Rigid-Body Equilibrium, CoG & Moments

The Forth Bridge has carried trains over Scottish water since 1890 on 53,000 tonnes of steel that goes nowhere. Most of the mechanical world is like that — bridges, cranes, shelves, the bones of your forearm holding this screen — not moving, and working hard at it. Staying put is a genuine physical achievement: inside every stationary structure, enormous forces are locked in perfect cancellation, and the price of a bookkeeping error is collapse. Here is the discovery that makes statics a subject: force balance alone is not enough. Push a book's opposite corners with equal and opposite forces — net force zero — and it spins on the spot. A rigid body has two independent things it can do, translate and rotate, so rest costs two separate conditions: $\sum F = 0$ and $\sum \tau = 0$ . Out of the second grows the entire craft of this module — the principle of moments (the see-saw law, humanity's oldest quantitative physics), the centre of gravity where a body's whole weight acts, the tipping rule that decides when a leaning tower falls, and the pivot-choice trick that makes unknown forces vanish from the ledger. This is a rare all-boards module, and each board speaks its own dialect: IGCSE and A-Level examine it as "moments" — the principle of moments, CoG experiments, stability — and it is core material there; CBSE, ICSE and NEET want both conditions and beam problems; JEE and AP add hinged rods, wire tensions and reaction forces at supports. The famous ladder, and the topple-or-slide contest, each get their own archetype module ahead; here we build the machinery they lean on.

Key takeaways

A rigid body is in equilibrium only when both books close: $\sum F = 0$ (no translation) and $\sum \tau = 0$ (no rotation) — about any point.
Neither implies the other: a couple (equal, opposite, offset forces) has zero net force yet spins the body — the standing counterexample.
Once forces balance, the net torque is the same about every point — so choose the pivot where the most unknown forces act, and they vanish from the torque equation.
Principle of moments: about the pivot, total clockwise moment = total anticlockwise moment; $W_1 d_1 = W_2 d_2$ for the two-weight balance.
Gravity on an extended body acts, for all torque purposes, at the centre of gravity — the centre of mass in uniform $g$ ; a uniform beam's weight acts at its midpoint.
Tipping rule: a body on a surface stays up while the vertical line through its CoG pierces its base of support; a box on a tilting ramp tips past $\tan\theta_c = w/h$ .
Key idea: standing still is a two-signature contract — forces cancel and moments cancel — and a well-chosen pivot turns most statics problems into one-line calculations.

The core insight

A rigid body rests only if $\sum F = 0$ and $\sum \tau = 0$ ; once the forces balance, torque balance about one point guarantees it about all points — so pivots are chosen to erase unknown forces — with gravity entering the ledger as a single weight acting at the centre of gravity, and stability decided by whether the CoG's plumb line stays inside the base of support.

Where this lives today

Statics is the most heavily exercised physics on Earth by sheer volume: every structural engineering office runs $\sum F = 0$ , $\sum \tau = 0$ continuously, with software decomposing buildings, bridges and aircraft frames into millions of elements and closing both books for each one simultaneously — building codes are, at bottom, statutory safety margins on this module's equations. The liveliest frontier is robotics: a walking humanoid stays up only while its ground-reaction resultant remains inside the support polygon of its feet — the tipping criterion (4), generalized and recomputed hundreds of times per second in every stumble recovery. Biomechanics applies the lever law to your frame: the forearm is a speed-multiplying lever with a roughly 1:7 arm ratio, which is why physiotherapists compute hundreds of newtons of biceps tension for a shopping bag — Archimedes' equation (2), billed to your elbow. And the tipping rule quietly polices daily life, from the anti-topple wall straps sold with every tall bookcase to the load–radius charts hanging in every crane cab.

