Weight, Normal Force & Apparent Weight

Step on a bathroom scale in a lift and watch the needle as the car starts to rise: for a second or two you apparently gain three or four kilograms, then the reading settles. Ride down and you briefly "lose" them. Your body never changed — so what was the scale measuring? The uncomfortable truth this module is built on: a scale cannot feel gravity at all. It can only report how hard its own surface pushes on your feet. That push — the normal force — is a different force from your weight, obeys different rules, and only pretends to equal $mg$ while nothing accelerates. Every board tests this triad. CBSE, ISC and IGCSE open with the mass–weight distinction (a guaranteed 2-marker) and the spring-versus-beam-balance question. JEE-Main and NEET run the apparent-weight machinery hard — lift readings, ratios with $g$ cancelling — and JEE-Advanced hides the constraint nature of the normal force inside every contact problem: the moment your algebra returns $N < 0$ , the examiner is telling you contact has broken. AP Physics 1, IB and A-Level phrase the same physics through orbiting astronauts, planting "no gravity in space" among the options. This module gives you the three ideas cleanly: mass as the amount of matter, weight as the local pull $mg$ , the normal force as the self-adjusting push that scales actually read. The full lift drill-family lives in the Lifts & Apparent Weight archetype; here you learn the law it runs on.

Key takeaways

Mass ( $m$ , kg): the amount of matter — a scalar, the same everywhere. Weight ( $W = mg$ , newtons): the gravitational force on it — on the Moon ( $g \approx 1.62\ \mathrm{m/s^2}$ ), one-sixth the weight, same mass.
The normal force $N$ is the perpendicular contact push of a surface: no formula of its own — it self-adjusts to whatever the second law demands, and it pushes but never pulls ( $N \ge 0$ ).
$N = mg$ only at rest on a horizontal surface with nothing else vertical. Press down: $N$ grows. Pull up: it shrinks. Incline: $mg\cos\theta$ . Accelerating lift: $m(g + a)$ .
Apparent weight = the scale reading = $N$ . Lift accelerating up: heavier. Down: lighter. Free fall: $N = 0$ — weightless with gravity fully on.
Orbiting astronauts float because the station falls around the Earth with them (gravity there is ~89% of the surface value), not because gravity is absent.
Key idea: gravity acts on every atom of you equally and cannot be felt — what you feel and what scales measure is the normal force, which tracks $mg$ only when nothing accelerates.

The core insight

Weight $W = mg$ belongs to the place; mass $m$ is the amount of matter and travels with you unchanged; the normal force $N$ is the self-adjusting perpendicular contact push a scale actually reads — found from Newton's second law, equal to $mg$ only in unaccelerated, horizontal, unpushed rest.

Where this lives today

On the International Space Station a bathroom scale is useless — everything aboard reads $N = 0$ forever — so crew "weigh-ins" shake the astronaut on a calibrated spring and extract mass from the oscillation period: inertia stepping in where weight has resigned. Parabolic "zero-g" flights sell the same physics — free-fall arcs buying twenty seconds of genuine $N = 0$ per parabola — with gravity at full strength throughout. Down here, the split runs the other way. Since 2019 the kilogram has been defined through Planck's constant and realized by Kibble balances, which weigh a mass against a precisely measured electromagnetic force. And every g-suit, crash-test standard and roller-coaster restraint is engineered around apparent weight, because $N = m(g+a)$ , not $mg$ , is what stresses a body: a pilot pulling $8g$ carries eight times her spine's usual load while her weight never changes by a newton.

The physical picture

Hold a brick and you "feel its weight" — but be precise about what your nerves report. Gravity pulls every atom of the brick, and of you, in exact proportion to its mass, all identically. A pull that moves every part of you in unison compresses nothing, strains nothing, and feels like nothing. What you actually sense is your hand's upward push on the brick and the strain in your muscles. Cut the imaginary lift cable and gravity carries on at full strength, yet every sensation of weight vanishes, because the push has vanished. The feeling of weight is a story about support, never about gravity directly.

Figure

Mass travels with you — weight belongs to the place, and what you feel is the support

The second character deserves the same respect. Put a book on a table: the table pushes up. Press down on the book and the table instantly pushes harder; tilt the table and the push weakens, staying perpendicular to the surface. No formula told the table what to do — microscopically its surface atoms compress like impossibly stiff springs, by exactly the amount needed to stop the book sinking in. That is why the normal force is a constraint force: its magnitude is demanded by the situation, and your only way to find it is Newton's second law. One rule it never breaks: a surface can push but cannot grab and pull — the moment physics would need $N < 0$ , contact simply ends.

