Statics and Rigid-Body Equilibrium

The Forth Bridge has stood over Scottish water since 1890, holding up trains with 53,000 tonnes of steel that goes nowhere. The ladder against your wall, the crane over a building site, the bones of your forearm as you hold a coffee cup — most of the mechanical world is not moving, and staying put turns out to be a demanding physical achievement. Every one of those structures is a battlefield of enormous forces in perfect cancellation: the bridge's members carry meganewtons of tension and compression, your biceps hauls with several times the weight of the cup, and the price of miscalculation is collapse. Statics is the physics of that cancellation, and its central discovery is that force balance alone is not enough. A rigid body can have zero net force on it and still whirl into motion, spun by a torque — two equal and opposite thumb-pushes on a steering wheel move nothing and yet turn everything. Keeping an extended object truly still requires two independent conditions: the forces must cancel, and the torques must cancel too. Out of that second condition grows the entire craft of structural reasoning — the principle of moments, the lever, the centre of gravity, the tipping criterion. You arrive with exactly the right tools: free-body diagrams, Newton's laws, and the torque concept from the last article. What you gain is the engineer's oldest superpower — the ability to compute forces you cannot see, hiding inside ladders, beams, and bridges — plus a set of classic problems (the leaning ladder above all) that every physicist is expected to have fought and won.

Key takeaways

A rigid body is in static equilibrium only when BOTH conditions hold: $\sum \vec F = 0$ (no translation) and $\sum \vec\tau = 0$ (no rotation).
Neither condition implies the other: a couple — two equal, opposite, offset forces — has zero net force yet nonzero torque, and spins the body.
If $\sum \vec F = 0$ , the net torque is the same about every point — so you may choose the pivot that kills the most unknowns.
Principle of moments: about any pivot, clockwise moments balance counterclockwise moments; a small force on a long lever arm matches a large force on a short one.
Gravity on an extended body acts, for torque purposes, entirely at the centre of gravity (the centre of mass in uniform $g$ ); a body tips when its centre of gravity passes outside its base of support.
The classic problems — see-saw, beam on two supports, leaning ladder — are all solved by the same three-line recipe: $\sum F_x = 0$ , $\sum F_y = 0$ , $\sum \tau = 0$ about a cleverly chosen point.
Key idea: standing still is a two-part contract — forces cancel and torques cancel — and choosing the right pivot point turns most statics problems into one-line calculations.

The core insight

A rigid body remains at rest only if the vector sum of all external forces vanishes AND the sum of all external torques vanishes about a point — and once the forces balance, torque balance about one point guarantees it about every point, which is why a well-chosen pivot dissolves statics problems.

Where this lives today

Statics is the most heavily exercised physics on Earth by sheer volume of daily use. Every structural engineering office runs it continuously: finite-element software decomposes buildings, bridges, and aircraft frames into millions of elements and solves exactly the equations of this article — $\sum\vec F = 0$ , $\sum\vec\tau = 0$ — for every one of them simultaneously, with the load paths of a skyscraper or the wing spar of an A350 falling out as the solution. Building codes are, at bottom, statutory statements about statics margins: the requirement that demand stay below capacity for every beam, column, and connection under worst-case loading is equations (1) with legally mandated safety factors. The liveliest modern frontier is robotics. A walking humanoid is a statics problem that re-solves itself continuously: the robot stays up only while the ground-reaction resultant stays inside the support polygon of its feet — the dynamic generalization of this article's tipping criterion (5), known as the zero-moment point, is the organizing concept of bipedal locomotion control, computed hundreds of times per second in every Atlas backflip and delivery-robot stumble recovery. Grasp planning for robotic hands runs the same physics in miniature — a stable grasp is one whose contact forces can close the force and torque ledgers against any small disturbance (the "force-closure" condition). And biomechanics applies the lever law to your own frame: the forearm is a third-class lever with a lever-arm ratio near 1:7, which is why holding 5 kg at arm's length demands hundreds of newtons of biceps tension — a number your physiotherapist computes with Archimedes' equation (2).

The physical picture

Why should keeping a rigid body still require two separate conditions? Because a rigid body can do two separate things: translate and rotate. Newton's second law, $\sum \vec F = M\vec a_{\rm cm}$ , controls only the centre of mass. Kill the net force and the centre of mass stays put — but the body is perfectly free to turn around that stationary centre. To see it, lay a book on a slick table and push its opposite corners with equal and opposite forces. The net force is exactly zero; the book's centre goes nowhere; and the book spins on the spot. That force pair — equal magnitude, opposite direction, offset lines of action — is called a couple, and it is the standing counterexample to every temptation to declare victory after checking forces alone. Torque, the rotational analogue of force from the previous article, has its own balance sheet, and it must independently close.

