A lunch tray is being held in one hand, as the drawing illustrates. The mass of the tray itself is 0.180 kg, and its center of gravity is located at its geometrical center. On the tray is a 1-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. I don't even have a clue where to start?? Here is what I tried though, but I'm not sure: Sum of Counter-clockwise Torques = Sum of Clockwise Torques Tf = Tplate + Tcup + Ttray then: Ff(.1m) = (1kg*9.80m/s^2)(.24m)+(.35kg*9.80m/s^2*.38m)+(.18kg*9.80m/s^2*.2m) and ended up with F being equal to: 40.1... but I'm not sure?