The physical picture

Why two conditions? Count what a rigid body can do. Newton's second law, $\sum F = Ma_{\text{cm}}$ , governs only the centre of mass: kill the net force and the CM stays put — but the body remains free to turn around its stationary centre. See it on a slick table: push a book's opposite corners with equal and opposite fingers. Forces cancel; the centre goes nowhere; the book spins up. That force pair — equal magnitude, opposite direction, offset lines of action — is the couple from the torque module, and it is the permanent proof that the two books of mechanics are audited separately. Concurrent-force equilibrium (the Laws-of-Motion chapter) closed only the first book, because forces through one point cannot turn anything; extended bodies open the second.

Figure

Two freedoms, two conditions — the couple is the proof they are independent

The second condition is where the craft lives, because torque cares where forces act. That is the principle of moments: a child far from the see-saw's pivot balances a parent sitting close in, force times lever arm against force times lever arm — Archimedes' law of the lever, twenty-two centuries old and unmodified. Every crowbar, wheelbarrow, bottle-opener and forearm is this principle with a job: levers that put the load close to the pivot and the effort far away multiply force; your forearm does the opposite (biceps attached a few centimetres from the elbow, load in the hand at ~35 cm) and multiplies speed at the price of force — holding 5 kg at the hand costs the muscle several hundred newtons. Two more ideas complete the toolkit. First: gravity pulls on every particle of a body, yet for torque bookkeeping the entire weight acts at one point — the centre of gravity, which in uniform $g$ is the centre of mass (derived below, and the licence behind every FBD you have drawn with a single $Mg$ arrow at the middle). Locate it experimentally in two minutes: hang a cardboard cutout from a pin, drop a plumb line, repeat from another corner — the lines cross at the CoG. Second: the CoG decides stability. A body on a surface stays up while the vertical line through its CoG pierces its base of support — the region spanned by the contact points — and tips the instant that plumb line escapes. That single criterion explains the wide stance on a lurching bus, the SUV's rollover reputation, the crane's counterweight, and why the Leaning Tower of Pisa leans but does not fall: its plumb line still lands well inside the foundation ring.

The math: two conditions and one free choice

The conditions of equilibrium, straight from the two laws of rigid-body motion ( $\sum F = Ma_{\text{cm}}$ and $\sum \tau = I\alpha$ , both zeroed):

\sum F_x = 0, \qquad \sum F_y = 0, \qquad \sum \tau_P = 0 \ \ \text{(about a point } P\text{)} \qquad (1)

where: the force sums run over all external forces (internal ones cancel in third-law pairs), and each torque is force times perpendicular lever arm about $P$ , anticlockwise counted positive. Three scalar equations for planar problems — which is why a 2-D statics question can determine at most three unknown forces. The quiet gift is the subscript $P$ : the derivation below proves that once the forces balance, the torque balance holds about every point if it holds about one — so you may put $P$ wherever it kills the most unknowns. For two weights on a lever, condition (1) collapses to the principle of moments:

W_1\,d_1 = W_2\,d_2 \qquad (2)

where: $d_1, d_2$ — perpendicular distances from the pivot to each weight's line of action. Spoken as every IGCSE and A-Level mark scheme wants it: total clockwise moment about the pivot = total anticlockwise moment. A $25\ \mathrm{kg}$ child at $2.4\ \mathrm{m}$ balances a $60\ \mathrm{kg}$ parent at $1.0\ \mathrm{m}$ ( $25 \times 2.4 = 60 \times 1.0$ ; the $g$ 's cancel, as they always do in pure balance). Gravity enters the ledger through one point:

\tau_{\text{grav}} = (\text{weight } Mg) \times (\text{lever arm of the CoG}) \qquad (3)

where: the centre of gravity coincides with the centre of mass whenever $g$ is uniform across the body — every laboratory and exam situation. For a uniform beam or rule, that is the geometric midpoint; for a non-uniform one, wherever the balance experiment says (a favourite IGCSE practical: the off-centre balance point of a loaded metre rule measures its CoG). And the stability criterion:

