fingers are .1m from the end and the thumb is .06m from the end. The force of the thumb is down and the force of the fingers is up

Answer 1

The tray is at rest and staying that way so the linear and the angular accelerations are both 0.

For the linear acceleration we have the sum of the forces up must equal the sum of the forces down, so

Wt + Wp + Wc + Ft = Ff where:
Wt is the weight of the tray
Wp is the weight of the plate of food
Wc is the weight of the cup of coffee
Ft is the force of the thumb
Ff is the force of the four fingers

To compute weight given mass, multiply by the gravitational acceleration (abut 9.8 m/s^2 at the surface of the Earth).

For the angular acceleration we can choose any point we want, but since we already have the lengths from "the end", we'll use that as our point. That means that all the forces are on the same side, so we have the same split into up and down.

But the sum is of the moments: forces times distances to the axis:

Wt x Ltray + Wp x Lp + Wc x Lc + Ft x Lthumb= Ff x Lf where:

Lt is the distance from the geometric center of the tray to the axis
Lp is the distance from the plate of food to the end of the tray
Lc is the distance from the cup of coffee to the end
Lthumb is the distance from the thumb to the end
Lf is the distance from the four fingers to the end

We know all the distances and all the weights, so we have two equations with two unknowns: Ft and Ff