Elevator Physics Explained: Tension and Normal Force Secrets
JEE/NEET Physics · Laws of Motion series · Part 2 of 8 · All parts →
- Tension: a rope pulls along its length, same strength at both ends (ideal rope, massless)
- Normal force: a surface pushes perpendicular to itself — its SIZE adjusts as needed
- Elevator going up: N = m(g + a) — you feel heavier; coming down: N = m(g − a)
- Free fall (a = g): N = 0 — weightlessness is just no floor-push
- Apparent weight = the normal force — what the weighing scale actually reads
Stand on a scale in a lift going up and you gain weight; going down you lose it; if the cable snaps the scale reads zero. Your mass never changed — the forces did. This single experiment contains most of what JEE and NEET ask from Newton’s second law. Part 2 of the Laws of Motion series shows you exactly why it happens and how to solve every version of it.
- Tension: the rope’s message
- Normal force: the floor’s argument
- The elevator problem
- Apparent weight and weightlessness
- Solved examples
- Common mistakes
- This physics in your daily life
- Practice set
- Frequently asked questions
- Recap
Tension: The Rope’s Message
A rope can pull, never push (try pushing a dog with a leash). The force a rope transmits is called tension (T), and it always acts along the rope, away from the body it is attached to. That direction rule matters enormously when you draw free-body diagrams: on every FBD, the tension arrow points from the object outward, along the string.
In an ideal (massless, inextensible) rope — the default assumption in every JEE/NEET problem unless stated otherwise — the tension is the same at every point along the rope and therefore the same at both ends. The rope faithfully transmits a force around corners via pulleys without weakening it. This is why a pulley simply “bends” the line of action of a force without changing its magnitude (again, assuming a frictionless, massless pulley).
The most important conceptual point: tension is not a fixed number. Its size is whatever the situation demands; you solve for it using Newton’s second law. If a 2 kg mass hangs at rest, tension is 20 N (with g = 10). If the same mass accelerates upward at 5 m/s², the rope must pull harder — 30 N. The rope is a messenger: it delivers exactly the force the physics requires, nothing more, nothing less.
One more subtlety worth memorising: a rope over two different sections can carry different tensions only if the rope or pulley has mass, or if the pulley has friction. If a problem says “light frictionless pulley,” you are licensed to write a single T throughout.
Normal Force: The Floor’s Argument
Press on a table; it pushes back perpendicular (‘normal’ means perpendicular in mathematics, which is where the name comes from) to its surface. Press harder, it pushes harder — a self-adjusting contact force that grows exactly as needed to prevent you passing through, and no more. It is the “argument” the surface puts up against your weight: firm enough to win, never stronger than required.
Because it self-adjusts, the normal force cannot be written down from a formula alone at the start of a problem — it is usually an unknown you solve for. That said, two standard cases cover most exam questions:
- Horizontal surface, no vertical acceleration: N = mg. The floor matches your weight exactly.
- Inclined plane (frictionless): only the component of gravity pressing into the surface matters, so N = mg cos θ. The steeper the incline, the weaker the normal force — which is why you feel “lighter” on a slide.
Keep in mind that N always acts perpendicular to the contact surface, not vertically. On a wall, normal force is horizontal; on an incline, it is tilted. Misdirecting this arrow is one of the most common FBD errors examiners love to punish.
The Elevator Problem
Now combine the two ideas. You stand on a scale inside a lift. Two vertical forces act on you: your weight mg downward and the normal force N (which is exactly what the scale reads) upward. Newton’s second law, taking upward as positive, gives:
Writing a as a signed quantity (positive up, negative down) collapses the three famous cases into a single line:
| Letter | What it means (plain words) | Value / unit |
|---|---|---|
| N | normal force — the scale’s reading, the ‘apparent weight’ | N |
| m | your mass — never changes in any of this | kg |
| a | the lift’s acceleration (signed: + up, − down) | m/s² |
| g | gravity’s pull per kg | 9.8 m/s² down (use 10 in problems) |
Notice the pattern: only acceleration changes the reading. A lift cruising upward at a smooth 5 m/s reads exactly the same as a stationary lift, because a = 0. Starting the ride (speeding up) and ending it (slowing down) are the only moments the scale misbehaves — speeding up while moving down feels the same as slowing down while moving up, since both share downward acceleration. This is the single most-tested twist in elevator problems.
Apparent Weight and Weightlessness
The scale never measures mg — it measures the force it must supply: N. Accelerating up, it must supply extra (heavier); accelerating down, it may supply less (lighter); falling with you at g, it supplies nothing at all. Physicists call N your apparent weight, to distinguish it from the true weight mg, which never changes.
This reframes weightlessness entirely. Astronauts’ ‘zero gravity’ is really ‘zero normal force’ — the International Space Station and everything in it is in perpetual free fall around the Earth, so no surface pushes on the astronauts and they feel nothing. Gravity there is still about 90% of its surface value; what is missing is not gravity but the floor. The same logic, as the Gravitation series explains, governs orbits and tides.
Solved Examples
Step 1: FBD — N up, mg down. Step 2: N − mg = ma. Step 3: N = 50 × (10 + 2) = 600 N (vs 500 N at rest — a 20% jump from a modest acceleration).
✔
Answer: 600 N
Downward acceleration means a = −3 in our sign convention: N = 50 × (10 − 3) = 350 N. Alternatively, write mg − N = ma directly and solve.