\text{stable rest} \iff \text{the plumb line from the CoG pierces the base of support} \qquad (4)

where: the base of support is the smallest convex region containing all contact points (the rectangle under a cabinet's feet, the line under a book's tilted edge). A uniform box of width $w$ and height $h$ on a tilting ramp keeps its plumb line inside the base up to $\tan\theta_c = w/h$ — squat and wide survives; tall and narrow topples early.

you want	use	notes
is it in equilibrium at all?	both conditions of (1)	the couple trap: forces alone prove nothing
balance two weights	moments, eq. (2)	clockwise = anticlockwise, $g$ cancels
unknown support/hinge forces	$\sum\tau$ about a point where an unknown acts, then $\sum F$	pivot choice = the free gift
where does gravity act?	CoG = CM (uniform $g$ ), eq. (3)	uniform beam: midpoint — always draw it there
will it tip?	plumb-line criterion, eq. (4)	box on ramp: $\tan\theta_c = w/h$
ladder, toppling-vs-sliding contests	the archetype modules ahead	this module supplies their machinery

Figure

The tipping rule — stability is the CoG plumb line against the base of support

Derivation AWhy one pivot is enough — torque balance is pivot-independent once forces balance

We derive the statement that makes statics solvable: if $\sum F = 0$ , the total torque about any point $P'$ equals the total torque about any other point $P$ — so checking rotation about one well-chosen pivot settles it everywhere, and the pivot is purely a convenience. The strategy is direct bookkeeping with signed lever arms along a line (the planar argument is identical with vectors); watch exactly where the force balance gets spent — the result is false for unbalanced forces, and that failure is itself exam material.

Step 1. Set up. Forces $F_1, F_2, \ldots$ act at positions $x_1, x_2, \ldots$ measured from $P$ (each torque about $P$ is $\tau_i = F_i \times x_i$ with the appropriate sign for its turning sense). Move the reference to $P'$ , a distance $D$ away: the same forces now act at positions $x_i - D$ .

Step 2. Write the new total torque and expand:

\sum_i F_i\,(x_i - D) = \sum_i F_i\,x_i \;-\; D\sum_i F_i

The shift $D$ is the same for every force, so it factors out onto the total force.

Step 3. Spend the force balance. If $\sum F_i = 0$ , the correction term dies:

\tau_{\text{about }P'} = \tau_{\text{about }P}

One number, every pivot. So in an equilibrium problem you may take moments about any point — and the professional choice is the point where the most unknown forces act, since a force through the pivot has zero lever arm and exits the ledger. In the beam example below, each support in turn erases the other's reaction.

Step 4. Read the two corollaries. First, the warning: if $\sum F \neq 0$ (anything accelerating), torques about different points genuinely differ by $D\sum F$ — "the torque happens to vanish about my favourite point" proves nothing then. Second, the couple's signature: a couple has $\sum F = 0$ , so by Step 3 its torque is the same about every point in space — $Fd$ , reference-free, the irreducible twist that no single force can balance (a single force always has a nonzero net force, so it cannot cancel a couple's zero-force turning; only another couple can). Verified: the see-saw check — $W_1 d_1 = W_2 d_2$ about the pivot; recomputing about the left seat gives pivot reaction $(W_1 + W_2)$ at arm $d_1$ against $W_2(d_1 + d_2)$ , equal exactly when (2) holds ✓. Standard treatment: NCERT Ch. 7, Halliday Ch. 12.

Go deeper · Derivation B · 2 minThe centre of gravity and the tipping rule

Derivation BThe centre of gravity and the tipping rule

We derive the licence used in every free-body diagram — that gravity's torque on an extended body equals that of the whole weight $Mg$ acting at the centre of mass — and then convert it into the tipping criterion (4). The route is a two-line sum plus a support-region argument; the subtlety worth owning is that everything hangs on $g$ being the same for every particle.

Step 1. Sum gravity's torques particle by particle. About any pivot, each particle $m_i$ at horizontal distance $x_i$ contributes torque $m_i g\,x_i$ (weight times horizontal lever arm, since weight is vertical). Total: $\sum_i m_i g\,x_i = g\sum_i m_i x_i$ — the $g$ factors out because it is the same for all particles.

