ARCHETYPE: Topple or Slide?

Brake hard with a tall box in the back of a truck and one of two things happens: the box skids forward across the bed, or it pitches over onto its face. Push a wardrobe across the floor and the same fork appears — shove it low and it slides; shove it high and it starts to tip. Two completely different failure modes, one object, and which one wins is decided by a crisp little race between two thresholds: friction's ceiling on one side, geometry's lever arms on the other. JEE has turned this race into an examination industry. "At what deceleration does the block topple rather than slide?" "At what height may the force be applied so the crate slides without tipping?" "As the incline is raised, does the cube slip first or fall over first?" Every one of these is the same computation — compare $\mu$ with a ratio of the object's dimensions — but the setup wears enough disguises (trucks, inclines, pushed cabinets) that unprepared students treat each as a new problem. This article makes the race itself the concept. You will learn where the normal force really acts on a tipping body — the one idea that unlocks everything — then run the slide-versus-topple comparison in its three exam costumes: the pushed block, the tilted incline, and the accelerating truck bed. One inequality, memorized honestly, converts a whole family of JEE-Advanced problems into ninety-second arithmetic.

Key takeaways

A rigid body on a surface fails in one of two ways: sliding when the needed friction exceeds $\mu N$ , or toppling when the needed normal-force position moves beyond the base's edge.
The secret character is the normal force's location: it shifts toward the leading edge to balance torques, and toppling begins the instant it reaches the edge.
Pushed block (force $F$ at height $y$ , base width $b$ ): slides at $F = \mu mg$ , topples at $F = \dfrac{mgb}{2y}$ — it slides first iff $\mu < \dfrac{b}{2y}$ .
Critical push height $y_c = \dfrac{b}{2\mu}$ : push below it and the block can only slide; above it, it topples before it ever slides.
On a slowly tilted incline: slides first if $\mu < \dfrac{b}{h}$ , topples first if $\mu > \dfrac{b}{h}$ (base $b$ along the slope, height $h$ ).
Accelerating/braking platform: slides at $a > \mu g$ , topples at $a > \dfrac{gb}{h}$ — same race, pseudo-force costume.
Key idea: compare $\mu$ against the geometry ratio — friction's ceiling versus the base's lever arm — and the smaller threshold wins the race.

The core insight

Whether a body slides or topples is a race between two thresholds — friction's limit $\mu N$ for sliding, and the normal force reaching the base's edge for toppling — and the outcome is decided by comparing $\mu$ with a pure geometry ratio of the body's base to the relevant lever height.

Where this lives today

Every SUV on sale carries this article's ratio as a safety rating. The NHTSA's Static Stability Factor is $\mathrm{SSF} = t/2h$ — half the track width over the centre-of-mass height, precisely our topple threshold in cornering costume — and vehicles with low SSF earn poor rollover stars. Warehouse rules encoding "push pallets below waist height" are the critical height $y_c = b/2\mu$ in workwear. Earthquake engineers run the platform version at scale: unanchored equipment on a shaking floor obeys the same $a > gb/h$ criterion, which is why hospital cabinets and server racks are bolted at the top, not the bottom — anchoring converts a topple problem into a much more forgiving slide problem. And roboticists inverted the entire archetype: a walking robot deliberately lets its support point migrate to the foot's edge and catches the fall with the next step. Walking, seen from this chapter, is controlled serial toppling.

The setup

The one idea that unlocks this whole archetype: the normal force is not glued to the centre of the base. A surface pushes on a resting body wherever contact pressure demands, and the resultant normal force positions itself so that torques balance. Push a block near its top: the floor presses harder near the block's front edge, and the resultant $N$ migrates forward. The harder you push, the further it shifts — silently keeping the block in rotational equilibrium. But the migration has a boundary. $N$ can act only where there is base. The instant equilibrium would require $N$ to stand beyond the front edge, the surface can no longer supply the balancing torque — and the block begins to rotate about that edge. Toppling is the normal force running out of room.

Figure

The travelling normal force — equilibrium's silent bookkeeper runs out of base

Set the cast for the standard problems. A uniform block: base width $b$ measured along the direction of motion or push, height $h$ , centre of gravity at the geometric centre — height $h/2$ , midway across the base. Three costumes, one race:

The pushed block — horizontal force $F$ at height $y$ on a level floor. The tilted incline — angle $\theta$ raised slowly. The accelerating platform — truck bed or railway wagon with acceleration (or braking) $a$ . In each costume, sliding has a threshold and toppling has a threshold, and the smaller threshold is the failure you actually observe.

