Question

Two people carry a heavy electric motor by placing it on a light board 2.00 m long. One person lifts at one end with a force of 400 N, and the other lifts the opposite end with a force of 600 N. (a) What is the weight of the motor, and where along the board is its center of gravity located? (b) Suppose the board is not light but weighs 200 N, with its center of gravity at its center, and the two people exert the same forces as before. What is the weight of the motor in this case, and where is its center of gravity located?

Answer 1

a) W = 600 N + 400 N = 1000 N to balance upwards and downwards forces giving W (weight of motor) = 1000 N

Let X be the distance between 400 N and the weight of the motor

W X = 600 * 2 to balance torques

1000 X = 1200 so X = 1.2 m and the person at 400 N is 1.2 m from the center of mass

Check: 1000 * 1.2 = 600 * 2 and 1200 = 1200

(b) again taking torque about the left end

2 * 600 = W * X + 200 * 1 taking torque about the left end

1200 = 1000 X + 200 * 1 balancing torque about left end

X = 1000 / 1000 = 1 so the motor would be at 1 m from the left end

1000 * 1 + 200 * 1 = 600 * 2

1200 = 1200