Step 2. Recognize the centre of mass. By definition, $\sum_i m_i x_i = M x_{\text{cm}}$ . Hence

\tau_{\text{grav}} = Mg\,x_{\text{cm}}

— exactly the torque of one force $Mg$ applied at the centre of mass. That proves CoG = CM in uniform gravity: not a definition, a theorem, and the reason a single weight-arrow at the midpoint of a uniform beam is honest bookkeeping. (In a genuinely non-uniform field — a 100-km space tether — the lower parts feel stronger gravity, the factoring fails, and the CoG migrates away from the CM. No laboratory object notices.)

Step 3. Stand the body on a surface. The upward contact forces can act only within the base of support — pushes distributed among the contact points; their resultant is effectively a single upward force acting somewhere inside that region (contact forces push but cannot pull, so their resultant cannot act outside the region they occupy).

Step 4. Take torques about the base's nearest edge. If the CoG's plumb line lands inside the base, the normal resultant can position itself directly below the CoG: both books close, equilibrium available. If the plumb line falls outside, gravity's torque about the edge turns the body over, and no arrangement of pushes inside the base can generate a counter-torque — the body must rotate about the edge and fall. That is criterion (4). For the uniform box of width $w$ , height $h$ on a ramp at angle $\theta$ : the plumb line from the centre reaches the downhill edge of the base when $\tan\theta = \dfrac{w/2}{h/2} = \dfrac{w}{h}$ — the tipping threshold, wider-or-squatter always safer. Verified: a cabinet $1.8\ \mathrm{m}$ tall, $0.6\ \mathrm{m}$ deep tips past $\tan\theta_c = 1/3$ , $\theta_c = 18.4^\circ$ — matching the energy view (its CoG is highest exactly when the plumb line crosses the pivot edge) ✓. Treatment as in Halliday Ch. 12; the plumb-line/hanging-cutout experiment is the standard IGCSE practical.

Limiting cases

case	result	meaning
point particle	$\sum\tau = 0$ automatic	torque balance is the extended-body condition
couple	$\tau = Fd$ about every point	zero net force, unbalanceable by any single force
forces unbalanced	torques differ from pivot to pivot	pivot freedom exists only in equilibrium
uniform beam, central pivot	own weight exits the ledger	zero lever arm — the see-saw simplification
box on ramp at $\tan\theta = w/h$	plumb line at the edge	tipping threshold; beyond it, no statics
support directly under the CoG	single support suffices	the waiter's tray, the balancing stone
non-uniform $g$	CoG ≠ CM	space tethers, not lab benches

Common trap"The forces on it cancel, so it is in equilibrium — and anyway the beam's weight acts at its end"

Two of the most reliably mark-losing sentences in statics. The first dies by the couple: equal and opposite forces on offset lines produce zero net force and a net torque — the body spins while its centre of mass politely stays put. Equilibrium is a two-signature contract, and $\sum\tau = 0$ must be checked separately, about some (any, once forces balance) point. The reverse omission is just as fatal: torque balance about one special point with unbalanced forces proves nothing, because then the torque is pivot-dependent and you may simply have chosen the one point where it vanishes. The second sentence corrupts the ledger at the source: the weight of a uniform beam acts at its midpoint — its centre of gravity — never at an end and never at the support. Hang it at the end and every reaction force you compute is wrong, typically by a factor of two. The discipline: draw the FBD with gravity's arrow at the CoG, write both force equations, then take moments about the busiest unknown. Forces, then torques, gravity at the centre — all three, every time.

Worked examples

1 · Board level — the bridge beam (CBSE/ICSE/A-Level pattern, 3 marks)

A uniform beam of length $8.0\ \mathrm{m}$ and weight $4000\ \mathrm{N}$ rests on supports at its ends A and B. A truck of weight $12{,}000\ \mathrm{N}$ stands with its centre $6.0\ \mathrm{m}$ from A. Find the load on each support.