The math: two thresholds, one race

Start with the pushed block. The sliding threshold is pure friction: the push must beat the ceiling.

F_{\text{slide}} = \mu\,mg \qquad (1)

where: $\mu$ — the coefficient of static friction; $N = mg$ on level ground. Below this force the block does not move at all; at it, the block begins to skid while remaining upright — provided toppling has not intervened first. The toppling threshold is pure torque. Take moments about the front bottom edge — the pivot-to-be. The push $F$ at height $y$ tips the block forward with torque $Fy$ ; the weight, acting at the centre, restores with lever arm $b/2$ :

F_{\text{topple}} = \frac{mg\,b}{2y} \qquad (2)

where: $b/2$ — the horizontal distance from the front edge back to the weight's line of action; $y$ — the push height. Read it physically: pushing twice as high halves the force needed to tip. This is why removal men push furniture low — they are keeping $F_{\text{topple}}$ safely above $F_{\text{slide}}$ . The race is now one line. The block slides (rather than topples) if $F_{\text{slide}} < F_{\text{topple}}$ :

\mu < \frac{b}{2y} \qquad\Longleftrightarrow\qquad y < y_c = \frac{b}{2\mu} \qquad (3)

where: $y_c$ — the critical push height. Push below $y_c$ : friction gives way first and the block skids upright. Push above $y_c$ : the tipping torque wins before friction's ceiling is ever reached, and the block pitches over — even though it never slid. Note what is absent: the mass. Both thresholds scale with $mg$ , so the verdict is geometry versus surface, never weight.

costume	slides when	topples when	topples first iff
pushed at height $y$	$F > \mu mg$	$F > \dfrac{mgb}{2y}$	$y > \dfrac{b}{2\mu}$
incline raised to $\theta$	$\tan\theta > \mu$	$\tan\theta > \dfrac{b}{h}$	$\mu > \dfrac{b}{h}$
platform acceleration $a$	$a > \mu g$	$a > \dfrac{gb}{h}$	$\mu > \dfrac{b}{h}$

Derivation AWhere the normal force stands — and the moment it runs out of base

We derive the position of the resultant normal force on a pushed block, and from it the toppling threshold, equation (2). The strategy: the block, while static, is in full equilibrium — forces and torques — so we let the normal force act at an unknown distance $x$ ahead of the centre-line and let the torque equation tell us where it must stand. Toppling is then read off as the moment $x$ would need to exceed $b/2$ , the physical edge of the base. Watch how this derivation replaces the vague idea "it tips when pushed too hard" with an exact bookkeeping statement.

Step 1. Forces on the block: push $F$ (horizontal, height $y$ ), weight $mg$ (down, at the centre), friction $f$ (backward, at the floor), normal $N$ (up, at unknown offset $x$ ahead of the centre-line). Equilibrium of forces: $N = mg$ and $f = F$ .

Step 2. Torques about the centre of gravity (any point works; the CoG kills the weight's term). The push turns the block forward: torque $F \times$ (its height relative to the CoG). The friction at the floor also turns it forward. The normal force, standing ahead of centre, turns it backward with torque $Nx$ . Rather than track each height separately, take torques about the bottom-centre point of the base — cleanest ledger: the push contributes $Fy$ forward, friction passes through the point (zero arm), the weight passes through it (zero arm), and $N$ at offset $x$ contributes $Nx$ backward:

F\,y = N\,x = mg\,x \quad\Longrightarrow\quad x = \frac{F\,y}{mg}

Why the normal force "knows" where to stand: contact pressure redistributes across the base — more squeeze at the front, less at the back — and $x$ is simply the centroid of that pressure. The floor is a self-adjusting torque supplier, exactly as static friction is a self-adjusting force supplier.

Step 3. Impose the geometric budget. The pressure centroid cannot leave the base: $x \le \dfrac{b}{2}$ . Substituting Step 2:

\frac{Fy}{mg} \le \frac{b}{2} \quad\Longrightarrow\quad F \le \frac{mgb}{2y}

Step 4. Read the boundary. At $F = \dfrac{mgb}{2y}$ the entire normal force stands on the front edge — the block is balanced on a knife's line, and any further push rotates it about that edge. This is equation (2), and the criterion "N reaches the edge" is the master key that unlocks every costume of this archetype. Verified: limits — $y \to 0$ (push at floor level): $F_{\text{topple}} \to \infty$ , you cannot tip a block by pushing its base ✓; $y = h$ (push at the top of a cube, $b = h$ ): $F_{\text{topple}} = mg/2$ , the standard cube result ✓. Cross-check: at threshold, torque about the front edge directly — $Fy$ tipping vs $mg(b/2)$ restoring — gives the same equation, confirming the pivot-free bookkeeping.

