Non-inertial reference frames — pseudo forces: Exercises

Before we start, five symbols you must own:

Definition The five characters in every problem

$m$ — the mass of the object (kilograms). Bigger $m$ = stronger pseudo force.
$a_{0}$ — the acceleration of the frame (the bus, lift, car, disc) measured by a ground observer who is not accelerating. The little arrow means it is a vector: it carries both a size and a direction.
$a_{0}$ (no arrow) — the magnitude (length) of that vector, a plain positive number in $m/s^{2}$ . So $a_{0}$ tells you how much and which way; $a_{0}$ tells you just how much. Whenever you see the arrow dropped, we are talking about the number only.
$F_{real}$ — the real force: the sum of every push and pull with a genuine physical source (gravity pulling down, a string tugging, a floor pushing, friction gripping). These are the forces a ground observer would list. They exist in every frame.
$F_{pseudo} = - m a_{0}$ — the fake force. The minus sign flips the arrow $a_{0}$ around: whichever way the frame speeds up, the fake force points the opposite way, and its size is $m a_{0}$ (mass times the magnitude). Unlike $F_{real}$ , this one has no physical source — it appears only because we chose an accelerating frame.

The two combine in the frame's working law $F_{real} + F_{pseudo} = m a_{in frame}$ , where $a_{in frame}$ is the acceleration measured inside the accelerating frame. Throughout, $g = 9.8 m/s^{2}$ is the strength of real gravity pulling everything straight down.

Intuition Figure 1 — the two arrows you always draw

Look at the figure below. It shows a boxed "frame" (think of it as a train car) rolling on two wheels. A blue arrow above it, labelled $a_{0}$ , points right — the direction the frame speeds up. Inside sits a yellow block of mass $m$ , and a pink arrow on that block, labelled $F_{pseudo} = - m a_{0}$ , points left — exactly opposite the blue one. The single lesson of the picture: the fake force (pink) is always the mirror-flip of the frame's acceleration (blue). Keep this pink-opposite-blue image in mind for every problem below.

Alt text: a rectangular train car on two wheels; a blue rightward arrow labelled frame acceleration a-zero sits above it; a yellow block labelled m sits inside with a pink leftward arrow labelled pseudo force equals minus m a-zero, pointing opposite the blue arrow.

Figure — Non-inertial reference frames — pseudo forces

Level 1 — Recognition

Problem 1.1

A train accelerates forward with $a_{0}$ of magnitude $a_{0} = 3 m/s^{2}$ . A suitcase of mass $m = 8 kg$ rests on the smooth luggage rack. In the train's frame, state the direction and size of the pseudo force on the suitcase.

Recall Solution 1.1

WHAT we want: the fake force $F_{pseudo} = - m a_{0}$ . WHY: we are asked to describe motion inside the train, an accelerating (non-inertial) frame, so we must add the fake force. Size: $∣ F_{pseudo} ∣ = m a_{0} = 8 \times 3 = 24 N$ (using the magnitude $a_{0}$ ). Direction: the minus sign flips $a_{0}$ . Train speeds up forward, so the fake force points backward (toward the rear of the train). Answer: $24 N$ , backward.

Problem 1.2

A lift descends, speeding up downward at magnitude $a_{0} = 2 m/s^{2}$ . A person of mass $m = 60 kg$ stands inside. In which direction does the pseudo force on the person point, and how big is it?

Recall Solution 1.2

$a_{0}$ points down (the lift is gaining downward speed). Flip it: $F_{pseudo} = - m a_{0}$ points up. Size $= m a_{0} = 60 \times 2 = 120 N$ . Answer: $120 N$ , upward. (This is why you feel briefly lighter as a lift starts going down — the upward fake force partly cancels your weight.)

Level 2 — Application

Problem 2.1

The smooth-rack suitcase of Problem 1.1 ( $m = 8 kg$ , train magnitude $a_{0} = 3 m/s^{2}$ forward) is free to slide. What is its acceleration as seen by a passenger inside the train?