Step 1. FBD: reactions $R_A$ , $R_B$ up at the ends; beam's weight $4000\ \mathrm{N}$ down at its midpoint ( $4.0$ m from A — uniform beam, CoG central); truck's $12{,}000\ \mathrm{N}$ down at $6.0$ m from A. Two unknowns: one torque equation plus one force equation.

Step 2. Moments about A (so the unknown $R_A$ exits): clockwise from the weights, anticlockwise from $R_B$ .

R_B \times 8.0 = 4000 \times 4.0 + 12{,}000 \times 6.0 = 88{,}000\ \mathrm{N\,m} \implies R_B = \boxed{11{,}000\ \mathrm{N}}

Step 3. Force balance: $R_A = 16{,}000 - 11{,}000 = \boxed{5000\ \mathrm{N}}$ .

Step 4. Audit with the other pivot — the freedom the derivation grants. Moments about B: $R_A \times 8.0 = 4000 \times 4.0 + 12{,}000 \times 2.0 = 40{,}000$ , so $R_A = 5000\ \mathrm{N}$ ✓. Physical read: the truck stands three-quarters of the way toward B, and B duly carries the lion's share — reading the split is the cheapest error check in statics.

2 · Board level — the metre rule and the principle of moments (IGCSE/ICSE pattern)

A uniform metre rule is pivoted at its 50 cm mark. A weight of $3.0\ \mathrm{N}$ hangs at the 15 cm mark. (a) Where must a $4.5\ \mathrm{N}$ weight hang to balance the rule? (b) Why does the rule's own weight not enter the calculation?

Step 1. (b) first, because it is the licence: the rule is uniform, so its CoG is at 50 cm — on the pivot. Zero lever arm, zero moment: its weight exits the ledger. (This is also why the empty rule balances at 50 cm.)

Step 2. (a) Principle of moments about the pivot. The 3.0 N weight sits $35\ \mathrm{cm}$ on one side: anticlockwise moment $3.0 \times 0.35 = 1.05\ \mathrm{N\,m}$ .

Step 3. Balance: $4.5 \times d = 1.05 \Rightarrow d = 0.233\ \mathrm{m}$ on the other side — the $\boxed{73.3\ \mathrm{cm}}$ mark.

Step 4. Marking-scheme sentence: "for equilibrium, the total clockwise moment about the pivot equals the total anticlockwise moment." (Variant to expect: pivot not at 50 cm — then the rule's own weight, acting at 50 cm, re-enters with its own lever arm, and the same equation measures an unknown rule weight. That is the standard IGCSE practical.)

3 · Competitive level — the hinged rod and the wire (JEE-Main/AP pattern, ~3 minutes)

A uniform rod of mass $4.0\ \mathrm{kg}$ and length $L$ is hinged to a wall and held horizontal by a wire from its far end to the wall, making $30^\circ$ with the rod. Find (a) the wire's tension, (b) the magnitude of the hinge force. ( $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. (a) Moments about the hinge — the pivot that erases the two unknown hinge components. Weight $39.2\ \mathrm{N}$ at $L/2$ (clockwise); the wire's tension at the far end, perpendicular component $T\sin 30^\circ$ acting at $L$ (anticlockwise):

T\sin 30^\circ \times L = 39.2 \times \frac{L}{2} \implies T = \frac{39.2 \times 0.5}{0.5} \cdot \frac{1}{1} = \boxed{39.2\ \mathrm{N}}

(The $L$ cancels — statics answers rarely care how long the rod actually is.)

Step 2. (b) Force balance, both axes. Horizontal: the wire pulls its end toward the wall with $T\cos 30^\circ = 33.9\ \mathrm{N}$ , so the hinge pushes away with $H = 33.9\ \mathrm{N}$ . Vertical: $V + T\sin 30^\circ = 39.2$ , so $V = 39.2 - 19.6 = 19.6\ \mathrm{N}$ .