At a = g it would read zero — the cable-cut case. ✔
Answer: 350 N
FBD: T up, mg down. T − mg = ma → T = 2(10 + 5) = 30 N.
Same elevator logic, tension playing the normal’s role — hanging masses and standing people obey identical equations. ✔
Answer: 30 N
Phase 1: N = 50 × 12 = 600 N. Phase 2: constant velocity → a = 0 → N = mg = 500 N. The “weight gain” vanishes the instant the lift stops accelerating — even though it is now moving fastest.
Answers: 600 N, then 500 N
Common Mistakes (and How to Dodge Them)
- The scale reads mg always. It reads N — only equal to mg when a = 0.
- Constant upward VELOCITY. Cruise speed changes nothing: a = 0 → N = mg. Only acceleration shifts weight.
- Weightlessness = no gravity. Astronauts have ~90% of surface gravity — they feel nothing because nothing pushes them (N = 0 in free fall).
- Tension pointing wrong. Ropes pull ALONG themselves; on an FBD the arrow is away from the body, along the rope.
- Sign chaos. Fix one direction as positive before writing N − mg = ma and stick to it; most lost marks here are pure sign errors, not concept errors.
- Confusing velocity direction with acceleration direction. A downward-moving lift that is slowing down has upward acceleration — the scale reads heavy. Always ask “is the speed increasing or decreasing?”, not “which way is it going?”
This Physics in Your Daily Life
- Theme-park drop towers — the stomach-lift is N shrinking toward zero: your organs, like the scale, only feel pushes.
- Aircraft takeoff pushes you into the seat — the seat’s normal force exceeds mg by exactly ma: you weigh more while climbing.
- Flat-bottomed weighing scales lie politely — stand on one in any accelerating lift and you’ll change weight by kilograms without gaining a gram.
- Laundry lines and crane cables — tension is the invisible hand throughout; engineers size every cable by T = m(g+a) with the worst-case jerk, then add a generous safety factor.
- Doctors’ hospital bed-lifts are chosen for gentle accelerations: patients feel the g+a and g−a swings that healthy knees ignore.
- Roller-coaster hill crests — at the top of a convex hump the track’s normal force drops below mg (N = m(g − v²/r) over a curve of radius r), which is why you float off the seat at speed; over a valley bottom it rises above mg and you feel pressed down hard.
You don’t feel gravity — you feel the floor stopping you from falling. In free fall there is no floor-push and no feeling at all (astronauts float not because gravity vanished but because nothing resists it). Weight, as felt, is a contact force: the push of whatever holds you up.
Lift up at 2 m/s²: floor must supply mg + ma = 500 + 100 = 600 N — a 20% weight gain you’d swear was real. Down at 5: 250 N — half weight. Cut the cable: 0 N, full weightlessness with gravity still at 100%. Three numbers, one formula, every ride.
Draw the FBD of the lift-rider: weight arrow (mg) down, normal arrow (N) up. Accelerating up → N must out-tug mg: draw N longer. Accelerating down → N shorter. Free fall → N vanishes. The relative lengths of two arrows ARE the felt experience.
Practice Set (Answers Hidden — Try First)
(NEET-level) 60 kg, lift up at 1.5 m/s² (g=10): N =
(JEE Main-level) Lift down at g/2: scale reads what fraction of mg?
(NEET-level) Astronauts in the ISS feel weightless because:
(Concept) Lift moving UP at constant 5 m/s: reading =
(JEE Main-level) 5 kg on a rope in lift up at 4 (g=10): T =
(JEE Advanced-level) A 50 kg rider stands on a scale in a lift. The scale reads 650 N. Find the lift’s acceleration and state its direction (g = 10).
(NEET-level) A man of mass m stands on a weighing scale in a lift whose cable has snapped (all resistance neglected). What does the scale read, and what is the man’s true weight?
Frequently Asked Questions
Is apparent weight a real force?
Can the normal force ever exceed mg by a large amount?
What if the lift accelerates downward with a > g?
Do I use g = 10 or g = 9.8 in exams?
- tension pulls along the rope, equal at both ends
- normal ⊥ surface, self-adjusting
- lift up: N = m(g+a); down: m(g−a)
- a = g → N = 0 → weightlessness
- scale reads N, not mg
- velocity direction never matters — only acceleration does
- 🔁 tension along rope, normal ⊥ surface
- 🔁 N = m(g ± a)
- 🔁 apparent weight = N
- 🧠 Chant: ‘heavier up, lighter down, nothing falling’.
- 🧠 Cruise doesn’t count: constant velocity = normal weight.
- 🧠 One equation rules them all: N = m(g + a) with signed a — the three cases are one formula.
- 🏠 Daily: drop-tower stomach lift = N shrinking.
- 🏠 Daily: takeoff pushback into seat = m(g+a).
Quick revision
- Tension: a rope pulls along its length, same strength at both ends (ideal rope, massless)
- Normal force: a surface pushes perpendicular to itself — its SIZE adjusts as needed
- Elevator going up: N = m(g + a) — you feel heavier; coming down: N = m(g − a)
- Free fall (a = g): N = 0 — weightlessness is just no floor-push
- Apparent weight = the normal force — what the weighing scale actually reads
- Tension: the rope’s message
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