## Motion in a Lift

The weight of a body is simply the force exerted by the earth on the body. If the body is on an accelerated platform, the body experiences a fictitious force, so the weight of the body appears changed and this new weight is called apparent weight. Let a man of weight W = Mg be standing in a lift.We consider the following cases:

**Case (a)**

If the lift moving with constant velocity v upwards or downwards.

In this case, there is no accelerated motion hence no pseudo force experienced by observer $0^{\prime}$ inside the lift.

So apparent weight $\mathrm{W}^{\prime}$ = Actual weight W.

**Case (b)**

If the lift is accelerated (i.e. a = constant upward) : Then net forces acting on the man are

- weight W = Mg downward
- fictitious force $\mathrm{F}_{0}$ = Ma downward.

So apparent weight

$W^{\prime}=W+F_{0}$

Or

$\mathrm{W}^{\prime}=\mathrm{Mg}+\mathrm{Ma}$

$=M(g+a)$

Effective gravitational acceleration

$g^{\prime}=g+a$

**Case (c)**

If the lift is accelerated downward with acceleration $a<g$: Then fictitious force $\mathrm{F}_{0}$ = Ma acts upward while the weight of man W = Mg always acts downward, therefore

So apparent weight

$W^{\prime}=W+F_{0}$

Or

$\mathrm{W}^{\prime}=\mathrm{Mg}-\mathrm{Ma}$

$=M(g-a)$

Effective gravitational acceleration

$g^{\prime}=g-a$

**Special Case:**

If g = a then

$W^{\prime}=0$ cond $^{n}$. Of weightlessness.

Thus, in a freely falling lift the man will experience weightlessness.

**Case (d)**

If lift accelerates downward with acceleration a > g

Then as in Case c

Apparent weight $W^{\prime}=M(g-a)$ = $-M(a-g)$ is negative, i.e., the man will be accelerated upward and will stay at the ceiling of the lift.

**Ex.**A spring weighing machine inside a stationary lift reads 50 kg when a man stands on it. What would happen to the scale reading if the lift is moving upward with (i) constant velocity, and (ii) constant acceleration?

**Sol.**(i) In the case of a constant velocity of lift, there is no fictitious force; therefore, the apparent weight = actual weight. Hence the reading of the machine is 50 kg wt.

(ii) In this case the acceleration is upward, the fictitious force R = ma acts downward, therefore apparent weight is more than actual weight i.e. $\mathrm{W}^{\prime}$ = W + R = m (g + a).

Hence scale shows a reading = m (g + a) newton

$=\frac{m g\left(1+\frac{a}{g}\right)}{g} k g$

$=\left(50+\frac{50 a}{g}\right) k g.w t$

So, that’s all from this blog. I hope you get the idea about the Motion in a lift. If you found this article helpful then don’t forget to share it with your friends.

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Frame of reference

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