Recall Solution 2.1

WHAT: apply the modified law $F_{real} + F_{pseudo} = m a_{in frame}$ . WHY: the passenger's frame is non-inertial, so this is the law that works there. Horizontally the rack is smooth $\Rightarrow$ real horizontal force $= 0$ . The only horizontal "force" is the pseudo force $= 24 N$ backward. $a_{in frame} = \frac{F _{pseudo}}{m} = \frac{24}{8} = 3 m/s^{2} backward.$ Answer: $3 m/s^{2}$ toward the rear. (Sanity check with the ground view: ground sees no force on the suitcase, so it stays still while the train pulls forward at $3 m/s^{2}$ — relative to the train the suitcase recedes backward at exactly $3 m/s^{2}$ . ✓)

Problem 2.2

A person of mass $m = 70 kg$ stands on a scale in a lift accelerating upward at magnitude $a_{0} = 1.5 m/s^{2}$ . What does the scale read (the apparent weight $N$ )? Use $g = 9.8 m/s^{2}$ .

Recall Solution 2.2

WHAT: find the normal force $N$ (the up-push of the scale, which is what it displays). Forces in the lift frame: weight $m g$ down, normal $N$ up, pseudo force $m a_{0}$ down (frame accelerates up $\Rightarrow$ fake force down). Person is at rest in the lift $\Rightarrow$ net $= 0$ : $N - m g - m a_{0} = 0 \Rightarrow N = m (g + a_{0}) = 70 (9.8 + 1.5) = 70 \times 11.3 = 791 N .$ Answer: $791 N$ (heavier than the resting $70 \times 9.8 = 686 N$ ). See Apparent weight & normal force.

Level 3 — Analysis

Problem 3.1

A pendulum bob hangs from the roof of a car. The car accelerates horizontally and the string settles at a steady angle $θ = 1 2^{\circ}$ from vertical. What is the magnitude $a_{0}$ of the car's acceleration? ( $g = 9.8 m/s^{2}$ .)

Recall Solution 3.1

WHAT: in the car frame the bob is in equilibrium under tension $T$ (along string), weight $m g$ down, and pseudo force $m a_{0}$ horizontal (opposite the car's acceleration). Split the string tension into horizontal and vertical pieces: $T sin θ = m a_{0} (horizontal balance), T cos θ = m g (vertical balance) .$ WHY divide them: dividing kills the unknowns $T$ and $m$ at once, leaving only the angle and the two accelerations: $tan θ = \frac{a _{0}}{g} \Rightarrow a_{0} = g tan θ = 9.8 \times tan 1 2^{\circ} = 9.8 \times 0.2126 = 2.08 m/s^{2} .$ Answer: $a_{0} \approx 2.08 m/s^{2}$ . The string is a built-in accelerometer.

Problem 3.2

Same pendulum-in-a-car setup. The bob has mass $m = 0.30 kg$ and the car accelerates at magnitude $a_{0} = 4.0 m/s^{2}$ . Find (a) the tilt angle $θ$ and (b) the string tension $T$ .

Intuition Figure 2 — the pendulum's force triangle

Look at the figure below. A string hangs from a ceiling bracket; the yellow bob dangles at angle $θ$ (marked with a pale-yellow arc) from the dotted vertical. Three arrows leave the bob: a blue one straight down labelled $m g$ (real weight), a pink one pointing backward labelled $m a_{0}$ (the fake force, opposite the car's motion), and a white one up along the string labelled $T$ (tension). At the bottom a blue arrow labelled "car $a_{0}$ " shows the car speeding up forward, so the bob lags backward — which is exactly why the pink arrow points backward. The white tension is the diagonal that balances the blue-down and pink-back pair.

Alt text: a pendulum hanging from a ceiling, string tilted by angle theta from a dotted vertical line; from the bob a blue arrow points straight down labelled m g, a pink arrow points backward labelled m a-zero, and a white arrow points up along the string labelled T; a separate blue arrow at the bottom labelled car a-zero points forward.

Recall Solution 3.2

(a) $tan θ = \frac{a _{0}}{g} = \frac{4.0}{9.8} = 0.4082 \Rightarrow θ = arctan (0.4082) = 22. 2^{\circ} .$ WHY arctan: $tan θ$ gives the ratio; arctan is the inverse question "which angle has this tangent?", which recovers $θ$ . (b) The bob effectively hangs in a stronger, tilted gravity of size $g_{eff} = g^{2} + a_{0}^{2}$ (the diagonal of the down-weight and sideways pseudo force). The tension carries all of it: $T = m g^{2} + a_{0}^{2} = 0.30 9. 8^{2} + 4. 0^{2} = 0.30 96.04 + 16 = 0.30 112.04 = 0.30 \times 10.585 = 3.18 N .$ Answer: $θ \approx 22. 2^{\circ}$ , $T \approx 3.18 N$ .

