Two people carry a heavy electric motor by placing it on a light board 1.50{\ m

Question

Two people carry a heavy electric motor by placing it on a light board 1.50{ m m} long. One person lifts at one end with a force of 370{ m N} , and the other lifts the opposite end with a force of 680{ m N} . Part A What is the weight of the motor? Part B Where along the board is its center of gravity located? Answer includes " { m m} from the end where the 370-{ m N} force is applied" at the end of it Part C Suppose the board is not light but weighs 200{ m 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? Part D Where is its center of gravity located? Answer includes " { m m} from the end where the 370-{ m N} force is applied" at the end of it

Explanation / Answer

The weight of the motor is simply the sum off the two lifting forces: 630+450. The center of mass would be found by calculating the torques of the lifting forces on each end minus the motor's torque for each. Note that the torques cancel out and you are left with no net force, which is equilibrium. Drawing a diagram would help. When the board weighs 190 N, the motor's original weight is now reduced by that amount. This is probably confusing, but you'll get it.

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