Non-inertial reference frames — pseudo forces: Worked examples

Before anything, let's earn every symbol once, in plain words.

Definition The symbols we will reuse

$m$ — the mass of the object, in kilograms. "How much stuff there is."
$a_{0}$ — the acceleration of the frame (the bus, lift, car) relative to the ground. An arrow: it has a size and a direction. We always pick a positive direction first, then $a_{0}$ is a signed number: $+$ if it points that way, $-$ if it points the other.
$g = 9.8 m/s^{2}$ — the strength of Earth's gravity pull, downward.
$F_{pseudo} = - m a_{0}$ — the fake force we add inside the accelerating frame. The minus sign means exactly opposite to $a_{0}$ : if the frame speeds up forward, the fake force points backward.
$a_{in frame}$ — the acceleration the observer inside the accelerating frame measures (parent-note symbol). In each example we rename it to the frame we're in: $a_{bus}$ , $a_{lift}$ , etc. It is what the passenger actually sees the object do, not what the ground sees.
$N$ — the normal force: the push a surface (floor, scale) gives perpendicular to itself. A scale reads $N$ (see Apparent weight & normal force).
$T$ — the tension: the pull carried along a string, directed from the object toward whatever the string hangs from.

Cell	What varies	Example
A. $a_{0}$ forward (+)	horizontal, sign $+$	Ex 1 — bus, ball slides
B. $a_{0}$ backward (−)	horizontal, sign $-$	Ex 2 — braking bus
C. $a_{0}$ up (+)	vertical, sign $+$	Ex 3 — lift going up
D. $a_{0}$ down (−)	vertical, sign $-$	Ex 4 — lift going down
E. Degenerate: free fall	$a_{0} = g$ down	Ex 5 — weightlessness
E′. Beyond free-fall	$a_{0} > g$ down	Ex 6 — driven-down cage
F. Zero input	$a_{0} = 0$	Ex 7 — inertial trap
G. Tilted / 2-D	horizontal + gravity combine	Ex 8 — pendulum tilt-meter
H. Limit behaviour	$a_{0} \to \infty$ , $a_{0} \to 0$	Ex 9 — how far can the string tilt?
I. Word problem	real-world story	Ex 10 — coffee cup on a train
J. Exam twist	block on incline in accelerating truck	Ex 11 — the classic trap

Horizontal cases (A & B)

Figure — Non-inertial reference frames — pseudo forces

Figure s01 — A forward-accelerating bus (orange arrow, $a_{0}$ ) and the backward pseudo force (red arrow, $- m a_{0}$ ) on a blue ball resting on the smooth floor. Covers Ex 1 and Ex 2.

Worked example Ex 1 — Ball on a smooth bus floor · Cell A (

$a_{0}$ forward) A bus accelerates forward at $a_{0} = 3 m/s^{2}$ . A ball of mass $m = 0.4 kg$ rests on the smooth (frictionless) floor. What acceleration does a passenger see, and in which direction?

Forecast: guess now — does the ball slide forward or backward, and is its acceleration bigger or smaller than $3 m/s^{2}$ ?

Pick the frame: the bus (non-inertial). List real horizontal forces: none — the floor is smooth. Why this step? Choosing one frame stops us double-counting. Smooth floor ⇒ no friction ⇒ zero real horizontal force.
Add the pseudo force: $F_{pseudo} = - m a_{0} = - (0.4) (3) = - 1.2 N$ , i.e. backward (opposite the bus's forward $a_{0}$ ). Look at the red arrow in the figure. Why this step? The minus sign in $- m a_{0}$ makes it point backward — that's the whole trick.
Apply the frame law: $F_{real} + F_{pseudo} = m a_{bus} \Rightarrow 0 + (- 1.2) = 0.4 a_{bus}$ . Why this step? Inside the non-inertial bus, Newton's law only closes if we include the pseudo force alongside the real ones; here $a_{bus}$ is the parent note's $a_{in frame}$ — what the passenger actually measures.
Solve: $a_{bus} = - 1.2/0.4 = - 3 m/s^{2}$ — backward, magnitude $3 m/s^{2}$ . Why this step? The passenger genuinely sees the ball recede backward exactly this fast.

