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**Table Of Content**

## Definition

**Apparent weight** of a today is the weight which does not actually exist. But it appears when the body is kept in a lift or elevator.

## Introduction

We know that earth attracts everybody towards its centre with some force of attraction called Gravity force. Due to this force of gravity, all the bodies experience their weight. If ‘m’ is the mass of a body then its weight is ‘mg’ (vertically downwards) which is equal to the gravity force. But if the body is on a plane which is accelerated up or down, then the force exerted by the body on the plane (that is, weight of the body) changes while the gravity force remains the same. In this situation, person realizes his changed weight (**apparent weight**)

When we stand on the earth surface, we press the earth downwards due to the gravity force. This downward force is our weight. Earth also exerts an upward force in reaction on our feet. Due to this reactionary force ‘R’, we experience our weight. Because we are habitual of it we do not notice it but whenever there is a change in the reactionary force on our feet, we feel a change in our weight for example our weight appears to be increased(**apparent weight**) in a lift accelerated upwards and decreased in lift accelerated downwards.

## Apparent weight of a man in a lift or elevator

Suppose a person of mass ‘m’ is standing on a weighing machine placed in a lift/elevator. The actual weight of the man is ‘mg’. This weight presses the machine. The machine also exerts a reactionary force ‘R’ on the person in upward direction where ‘R = W’ (Newton’s Third Law). Thus, two forces are acting on the person; gravity force ‘mg’ and reactionary force ‘R’. Since the two forces are in opposite directions, the net force on the person is given by:

F = mg – R (downwards)

From fig.(a),but the person is at rest (no acceleration) hence the net force on him should be zero, that is

F = mg – R = 0

R = mg

But

R = W

W = mg

That is, that **apparent weight** is equal to the actual weight of the person.

Now, suppose the elevator is moving uniformly in the upward directions. The acceleration of the lift be ‘a’ (say). Again, the net force on the person is F = mg – R (downwards) .Now the person has acceleration ‘a’ in upward direction. Hence according to Newton’s Second Law of Motion, net force on him is given by F = ma (upwards) or F = -ma (downwards).Thus refer fig.(b)

-ma = mg – R

R = mg + ma

As

R = w

w = m.(g + a)

Thus, in this condition, the **apparent weight** ‘W’ of the person is greater than the gravity force ‘mg’. That is, the person will experience his weight to be increased than in the normal state.

Now, suppose the lift is coming down with acceleration ‘a’ (fig (c)). Again, the net force on the person is

F = mg – R (downwards)

As the acceleration ‘a’ is directed downwards, hence the net force F = ma should also be directed downwards. That is

-ma = mg – R

R = mg + ma

But

R = W

W = mg – ma

W = m(g – a)

In this situation, the person feels his **apparent weight** to be decreased than the usual weight.

If the string of the descending lift is broken then it will fall freely under gravity. In this situation,

a = g

Then from above equation

W = mg – ma

W = mg – mg = 0

Now the machine will read zero or the body becomes weightless. In this state, the person will not feel even the weight of a body placed on his head. If the person drops a ball from his hand, the ball will not fall down on the floor of the lift, but will appear floating in air. This is because the ball and the lift both are falling under the same acceleration g.

If the acceleration of the descending lift is greater than the acceleration due to gravity ‘g’ then according to eqn.

W = mg – ma

The **apparent weight** of the person will be negative. Under this condition, the person will rise from the floor of the lift and stick to the ceiling of the lift.

## References

- Nootan Physics by Kumar Mitlal
- Pradeep’s Fundamental Physics (XI)