Level 4 — Synthesis

Problem 4.1

A block of mass $m = 2.0 kg$ sits on the floor of a truck. The coefficient of static friction between block and floor is $μ_{s} = 0.30$ . The truck accelerates forward. What is the maximum forward acceleration magnitude $a_{0}$ the truck can have before the block starts sliding backward (relative to the truck)? ( $g = 9.8 m/s^{2}$ .)

Recall Solution 4.1

WHAT: work in the truck frame. The block is on the verge of sliding, so it is still at rest in the frame $\Rightarrow$ net force zero, with friction at its maximum. Define the friction symbol: let $f$ be the static friction force — the grip the floor exerts sideways on the block to stop it sliding. Static friction adjusts itself up to a ceiling; that ceiling is $f_{m a x} = μ_{s} N$ , where $μ_{s}$ is the coefficient of static friction and $N$ is the normal (perpendicular) push of the floor. Horizontal forces in the truck frame:

Pseudo force $m a_{0}$ backward (frame accelerates forward).
Static friction $f$ forward, holding the block in place, up to a maximum $f_{m a x} = μ_{s} N$ . Vertically nothing accelerates and there is no vertical pseudo force, so $N = m g$ . On the verge of slipping, friction is maxed and balances the pseudo force: $m a_{0} = f_{m a x} = μ_{s} N = μ_{s} m g \Rightarrow a_{0} = μ_{s} g = 0.30 \times 9.8 = 2.94 m/s^{2} .$ WHY the mass cancels: both the pseudo force ( $m a_{0}$ ) and the friction limit ( $μ_{s} m g$ ) scale with $m$ , so it drops out — a heavier block is no easier or harder to keep put. Answer: $a_{0}^{m a x} = 2.94 m/s^{2}$ .

Problem 4.2

A person ( $m = 65 kg$ ) stands on a scale inside a lift. The scale reads $520 N$ . Is the lift accelerating up or down, and what is the magnitude of $a_{0}$ ? ( $g = 9.8 m/s^{2}$ .)

Recall Solution 4.2

WHAT: the general lift result is $N = m (g \pm a_{0})$ , with $+$ for upward acceleration and $-$ for downward. The resting weight is $m g = 65 \times 9.8 = 637 N$ . The scale reads $520 N < 637 N$ , i.e. the person is lighter, so the pseudo force must point up, which happens when $a_{0}$ points down. Use the minus form: $N = m (g - a_{0}) \Rightarrow 520 = 65 (9.8 - a_{0}) \Rightarrow 9.8 - a_{0} = \frac{520}{65} = 8.0 \Rightarrow a_{0} = 1.8 m/s^{2} .$ Answer: the lift accelerates downward at magnitude $a_{0} = 1.8 m/s^{2}$ .

Level 5 — Mastery

Problem 5.1

A block of mass $m$ rests on a frictionless wedge of incline angle $α$ . The wedge is pushed horizontally so the block stays fixed relative to the wedge (no sliding). Find the required horizontal acceleration magnitude $a_{0}$ of the wedge in terms of $g$ and $α$ . Then evaluate for $α = 3 0^{\circ}$ ( $g = 9.8 m/s^{2}$ ). Finally, discuss what happens if the wedge accelerates faster or slower than this value.

Recall Solution 5.1

WHAT: go to the wedge's frame. The block is at rest there, so real forces + pseudo force = 0. Forces on the block:

Weight $m g$ straight down.
Normal force $N$ perpendicular to the incline surface.
Pseudo force $m a_{0}$ horizontal, opposite the wedge's acceleration. WHY use axes along/across the incline: the unknown $N$ is perpendicular to the surface, so choosing axes along and perpendicular to the incline lets one equation be $N$ -free.

Where the $cos α$ and $sin α$ come from. The incline surface tilts by $α$ from horizontal. Take the "along-slope" direction (pointing up the ramp). Project each horizontal or vertical vector onto it — this is exactly the "break the arrow into two perpendicular legs" move from the opening definition:

The pseudo force is horizontal. The angle between a horizontal arrow and the up-slope direction is $α$ , so its along-slope component is $m a_{0} cos α$ (adjacent-side projection). It points up the slope.
The weight is vertical. The angle between a vertical arrow and the up-slope direction is $9 0^{\circ} - α$ , so its along-slope component is $m g cos (9 0^{\circ} - α) = m g sin α$ , pointing down the slope.