Mass and weight — two different quantities

Start with clean definitions — every board's 2-marker depends on stating them sharply. Mass measures how much matter an object contains — equivalently, how much it resists acceleration (inertia, from Force, Inertia & Newton's First Law). Weight is the gravitational force the local planet exerts on it:

W = mg \qquad (1)

where: $W$ — weight in newtons (a force, never kilograms); $m$ — mass in kg; $g$ — the local field strength: $9.8\ \mathrm{m/s^2}$ near Earth's surface, $1.62$ on the Moon, about $3.7$ on Mars. $\vec W$ points toward the planet's centre — "down" is defined by it. Read the equation as a partnership: $m$ belongs to you, the same everywhere in the universe; $g$ belongs to the place. A $60\ \mathrm{kg}$ student weighs $588\ \mathrm{N}$ in Delhi and $97\ \mathrm{N}$ on the Moon.

	mass	weight
what it measures	amount of matter / inertia	gravitational force on the body
scalar or vector	scalar	vector (toward the planet's centre)
SI unit	kilogram (kg)	newton (N)
changes with location?	never	yes — proportional to local $g$
measured by	beam balance (compares masses — same reading on the Moon)	spring balance / weighing scale (reads a force)

The beam-balance row is a favourite trick: it compares the weights of two sides, but both sides sit in the same $g$ , so the comparison is really between masses — it reports the same value on the Moon. A spring balance reads a force, so it reads six times less there. "Which instrument gives a different reading on the Moon?" is a standard ICSE/IGCSE one-liner.

The normal force — the force with no formula

Everything the normal force is fits in one line: a direction and an inequality.

\vec N \perp \text{surface}, \qquad N \ge 0 \qquad (2)

where: $\vec N$ is the contact force the surface exerts on the body, perpendicular ("normal" is geometry-speak for perpendicular) to the touching surfaces, pointing away from the surface. The inequality encodes "surfaces push, never pull". Notice what is missing: any expression for the magnitude. $N$ is an output of Newton's second law, recomputed fresh in every problem — treating $N = mg$ as a law of nature is the most reliable way to lose marks in this chapter. Here is the honest version of that famous equality. For a body at rest on a horizontal surface with nothing else vertical going on, the vertical equation reads $N - mg = 0$ , so

N = mg \qquad (3)

where: the conditions do all the work — horizontal surface, zero vertical acceleration, no other vertical force. A $2.0\ \mathrm{kg}$ book on a level table gets $N = 19.6\ \mathrm{N}$ ; press down with $15\ \mathrm{N}$ and the table answers with $34.6\ \mathrm{N}$ ; pull up with $10\ \mathrm{N}$ and it relaxes to $9.6\ \mathrm{N}$ ; pull with $19.6\ \mathrm{N}$ and $N = 0$ — the book is on the verge of leaving the table. The same self-adjustment handles tilted geometry (the cheat-sheet card collects the full which- $N$ -when table). On a smooth incline at angle $\theta$ , gravity still points straight down, but the surface only has to prevent motion into itself — and only the perpendicular slice of the weight, $mg\cos\theta$ , threatens that:

N = mg\cos\theta \qquad (4)

where: $\theta$ is the incline angle above horizontal. The remaining slice $mg\sin\theta$ points along the slope — it tries to slide the body, which is Static Friction's department, not the surface's perpendicular push. Check the ends of the dial: $\theta = 0$ recovers the flat floor $N = mg$ ; $\theta = 90°$ gives $N = 0$ , because a vertical wall beside a freely falling block supports none of its weight. Those two limits are the fastest way to catch a $\sin$ – $\cos$ swap in an exam.

Figure

N is not a fixed number — four situations, four different normal forces on the same block

Derivation ANormal force on a tilted surface

We derive $N = mg\cos\theta$ for a block on a smooth incline — the geometry every board draws, and the cleanest proof that the normal force is computed, never assumed. The strategy: tilt the axes to match the surface, split the weight along them, then read Newton's second law perpendicular to the surface, where the acceleration is zero while contact holds. Watch the decomposition — the incline angle lands between the weight vector and the perpendicular axis, so that component carries $\cos\theta$ ; the $\theta \to 0$ limit catches the classic swap instantly.

Step 1. Choose axes glued to the geometry: $x$ along the slope, $y$ perpendicular. Two forces act — the weight $mg$ straight down, and $N$ along $+y$ . (A rough incline adds friction along $x$ ; it has no $y$ -component, so nothing below changes.)

