Motion in a Lift: Apparent Weight, Pseudo Force and All Cases (Class 11 Physics)

Motion in a lift is one of the most common applications of Newton's laws in Class 11 Physics. It belongs to the chapter Laws of Motion (Chapter 4 of the current NCERT book), and it asks one question: what does a weighing scale in a lift read? For the chapter questions, see eSaral's NCERT Solutions for Class 11 Physics Chapter 4.

What Is Apparent Weight in a Lift, and Why Does It Change?

Your actual weight is the gravitational force of the Earth on you:

$$W = Mg$$

It does not change as long as $M$ and $g$ are the same. Your apparent weight is the force that you feel, which is the normal reaction $N$ of the floor of the lift on you, and it is the value that a weighing scale shows. When the lift is at rest or moves at a constant velocity, $N = Mg$, so the apparent weight is equal to the actual weight.

When the lift accelerates, the lift is a non-inertial (accelerated) frame. To analyse the forces from inside the lift, we add a pseudo force

$$F_0 = Ma$$

on every body, in the direction opposite to the acceleration of the lift. This changes the net force on you, and so it changes your apparent weight. (In the ground frame, which is an inertial frame, there is no pseudo force: you use Newton's second law, $\Sigma F = Ma$, directly.)

Motion in a Lift: All the Cases

These are the cases of motion in a lift that you need for JEE and NEET.

Case 1: The lift moves with a constant velocity

When a lift moves up or down at a constant velocity, $a = 0$, and there is no pseudo force:

$$N - Mg = 0 \;\Rightarrow\; W' = N = Mg = W$$

The apparent weight equals the actual weight, as when the lift is at rest. Note that a constant speed means zero acceleration: a lift that moves upward at 5 m/s without any change in speed has no net force on the passenger.

Case 2: The lift accelerates upward

For an upward acceleration $a$, the pseudo force $F_0 = Ma$ acts downward. In the lift frame, the forces on the passenger are $Mg$ and $Ma$ downward, and $N$ upward:

$$N = Mg + Ma = M(g + a)$$ $$W' = M(g + a), \qquad g' = g + a$$

The passenger feels heavier. A 60 kg person in a lift that accelerates upward at $2\ \text{m/s}^2$ has an apparent weight of $60\times(10 + 2) = 720$ N, instead of 600 N. The same formula holds when the lift decelerates while it moves downward, because the acceleration is then upward.

Case 3: The lift accelerates downward ($a < g$)

For a downward acceleration $a$, the pseudo force acts upward:

$$N = Mg - Ma = M(g - a)$$ $$W' = M(g - a), \qquad g' = g - a$$

The passenger feels lighter. This is the sinking feeling when a lift starts to go down, or when a lift going up slows down for the top floor.

Special case: free fall. If $a = g$, then $W' = M(g - g) = 0$. The normal force is zero, and the passenger floats: this is weightlessness.

Case 4: The lift accelerates downward faster than free fall ($a > g$)

If $a > g$, then $M(g - a)$ is negative. The floor cannot pull the passenger down, so the passenger is pressed against the ceiling, and the ceiling exerts a downward force

$$N_{ceiling} = M(a - g)$$

This can happen only if something other than gravity, such as a cable, pulls the lift downward. JEE problems often ask: "What force does the ceiling exert on the passenger?"

Motion in a Lift: Quick Reference Table

How to Solve Motion in a Lift Problems Step by Step

1. Find the direction of the acceleration. A lift that moves upward but slows down has a downward acceleration. A lift that moves downward but slows down has an upward acceleration.
2. Draw the free-body diagram. Show $Mg$ downward and $N$ upward. In the lift frame, also show the pseudo force $Ma$ opposite to the acceleration.
3. Apply Newton's second law from the ground frame:
4. • Upward acceleration: $N - Mg = Ma \Rightarrow N = M(g + a)$
   • Downward acceleration: $Mg - N = Ma \Rightarrow N = M(g - a)$
4. Interpret the answer. $N$ is the reading of the scale in newtons. Divide by $g$ for the reading in kg.

Motion in a Lift: Effective Gravity

In motion in a lift problems, everything inside an accelerating lift behaves as if gravity were $g' = g \pm a$ ($+$ for an upward acceleration, $-$ for a downward acceleration). This gives quick answers to other questions:

• A body dropped inside the lift falls with an acceleration $g'$ relative to the lift, so the time to fall from a height $h$ is $t = \sqrt{\dfrac{2h}{g'}}$.
• A simple pendulum has the time period $T = 2\pi\sqrt{\dfrac{l}{g'}}$.
• The tension in the cable of a lift of mass $M$ with a load $m$ is $T = (M + m)\,g'$.
• A spring balance reads $m g'$.

Solved Examples on Motion in a Lift

These solved examples of motion in a lift use the formulas above.

Example 1: A spring balance in a lift

A spring balance in a stationary lift reads 50 kg when a man stands on it. What does it read if the lift moves upward with (i) a constant velocity, and (ii) a constant acceleration of $3\ \text{m/s}^2$? ($g = 10\ \text{m/s}^2$)

(i) Constant velocity: $a = 0$, so $N = mg = 50\times 10 = 500$ N, and the reading is $\dfrac{500}{10} = 50$ kg.

(ii) Upward acceleration $3\ \text{m/s}^2$:

$$N = m(g + a) = 50\times(10 + 3) = 650\ \text{N}, \qquad \text{reading} = \frac{650}{10} = 65\ \text{kg}$$

Check with the pseudo force: $ma = 50\times 3 = 150$ N, and $mg + ma = 500 + 150 = 650$ N, which is $N$.

Example 2: Apparent weight in a decelerating lift

A man of mass 60 kg is in a lift that moves downward and slows down at $2\ \text{m/s}^2$. Find his apparent weight. ($g = 10\ \text{m/s}^2$)

The lift moves down and slows down, so the acceleration is upward:

$$N = M(g + a) = 60\times(10 + 2) = 720\ \text{N}$$

Example 3: Tension in the cable

A lift of mass 500 kg carries a load of 200 kg. Find the tension in the cable when the lift accelerates upward at $2\ \text{m/s}^2$. ($g = 10\ \text{m/s}^2$)

$$T = (M + m)(g + a) = 700\times 12 = 8400\ \text{N}$$

Example 4: A pendulum in a lift

A simple pendulum of length 1 m is in a lift that accelerates upward at $3\ \text{m/s}^2$. Find its time period. ($g = 10\ \text{m/s}^2$)

$$T = 2\pi\sqrt{\frac{l}{g + a}} = 2\pi\sqrt{\frac{1}{13}} \approx 1.74\ \text{s}$$

In a stationary lift, it is $2\pi\sqrt{1/10} \approx 1.99$ s, so the time period is shorter when the lift accelerates upward.

Example 5: A ball dropped in a lift

A ball is dropped from a height of 2.5 m inside a lift that accelerates downward at $5\ \text{m/s}^2$. How long does it take to reach the floor? ($g = 10\ \text{m/s}^2$)

The acceleration of the ball relative to the lift is $g' = g - a = 5\ \text{m/s}^2$:

$$t = \sqrt{\frac{2h}{g'}} = \sqrt{\frac{2\times 2.5}{5}} = 1\ \text{s}$$

Common Mistakes in Motion in a Lift Problems

These mistakes cost marks in motion in a lift questions.

More on Laws of Motion

To go deeper into motion in a lift and accelerated frames, see eSaral's page on the frame of reference, and the complete Class 11 Physics notes. To watch free physics videos, install the eSaral App.