Apparent Weight Of A Man In A Lift

Let we consider a man of mass m is standing on a weighting machine placed in an elevator or lift. The actual weight mg of the man acts on the weighting machine and offers a reaction R given by the reading of the weighting machine.

This reaction R exerted by the surface of contact on the man is the apparent weight of the man.

Now, we consider how R is related to mg in the different conditions.

1. When the lift moves upwards with acceleration a as given, the net upward force on the man is

R – mg = ma ⇒ R = ma + mg

Apparent Weight R = m(a+g)

So, when a lift accelerate upward, the apparent weight of the man inside it increases.

2. When the lift moves downward with acceleration a as given, the net downward force on man is

mg – R = ma

Apparent Weight R = mg – ma ⇒ R = m(g-a)

So, when a lift accelerate downward, the apparent weight of the man inside it decreases.

3. When the lift is at rest or moving with uniform velocity v, downward or upward

The net force on man, R – mg = ma

R – mg = m×0

R = mg, Then

Apparent Weight = Actual Weight

So, when the lift is at rest then the apparent weight of the man is the actual weight of the man.

4. When the lift falls freely under gravity, if the supporting cable of the lift breaks.

Then, a = g, So the net downward force on the man is

mg – R = ma

R = mg – ma

R = mg – mg [Since, a = g], So R = 0

Thus, the appearantee of the man becomes zero.

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