Verify: the ground observer sees no force on the ball, so it stays still; the bus floor slides forward under it at $3 m/s^{2}$ , so relative to the bus the ball moves back at $3 m/s^{2}$ . Same number ✓. Units: $N / kg = m/s^{2}$ ✓.

Worked example Ex 2 — Braking bus · Cell B (

$a_{0}$ backward) The same bus now brakes, decelerating at $4 m/s^{2}$ . Take forward as positive, so the frame's acceleration is the signed value $a_{0} = - 4 m/s^{2}$ (pointing backward). Same $m = 0.4 kg$ ball on smooth floor. Which way does it slide?

Forecast: you're in a braking bus — which way do loose things on the floor go?

Frame's signed acceleration: $a_{0} = - 4 m/s^{2}$ (backward).
Pseudo force: $F_{pseudo} = - m a_{0} = - (0.4) (- 4) = + 1.6 N$ — forward (positive). Why this step? Two minus signs make a plus: opposite of "backward" is "forward". The ball lurches toward the front.
Frame law: $0 + 1.6 = 0.4 a_{bus} \Rightarrow a_{bus} = + 4 m/s^{2}$ forward. Why this step? Same closure as Ex 1 — the only horizontal "force" is the pseudo force, so it sets $a_{bus}$ .

Verify: everyday experience — when a bus brakes, unsecured things fly forward. Magnitude equals $∣ a_{0} ∣$ on a smooth floor ✓. The sign flipped correctly versus Ex 1 ✓.

Vertical cases (C, D, E, E′)

Figure s02 — Free-body of a person on a scale in a lift accelerating up (orange). Weight $m g$ down (gray), normal $N$ up (green, the scale reading), pseudo force $m a_{0}$ down (red). Covers Ex 3–6.

Here gravity $m g$ (always down) shares the stage with the pseudo force. A person of mass $m = 60 kg$ stands on a bathroom scale; the scale reads the normal force $N$ (defined above). Take up as positive.

Worked example Ex 4 — Lift accelerating down · Cell D (watch the signs!)

Same lift now accelerates down at rate $2 m/s^{2}$ . Keeping up positive, the signed frame acceleration is $a_{0} = - 2 m/s^{2}$ . Scale reading?

Forecast: lighter or heavier this time?

Substitute the signed value into $F_{pseudo} = - m a_{0}$ : $F_{pseudo} = - m a_{0} = - (60) (- 2) = + 120 N$ — a positive (upward) force. As a magnitude that is $120 N$ , and its direction is up. Why this step? This is exactly the signed-vs-magnitude trap flagged earlier: because $a_{0}$ is already $- 2$ , the formula $- m a_{0}$ produces $+ 120 N$ on its own — we do not also flip a sign by hand. The fake force lifts you slightly off the scale.
Equilibrium in lift frame ( $a_{lift} = 0$ ), up positive: $N - m g + 120 = 0$ . Why this step? Sum of signed forces is zero because the person is at rest in the lift. Here $+ 120$ is the signed pseudo force from step 1, not a magnitude we chose the sign of.
Solve: $N = m g - 120 = 60 (9.8) - 120 = 588 - 120 = 468 N$ . Equivalently $N = m (g - 2) = 60 (7.8) = 468 N$ .

Verify: $N < 588 N$ — you feel lighter, correct for a downward-accelerating lift ✓. Cross-check with Ex 3: the compact formula $N = m (g + a_{0})$ with signed $a_{0} = - 2$ gives $60 (9.8 - 2) = 468$ ✓ — the signed substitution and the by-hand version agree.

Worked example Ex 6 — Cage driven down

faster than gravity · Cell E′ (beyond free-fall) A motor drags the lift downward at rate $2 m/s^{2}$ past gravity, i.e. the signed frame acceleration is $a_{0} = - 12 m/s^{2}$ (up positive) — faster down than gravity. The person now stands on a scale that is on the ceiling of the cage (a spring pressing down on their head), which can only push, giving a downward normal force $N_{c} \geq 0$ . What does that ceiling-scale read?

