What does a weighing scale actually measure?
A bathroom scale does not measure gravity. It measures the force you press into its top surface — the spring (or load cell) inside compresses under that push, and the dial converts the deflection into kilograms or newtons. By Newton's third law, the force you exert downward on the scale equals the force the scale exerts upward on your feet — that upward force is the normal reaction \(N\).
The displayed number is the magnitude of \(N\). On a stationary floor, \(N\) happens to equal \(mg\), so the scale reads your true weight. The moment any vertical acceleration enters the picture, \(N\) and \(mg\) part ways. The scale faithfully reports \(N\) — your apparent weight — but stops reporting your true weight.
True weight \(W = mg\) is the gravitational force; it depends only on mass and the local gravitational field, both essentially unchanged inside a lift. Apparent weight depends on the lift's acceleration and can be larger, smaller or even zero.
The lift FBD
Draw a person of mass \(m\) standing on the scale inside a lift. Only two vertical forces act on the person:
- Weight \(mg\) downward (real gravitational force).
- Normal force \(N\) upward (contact force from the scale; the scale reading).
Choose upward as positive and let \(a\) denote the lift's acceleration (positive up, negative down). Newton's second law for the person:
\[N - mg = ma \quad\Longrightarrow\quad \boxed{\,N = m(g + a)\,}\]
This single formula handles every lift problem you will meet at NEET.
Trap 1 · True weight and apparent weight are not the same thing
"Weight" in everyday English means whatever the scale says. In physics, true weight \(W = mg\) is the gravitational force; apparent weight is the normal reaction \(N\). They coincide only when the vertical acceleration is zero. The scale reads kilograms by dividing \(N\) by \(g\) internally — so a "60 kg" reading in an accelerating lift does not mean your mass is 60 kg.
Read every lift question as: "find \(N\)", not "find weight". Use \(N = m(g + a)\) consistently.
The four canonical cases (NCERT Exercise 4.13)
NCERT's worked exercise puts a 70 kg passenger on a scale and runs the lift through four conditions. Take \(g = 10\) m/s\(^2\).
Case (a): Uniform upward velocity of 10 m/s
Constant velocity means \(a = 0\), so \(N = 70 \times 10 = 700\) N. Apparent weight equals true weight. The fact that the lift is moving upward is irrelevant — only acceleration matters.
Case (b): Downward acceleration of 5 m/s\(^2\)
Here \(a = -5\) m/s\(^2\), so \(N = 70 \times (10 - 5) = 350\) N — half the true weight. The passenger feels lighter because the floor is dropping away beneath them.
Case (c): Upward acceleration of 5 m/s\(^2\)
\(a = +5\), so \(N = 70 \times 15 = 1050\) N. The passenger feels 50% heavier; the floor must push hard enough to both support the weight and accelerate it upward at 5 m/s\(^2\).
Case (d): The cable snaps — free fall
Now \(a = -g = -10\) m/s\(^2\), so \(N = 70 \times 0 = 0\) N. The scale reads zero. The passenger is in apparent weightlessness — the most NEET-relevant phenomenon in this topic.
| Lift motion | Acceleration \(a\) | Scale reading \(N\) | Sensation |
|---|---|---|---|
| Stationary or uniform velocity (up or down) | \(0\) | \(mg\) | Normal weight |
| Accelerating upward (or decelerating while descending) | \(+a\) | \(m(g + a) > mg\) | Heavier |
| Accelerating downward (or decelerating while ascending) | \(-a\) | \(m(g - a) < mg\) | Lighter |
| Free fall (cable cut) | \(-g\) | \(0\) | Weightless |
Why uniform velocity gives no change
A lift cruising upward at 10 m/s feels exactly like a lift cruising downward at 10 m/s feels exactly like a stationary lift — provided each holds its speed steadily. Newton's first law: zero net force implies zero acceleration, which implies \(N = mg\). Inside a constant-velocity lift you are in an inertial frame; acceleration is the only thing your body and the scale can detect.
Trap 2 · Uniform velocity in a lift means apparent weight equals true weight
A favourite distractor stem reads: "The lift is moving upward with velocity 10 m/s. What does the scale read?" Students see "upward" and pick the heavier option \(m(g+a)\). But there is no \(a\) — just \(v\). The correct answer is \(mg\). Always extract the acceleration value from the stem before substituting; if only velocity is given (and it is uniform), \(a = 0\).
Velocity has no place in \(N = m(g + a)\). Only acceleration enters.
Why free fall produces weightlessness
When the cable snaps and the brakes fail, gravity is the only force on the cabin. The lift accelerates downward at \(g\), and the person inside also accelerates downward at \(g\). Both share the same acceleration, so there is no relative push between them. The scale's spring does not compress; the dial reads zero. Substituting \(a = -g\) into \(N = m(g + a)\) gives \(N = 0\).