Go deeper · Derivation B · 2 minThe incline race — slide angle versus topple angle

Derivation BThe incline race — slide angle versus topple angle

We derive both failure angles for a uniform block (base $b$ along the slope, height $h$ ) on an incline raised slowly from horizontal, and the comparison that decides which failure occurs. The strategy: for sliding, resolve gravity along and into the incline and apply the friction ceiling; for toppling, use the plumb-line criterion — the block tips when the vertical line through its centre of gravity leaves the downhill edge of its base. Watch that the same two numbers ( $b$ and $h$ ) that set the moment of inertia of exam nightmares here appear as a simple ratio.

Step 1. Sliding threshold. On the incline, gravity's component along the slope is $mg\sin\theta$ ; the normal force is $N = mg\cos\theta$ . The block slips when the demand exceeds the ceiling:

mg\sin\theta > \mu\,mg\cos\theta \quad\Longrightarrow\quad \tan\theta > \mu

Step 2. Toppling threshold, by the plumb line. The centre of gravity sits at the block's geometric centre — $b/2$ up-slope from the downhill edge, $h/2$ above the base. The block remains torque-balanced while the vertical through the CoG lands inside the base; it reaches the downhill edge when the incline satisfies:

\tan\theta = \frac{b/2}{h/2} = \frac{b}{h}

Why the plumb-line works: at the tipping angle, the weight's line of action passes exactly through the pivot edge, so gravity's restoring torque about that edge has shrunk to zero — the same "normal force reaches the edge" statement, seen from the geometry side.

Step 3. The race. As $\theta$ rises from zero, the block fails at whichever threshold comes first:

\mu < \frac{b}{h}: \text{slides first} \qquad\qquad \mu > \frac{b}{h}: \text{topples first}

Why squat shapes slide and tall shapes fall: for a wide, low crate $b/h$ is large, and almost no realistic $\mu$ can beat it — it will always skid. For a tall bookcase $b/h$ is small and even modest friction holds it back from sliding long enough for the plumb line to escape — it goes over. *Verified: cabinet numbers — $b = 1.0$ m, $h = 2.0$ m gives $\tan\theta_c = 0.5$ , $\theta_c = 26.6^\circ$ ; with $\mu = 0.4 < 0.5$ it slides at $\tan\theta = 0.4$ ( $21.8^\circ$ ) before it could tip ✓. A cube ( $b = h$ ) topples first only if $\mu > 1$ — rare — which is why dice slide, not somersault, when you tilt a tray.*

Figure

The incline race — friction threshold versus plumb-line threshold

The truck-bed costume needs one honest remark. In the truck's (accelerating) frame, add the pseudo-force $ma$ acting at the centre of mass, horizontally, opposite the acceleration. The ledgers then read exactly like the incline's: sliding when $ma > \mu mg$ , i.e. $a > \mu g$ ; toppling about the leading bottom edge when the pseudo-force's torque beats the weight's, $ma\cdot\frac{h}{2} > mg\cdot\frac{b}{2}$ , i.e. $a > \frac{gb}{h}$ . Same race, same referee: compare $\mu$ with $b/h$ . (Ground-frame purists get identical thresholds from "friction must supply $ma$ " and "N at the edge" — the pseudo-force is bookkeeping, not new physics.)

Limiting cases

case	result	meaning
push at floor level ( $y \to 0$ )	$F_{\text{topple}} \to \infty$	can't tip a block through its base — it can only slide
very tall body ( $h \gg b$ )	topple threshold $\to 0$	bookcases fall over; they almost never skid
very squat body ( $b \gg h$ )	slide threshold reached first for any real $\mu$	crates skid; they almost never tip
$\mu \to 0$ (ice)	slides at the faintest provocation	no friction, no toppling: the race is forfeit
$\mu = \dfrac{b}{h}$ exactly	both failures at once	the block slides and tips simultaneously — the examiner's boundary case
cube pushed at its top	$F_{\text{topple}} = \dfrac{mg}{2}$	topples first only if $\mu > 0.5$

Worked examples

1 · Board level — cube pushed at the top (JEE-Main entry pattern)

A uniform cube of side $0.50\ \mathrm{m}$ and mass $40\ \mathrm{kg}$ rests on a floor with $\mu_s = 0.40$ . A horizontal force is applied at the top edge, gradually increasing. Does the cube slide or topple first, and at what force does it happen? ( $g = 10\ \mathrm{m/s^2}$ .)