Along the incline (up-slope positive), balance: $m a_{0} cos α = m g sin α \Rightarrow a_{0} = g tan α .$ Perpendicular to the incline (for completeness — this fixes $N$ ): the perpendicular components must also sum to zero. Weight contributes $m g cos α$ into the surface; the pseudo force contributes $m a_{0} sin α$ into the surface; the normal $N$ pushes out. So $N = m g cos α + m a_{0} sin α .$ This confirms the surface really can supply the needed push, and shows $N$ grows as the wedge accelerates harder. For $α = 3 0^{\circ}$ : $a_{0} = 9.8 \times tan 3 0^{\circ} = 9.8 \times 0.5774 = 5.66 m/s^{2} .$

Edge cases — what if $a_{0} \neq = g tan α$ ?

If $a_{0} = g tan α$ exactly, the along-slope forces cancel and the block stays put. ✓
If $a_{0} > g tan α$ , the pseudo force's up-slope push wins, so in the wedge frame the block slides up the ramp.
If $a_{0} < g tan α$ (including $a_{0} = 0$ , an unaccelerated wedge), gravity's down-slope pull wins and the block slides down — the ordinary "block slips on a frictionless ramp" case. Answer: $a_{0} = g tan α$ ; for $3 0^{\circ}$ , $a_{0} \approx 5.66 m/s^{2}$ . Notice this is the same law as the pendulum ( $tan θ = a_{0} / g$ ) — the incline surface plays the role of the tilted "effective vertical".

Problem 5.2

A stone hangs by a string from the ceiling of a lift. The lift accelerates upward at magnitude $a_{0} = 3.0 m/s^{2}$ while the stone (mass $m = 0.50 kg$ ) is simultaneously held so the string makes $θ = 2 0^{\circ}$ with the vertical and the stone stays fixed in the lift frame. Treating this as motion under an effective gravity, find (a) the effective gravity $g_{eff}$ felt inside the lift and (b) the horizontal hold force $F_{h}$ needed to keep the stone at $2 0^{\circ}$ . ( $g = 9.8 m/s^{2}$ .)

Recall Solution 5.2

Define the symbol: $F_{h}$ is the horizontal holding force — a real sideways push (imagine a light spring or finger from the lift wall) that keeps the tilted stone from swinging back to vertical. It is horizontal, hence the subscript $h$ . (a) Effective gravity. In the upward-accelerating lift the downward pull is boosted: the pseudo force adds $m a_{0}$ downward, so per unit mass the effective downward gravity is $g_{eff} = g + a_{0} = 9.8 + 3.0 = 12.8 m/s^{2} .$ WHY add: frame accelerates up $\Rightarrow$ pseudo force down $\Rightarrow$ stacks on real gravity. (b) Sideways hold. In the lift frame the stone is in equilibrium under: tension $T$ (along string), effective weight $m g_{eff}$ down, and the horizontal hold $F_{h}$ . Break the tension into legs: Vertical: $T cos θ = m g_{eff}$ . Horizontal: $T sin θ = F_{h}$ . Divide to remove $T$ : $F_{h} = m g_{eff} tan θ = 0.50 \times 12.8 \times tan 2 0^{\circ} = 0.50 \times 12.8 \times 0.36397 = 2.33 N .$ Answer: $g_{eff} = 12.8 m/s^{2}$ , $F_{h} \approx 2.33 N$ .

Active recall

Recall Quick self-check

Fake-force size in Problem 1.1 ::: $24 N$ , backward. Truck's max acceleration before block slides (Problem 4.1) ::: $a_{0} = μ_{s} g = 2.94 m/s^{2}$ . Wedge acceleration to freeze the block (Problem 5.1) ::: $a_{0} = g tan α$ . Wedge accelerates faster than $g tan α$ — which way does the block slide? ::: Up the ramp (pseudo force's up-slope push wins). Lift in Problem 4.2 accelerates which way? ::: Downward, $1.8 m/s^{2}$ (scale reads light).

Connections

Newton's Second Law — every solution is $F = m a$ with the fake term added.
Apparent weight & normal force — Problems 2.2 and 4.2 are apparent-weight questions.
Centrifugal force · Coriolis force — the rotating-frame cousins of these fake forces.
Galilean relativity — reminds you which frames need no pseudo force.