Weightlessness — true (free fall) vs apparent: Worked examples

Before anything, let us re-earn the two symbols we will use on every line:

$N = m (g + a) (up positive)$

The scenario matrix

Every problem this topic can throw at you lands in exactly one of these cells. The examples below are labelled by cell so you can see the whole board is covered.

#	Cell (case class)	Sign / value of $a$	Predicted feel	Example
A	At rest / constant velocity	$a = 0$	Normal ( $N = m g$ )	Ex 1
B	Accelerating up	$a > 0$	Heavy ( $N > m g$ )	Ex 2
C	Accelerating down (not free)	$- g < a < 0$	Light ( $N < m g$ )	Ex 3
D	Free fall (degenerate)	$a = - g$	Weightless ( $N = 0$ )	Ex 4
E	"Trap" — moving down, but decelerating	$a > 0$	Heavy	Ex 5
F	Beyond free fall (pushed down harder than $g$ )	$a < - g$	$N = 0$ , body leaves floor	Ex 6
G	Orbit / circular free fall	$a = g$ inward = " $- g$ " locally	Weightless	Ex 7
H	Different planet / altitude ( $g$ changes)	any $a$ , new $g$	recompute	Ex 8

Notice the two special boundaries: $a = 0$ separates "normal" from "heavy/light," and $a = - g$ is the exact edge where $N$ hits zero. Everything below $a = - g$ (cell F) is qualitatively new — the floor can no longer keep up with the body, so contact breaks entirely.

Figure — Weightlessness — true (free fall) vs apparent

Worked examples

Take $g = 10 m/s^{2}$ and $m = 60 kg$ unless a problem says otherwise.

Worked example

Ex 5 — Cell E: the trap (moving down while decelerating) Elevator is moving downward at $6 m/s$ and slowing at $3 m/s^{2}$ (about to stop). Reading?

Forecast: "It's going down, so I feel light" — is that right?

Velocity vs acceleration. Velocity is downward, but it's decreasing in magnitude. To slow a downward motion, acceleration must point up: $a = + 3$ . Why this step? "Slowing down while moving down" = acceleration opposite to velocity = upward. The moving direction is a decoy.
Apply. $N = 60 (10 + 3) = 780 N$ . Why this step? Upward acceleration means the floor pushes harder → you feel heavier, even though you're descending.

Verify: $780 > 600$ ✓ — heavier despite going down. This is exactly the Non-inertial Frames & Pseudo-forces gotcha the parent warns about: velocity direction is irrelevant.

Worked example

Ex 6 — Cell F: pushed down harder than gravity A rocket floor is fired downward so the whole cabin accelerates downward at $15 m/s^{2}$ (faster than free fall). What does the scale read?

Forecast: The formula gives a negative number — what does that mean?

Sign of $a$ . Downward $15 \Rightarrow a = - 15$ , and $∣ a ∣ = 15 > g = 10$ . Why this step? We've crossed below the free-fall boundary $a = - g$ .
Apply naively. $N = 60 (10 - 15) = 60 (- 5) = - 300 N$ . Why this step? To see that the maths flags a problem — a floor can't pull your feet, so a negative $N$ is physically impossible.
Interpret. A surface can only push ( $N \geq 0$ ). So the true reading is $N = 0$ : your feet leave the floor. You'd float against the ceiling (or a ceiling would have to push you down instead). Why this step? When the demanded $N$ goes negative, contact simply breaks — the body separates from that surface.

Verify: Boundary check: $N = 0$ first occurs at $a = - g = - 10$ ; for any $a < - 10$ the floor can't keep up, so $N = 0$ and the body separates. ✓ (The $- 300$ N is a signal, not a real force.)

Worked example

Ex 7 — Cell G: orbit (curved free fall) An astronaut of mass $70 kg$ orbits at altitude where the local gravitational acceleration is $g^{'} = 8.7 m/s^{2}$ . The station is in a circular orbit whose centripetal acceleration equals $g^{'}$ (that's what "orbit" means). Apparent weight?

Forecast: With $g^{'} = 8.7$ , does she weigh $70 \times 8.7 = 609 N$ , or zero?

What is the acceleration of the system? For a circular orbit, gravity is the centripetal force, so $a = g^{'} = 8.7$ directed toward Earth — the same direction gravity pulls. Why this step? Circular Motion & Centripetal Acceleration: the net inward acceleration is supplied entirely by gravity, nothing else.
Go into the falling frame. Relative to the station, the astronaut accelerates at $g^{'}$ in the exact direction gravity pulls — this is free fall along a curve. Locally "down" points to Earth, so $a = - g^{'}$ in the up-positive local frame. Why this step? Same setup as the snapped cable (Ex 4), just curved instead of straight.
Apply. $N = m (g^{'} + a) = 70 (8.7 - 8.7) = 0 N$ . Why this step? Station and astronaut share the same acceleration, so neither presses on the other.

Verify: $N = 0$ ✓ despite $g^{'} = 8.7$ (about 89% of surface gravity, from Gravitation — Variation of g with altitude). Floating comes from shared acceleration, not absent gravity.

Worked example

Ex 8 — Cell H: different planet, elevator accelerating On the Moon ( $g_{M} = 1.6 m/s^{2}$ ) an astronaut of mass $80 kg$ rides a lander accelerating upward at $2.0 m/s^{2}$ . Scale reading? Then: what upward-or-downward $a$ would make him weightless?

Forecast: Will the reading be near his Earth weight or much smaller?

Use the local $g$ . Replace $g$ by $g_{M} = 1.6$ . Up is positive, $a = + 2.0$ . Why this step? The formula's $g$ is always the local gravitational acceleration — nothing about it is Earth-specific.
Apply. $N = 80 (1.6 + 2.0) = 80 (3.6) = 288 N$ . Why this step? Heavy relative to his standing Moon weight $80 \times 1.6 = 128 N$ , because of the upward acceleration.
Weightless condition. Need $N = 0 \Rightarrow g_{M} + a = 0 \Rightarrow a = - g_{M} = - 1.6 m/s^{2}$ (lander in free fall toward the Moon). Why this step? The weightless edge is always $a = - g$ with the local $g$ — here $- 1.6$ , not $- 10$ .

Verify: Standing Moon weight $80 \times 1.6 = 128 N$ ; accelerating up gives $288 > 128$ ✓. Free-fall $a = - 1.6$ gives $N = 80 (1.6 - 1.6) = 0$ ✓.

Recall

Recall Which cell am I in?

Moving up at constant speed — heavy, light, or normal? ::: Normal ( $a = 0$ , so $N = m g$ ). Slowing down while descending — heavy or light? ::: Heavy (acceleration points up, $a > 0$ ). Formula gives $N = - 200$ N — physical reading? ::: $0$ N; the body leaves the surface (a floor can't pull). On the Moon, what $a$ gives weightlessness? ::: $a = - g_{M} = - 1.6 m/s^{2}$ (free fall toward the Moon). In orbit with $g^{'} = 8.7$ , apparent weight? ::: $0$ N — shared acceleration, not zero gravity.

Connections

Parent: Weightlessness — the master formula built here.
Newton's Second Law — every example is $F_{n e t} = ma$ vertical.
Normal Force — why $N \geq 0$ and what " $N = 0$ " means physically.
Newton's Third Law — why the scale reads your push.
Non-inertial Frames & Pseudo-forces — the "moving down feels light" trap.
Circular Motion & Centripetal Acceleration — orbit as curved free fall (Ex 7).
Gravitation — Variation of g with altitude — where $g^{'} = 8.7$ comes from.
Free Fall Kinematics — the $a = - g$ case in motion.