Step 2. Decompose the weight: it makes angle $\theta$ with the $-y$ axis, so the components are $mg\sin\theta$ along the slope and $mg\cos\theta$ into the surface. Why cosine goes with the perpendicular: tip the ramp flat — at $\theta = 0$ , $\cos\theta = 1$ keeps all of $mg$ perpendicular while $\sin\theta = 0$ kills the along-slope pull.

Step 3. Apply Newton's second law along $y$ . The block neither sinks into the ramp nor lifts off, so $a_y = 0$ :

N - mg\cos\theta = 0 \quad\Longrightarrow\quad N = mg\cos\theta

Why this is the whole trick: the constraint "stays on the surface" converts the unknown push into one solvable line. The surface supplies exactly $mg\cos\theta$ — less than the full weight, because the rest of the weight is trying to slide, not penetrate.

Step 4. Interrogate the limits: $\theta \to 0$ gives $N \to mg$ (flat table); $\theta \to 90°$ gives $N \to 0$ (a block falling beside a vertical wall it barely grazes). Both endpoints are the right physics. Verified: $\theta = 37°$ , $m = 5.0\ \mathrm{kg}$ , $g = 10$ : $N = 50 \times 0.80 = 40\ \mathrm{N}$ — the standard NCERT-pattern value; limits as above; agrees with HC Verma Ch. 6.

Apparent weight — what the scale actually reads

Now combine the two characters in the setting every exam loves: the lift. A scale reads the normal force its surface supplies. Standing still, $N = mg$ and the scale tells the truth. Let the floor accelerate and $N$ must both support you and accelerate you — the reading detaches from $mg$ . That reading is your apparent weight, and one derivation covers every case.

Go deeper · Derivation B · 1 minApparent weight in an accelerating lift

Derivation BApparent weight in an accelerating lift

We derive the scale reading $N = m(g + a)$ for a person on a scale in a lift with signed upward acceleration $a$ , then extract the three readings every exam asks for — heavier, lighter, weightless. The strategy: forces on the person alone, plus Newton's second law along the vertical. The discipline that earns the marks: use the acceleration, never the velocity — a lift cruising upward at steady speed has $a = 0$ and reads exactly $mg$ .

Step 1. Forces on the person: weight $mg$ down, normal force $N$ up from the scale. Taking up as positive, Newton's second law reads

N - mg = ma

Why the person and not the lift: the scale presses on the person, so the person's equation is the one containing $N$ . Cables, motors and counterweights never enter.

Step 2. Solve for the reading:

N = m(g + a)

Why this is "apparent weight": every sensation of heaviness — soles, spine — is produced by $N$ . When the lift accelerates upward ( $a > 0$ , which includes braking during a descent) the floor pushes harder: heavier. Accelerating downward, it eases off: lighter.

Step 3. Take the free-fall limit $a = -g$ : $N = m(g - g) = 0$ . Why weightlessness is not "no gravity": gravity is the only force left — exactly why the floor no longer needs to push. Person, scale and lift fall together; every contact force relaxes to zero. An orbiting station is this lift falling forever: at ISS altitude $g \approx 8.7\ \mathrm{m/s^2}$ (89% of surface), yet astronauts float because nothing needs to hold them up. *Verified: $a = 0$ recovers $N = mg$ ; $a = +g$ doubles the reading; $a = -g$ zeroes it. With $m = 60$ , $g = 9.8$ , $a = +1.2$ : $N = 660\ \mathrm{N}$ vs $588\ \mathrm{N}$ at rest — a 12% gain, the familiar lift-start heaviness. NCERT Ch. 5, HC Verma Ch. 5.*

The Lifts & Apparent Weight archetype module runs this formula through the full exam drill-family — multi-stage lift journeys, breaking cables, scales under blocks under blocks. Here, keep the law and its three faces (panel 4 of the gallery figure above shows the geometry): $a$ up means heavy, $a$ down means light, $a = -g$ means zero.