Step 1. Sliding threshold: $F_{\text{slide}} = \mu mg = 0.40 \times 40 \times 10 = 160\ \mathrm{N}$ .

Step 2. Toppling threshold — push height $y = 0.50$ m, base $b = 0.50$ m: $F_{\text{topple}} = \dfrac{mgb}{2y} = \dfrac{400 \times 0.50}{2 \times 0.50} = 200\ \mathrm{N}$ .

Step 3. Rank them: $160 < 200$ , so the cube slides first, at $F = \boxed{160\ \mathrm{N}}$ , and never topples.

Step 4. State the general rule for the mark: a cube pushed at its top topples first only when $\mu > b/2y = 0.5$ . Ordinary surfaces lose the race; only a very grippy floor lets a cube go over.

2 · Board level — the cabinet on a tilting ramp

A cabinet ( $b = 1.0\ \mathrm{m}$ base along the slope, $h = 2.0\ \mathrm{m}$ tall, uniform) stands on a ramp whose angle is slowly increased. Take $\mu_s = 0.40$ . (a) At what angle would it slide? (b) At what angle would it topple? (c) Which happens?

Step 1. Slide angle: $\tan\theta_s = \mu = 0.40 \Rightarrow \theta_s = 21.8^\circ$ .

Step 2. Topple angle, plumb-line criterion: $\tan\theta_t = \dfrac{b}{h} = \dfrac{1.0}{2.0} = 0.50 \Rightarrow \theta_t = 26.6^\circ$ .

Step 3. The smaller angle arrives first as the ramp rises: $21.8^\circ < 26.6^\circ$ , so the cabinet slides at $\boxed{21.8^\circ}$ and the topple angle is never reached.

Step 4. Flip the scenario for understanding: on a rubber-coated ramp with $\mu = 0.7 > 0.5$ , friction would hold the cabinet past $26.6^\circ$ and it would topple there instead. Same object, different floor, different failure.

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

A tall box ( $b = 0.80\ \mathrm{m}$ along the truck's direction of motion, $h = 1.6\ \mathrm{m}$ , uniform) stands on a truck bed with $\mu_s = 0.60$ . Find the maximum braking deceleration that leaves the box undisturbed, and state which failure occurs beyond it. ( $g = 10\ \mathrm{m/s^2}$ .)

Step 1. Sliding threshold: friction must supply the box's deceleration, so $a_{\text{slide}} = \mu g = 6.0\ \mathrm{m/s^2}$ .

Step 2. Toppling threshold (truck frame: pseudo-force $ma$ forward at the CoM, height $h/2$ ; pivot at the front bottom edge): $ma\cdot\dfrac{h}{2} = mg\cdot\dfrac{b}{2} \Rightarrow a_{\text{topple}} = \dfrac{gb}{h} = \dfrac{10 \times 0.8}{1.6} = 5.0\ \mathrm{m/s^2}$ .

Step 3. Rank: $5.0 < 6.0$ — the box topples first. Maximum safe deceleration $= \boxed{5.0\ \mathrm{m/s^2}}$ .

Step 4. Sanity: $b/h = 0.5 < \mu = 0.6$ — tall-ish box on a grippy bed, so geometry loses before friction does, confirming the topple verdict. Timing note: both thresholds are one-line formulas; the entire question is computing $\mu g$ and $gb/h$ and circling the smaller — 2.5 minutes with the audit. The distractor options will be $6.0$ (sliding, wrong failure) and $10$ ( $gb/h$ computed with $h/2$ misused).

4 · Competitive level — JEE-Advanced stretch: the safe pushing window (~4 minutes)

A crate ( $b = 0.60\ \mathrm{m}$ , $h = 1.8\ \mathrm{m}$ , mass $50\ \mathrm{kg}$ , uniform) must be slid across a floor with $\mu_s = 0.30$ without tipping. ( $g = 10\ \mathrm{m/s^2}$ .) (a) Find the minimum push force and the maximum height at which it may be applied. (b) A worker pushes at $y = 1.5\ \mathrm{m}$ . What happens, and at what force?

