Normal force — reaction force, not always = mg: Worked examples

Before anything, one reminder of the two symbols we will use everywhere:

The scenario matrix

Every normal-force problem is one row of this table. The whole point of the page is that each row gets at least one fully worked example so you never meet an unshown case.

Cell	Situation class	What changes about $N$	Expected result	Example
A	Flat floor, nothing else	baseline	$N = m g$	Ex 1
B	Flat floor + push down	more to hold	$N = m g + F$	Ex 2
C	Flat floor + pull up (with the degenerate "lifts off" case)	less to hold, can hit $0$	$N = m g - F$ , floored at $0$	Ex 3
D	Elevator accelerating up	floor pushes harder	$N = m (g + a)$	Ex 4
E	Elevator accelerating down / free fall ( $a = g$ , degenerate)	floor pushes softer, can hit $0$	$N = m (g - a)$	Ex 5
F	Incline, angle $θ$ (general + limits $θ \to 0, θ \to 9 0^{\circ}$ )	only $⊥$ part of gravity	$N = m g cos θ$	Ex 6
G	Vertical wall, held by a horizontal push (word problem)	gravity is along the surface, not into it	$N = F_{push}$	Ex 7
H	Incline accelerated horizontally (exam twist)	$a_{⊥} \neq = 0$	$N = m (g cos θ + a sin θ)$	Ex 8

Prerequisite tools used below: Newton's Second Law, Free Body Diagrams, and for rows F/H the geometry from Inclined Plane Problems; row G touches Friction ideas; rows D/E are the heart of Apparent Weight & Elevators.

Cell A — Flat floor, baseline

Worked example Ex 1 — Book on a table

A $2 kg$ book lies still on a flat table. Find $N$ . ( $g = 10$ .)

Forecast: Guess before reading — is $N$ bigger, smaller, or equal to the $20 N$ weight?

List the forces. Weight $m g = 2 \times 10 = 20 N$ down; normal $N$ up. Why this step? A free body diagram must include every touching or pulling agent — here only Earth (gravity) and the table (normal).
Write Newton's 2nd law, up as positive. $N - m g = m a_{y}$ . Why this step? Newton's Second Law says net force $=$ mass $\times$ acceleration, and we resolve it along the interesting axis (vertical).
Insert $a_{y} = 0$ . The book is at rest and stays at rest, so it does not accelerate. $N - 20 = 0 \Rightarrow N = 20 N$ . Why this step? No vertical acceleration is the exact special condition that makes $N = m g$ .

Verify: $N = m g = 20 N$ . Units: $kg \cdot m/s^{2} = N$ ✔. Equal to weight, as the baseline row predicts. ✔

Cell B — Extra push down

Cell C — Pull up, including the lift-off degenerate case

Worked example Ex 3 — Pulling up, two sub-cases

The $2 kg$ book on the table. (a) You pull up with $F = 8 N$ . (b) You pull up with $F = 25 N$ . Find $N$ in each.

Forecast: In part (b) can $N$ go negative? Think about what a negative "push" would even mean.

Newton's 2nd law, up positive, $a_{y} = 0$ (while it stays on the table). $N + F - m g = 0 \Rightarrow N = m g - F$ . Why this step? The upward pull helps hold the book, so the table has less left to do.
Part (a): $N = 20 - 8 = 12 N$ . Why this step? $F < m g$ , so the surface is still needed. Fine.
Part (b): the formula gives $20 - 25 = - 5 N$ . Why this step and the catch: a surface can only push, never pull. A negative $N$ is impossible, so it means the book has left the surface: the true answer is $N = 0$ . Why: once $F \geq m g$ the object lifts off and there is no contact at all.

Verify: (a) $N = 12 N$ , in $[0, m g]$ ✔. (b) formula $= - 5$ ⇒ clamp to $N = 0$ ✔ (degenerate lift-off cell handled).

Cell D — Elevator accelerating up

Worked example Ex 4 — Heavy feeling in a rising lift

A $60 kg$ person stands in an elevator accelerating upward at $a = 2 m/s^{2}$ . Find the floor's push $N$ (their apparent weight).