Step 3. Magnitude: $F_{\text{hinge}} = \sqrt{33.9^2 + 19.6^2} = \sqrt{1149 + 384} = \boxed{39.2\ \mathrm{N}}$ — coincidentally equal to $T$ here (the $30^\circ$ –half-length geometry), a symmetry the examiner may probe with "compare $T$ and the hinge force." Timing note: ~2.5 minutes. The whole problem is pivot choice: moments about the hinge in one line, then forces. Starting with force equations strands you with three unknowns in two equations. And the perpendicular-component discipline ( $T\sin 30^\circ$ , not $T$ ) at the moment step is where the marks bleed.

4 · Competitive level — tipping thresholds, from cabinets to cranes (NEET/JEE stretch, ~3 minutes)

(a) A cabinet $1.8\ \mathrm{m}$ tall and $0.60\ \mathrm{m}$ deep stands on a ramp. At what ramp angle does it tip? (b) A tower crane's horizontal jib carries a $2000\ \mathrm{kg}$ load at $30\ \mathrm{m}$ from the mast; the counterweight of $8000\ \mathrm{kg}$ sits on the other side. Where must it sit for the mast to feel no net moment?

Step 1. (a) Tipping when the CoG's plumb line reaches the downhill base edge: $\tan\theta_c = \dfrac{w}{h} = \dfrac{0.60}{1.8} = \dfrac13$ , so $\theta_c = \boxed{18.4^\circ}$ . Lower or wider furniture survives steeper floors — the whole geometry of stability in one fraction.

Step 2. (b) Moments about the mast: $8000g \times x = 2000g \times 30$ , so $x = \boxed{7.5\ \mathrm{m}}$ — the $g$ 's cancel as always in pure balance.

Step 3. Read the engineering: the crane's counterweight is fixed, the load is not — so real cranes publish a load–radius chart, which is equation (2) solved for the maximum load at each jib distance. Every "how far out may the crane carry this?" question is the see-saw with a safety factor. Timing note: ~2 minutes total. In (a) the planted error is $h/w$ (giving $71.6^\circ$ ) — remember tall-and-narrow must tip early, so the small ratio is right. In (b), including $g$ on one side only is the sign of rushing; it cancels or it doesn't appear.

Exam traps

Both books, every time. $\sum F = 0$ and $\sum \tau = 0$ are independent signatures; the couple defeats anyone who checks only forces. On "is this in equilibrium?" questions, audit both explicitly.

Gravity acts at the CoG. Uniform beam: midpoint. Non-uniform (the loaded rule, the wheelbarrow): wherever the balance point says. Hanging the weight at an end doubles support-force errors.

Pivot freedom needs force balance first. For anything accelerating, torque is pivot-dependent — the "take moments about a clever point" move is a statics privilege.

Perpendicular lever arms. A wire at angle $\theta$ to the rod contributes $T\sin\theta \times L$ , not $TL$ ; a weight on a tilted beam acts at the horizontal distance. Extend lines of action and drop perpendiculars — the torque module's geometry, now with money on it.

State the convention. "Taking anticlockwise as positive" or "clockwise = anticlockwise" written once protects every sign that follows; unstated conventions produce self-contradicting equations under time pressure.

Three unknowns maximum (planar). Two force equations plus one torque equation. A fourth unknown (wall friction on a ladder, a fourth table leg) makes the problem statically indeterminate — the exam-honest answer is to say so, not to invent a fourth equation.

AL/IG scope line. A-Level (9702) and IGCSE examine this module as moments: principle of moments, CoG location experiments, stability and tipping — with perpendicular distances throughout. The vector torque form and hinge-reaction resolutions (Example 3) are the JEE/AP lane; don't import them into a moments answer that wants two products and a balance sentence.