Limiting cases

case	result	meaning
horizontal, at rest, unpushed	$N = mg$	the only case where the reflex is safe
incline, $\theta \to 0$	$N \to mg$	the ramp flattens into a floor — formulas must agree
incline, $\theta \to 90°$	$N \to 0$	a vertical wall supports none of the weight
lift, $a = +g$	$N = 2mg$	apparent weight doubles — rocket-launch territory
lift, $a = -g$ (free fall)	$N = 0$	weightless with gravity at full strength

Common trapThe normal force always equals mg

It equals $mg$ in exactly one situation: a body at rest (or unaccelerated) on a horizontal surface with no other vertical forces. Everywhere else the reflex quietly poisons the problem — and because $N$ feeds the friction ceiling $\mu N$ in the next module, one wrong $N$ corrupts every later line. On a $30°$ incline, $N \approx 0.87\,mg$ . Under a $15\ \mathrm{N}$ press, a $2\ \mathrm{kg}$ book gets $34.6\ \mathrm{N}$ , not $19.6$ . In a lift accelerating downward, $N < mg$ before you finish reading the question. The cure is mechanical: draw the forces, write the second law perpendicular to the surface with the actual perpendicular acceleration, and let the algebra hand you $N$ . The normal force is an output of the second law — never an input.

Worked examples

1 · Board level — Earth to Moon (CBSE/ICSE pattern, 3 marks)

A student has mass $60\ \mathrm{kg}$ ( $g_{\text{Earth}} = 9.8\ \mathrm{m/s^2}$ , $g_{\text{Moon}} = 1.62\ \mathrm{m/s^2}$ ). Find her weight on both worlds, and what a beam balance and a spring balance each read on the Moon if both read "60 kg" on Earth.

Step 1. Weight on Earth: $W_E = mg = 60 \times 9.8 = \boxed{588\ \mathrm{N}}$ .

Step 2. Weight on the Moon: $W_M = 60 \times 1.62 = \boxed{97.2\ \mathrm{N}}$ — one-sixth of $588$ (check: $588/97.2 \approx 6.0$ ✓). Her mass stays $60\ \mathrm{kg}$ in both places.

Step 3. The beam balance compares her against standard masses in the same lunar $g$ — still $\boxed{60\ \mathrm{kg}}$ . The spring balance reads the force, now $97.2\ \mathrm{N}$ , so its kg dial shows $97.2/9.8 \approx \boxed{9.9\ \mathrm{kg}}$ . One measures mass; the other measures a force wearing a kg costume.

2 · Board level — the pressed and pulled book (ISC pattern)

A $2.0\ \mathrm{kg}$ book lies on a horizontal table ( $g = 9.8\ \mathrm{m/s^2}$ ). Find the normal force when (a) nothing else acts, (b) a hand presses down on it with $15\ \mathrm{N}$ , (c) a string pulls up on it with $10\ \mathrm{N}$ . (d) At what upward pull does the book leave the table?

Step 1. (a) Vertical balance, two forces: $N = mg = 2.0 \times 9.8 = \boxed{19.6\ \mathrm{N}}$ .

Step 2. (b) Three forces, still $a = 0$ : $N - mg - 15 = 0 \Rightarrow N = \boxed{34.6\ \mathrm{N}}$ . The table answers the squeeze instantly.

Step 3. (c) The pull helps support the book: $N + 10 - mg = 0 \Rightarrow N = \boxed{9.6\ \mathrm{N}}$ .

Step 4. (d) Contact ends at $N = 0$ : pull $= mg = \boxed{19.6\ \mathrm{N}}$ . Any harder and the book lifts off — the table cannot pull it back, because $N \ge 0$ always.

3 · Competitive level — the lift reading (JEE-Main pattern, ~2.5 minutes)

A $10\ \mathrm{kg}$ block rests on a weighing scale on the floor of a lift ( $g = 10\ \mathrm{m/s^2}$ ). Find the scale reading (in kg) when the lift (a) accelerates upward at $2\ \mathrm{m/s^2}$ , (b) moves upward but is slowing at $3\ \mathrm{m/s^2}$ , (c) moves downward at a constant $4\ \mathrm{m/s}$ .

Step 1. The scale reads $N = m(g + a)$ , $a$ the signed upward acceleration. Translate each story into an $a$ — that translation is the entire question.

Step 2. (a) Accelerating up, $a = +2$ : $N = 10 \times 12 = 120\ \mathrm{N}$ , displayed as $\boxed{12\ \mathrm{kg}}$ .

Step 3. (b) Moving up but slowing: the acceleration points down, $a = -3$ : $N = 10 \times 7 = 70\ \mathrm{N} \to \boxed{7\ \mathrm{kg}}$ . Direction of motion never entered.

Step 4. (c) Constant velocity, any direction: $a = 0$ , $N = 100\ \mathrm{N} \to \boxed{10\ \mathrm{kg}}$ , the true value. Timing note: ~2.5 min, almost all of it Step 1. The planted distractor is the velocity — "moving downward" tempts a lighter reading in (c). Write the signed $a$ next to each part first and the three answers fall out in thirty seconds.

