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

Question

Two people carry a heavy electric motor by placing it on a light board 1.50 m long. One person lifts at one end with a force of 440 N , and the other lifts the opposite end with a force of 670 N .

1.What is the weight of the motor?

2.Where along the board is its center of gravity located?

3.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

4.Where is its center of gravity located?

Explanation / Answer

Key concepts;

1. We have to balance all the force on the board

2 . since board is in equilibrium so net torque must be zero , so we will balance all the torque.

3. We don't know the center of gravity so we will take torque of all forces about any one end.

4 . all the forces on board are in vertical direction

1.)

Net upward force = Nb + Na. = 670+ 440 = 1110 N

( I have used name B for right end and name A for left end)

Net downward force = weight of motor = w

( in part 1 and 2 board is assumed massless according to question)

So net force is zero means total upward force must be same in magnitude of total downward force

Thus w = 1110N

2.)

Let's take torque of all the three forces about left end A ( where 440 N normal is acting)

Also let's assume the distance of center of gravity from left end A ( where 440N force is acting ) is X.

There fore

Both force Nb and w are on the same side of end A but their directions are opposite so their torque will be in opposite directions.

And are taking torque about A so any force acting at A are passing through point A and hence their torque will be zero.

Thus

Torque of Nb + torque of w = 0

670×1.5 - 1110× X = 0

(670 N forceis acting at a perendired distance of 1.5 m from end A)

So X = 0.905m ( from that end where 440 N force is acting)

3.

. Now if board has mass of 180 N

It means net downward force = 180+w

But net upward force is same = Na +Nb = 1110N

So in this case weight of motor will be lesser

And equating both we get

Weight w= 830 N

4.

We know that Wight of any extended object acts on center on mass .and center of mass of the symmetrical board will be at center of board I.e. at a distance of 0.75 m from each end

Now again taking torque of all forces about end A ( where 440 N force is applied)

Torque of. Nb will be same but torque of w and weight of board will be different

Both weight of board and w are in same direction and on same side of A so they will exert torque in same direction ( which are opposite to the torque of Nb)

Torque of weight of board = 180 ×0.75

Torque of w = w×X

Balancing all torques we get

670×1.5 - 180×0.75 - w× X = 0

W= 830 N in this case ( we derived in questiOn 3)

Thus

X = 870/ 830 = 1.048 m from end A ( where force 440N is applied)

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