Two people carry a heavy electric motor by placing it on a light board 3.00 m lo

Question

Two people carry a heavy electric motor by placing it on a light board 3.00 m long. One person lifts at one end with a force of 250 N, and the other lifts the opposite end with a force of 900 N.

(a)What is the weight of the motor?
N

Where along the board is its center of gravity located?
m (from the end where the 250 N force is applied)

(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 force as before. What is the weight of the motor in this case?
N

Where is its center of gravity located?
m (from the end where the 250 N force is applied)

Explanation / Answer

part a)

weight of the motor is equal to applied forces

weight = 250 + 900

weight = 1150 N

x = lF2 /(F1+F2)

x = 3*900 / (250+900)

x = 2.35 m from the end where 250 N force is applied

part (b)

if board is not light

then weight = F1 + F2 - Fo

weight = 250 + 900 - 200 = 950 N

here sum of all torque is zero at the end of board where 250 N acts

we have to consider the torque due to the weight of the board, this will act at the midpoint of the board

(900)3 = (200)*1.5 + 950*x

x = 2.53 m from the end where 250 N force applied

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