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Stand on a weighing machine in a lift and watch the dial. It jumps up as the lift starts rising, settles back while the lift cruises, and dips as the lift slows near the top. Questions built on this test one idea: what the machine actually measures. This post follows a full lift ride with numbers, shows how to work backwards from a reading to the acceleration, and lists the traps.
What does a weighing machine actually measure?
It measures how hard you press on it, which by Newton's third law equals how hard it pushes back on you: the normal force N. OpenStax University Physics calls the bathroom scale an excellent example of a normal force, giving a reading of how much it must push upward to support an object. The dial is marked in kilograms only because the maker divided the force by g.
The NCERT Class 11 chapter on laws of motion warns, in its Points to Ponder, that the familiar mg = R holds only if the body is in equilibrium, and gives a body in an accelerated lift as the case where they differ. Its Exercise 4.13 asks for scale readings for a man in a lift, so this is squarely school material too. The JEE (Main) 2026 syllabus lists Newton's laws under Unit 3; check the syllabus for your exam year.
What is the one equation?
Take upward as positive. For a person of mass m in a lift with acceleration a:
N − mg = ma, so N = m(g + a).
The scale reads N/g in kilograms. If a is upward, the reading is high. If a is downward (a negative), it is low. If a = 0, it is normal, whatever the speed. OpenStax's own example works this for a 75.0 kg man, both while the lift accelerates upward and while it moves up at constant speed.
Worked example 1: one complete ride
A 60 kg student rides a lift from the ground floor upwards. Take g = 9.8 m/s². The lift's motion uses sample numbers:
0 to 2 s: speeds up at 1.5 m/s², from rest to 3 m/s.
2 to 10 s: moves up at a steady 3 m/s.
10 to 12 s: slows at 1.5 m/s² to rest.
Speeding up (a = +1.5 m/s²): N = 60 × (9.8 + 1.5) = 60 × 11.3 = 678 N. Reading = 678/9.8 ≈ 69.2 kg.
Steady speed (a = 0): N = 60 × 9.8 = 588 N. Reading = 60 kg.
Slowing down while still moving up (a = −1.5 m/s²): N = 60 × 8.3 = 498 N. Reading ≈ 50.8 kg.
The lift climbs ½ × 1.5 × 2² = 3 m, then 3 × 8 = 24 m, then another 3 m, a total of 30 m.
On the way down the pattern reverses. Starting downward means the acceleration points down, so the reading drops first. Stopping at the bottom means the acceleration points up, so the reading rises last.
Worked example 2: from the reading back to the motion
The same student sees the scale read 54 kg. What is the lift doing?
Normal force: N = 54 × 9.8 = 529.2 N.
a = N/m − g = 529.2/60 − 9.8 = 8.82 − 9.8 = −0.98 m/s².
The acceleration is 0.98 m/s² downward, one tenth of g.
There are two possible motions: moving down and speeding up, or moving up and slowing down. The reading alone cannot tell them apart. A question that asks "is the lift going up or down?" from a single reading has no single answer; a well-set question will also tell you the direction of motion.
This backwards task is harder than it looks. In one study, upper secondary students in groups tried to calculate a lift's acceleration from a scale, and the researchers reported multiple difficulties, even among high-achieving students, linked to weak grasp of the basic concepts.
Worked example 3: free fall and beyond
If the cable breaks and the lift falls freely, a = −g and N = m(g − g) = 0. The scale reads zero; the student feels weightless. This does not mean gravity has switched off. Gravity is the only force left, which is why everything falls together.
If the lift were pulled down faster than g, the formula gives a negative N. A scale cannot pull, so the reading stays at zero.
Why is "weight" such a slippery word?
Part of the confusion is the word itself. Igal Galili's review, Weight versus gravitational force (2001), describes two definitions in textbooks: weight as the gravitational force mg, and an operational weight, what a scale reads. He argues that keeping the two apart may improve science teaching. In a lift, mg never changes; the scale reading does. When a question says "apparent weight", it means the scale reading, N.
Common mistakes and how to avoid them
Using velocity to decide the reading. Use acceleration only.
Writing N = m(g − a) with a sign already inside a. Choose one convention, upward positive, and stick to it.
Thinking the reading changes at constant speed. It does not.
Adding a pseudo force while working from the ground. If you prefer the lift frame, our post on pseudo forces shows how, but never mix the two frames.
A practice routine
Day 1: Redo the full ride for a descending lift with the same numbers. Write the reading for each stage.
Day 2: Four "reading to acceleration" problems, giving both possible motions each time.
Day 3: Two spring-balance versions, with a bag hanging from a balance in a lift. The tension plays the role of N.
Day 4: Solve NCERT Exercise 4.13 and check each answer against N = m(g + a).
If the free-body diagram is the part that goes wrong, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning, so you can check each force and sign. Our guide to free-body diagrams covers the drawing habit.
Next time you ride a lift, predict what a scale would read at each stage before the doors close.
