Gtwo 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

(A) 1000 N

(B) 1.2 m from the 400 N force

Explanation:

length of the board (R) = 2 m

force at one end (F1) = 400 N

force at other end (F2) = 600 N

let the distance of force F1 from the motor be = r

let the distance of force F2 from the motor be = r

therefore r + r' = 2 m

(A) what is the weight of the motor?

    since the system is in equilibrium, the weight of the motor will be equal to the force applied by the two persons to lift the motor. therefore

weight of the motor = 600 + 400 = 1000 N

(B) where along the board is the center of gravity located?

 since the system is in equilibrium, the summation of the torque exerted by the 400 N (F1) will be equal to the torque exerted by the 600 N (F2)

torque exerted by the 400 N (F1) = torque exerted by the 600 N (F2)

 F1 x r = F2 x r'

 F1 x r = F2 x (2 - r)

400r = 600(2-r)

400r = 1200 - 600r

400r + 600r = 1200

 1000r = 1200

  r = 1200/1000

  r = 1.2 m

therefore the motor is 1.2 m from the 400 N force