Forecast: the floor can't hold them anymore — what pushes them, and how hard?

First test the floor scale with $N = m (g + a_{0})$ , signed $a_{0} = - 12$ : $N = 60 (9.8 - 12) = 60 (- 2.2) = - 132 N$ . Why this step? A floor can only push up ( $N \geq 0$ ). A negative answer means the floor would have to pull — impossible. So the person leaves the floor and presses the ceiling instead.
Redraw the free body with the ceiling in contact. Keeping up positive, list every signed force:
- weight: $- m g$
- pseudo force from the formula: $- m a_{0} = - (60) (- 12) = + 720$ (up)
- ceiling push: $- N_{c}$ (the ceiling can only push the head down, so its signed value is negative) Why this step? We substitute the signed $a_{0} = - 12$ straight into $- m a_{0}$ (no by-hand sign flip), keeping the same convention as Ex 4. This gives an upward pseudo force of $+ 720 N$ .
Set the signed sum to zero (person at rest in the cage, $a_{cage} = 0$ ): $- m g + (- m a_{0}) - N_{c} = 0 \Rightarrow - 588 + 720 - N_{c} = 0.$ Why this step? All signed forces cancel because the person doesn't move relative to the cage; this is the same equilibrium closure used in Ex 3–5, now with the ceiling term added.
Solve for the reading: $N_{c} = 720 - 588 = 132 N$ — the ceiling presses on their head with $132 N$ .

Verify: magnitude matches the impossible floor reading from step 1 ( $∣ - 132∣ = 132$ ) ✓ — physics simply moved the contact from floor to ceiling. This is the natural case just beyond free-fall (Ex 5's $N = 0$ is the crossover at $a_{0} = - g$ ).

The zero-input trap (F)

Worked example Ex 7 — No acceleration · Cell F

A train moves at a constant $20 m/s$ (so $a_{0} = 0$ ). A student "helpfully" adds a pseudo force to explain why their coffee stays put. What is the pseudo force?

Forecast: should there be any pseudo force at all?

$F_{pseudo} = - m a_{0} = - m \cdot 0 = 0 N$ . Why this step? Constant velocity ⇒ $a_{0} = 0$ ⇒ the frame is inertial (see Galilean relativity). No correction is needed.
If you add a nonzero pseudo force here, you break Newton's law that was already fine — a double count. Why this step? An inertial frame already satisfies $F = m a$ ; an extra fake force would over-count and give wrong motion.

Verify: the coffee stays still because no real force acts and no fake one is needed; $F = m a$ works as-is ✓. This is exactly the third [!mistake] warned about in the parent note.

Tilted / two-dimensional case (G)

Figure s03 — Pendulum bob (blue) in a forward-accelerating car (orange). String tension along the string, weight $m g$ down (gray), pseudo force $m a_{0}$ backward (red); the tilt angle $θ$ from vertical (red arc) satisfies $tan θ = a_{0} / g$ .

Worked example Ex 8 — Pendulum accelerometer · Cell G

A pendulum bob hangs from the roof of a car. The car accelerates horizontally at $a_{0} = 4.9 m/s^{2}$ . The bob settles at a steady angle $θ$ from vertical. Find $θ$ .

Forecast: guess $θ$ — bigger or smaller than $4 5^{\circ}$ ? (Hint: compare $a_{0}$ with $g$ .)

