# Elevator Problem

The weight of a book was measured using a scale. The scale showed 15N. Next, the book and the scale were placed in an elevator that does not show the floor number. When the elevator started moving, the reading of the scale has changed, but later was reading 15N again. Using the fact that the scale shows the same value in the elevator, can you tell if the elevator is moving upward or downward? What kind of information would you need to make that assessment? Explain

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

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Let us consider an object with mass m.

The weight of the object is given as

W=mg

If the object is placed inside an elevator, then the apparent weight of the object is given as follows:

N=W+W^'

where,W=mg,the actual weight of the body

and W^'=ma,where a=acceleration of the elevator

Case -I: When the elevator is at rest or is moving with constant velocity

When the elevator is at rest or is moving with constant velocity, the acceleration of the elevator is zero.

The normal force acting on the object placed inside the elevator is given as,

N=W+W^'

Or,N=mg+ma [since a=0]

Or,N=mg+0

Or,N=mg

The normal force is equal to the apparent weight of the object.

Therefore, the apparent weight of the object placed in an elevator, which is either is at rest or is moving with constant velocity, is same as its actual weight.

Case -II: When the elevator is moving upward with a constant acceleration

The normal force acting on the object placed inside the elevator is given as follows:

N=W+W^'

Or,N=mg+ma

Or,N=m(g+a)

Therefore, the apparent weight of the object placed in an elevator, which is moving upward with a constant acceleration, is greater than the actual weight of the object.

Case -III: When the elevator is moving downward with a constant acceleration

The normal force acting on the object placed inside the elevator is given as follows:

N=W+W^'

Or,N=mg+m(-a)

Or,N=mg-ma

Or,N=m(g-a)

Therefore, the apparent weight of the object placed in an elevator, which is moving downward with a constant acceleration, is less than the actual weight of the object.

In all the above conditions, once the object is taken outside the elevator, its weight will again be equal to the actual weight.

In the given problem, we have been told that the weight of the object changes as the elevator starts moving. Therefore, we can conclude that the elevator is moving with a constant acceleration. Since, had the elevator been moving with a constant velocity or zero acceleration, the apparent weight of the object would have remained the same. But based on the given information, we cannot conclude whether the elevator is moving upward or downward. We need to know whether the weight of the object increased or decreased inside the moving elevator, in order to access whether the elevator is moving upward or downward.

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