Origins · 1 minArchimedes and how this was discovered

Archimedes

c. 287–212 BC · Syracuse

The centre of gravity is Archimedes' personal invention. In On the Equilibrium of Planes he did not merely state the law of the lever — he built, from symmetry axioms, the first theory of where a body's weight effectively acts, and then computed that point by pure geometry for triangles, parallelograms, trapezia, and (his masterpiece, via The Method) segments of parabolas. That is this module's equation (3), executed twenty-two centuries early without algebra. He treated the theory as working engineering: witnesses describe him single-handedly drawing a laden ship across the beach with compound pulleys — a demonstration of leverage staged for King Hiero — and the cranes and stone-throwers he built held the Roman siege of Syracuse at bay for two years. He died at its fall, by tradition absorbed in a diagram. His statics is unusual among ancient physics in never having been corrected: every FBD with one weight-arrow at the CoG, every moment balance in every structural calculation on Earth, still runs on the equilibrium of planes.

History

250

BC — Archimedes' On the Equilibrium of Planes derives the lever law and computes centres of gravity — quantitative statics born two millennia before dynamics.

1586

— Simon Stevin's Weeghconst settles the inclined plane and composes forces by the parallelogram — vector statics before vectors.

1687

— Varignon's theorem (the moment of a resultant equals the sum of the moments) arrives the same year as Newton's Principia, welding the lever tradition to the new mechanics.

1773

— Coulomb's Essai applies equilibrium-with-friction to arches, beams and retaining walls: statics becomes civil engineering's working language.

1890

— the Forth Bridge opens — 53,000 tonnes in calculated cantilever balance, celebrated by the famous human-cantilever photograph of its moment diagram.

Going further

Texts. NCERT Physics Class 11, Ch. 7 (equilibrium of rigid bodies) · HC Verma Vol. 1, Ch. 10 · Halliday–Resnick–Walker Ch. 12 · any CIE 9702/0625 moments chapter for the AL/IG dialect. For the ladder, indeterminacy, and the full craft: Statics & Rigid-Body Equilibrium (UG deep dive).

Provenance. All results are standard. A verifier should recompute: the bridge reactions ( $R_B = 88{,}000/8 = 11{,}000$ N, $R_A = 5000$ N, audited from pivot B: $40{,}000/8$ ); the metre rule ( $3.0 \times 35 = 4.5 \times d \Rightarrow d = 23.3$ cm → 73.3 cm mark); the hinged rod ( $T = 39.2$ N, $H = 33.9$ N, $V = 19.6$ N, $F = \sqrt{1149+384} = 39.2$ N); the cabinet $\theta_c = \arctan(1/3) = 18.4^\circ$ ; the crane $x = 60{,}000/8000 = 7.5$ m.

Prerequisites. Torque & the Turning Effect · Equilibrium of Concurrent Forces · Locating the Centre of Mass. Leads to. The Ladder Against the Wall (archetype) · Topple or Slide? (archetype) · Statics & Rigid-Body Equilibrium (UG deep dive).

Why exactly two conditions — why not one, or three?

Count the freedoms. A rigid body in a plane can do three independent things: translate along $x$ , translate along $y$ , and rotate. Newton's second law (two components) governs the translations; the rotational law governs the turning — three scalar conditions, packaged as " $\sum F = 0$ and $\sum\tau = 0$ ." A point particle, having no size, cannot meaningfully rotate: for it, torque balance is automatic and force balance alone defines equilibrium — which is why the Laws-of-Motion chapter never needed moments. Torque balance is precisely the new physics of extended bodies. In full 3-D the count rises to six (three translations, three rotation axes), which is the version structural engineers solve; the planar three is the exam's world, and it caps solvable unknowns at three.

Does my choice of pivot change the answer? Different pivots give different-looking equations.