In the car frame the bob is in equilibrium. Three forces (see figure): tension $T$ along the string, weight $m g$ down, pseudo force $m a_{0}$ backward (opposite the car's forward $a_{0}$ ). Why this step? Steady angle ⇒ zero acceleration in the car frame ( $a_{car} = 0$ ) ⇒ forces sum to zero.
Split the tension into pieces. Why tan and not sin or cos? We want the angle from two known perpendicular forces — the horizontal pull ( $m a_{0}$ ) and the vertical pull ( $m g$ ). The ratio of the side opposite $θ$ to the side next to it is $tan θ$ ; that ratio is exactly "how tilted," so tan is the right tool.
Horizontal balance: $T sin θ = m a_{0}$ . Vertical balance: $T cos θ = m g$ . Why this step? The string tilts by $θ$ from vertical, so its tension has a horizontal part $T sin θ$ (holding back the pseudo force) and a vertical part $T cos θ$ (holding up the weight). At equilibrium each direction balances separately.
Divide the two equations to eliminate the unknown $T$ : $\frac{T sin θ}{T cos θ} = \frac{m a _{0}}{m g} \Rightarrow tan θ = \frac{a _{0}}{g} = \frac{4.9}{9.8} = 0.5$ . Why this step? We don't know $T$ and don't care about it, so dividing cancels it and also cancels $m$ — leaving a pure relation between the angle and the two accelerations.
Undo the tan with arctan (the question "which angle has this tan?"): $θ = arctan (0.5) \approx 26.5 7^{\circ}$ . Why this step? $tan$ turned the angle into a ratio; $arctan$ is its inverse, turning the ratio back into the angle we actually want.

Verify: $a_{0} < g$ ⇒ $tan θ < 1$ ⇒ $θ < 4 5^{\circ}$ ✓ (matches forecast). The bob leans backward, opposite the car's motion, as a real accelerometer does ✓. Effective gravity has magnitude $g^{2} + a_{0}^{2} = 9. 8^{2} + 4. 9^{2} \approx 10.96 m/s^{2}$ , pointing along the string ✓.

Limit behaviour (H)

Worked example Ex 9 — How far can the string tilt? · Cell H (limits)

Using $tan θ = a_{0} / g$ from Ex 8, examine two extremes: (a) $a_{0} \to 0$ , (b) $a_{0} \to \infty$ .

Forecast: at huge acceleration, does the string reach $9 0^{\circ}$ ?

Small limit: as $a_{0} \to 0$ , $tan θ \to 0 \Rightarrow θ \to 0^{\circ}$ . Why this step? No horizontal fake force ⇒ string hangs straight down. Sanity anchor at the bottom of the range.
Large limit: as $a_{0} \to \infty$ , $tan θ \to \infty \Rightarrow θ \to 9 0^{\circ}$ but never reaches it. Why this step? $tan (9 0^{\circ})$ is undefined (the vertical line has "infinite steepness"); the string approaches horizontal but any finite $a_{0}$ leaves $θ$ a hair below $9 0^{\circ}$ .
Concrete midpoint: $a_{0} = g = 9.8 \Rightarrow tan θ = 1 \Rightarrow θ = arctan (1) = 4 5^{\circ}$ exactly. Why this step? Equal horizontal and vertical pulls give a $4 5^{\circ}$ tilt — a memorable landmark between the two limits.

Verify: monotone rise from $0^{\circ}$ (at rest) through $4 5^{\circ}$ (at $a_{0} = g$ ) toward $9 0^{\circ}$ (infinite $a_{0}$ ). At $a_{0} = g$ : $arctan (1) = 4 5^{\circ}$ ✓; $tan (0) = 0$ anchors the low end ✓.

Word problem (I)

Worked example Ex 10 — Coffee cup on a starting train · Cell I

A train starts and reaches $15 m/s$ in $5 s$ (uniform acceleration). A coffee cup ( $m = 0.3 kg$ ) sits on a table with friction coefficient $μ = 0.15$ between cup and table. Does the cup slide backward? (Take $g = 9.8$ .)

Forecast: does the fake force beat friction?

Frame acceleration: $a_{0} = Δ v /Δ t = 15/5 = 3 m/s^{2}$ forward. Why this step? We need $a_{0}$ before any pseudo force; it comes straight from "change in speed over time."
Pseudo force on cup: $F_{pseudo} = m a_{0} = 0.3 \times 3 = 0.9 N$ backward. Why this step? Inside the train frame this backward ghost-shove is what tries to slide the cup.
Maximum static friction available: $f_{m a x} = μ m g = 0.15 \times 0.3 \times 9.8 = 0.441 N$ . Why this step? Friction can only supply up to this much grip; if the pseudo force exceeds it, the cup slips.
Compare: $0.9 N > 0.441 N$ ⇒ the cup slides backward. Why this step? The demand ( $0.9$ N) beats the supply ( $0.441$ N), so static friction is overwhelmed.

