[SOLVED] Rigid Objects in Equilibrium & Center of Gravity 1. The problem statement, all variables and given/known data http://img149.imageshack.us/img149/8568/0968lr6sc4.gif [Broken] A lunch tray is being held in one hand, as the drawing illustrates. The mass of the tray itself is 0.220 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 0.350 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. T = ___N (downward) F = ___N (upward) 2. Relevant equations Torque = force * distance 3. The attempt at a solution Mug 0.350*9.8* 0.300= 1.029 Nm Plate 1.00*9.8*0.140= 1.372 Nm Tray 0.220*9.8*0.100= 0.2156 Nm Total Clockwise torque = 2.6166 Nm 2.6166 = 0.04T T = 65.415 N - wrong answer And I have no idea how to solve for F. Please help me.