Step 1. To slide at all: $F \ge \mu mg = 0.30 \times 500 = 150\ \mathrm{N}$ .

Step 2. For sliding to win the race, the push height must satisfy $y \le y_c = \dfrac{b}{2\mu} = \dfrac{0.60}{0.60} = \boxed{1.0\ \mathrm{m}}$ . Push anywhere in the bottom metre with $150$ N: the crate skids upright.

Step 3. For (b), at $y = 1.5 > y_c$ the topple threshold sits below the slide threshold: $F_{\text{topple}} = \dfrac{mgb}{2y} = \dfrac{500 \times 0.60}{3.0} = \boxed{100\ \mathrm{N}}$ , versus $F_{\text{slide}} = 150$ N.

Step 4. So as the worker's push grows past $100$ N the crate begins to tip about its front edge — it never slides. The friction it experiences at that moment is only $100$ N, well under the $150$ N ceiling: friction is fine; geometry failed. Practical translation: push at waist height, not shoulder height. Timing note: compute $y_c$ first — it instantly routes the question into "sliding physics" or "toppling physics" and halves the remaining work. The classic error is reporting 150 N for part (b): check the height against $y_c$ before quoting any sliding answer.

Exam traps

Using $b$ instead of $b/2$ . The weight's restoring lever arm about the edge is half the base — the CoG stands at the centre, not the far edge. Every threshold formula carries that factor; dropping it doubles $F_{\text{topple}}$ and flips verdicts.

Which dimension is the "base"? $b$ is the base extent along the push or motion direction; the perpendicular width is a spectator. A tall thin bookcase pushed side-on and face-on has two different $b$ 's and two different fates. Read the geometry from the figure, not from habit.

Announcing the wrong failure. Computing one threshold is half the question. Always compute both, rank them, and report the smaller — a block that topples at $100$ N will never exhibit its $150$ N sliding threshold (worked example 4).

Pseudo-force at the wrong height. In platform problems the pseudo-force $ma$ acts at the centre of mass ( $h/2$ for a uniform block), not at the base or the top. Its torque about the edge is $ma \cdot h/2$ — the $h/2$ is where the $b/h$ ratio comes from.

Forgetting friction's role in toppling. The tipping analysis assumed the block doesn't slide (else the static ledger breaks). If a question declares the floor smooth, the block can never topple from a horizontal push — it simply accelerates away. Toppling needs grip to brace against.

"At the point of toppling, N acts at the edge." Not a trap but a mark-earner: state it explicitly. Many JEE-Advanced parts ask for the normal force's position at a given push — use $x = Fy/mg$ from the derivation.

Which $g$ ? JEE-Main/Advanced almost always declare $g = 10\ \mathrm{m/s^2}$ — use exactly what is stated; boards using $9.8$ will say so.

Origins · 1 minGalileo Galilei and how this was discovered

Galileo Galilei

1564–1642 · Pisa & Florence

Galileo grew up in the shadow of the most famous stability problem in the world: the campanile of Pisa, already leaning four degrees when he (by legend) dropped weights from it. His last and greatest book, the Two New Sciences of 1638, opened with exactly this article's questions — why do large structures fail where small models stand, when do columns tip, how does a beam's own weight conspire against it? Galileo's scaling insight was that weight grows with volume while lever-arm geometry grows only with length: big things live closer to their toppling thresholds. The Tower itself vindicates the plumb-line criterion — its centre of gravity's vertical still lands metres inside the foundation ring, which is why eight centuries of lean have not yet become a fall.

History

1173

— construction begins on Pisa's campanile; it starts leaning before the third storey is finished, becoming civilization's longest-running toppling experiment.

1638

— Galileo's Two New Sciences founds the mechanics of structures, asking when beams break and bodies overturn as questions of geometry and load.

1781

— Coulomb's friction memoir supplies the other contestant in the race: the ceiling

\mu N

that decides whether a body slips before it tips.

1990

— the Pisa tower is closed to the public as its lean approaches the danger zone; engineers extract soil to rotate it back — torque bookkeeping at heritage scale.

2000

s — the US NHTSA begins publishing the Static Stability Factor for every car sold — the topple criterion

b/2h

, wearing a regulatory badge.