4 · Competitive level — the ratio that drops g (NEET pattern, ~2 minutes)

A person stands on a weighing machine in a lift. The reading when the lift accelerates upward at $g/3$ is $R_1$ ; when it accelerates downward at $g/4$ the reading is $R_2$ . Find $R_1 : R_2$ .

Step 1. Upward case: $N_1 = m\left(g + \dfrac{g}{3}\right) = \dfrac{4mg}{3}$ .

Step 2. Downward case: $a = -g/4$ , so $N_2 = m\left(g - \dfrac{g}{4}\right) = \dfrac{3mg}{4}$ .

Step 3. Ratio: $\dfrac{R_1}{R_2} = \dfrac{4/3}{3/4} = \dfrac{16}{9}$ , so $R_1 : R_2 = \boxed{16 : 9}$ . Both $m$ and $g$ cancel — no numbers needed. Timing note: ~2 min ceiling, under 1 min once you spot that fractions of $g$ guarantee cancellation. If your ratio still contains $m$ or $g$ , you slipped — the question was engineered for them to vanish.

Exam traps

Quoting weight in kilograms. "My weight is 60 kg" is everyday speech, not physics: weight is a force, $588\ \mathrm{N}$ at $g = 9.8$ . If a paper uses kg-wt or kgf: $1\ \mathrm{kgf} = 9.8\ \mathrm{N}$ . Writing " $W = 60\ \mathrm{kg}$ " in a CBSE/ISC answer costs the unit mark.

The $N = mg$ reflex. Valid only flat, unaccelerated, unpushed. Incline: $\cos\theta$ (check the $\theta = 0$ limit to catch a $\sin$ – $\cos$ swap: you must recover $N = mg$ ). Press: add it. Pull: subtract it. Lift: $m(g+a)$ . Before writing $N$ , ask what the perpendicular equation says here.

Velocity smuggled into the lift formula. "Moving upward at $5\ \mathrm{m/s}$ " contributes nothing; "slowing while moving up" means $a$ points down. Translate every lift story into a signed $a$ first — Examples 3(b) and 3(c) are the standard traps.

Calling $N$ and $mg$ a third-law pair. They act on the same body from two different agents (table, Earth). Partners act on different bodies: "Earth pulls you" pairs with "you pull the Earth". Proof they are not partners: in an accelerating lift $N \ne mg$ , while genuine pairs can never disagree. Newton's Third Law & Interaction Pairs drills this.

Beam vs spring balance on the Moon. Beam: unchanged (compares masses). Spring: one-sixth. A one-line justification is expected, not just the answer.

Which $g$ ? CBSE/ISC use $9.8\ \mathrm{m/s^2}$ ; JEE/NEET nearly always say "take $g = 10$ " — obey the paper, not your habit; A-Level uses $9.81$ ; AP-1 accepts $10$ for estimation. In ratio questions (Example 4), $g$ cancels — substitute nothing.

Origins · 1 minAlbert Einstein and how this was discovered

Albert Einstein

1879–1955 · Bern patent office, Berlin, Princeton

In 1907, still examining patents in Bern, Einstein had what he later called "the happiest thought of my life": a person falling freely does not feel their own weight. Everything in this module points at that sentence — gravity cannot be felt because it accelerates every part of you identically; what you feel is the normal force; remove the support and the sensation vanishes though gravity has not weakened. Einstein promoted the bookkeeping into a principle — free fall is locally indistinguishable from gravity-free space, a rocket at $g$ from standing on Earth — and rode it to general relativity, where gravity is not a force but the shape of spacetime. Every lift question you solve with $N = m(g+a)$ is a small rehearsal of that idea.

History

1589

— Galileo argues that heavy and light bodies fall alike, splitting mass (how much matter) from weight (how hard it is pulled) in all but name.

1687

— Newton's Principia makes weight one instance of universal gravitation: the same pull on the apple and on the Moon, proportional to mass.

1901

— The 3rd CGPM separates the kilogram (mass) from the kilogram-force, fixing standard gravity at 9.80665 m/s².

1907

— Einstein's "happiest thought": a freely falling person feels no weight — apparent weight, not weight, is what the body senses.

1971

— Apollo 15's David Scott drops a hammer and a feather on the Moon; they land together, on camera, in one-sixth g.

2019

— The kilogram is redefined via Planck's constant — no reference lump, and no role for g.

