Problem 10.50 Two people carry a heavy electric motor by placing it on a light b

Question

Problem 10.50

Two people carry a heavy electric motor by placing it on a light board 1.90m long. One person lifts at one end with a force of 420N , and the other lifts the opposite end with a force of 620N

Part A

What is the weight of the motor?

Part B

Where along the board is its center of gravity located?

Part C

Suppose the board is not light but weighs 180N , 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?

Part D

Where is its center of gravity located?

Explanation / Answer

appy Sum of all the forces along Y axis is Zero

forces on to ends ( F1 and F2) acts upwards while force at centre (due to weigth) in downward direction

sp

F1 - W   +F2 = 0

W = F1+F2

W = 420 + 620

W = 1040 N -----------------<<<<<<<<<<<<<<<Answer to part A

---------------------------------

let x be the location of CG

so

use net torque T = 0

T1x1 = T2 (Y-x)

420 x = 620( 1.9-x)

420 x = 1178 - 620 x

1040 x = 1178

x = 1178/1040

x = 1.132 m

-----------------------------------------

------------------------------------

if Borad weighs 180 N

net W = 420 + 620 -180

W = 860 N

-----------------------------------

let x be the CG

then using sum of torques = 0

we have

420 * 0    - 180 *1    + 860 * x   + 620 (1.9) = 0

860 x = 1178 + 180

x = 1.58 m   is the new CG

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