Forecast: Heavier or lighter than their true weight $600 N$ ?

Free body: only $N$ up, $m g$ down. Why this step? Nothing touches the person except the floor; the elevator's motion enters through acceleration, not a new force.
Newton's 2nd law, up positive, $a_{y} = + a$ . $N - m g = ma$ . Why this step? The person really is accelerating upward with the cabin, so the right side is not zero — this is the core Apparent Weight & Elevators idea.
Solve. $N = m (g + a) = 60 \times (10 + 2) = 720 N$ . Why: the floor must both hold the weight and supply the extra push to speed the person up.

Verify: $N = 720 N > 600 N$ ✔ — you feel heavier accelerating upward, matching row D.

Cell E — Accelerating down, and the free-fall degenerate case

Worked example Ex 5 — Descending lift and a snapped cable

Same $60 kg$ person. (a) Elevator accelerates downward at $a = 4 m/s^{2}$ . (b) The cable snaps and the cabin is in free fall ( $a = g = 10$ ). Find $N$ each.

Forecast: In part (b), what does the floor push equal when everything falls together?

Newton's 2nd law, up positive, downward acceleration means $a_{y} = - a$ . $N - m g = m (- a) \Rightarrow N = m (g - a)$ . Why this step? Downward acceleration reduces the net upward force needed, so the floor relaxes.
Part (a): $N = 60 \times (10 - 4) = 360 N$ . Why: less than $600 N$ — you feel lighter descending. ✔
Part (b), free fall: $N = 60 \times (10 - 10) = 0 N$ . Why: person and floor fall at the same rate, so nothing presses on anything — weightlessness, the degenerate limit of this row.

Verify: (a) $N = 360 N$ (in $[0, m g]$ ) ✔; (b) $N = 0 N$ ✔ — free-fall weightlessness confirmed.

Cell F — Incline, general angle plus both limits

Here the surface is tilted, so gravity no longer points straight into it. We split the weight into two arrows: one into the slope and one along the slope. Only the "into the slope" part must be balanced by $N$ .

Figure — Normal force — reaction force, not always = mg

Worked example Ex 6 — Block on a

$3 7^{\circ}$ frictionless ramp, plus limits A $4 kg$ block sits on a frictionless incline at $θ = 3 7^{\circ}$ . Find $N$ . Then check the two extreme angles $θ = 0^{\circ}$ and $θ = 9 0^{\circ}$ . (Use $cos 3 7^{\circ} = 0.8$ .)

Forecast: More or less than $m g = 40 N$ ? And at a vertical cliff ( $9 0^{\circ}$ ), what should $N$ become?

Choose axes along/perpendicular to the surface (see figure). Why this step? The block can only slide along the ramp, so along that axis is where motion lives, and perpendicular is where the constraint lives. Tilting the axes makes the constraint one clean equation.
Resolve gravity. Perpendicular part $= m g cos θ$ (pressing into the ramp); along-slope part $= m g sin θ$ (pulling it down the ramp). Why $cos θ$ for the perpendicular part? As the ramp tilts by $θ$ , the weight arrow tilts $θ$ away from the surface normal, and the adjacent (perpendicular) side of that little right triangle scales as $cos θ$ — see the red/mint split in the figure.
Newton's 2nd law perpendicular, $a_{⊥} = 0$ (block stays on the surface). $N - m g cos θ = 0 \Rightarrow N = m g cos θ$ . Why $a_{⊥} = 0$ ? The block never sinks into or leaps off the ramp, so its acceleration into the surface is zero.
Number: $N = 4 \times 10 \times 0.8 = 32 N$ .
Limit $θ = 0^{\circ}$ : $cos 0 = 1 \Rightarrow N = 40 N = m g$ — flat ground, back to Cell A. ✔
Limit $θ = 9 0^{\circ}$ : $cos 9 0^{\circ} = 0 \Rightarrow N = 0$ — a vertical wall the block just slides straight down, pressing nothing into the surface. Why: gravity is now entirely along the surface, none into it.