Going further

Texts. HC Verma, Concepts of Physics Vol. 1, Ch. 10 (friction and rotation problems) · Irodov, Problems in General Physics 1.7 (equilibrium race problems) · Halliday–Resnick–Walker Ch. 12 for the plumb-line criterion.

Provenance. All results are standard rigid-body statics canon. A verifier should recompute: $x = Fy/mg$ and the edge condition $F = mgb/2y$ ; the incline pair $\tan\theta = \mu$ vs $\tan\theta = b/h$ (cabinet: $21.8^\circ$ vs $26.6^\circ$ at $\mu = 0.4$ , $b/h = 0.5$ ); example 1's $160$ vs $200$ N; example 3's $5.0$ vs $6.0\ \mathrm{m/s^2}$ ; example 4's $y_c = 1.0$ m and $100$ vs $150$ N.

Prerequisites. Rigid-Body Equilibrium, CoG & Moments · Static Friction · Locating the Centre of Mass. Leads to. Non-Inertial Frames & Pseudo-Forces (truck-bed frame) · Statics & Rigid-Body Equilibrium (UG deep dive).

Where exactly does the normal force act, physically?

Across the whole contact patch — the floor pushes up a little everywhere the base touches. The single arrow $N$ we draw is the resultant of that pressure distribution, positioned at its centroid. When you push the block near the top, the front of the base presses harder, the pressure centroid migrates forward, and the resultant migrates with it. The migration is what keeps torques balanced without any new force appearing. Toppling is simply the centroid needing to stand outside the base — a physical impossibility, so rotation begins instead.

Why does the weight's restoring lever arm use b/2 and not b?

Because the weight acts at the centre of gravity, which for a uniform block stands halfway across the base — a horizontal distance $b/2$ from the pivot edge. The full width $b$ would be right only if all the mass sat at the far edge. Loaded shelves and asymmetric objects shift the CoG, and good JEE questions exploit exactly that: always measure the restoring arm from the pivot edge to the CoG's vertical line, wherever the question puts it.

Does the slide-or-topple verdict depend on how heavy the object is?

No — and this surprises everyone moving furniture. Both thresholds are proportional to $mg$ : doubling the mass doubles the force needed to slide and the force needed to tip, leaving the race's outcome untouched. What weight changes is the effort, not the failure mode. The verdict belongs to two dimensionless numbers alone: $\mu$ and the geometry ratio ( $b/2y$ for a push, $b/h$ for inclines and platforms).

Why do we compare μ with b/h on the incline but with b/2y for a push?

Same referee, different lever arms. In each costume the toppling threshold is "tipping torque = restoring torque about the edge," but what provides the tipping torque differs: a push at height $y$ (arm $y$ ), or gravity's downhill component acting at the CoM height $h/2$ (which, against the restoring arm $b/2$ , produces the ratio $b/h$ ). Rather than memorizing three formulas, re-derive in one line: identify the tipping force, its lever arm about the edge, and set the torque against $mg \times (\text{CoG's horizontal offset from the edge})$ .

Can a block slide and topple at the same time?

At the exact boundary $\mu = b/h$ (or $y = y_c$ ), the two thresholds coincide and both failures onset together — an examiner's favourite boundary case. Off the boundary, one strictly precedes the other. But note a real-world subtlety the static model hides: once sliding starts, kinetic friction (usually below $\mu_s N$ ) takes over and the torque books change; a block skidding to a sudden stop against a floor ridge can convert its motion into a topple. Exams stay within the static race unless they say otherwise.

In the truck problem, is the pseudo-force method compulsory?

No — it is a convenience, not a necessity. In the ground frame: the box accelerates at $a$ , so friction must supply $f = ma$ (sliding threshold $a = \mu g$ appears immediately), and the torque balance about the CoM with the normal force offset $x$ gives $N x = f \cdot h/2$ , running out of base at the same $a = gb/h$ . The truck-frame pseudo-force $ma$ at the CoM merely converts the problem into a statics question, where the plumb-line and edge-pivot habits work unchanged. Same thresholds, pick the frame you can audit fastest.

Why are cupboards anchored at the top rather than the bottom?

An earthquake shakes the floor sideways — the platform costume. A top anchor has the longest possible lever arm about the base edge, so a small anchor force cancels a large tipping torque; a base anchor sits at the pivot with almost no arm at all. The same logic in reverse explains why you push furniture low: force applied near the pivot line has the least tipping effect. Lever arms, not force sizes, run this entire chapter.