Gravity has not switched off — your true weight \(mg\) is unchanged — but the scale cannot detect it because the scale is in free fall too. This is the definition of apparent weightlessness. You can demonstrate the same effect by dropping a half-full water bottle: while falling, the water does not press on the cap, because both bottle and water share the same \(g\).
Trap 3 · "Weightless" never means "no gravity"
A common wrong claim: "Astronauts are weightless because there is no gravity in space." False on two counts. (i) Gravity is the dominant force on the ISS — about 89% of its surface value. (ii) Apparent weightlessness happens because spacecraft and astronaut are in continuous free fall together, not because gravity is absent. Same logic for the person in a falling lift.
Free fall and apparent weightlessness are the same phenomenon. Gravity is fully present in both.
Effective gravity inside an accelerating lift
For an observer inside the lift, every object behaves as if gravity had a new value. Define an effective gravitational acceleration \(g_{\text{eff}} = g + a\) (with \(a\) signed). The apparent weight of any object of mass \(M\) inside the lift is then \(M\,g_{\text{eff}}\). A simple pendulum of length \(L\) has period
\[T = 2\pi\sqrt{\dfrac{L}{g_{\text{eff}}}} = 2\pi\sqrt{\dfrac{L}{g + a}}\]
If the lift accelerates upward, \(g_{\text{eff}} > g\) and the pendulum swings faster. If it accelerates downward, the period grows. In free fall, \(g_{\text{eff}} = 0\), the restoring torque vanishes, and the pendulum stops oscillating — it simply floats.
Astronauts in orbit: free fall around the Earth
The ISS orbits at about 400 km altitude where gravity is still ~8.7 m/s\(^2\) (only 11% weaker than at sea level). Astronauts feel no weight because the station is continuously falling toward Earth while moving sideways fast enough (~7.7 km/s) that the surface keeps curving away beneath it. Astronaut and station share the same gravitational acceleration, so the normal force between them is zero — just as in a falling lift. "Microgravity" is a more honest label than "zero gravity": gravity is doing all the work of keeping the station in orbit.
Worked examples
A 70 kg man stands on a scale in a lift. Find the reading in each case (g = 10): (a) uniform speed 10 m/s upward; (b) downward acceleration 5 m/s\(^2\); (c) upward acceleration 5 m/s\(^2\); (d) free fall.
Method. Use \(N = m(g + a)\) with \(a\) signed.
(a) \(a = 0 \Rightarrow N = 700\) N.
(b) \(a = -5 \Rightarrow N = 350\) N.
(c) \(a = +5 \Rightarrow N = 1050\) N.
(d) Free fall: \(a = -g \Rightarrow N = 0\). Apparent weightlessness.
A spring balance in a stationary lift reads 600 N for a passenger. What does it read if the lift accelerates downward at 2 m/s\(^2\)? (g = 10)
Step 1. Mass \(m = 600/10 = 60\) kg.
Step 2. With \(a = -2\), \(N = 60 \times 8 = 480\) N. The passenger feels 20% lighter.
A 50 kg person on a scale in a lift reads 600 N. Find the lift's acceleration. (g = 10)
Step 1. \(mg = 500\) N; the scale shows 600 N (greater), so acceleration is upward.
Step 2. \(600 = 50(10 + a) \Rightarrow a = +2\) m/s\(^2\) upward.
Trap 4 · Apparent weight cannot be negative
Suppose a question states that the lift accelerates downward at \(2g\), i.e. \(a = -2g\). Plugging into the formula gives \(N = m(g - 2g) = -mg\). A naive answer would be "apparent weight equals \(-mg\)". But a scale cannot pull on you — normal contact forces are always \(\geq 0\). Physically, the moment the cabin accelerates downward faster than \(g\), the floor falls away from the passenger, contact is lost, and \(N = 0\). The person floats inside the cabin until the cabin slows or the ceiling stops them.
The formula \(N = m(g + a)\) is valid only while contact holds. If it predicts \(N < 0\), the correct physical answer is \(N = 0\) (and the object has separated from the surface).
The whole topic on one card
- Scale reading = normal force = apparent weight; true weight \(mg\) is unchanged inside a lift.
- Master equation: \(N = m(g + a)\), with \(a\) positive upward.
- Uniform velocity: \(a = 0\), so \(N = mg\). Velocity never enters the formula.
- Free fall: \(a = -g\), so \(N = 0\) — apparent weightlessness; gravity is still acting.
- Astronauts in orbit are in continuous free fall around Earth.
- Effective gravity \(g_{\text{eff}} = g + a\); pendulum period uses \(g_{\text{eff}}\) in place of \(g\).
- Apparent weight cannot go negative; if \(N\) would be negative, contact is lost and \(N = 0\).