Do Scales Work In Elevators?

Classical Mechanics Level 1

A boy stands on a scale in an elevator (lift) while it is stopped and the scale reads 40 kg. What will the scale display when the elevator begins to ascend?

More than 40 kg Less than 40 kg Exactly 40 kg

Krishna Kanth Dasari by Krishna Kanth Dasari , 29, India

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2 solutions

When the boy is on the ground only 'acceleration due to gravity' is acting on him so his weight is simply

Mass * Acceleration due to gravity

But when the lift started moving up there is one more acceleration acting on the body due to his inertia, in the direction opposite to the motion. so now the new weight is

Mass * (Acceleration due to gravity + Acceleration due to inertia)

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what if the elevator is moving at a constant velocity?

Paris Patel - 5 years, 8 months ago

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Thanks Patel for your comment. Think the elevator is on ground level with velocity zero. It started moving up with some constant velocity. You see there is a change in velocity. This means that the elevator has accelerated at least for some time. During this time the weight shown in the weighing machine will reduce. this new weight will remain constant once the acceleration is stopped (constant velocity is achieved). If we stop the elevator from that constant velocity, then there is a sure deceleration. In this case the weight starts increasing and comes original value when the elevator is stopped.

Now you see there is only change in weight only if acceleration or deceleration took place. The weight will increase if we accelerate and reduce if we decelerate...

Krishna Kanth Dasari - 5 years, 8 months ago

But the scale is also going up with the elevator (with same acceleration) so how will the scale calculate the extra acceleration?

Ahmad Sofiullah - 5 years, 8 months ago

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Scale shows the reading of net forces externally applied on it. So when the body is in upward motion with constantly increasing velocity, due to inertia and the body's own weight, force on the scale will be m(a+g)

Shreyansh Ajmera - 5 years, 8 months ago

I think the answer to this question is still 40 kg. Remember that the gravity only affects the weight of an object, not its mass.

Reynald Jose - 5 years, 8 months ago

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Mass is constant, no doubt but the forces are changing.

Shreyansh Ajmera - 5 years, 8 months ago

Correct, but a scale doesn't truly measure mass; it measures weight. Granted it would me more precise to convert to newtons or use a unit with slightly a slightly less confusing connotation surrounding it such as lbf.

Adam Bachmann - 5 years, 8 months ago

The weight of the body is nothing but product of mass to acceleration (Acceleration due to gravity)... As all of us know the mass is constant. But there is a change in the acceleration wright. this will change ur weight...

Krishna Kanth Dasari - 5 years, 8 months ago

Since the answer is in kilograms, you are actually asking about what happens to the mass. Since mass doesn't change, the answer correct answer should be "exactly 40 kg".

Brady Hornstra - 5 years, 8 months ago

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It is not asking about mass, nor is it asking about weight, it is asking about what the scale displays. And the scale incorrectly displays weight in kilograms.

Henk Elemans - 5 years, 8 months ago
Fhil Caballar
Oct 4, 2015

When at rest, the actual weight and the apparent weight are equal. AW = W = mg.
During this state, where the elevator is not moving, we can draw a free-body diagram where only two forces are involved. Normal Force and Weight. Since it is at rest. Fn-mg=0, therefore, Fn = mg

When the elevator is moving upward, so there is motion. F = ma. In our free-body diagram, only two forces are involved, hence, ma = Fn - mg, Fn = ma + mg, Fn = m( a + g ); a is positive because elevator is moving up.

Therefore, when the elevator is moving upward, the Apparent weight is more than our Actual weight.

Using Fn = m( a + g ) We can also deduce that when moving downward, where a is negative, the Apparent weight will be less than the actual weight.

But when the elevator is in a free fall, where a = g, the person inside will appear to be weightless.

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