Different-looking, same solution set — that is the pivot-independence theorem. Once $\sum F = 0$ holds, the net torque is one number regardless of reference point, so any pivot yields a valid equation and all of them, combined with the force equations, pin down the same unknowns. The craft is efficiency: put the pivot where the most annoying unknowns act (the hinge with its two components, the support whose reaction you don't want yet) and they vanish, having zero lever arm. Example 1 solves each support from the other's pivot and uses the second as a free audit. One caution: this freedom is a consequence of force balance — for accelerating bodies, torques genuinely differ from point to point, and pivot games prove nothing.

Is the centre of gravity the same as the centre of mass?

On every lab bench and exam paper, yes — and the derivation shows why: when $g$ is the same vector for every particle, gravity's total torque reorganizes into (total weight) acting at the mass-weighted average position, i.e. the centre of mass. The names part company only when $g$ varies noticeably across the body: then lower portions, feeling stronger gravity, carry extra torque-weight, and the effective point of action shifts away from the CM. For a metre-long object near Earth the difference is parts in $10^{-7}$ — hopeless to measure. For a 100-km orbiting tether it is real engineering: gravity-gradient torques orient satellites precisely because CoG and CM separate. Use the terms interchangeably at this level, but define whichever you write — the one-mark definition question loves this pair.

How do I stop getting the signs of moments wrong?

Adopt one convention per problem and let it carry the signs. Write "taking anticlockwise as positive" (or the mark-scheme's "clockwise = anticlockwise") at the top. Then, for each force, ask which way it would rotate the body about your pivot if it acted alone — that turning sense, not the force's up/down direction, fixes its sign (a downward force can turn either way depending on which side of the pivot it acts). Compute each magnitude as force × perpendicular lever arm, extending lines of action where needed. Two habits kill most sign errors: draw the FBD with forces at their true application points (gravity at the CoG), and when a spare minute exists, re-derive one answer from a different pivot — a sign slip almost never survives two pivots, as the bridge example shows.

What makes an equilibrium stable rather than just balanced?

Equilibrium says: placed exactly so, it stays. Stability asks: nudged slightly, does it return? A marble in a bowl, on a dome, and on a table all satisfy both conditions of (1), yet a nudge produces a restoring push in the first (stable), a runaway in the second (unstable), indifference in the third (neutral). The energy diagnosis you met in the potential-energy module carries over: stable = potential-energy minimum. For rigid bodies on surfaces it usually reduces to watching the CoG: if a small tilt raises the centre of gravity, gravity fights the tilt — stable; the tipping cabinet resists until its CoG crests the pivot edge (the energy hilltop), and beyond that the same gravity finishes the job. That crest is exactly the plumb-line-at-the-edge geometry of the tipping rule.

Why can a couple never be balanced by a single force?

Bookkeeping both conditions shows it in two lines. A couple contributes zero net force and torque $Fd$ (the same about every point). To balance it, a candidate single force $F'$ must close both books: the force book demands $F' = 0$ ... which closes the torque book at zero contribution, leaving $Fd$ unbalanced. Contradiction — only another couple (zero force, opposite $Fd$ ) can do it. This is why steering columns transmit couples, why two-pin supports exist in engineering, and why "a single well-placed force keeps the spinning plate in equilibrium" is a planted false option. The reverse insight is also examinable: a single force through the CoG shoves without turning — force book open, torque book closed — the other half of the couple's lesson that the two books are independent.

Why does a three-legged stool never rock, while a four-legged table often does?

Count equations against unknowns. A rigid body on the floor has three equilibrium equations available (planar-vertical version: one vertical force balance, two moment balances about perpendicular horizontal axes). Three legs = three unknown leg forces: exactly determined — whatever the floor's bumps, the three contact forces adjust to a unique solution, and three points always define a plane, so all three legs touch. Four legs = four unknowns with three equations: statically indeterminate. On a perfect floor the extra freedom is settled by tiny elastic deformations; on a real floor, one diagonal pair often carries everything while the other rocks. Engineers resolve indeterminacy by adding elasticity equations (much of a structures degree); milkmaids and camera tripods resolve it by using three legs. It is the everyday face of the "three unknowns maximum" rule above.