Verify: $N = 32 N < 40 N$ ✔; flat limit $= 40$ ✔; vertical limit $= 0$ ✔.

Cell G — Vertical wall (word problem)

Now the surface's normal is horizontal, so gravity plays no part in $N$ . This is the case people forget exists.

Worked example Ex 7 — Pinning a poster to a wall

You press a $0.5 kg$ poster flat against a vertical wall with a horizontal push $F = 6 N$ straight into the wall. Find the wall's normal force $N$ on the poster.

Forecast: Does $N$ depend on the poster's weight at all here?

Identify the normal direction. The wall is vertical, so its normal points horizontally outward (see figure). Why this step? Normal force is always $⊥$ to the surface — for a wall that means sideways, not up.
Newton's 2nd law along the horizontal (into-wall) axis, $a_{x} = 0$ (poster not accelerating into/out of the wall). $F - N = 0 \Rightarrow N = F$ . Why this step? Your push and the wall's push are the only horizontal forces; gravity is vertical and lives in a different equation entirely.
Number: $N = 6 N$ . Gravity ( $m g = 5 N$ ) never appears in the horizontal balance. Why: it is perpendicular to the wall's normal, so it cannot affect $N$ .

Verify: $N = F = 6 N$ , independent of $m g$ ✔ — the "normal $\neq = m g$ " lesson at its most extreme: here $N$ has nothing to do with weight.

Cell H — Incline that itself accelerates (exam twist)

Worked example Ex 8 — Block on a ramp pushed sideways

A $2 kg$ block rides a frictionless wedge of angle $θ = 3 0^{\circ}$ . The whole wedge is pushed horizontally so the block accelerates with the wedge at $a = 5 m/s^{2}$ , staying pressed to the slope. Find $N$ . (Use $cos 3 0^{\circ} = 0.866$ , $sin 3 0^{\circ} = 0.5$ .)

Forecast: More or less than the static-ramp answer $m g cos θ = 17.3 N$ ?

Keep the along/perpendicular axes, but $a_{⊥}$ is no longer zero. The horizontal acceleration $a$ has a component into the surface. Why this step? The block moves with the wedge horizontally; part of that horizontal motion is directed into the slope, so $a_{⊥} \neq = 0$ — the trap the parent note warned about.
Perpendicular components. Into-surface gravity part $= m g cos θ$ . The horizontal acceleration, projected onto the inward normal, contributes $m a sin θ$ toward the surface. Why $sin θ$ ? The normal points at angle $θ$ from vertical, so a horizontal push lands on it with a $sin θ$ factor.
Newton's 2nd law perpendicular. Taking "out of surface" positive: $N - m g cos θ - ma sin θ = 0$ , so $N = m (g cos θ + a sin θ)$ . Why this step: $N$ must supply both the gravity balance and the perpendicular part of the required acceleration.
Number: $N = 2 \times (10 \times 0.866 + 5 \times 0.5) = 2 \times (8.66 + 2.5) = 2 \times 11.16 = 22.32 N$ .

Verify: $N = 22.32 N > 17.3 N$ ✔ — accelerating the ramp into the block presses harder, and setting $a = 0$ recovers $m g cos θ = 17.32 N$ ✔.

Recall Did every cell get covered?

Row A → Ex 1; B → Ex 2; C (with lift-off) → Ex 3; D → Ex 4; E (with free fall) → Ex 5; F (with $0^{\circ}$ / $9 0^{\circ}$ limits) → Ex 6; G (wall) → Ex 7; H (accelerated ramp) → Ex 8. Every sign, every zero, every limit is shown.

Active Recall

Connections

Newton's Second Law — every row is just $\sum F_{⊥} = m a_{⊥}$ .
Apparent Weight & Elevators — Cells D and E.
Inclined Plane Problems — Cells F and H.
Free Body Diagrams — the setup tool for all eight.
Newton's Third Law — the true reaction to each $N$ is the object pushing back on the surface.
Friction — once you have these $N$ values, $f = μ N$ follows.