Motion in a lift - Laws of Motion, Physics, Class 11 - eSaral

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

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

$W = Mg$

This never changes as long as your mass and $g$ remain constant.

Your apparent weight, however, is the force you feel — the normal reaction $N$ that the floor of the lift pushes back on you. When the lift is stationary or moving at constant velocity, $N = Mg$, so apparent weight equals actual weight.

The moment the lift accelerates, you are in a non-inertial (accelerated) reference frame. To analyse forces from inside the lift, we introduce a pseudo force (also called fictitious force) equal to:

$F_0 = Ma$

directed opposite to the direction of acceleration of the lift. This pseudo force changes the net force on you and therefore changes your apparent weight.

This concept falls under Chapter 5 — Laws of Motion in the Class 11 Physics NCERT syllabus, and it is one of the most frequently tested topics in JEE Main and NEET. For the full NCERT treatment, check the NCERT Solutions for Class 11 Physics on eSaral, where IIT Bombay faculty have solved every problem with detailed explanations.

All 4 Cases of Motion in a Lift — Explained

Case 1: What happens when the lift moves with constant velocity?

When a lift moves upward or downward at constant velocity, its acceleration is zero:

$a = 0$

There is no pseudo force. The net force equation along the vertical direction gives:

$N - Mg = 0$

Therefore:

$W' = N = Mg = W$

Apparent weight = Actual weight. The weighing scale shows your true mass. This is identical to the situation when the lift is stationary.

Key point for JEE: Constant velocity means zero acceleration — do not confuse speed with acceleration. A lift moving at 5 m/s upward with no change in speed has zero net force on the passenger.

Case 2: What happens when the lift accelerates upward?

When the lift accelerates upward with acceleration $a$, the passenger is in a non-inertial frame. The pseudo force:

$F_0 = Ma$

acts downward (opposite to the upward acceleration of the lift).

Forces on the passenger (taking downward as positive):

• Weight $W = Mg$ → downward
• Pseudo force $F_0 = Ma$ → downward
• Normal force $N$ → upward

For the passenger to be in equilibrium inside the accelerating lift frame:

$N = Mg + Ma = M(g+a)$

Therefore:

$W' = M(g+a)$

Effective gravitational acceleration:

$g' = g+a$

The passenger feels heavier. A 60 kg person in a lift accelerating upward at 2 m/s² would register an apparent weight of:

$60 \times (10+2) = 720\ N$

instead of 600 N.

This also applies when the lift decelerates while moving downward — the deceleration has an upward component, so the same formula holds.

Case 3: What happens when the lift accelerates downward?

When the lift accelerates downward with acceleration $a$ (where $a < g$), the pseudo force:

$F_0 = Ma$

acts upward (opposite to the downward acceleration).

Forces on the passenger:

• Weight $W = Mg$ → downward
• Pseudo force $F_0 = Ma$ → upward
• Normal force $N$ → upward

Equilibrium inside the lift frame:

$N = Mg - Ma = M(g-a)$

Therefore:

$W' = M(g-a)$

Effective gravitational acceleration:

$g' = g-a$

The passenger feels lighter. This is why you feel a sinking sensation in your stomach when a lift starts moving downward or when a fast lift decelerates at the top floor.

Special Case — Free Fall

$a = g$

$W' = M(g-g) = 0$

This is the condition of weightlessness. In a freely falling lift, the normal force between the passenger and the floor becomes zero. The passenger floats.

Case 4: What happens when the lift accelerates downward faster than free fall?

If the lift's downward acceleration exceeds $g$:

$a > g$

then:

$W' = M(g-a)$

Since $a > g$, $(g-a)$ becomes negative.

This means the normal force from the floor is negative. The passenger is pressed against the ceiling of the lift.

The ceiling exerts a downward normal force on the passenger, and the passenger experiences an upward apparent weight of:

$M(a-g)$

In JEE problems, this case is often presented as: "What force does the ceiling exert on the passenger?" The answer is:

$M(a-g)$

directed downward at the passenger.

Quick Reference Table: All Cases at a Glance

How Do You Solve Any Lift Problem Step by Step?

Step 1 — Identify the Direction of Acceleration

Is the lift speeding up or slowing down? Which direction is it moving? A lift moving upward but slowing down has a downward acceleration.

Step 2 — Draw the Free Body Diagram (FBD)

Show weight $Mg$ downward and normal force $N$ upward. If working in the non-inertial frame, also show the pseudo force $Ma$ opposite to the lift's acceleration.

Step 3 — Apply Newton's Second Law

From the ground (inertial) frame:

$\Sigma F = Ma$

For upward acceleration:

$N - Mg = Ma$

$N = M(g+a)$

For downward acceleration:

$Mg - N = Ma$

$N = M(g-a)$

Step 4 — Interpret the Answer

$N$ is what the weighing scale reads (in Newtons). Divide by $g$ to get the scale reading in kg.

Solved Example

Problem: 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 moves upward with (i) constant velocity, and (ii) constant acceleration of 3 m/s²? (Take $g = 10\ m/s^2$)

Solution

Given:

$m = 50\ kg$

$g = 10\ m/s^2$

(i) Constant Velocity ($a = 0$)

There is no net acceleration.

$N - mg = 0$

$N = mg = 50 \times 10 = 500\ N$

Scale reading:

$\frac{500}{10} = 50\ kg$

Answer: 50 kg (unchanged)

(ii) Constant Upward Acceleration ($a = 3\ m/s^2$)

Applying Newton's Second Law:

$N - mg = ma$

$N = m(g+a)$

$N = 50(10+3)$

$N = 50 \times 13$

$N = 650\ N$

Scale reading:

$\frac{650}{10}=65\ kg$

Answer: 65 kg

The scale reads 15 kg more than actual weight because the lift is accelerating upward.

Verification Using Pseudo Force Method

Pseudo force:

$ma = 50 \times 3 = 150\ N$

Total downward force:

$mg + ma = 500 + 150 = 650\ N$

$N = 650\ N$ ✓

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