Going further

Texts. NCERT Physics Class 11, Ch. 5 (Laws of Motion) · HC Verma, Concepts of Physics Vol. 1, Ch. 4–5 · Halliday, Resnick & Walker, Ch. 5.

Provenance. Standard Newtonian mechanics. A verifier should recompute: Earth/Moon weights ( $588\ \mathrm{N}$ , $97.2\ \mathrm{N}$ , ratio $\approx 6.0$ , spring dial $9.9\ \mathrm{kg}$ ); the pressed/pulled book ( $19.6$ , $34.6$ , $9.6$ , threshold $19.6\ \mathrm{N}$ ); the incline check ( $5.0\ \mathrm{kg}$ , $37°$ , $g=10$ : $N = 40\ \mathrm{N}$ ); the lift readings ( $120 \to 12\ \mathrm{kg}$ , $70 \to 7\ \mathrm{kg}$ , $100 \to 10\ \mathrm{kg}$ ); the NEET ratio $16/9$ ; the ISS figure ( $8.7\ \mathrm{m/s^2} \approx 89\%$ ).

Prerequisites. Newton's Second Law: F = ma · Free Fall & Motion Under Gravity. Leads to. Free-Body Diagrams & Problem-Solving Strategy · Lifts & Apparent Weight (the archetype drilling this module's lift law) · Static Friction — and, for the deep dive, the university-tier Normal Force and Gravitational Force Near Earth's Surface (Advanced Mechanics).

If gravity is pulling me right now, why can't I feel it?

Because it pulls every part of you identically. A force you can feel must squeeze or stretch you, which requires different parts of your body to be pushed differently. Gravity accelerates every atom in unison — zero internal strain. The floor is what you feel: it pushes only on your feet, and your skeleton transmits that push upward, part by part. Cutting the support (free fall) removes every sensation of weight instantly while gravity carries on at full strength — "what does the scale read?" is always secretly "what is $N$ ?".

My weight is "60 kg" — is that wrong in an exam?

In physics, yes. Kilograms measure mass; weight is a force in newtons: a $60\ \mathrm{kg}$ person weighs $588\ \mathrm{N}$ . Everyday scales blur this by measuring force but printing mass, silently dividing by Earth's $g$ . Older papers use the kilogram-force, $1\ \mathrm{kgf} = 9.8\ \mathrm{N}$ : if a question says "a body weighs 5 kgf", translate immediately — $W = 49\ \mathrm{N}$ , $m = 5\ \mathrm{kg}$ . State the conversion once and the examiner knows you know.

Why do scales show kilograms if they really measure newtons?

Convenience plus a hidden assumption. A bathroom scale is a stiff spring measuring the normal force on its surface, in newtons; the maker divides by $9.8\ \mathrm{m/s^2}$ and prints kilograms, assuming you are at rest on Earth. Both assumptions can fail: in an accelerating lift the reading changes though your mass has not, and on the Moon the same scale reports a sixth of your mass. A beam balance dodges this by comparing you against standard masses in the same $g$ — which is why it tells the truth everywhere.

Why can a normal force only push and never pull?

Because of where it comes from: contact is atoms being squeezed, and squeezed bonds push back like compressed springs. Pulling would require the surfaces to be bonded — glue, suction, magnetism — a different force with its own name. The rule $N \ge 0$ is a prediction machine: if the algebra returns $N < 0$ , the contact assumption was false — the body has already left the surface, and the physics from that instant is free flight with $N = 0$ . "Verge of leaving contact" always translates to "set $N = 0$ ".

The scale under me reads N and my weight is mg — in a stationary room they're equal. Doesn't that make them action and reaction?

No — the most-tested subtlety in the topic. Third-law partners are the two ends of one interaction: swap the labels in "Earth pulls you" and you get "you pull the Earth" — a force on the planet. $N$ and $mg$ are two different interactions (table contact, Earth's gravity) that both act on you, balancing only because your acceleration is zero. In a lift accelerating downward, $N < mg$ immediately — genuine pairs can never disagree. Equality by equilibrium is the second law, not the third.

If the lift accelerates downward faster than g, what does the scale read?

Zero — and you leave the floor. $N = m(g + a)$ with $a < -g$ would demand a negative $N$ , which a floor cannot supply: it cannot pull you down. Contact breaks, you become a freely falling body inside a faster-falling box, and you drift toward the ceiling until it pushes you (a downward normal force — perpendicular contact push, as always). Scales read zero throughout. This is the $N \ge 0$ constraint doing physics: the algebra's "impossible" answer signals that the geometry of contact has changed.