Two people carry a heavy electric motor by placing it on a light board 1.80 m lo

Question

Two people carry a heavy electric motor by placing it on a light board 1.80 m long. One person lifts at one end with a force of 420 N , and the other lifts the opposite end with a force of 700 N . What is the weight of the motor? Where along the board is its center of gravity located? Suppose the board is not light but weighs 210 N , with its center of gravity at its center, and the two people each exert the same forces as before. What is the weight of the motor in this case? Where is its center of gravity located?

Explanation / Answer

F1 = 420 N

F2 = 700 N

Fo = 210 N

l = 1.80 m

Wm = 420 + 700 = 1120 N (F = 0)

x = lF2 / (F1 + F2)

x = (1.80 x 700) / 1120

x = 1.125 m

from the end where the 420­N force is applied

w = F1 + F2 - Fo = 1120 - 210 = 910 N

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