Verify: the ghost-shove ( $0.9$ N) outmuscles the grip ( $0.441$ N), so coffee slides — exactly why it spills toward you when a train jerks forward ✓. Units all newtons ✓.

Exam twist (J)

Figure s04 — A block (blue) on a frictionless $3 0^{\circ}$ wedge inside a truck accelerating forward (orange). Weight $m g$ down (gray) and pseudo force $m a_{0}$ backward (red) balance along the slope when $a_{0} = g tan α$ .

Worked example Ex 11 — Block on a frictionless wedge inside an accelerating truck · Cell J

A wedge of incline angle $α = 3 0^{\circ}$ sits on a truck bed. A block rests on the frictionless incline. The truck accelerates forward (toward the high side of the wedge) at $a_{0}$ . Find the $a_{0}$ that keeps the block stationary relative to the wedge (no sliding).

Forecast: must $a_{0}$ be large or small to pin a block on a $3 0^{\circ}$ slope?

Frame: the truck. Forces on block: weight $m g$ down, normal $N$ perpendicular to incline, pseudo force $m a_{0}$ backward. Why this step? Frictionless ⇒ only gravity, normal, and the fake force act. No friction to help.
"Stationary relative to wedge" ⇒ net force along the incline surface is zero. Why this step? Only the along-slope direction can make it slide; balancing it is the condition.
Resolve along the incline (take up-slope positive). Gravity's along-slope pull: $- m g sin α$ . Pseudo force's along-slope component: the backward force, projected up the slope, gives $+ m a_{0} cos α$ . Why this step? We only need components parallel to the surface; the normal force has none.
Balance: $m a_{0} cos α - m g sin α = 0 \Rightarrow a_{0} = g tan α$ .
Number: $a_{0} = 9.8 \times tan 3 0^{\circ} = 9.8 \times 0.5774 \approx 5.66 m/s^{2}$ . Why this step? This is the only acceleration that pins the block; slower and it slides down, faster and it slides up.

Verify: identical in form to the pendulum, $tan α = a_{0} / g$ — the "effective gravity" $g^{2} + a_{0}^{2}$ must point perpendicular to the incline for the block not to slide. Consistent with Centrifugal force-style effective-gravity reasoning ✓. Number check: $9.8 tan 3 0^{\circ} \approx 5.66$ ✓.

Active recall

Recall Quick self-test

Ball on smooth bus floor, bus accelerates forward at $3 m/s^{2}$ — ball's acceleration in bus frame? ::: $3 m/s^{2}$ backward. Scale reading for $60 kg$ in a lift accelerating up at $2 m/s^{2}$ ? ::: $N = 60 (9.8 + 2) = 708 N$ . Same lift accelerating down at $2 m/s^{2}$ (signed $a_{0} = - 2$ )? ::: $N = 60 (9.8 - 2) = 468 N$ . Free-fall scale reading? ::: $0 N$ (weightless). Cage driven down at $12 m/s^{2}$ — floor "reading" and what really happens? ::: Floor formula gives $- 132 N$ (impossible); person presses the ceiling with $132 N$ . Pendulum tilt for $a_{0} = 4.9$ , $g = 9.8$ ? ::: $θ = arctan 0.5 \approx 26.5 7^{\circ}$ . $a_{0}$ to pin a block on a frictionless $3 0^{\circ}$ wedge? ::: $a_{0} = g tan 3 0^{\circ} \approx 5.66 m/s^{2}$ . Pseudo force when the frame moves at constant velocity? ::: Zero — the frame is inertial.

Connections

Parent topic — this page fills in every case class.
Newton's Second Law — every cell reduces to force-balance once the ghost force is added.
Apparent weight & normal force — Ex 3–6 are the lift family.
Newton's Third Law — recall the pseudo force never has a reaction partner.
Galilean relativity — Ex 7's constant-velocity frame is inertial.
Centrifugal force · Coriolis force — rotating-frame pseudo forces build on the same $- m a_{0}$ idea.

Worked examples — Non-inertial reference frames — pseudo forces

The scenario matrix