Figure

The two conditions of equilibrium — force balance kills translation, torque balance kills rotation

The second condition is where all the craft lives, because torque depends on where forces act, not just how big they are. That is the content of the principle of moments, humanity's oldest quantitative law of mechanics: a force's turning effect is force times lever arm, so a child far from a see-saw's pivot balances a parent sitting close in. Archimedes formalized it twenty-two centuries ago and boasted that with a long enough lever he could move the Earth — the boast is arithmetically sound, and every wheelbarrow, crowbar, bottle opener and forearm since has been cashing it in. Two more ideas complete the toolkit. First, gravity pulls on every particle of an extended body, yet for torque bookkeeping the entire weight can be treated as acting at a single point — the centre of gravity, which in the uniform gravity of a lab bench coincides with the centre of mass (a happy fact derived below). Hang a cardboard cutout from a pin and drop a plumb line; do it again from another corner; the lines cross at the centre of gravity — a two-minute experiment that locates the one point where the whole weight story is told. Second, the centre of gravity decides stability: a body resting on a surface stays up while the vertical line through its centre of gravity pierces the base of support, and tips the moment it escapes. That single criterion explains why you widen your stance on a bus, why SUVs roll over more readily than sports cars, and why the Leaning Tower of Pisa — its centre of gravity still safely above its foundation ring — leans but does not fall.

The math: two conditions and a free choice

The conditions of equilibrium come straight from the two laws of rigid-body motion. Translation is governed by $\sum \vec F = M\vec a_{\rm cm}$ and rotation by $\sum \vec\tau = I\vec\alpha$ ; a body at rest and staying at rest needs $\vec a_{\rm cm} = 0$ and $\vec\alpha = 0$ , so

\sum \vec F = 0 \qquad\text{and}\qquad \sum \vec\tau_P = 0 \quad\text{(about a point } P\text{)} \qquad (1)

where: the force sum runs over all external forces (internal ones cancel in pairs by Newton's third law), and $\vec\tau_P = \vec r \times \vec F$ is each force's torque about the chosen reference point $P$ , with $\vec r$ running from $P$ to the force's point of application. For the flat, two-dimensional problems that dominate practice, (1) is three scalar equations — $\sum F_x = 0$ , $\sum F_y = 0$ , $\sum \tau = 0$ with counterclockwise counted positive — which is why a planar statics problem can determine at most three unknown forces. The quiet miracle of equation (1) is the subscript $P$ : it looks like a new condition for every possible pivot, infinitely many equations. The first derivation shows that once the forces balance, one pivot's torque balance buys them all — and that freedom to choose $P$ is the principal tool of the trade: put the pivot where the most annoying unknown force acts, and that force vanishes from the torque equation. For a pair of forces on a lever or see-saw, the torque condition alone becomes the principle of moments. With weights $W_1$ and $W_2$ at distances $d_1$ and $d_2$ on opposite sides of a pivot,

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

where: each product $W\,d$ is a moment — force times perpendicular lever arm. Equation (2) is the see-saw, the balance scale, the crowbar, and the reason a 25 kg child at 2.4 m balances a 60 kg parent at 1.0 m ( $25 \times 2.4 = 60 \times 1.0$ , in consistent units). Multiplied through by $g$ it is Archimedes' law of the lever, unmodified since antiquity. Gravity itself enters the torque ledger through one point. For a body of mass $M$ in a uniform field,

\vec\tau_{\rm grav} = \vec r_{\rm cm} \times M\vec g \qquad (3)

where: $\vec r_{\rm cm}$ locates the centre of mass. Equation (3) — derived below — says the distributed pull of gravity on every particle exerts exactly the torque of a single force $M\vec g$ applied at the centre of mass, which is therefore also the centre of gravity whenever $\vec g$ is uniform (for any laboratory object the distinction is utterly negligible; for a 100-km space tether it is real). This is the licence behind every free-body diagram you have ever drawn with a single $Mg$ arrow at the middle.

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

We are deriving the statement that makes statics solvable: if $\sum \vec F = 0$ , then 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 for all pivots, and the pivot choice is purely a matter of convenience. The strategy is direct bookkeeping: write every torque about the new point, split each position vector into "old point plus shift," and watch the shift term die against the force balance. The thing to watch is exactly where the force sum gets used — the result is false for a body with unbalanced forces, and the failure mode is instructive.

Step 1. Let forces $\vec F_i$ act at positions $\vec r_i$ measured from $P$ , and let $P'$ sit at displacement $\vec d$ from $P$ , so the same application points measured from $P'$ are $\vec r_i' = \vec r_i - \vec d$ . The torque about $P'$ is $\sum_i \vec r_i' \times \vec F_i$ . Why this decomposition: torque is the only player here that cares about the reference point, and this substitution isolates that dependence in a single term.

Step 2. Expand: $\sum_i (\vec r_i - \vec d) \times \vec F_i = \sum_i \vec r_i \times \vec F_i \;-\; \vec d \times \sum_i \vec F_i$ . The first sum is the torque about $P$ ; the second contains the total force. Why $\vec d$ factors out: it is the same shift for every force, so the cross product distributes it onto the force sum.

Step 3. Impose the first equilibrium condition, $\sum_i \vec F_i = 0$ . The correction term vanishes identically, leaving

\sum_i \vec\tau_{P'} = \sum_i \vec\tau_{P} \qquad (4)

Why this is the working tool of statics: since all pivots give the same total torque, you may put the pivot on top of an unknown force — the ladder's floor contact, a beam's support — and that force, having zero lever arm, disappears from the torque equation. One equation, one fewer unknown.

Step 4. Read the converse warning from the algebra. If $\sum \vec F \ne 0$ , the torques about different points genuinely differ by $-\vec d \times \sum\vec F$ — and for a couple the opposite happens: the force sum is zero, so by (4) a couple's torque $Fd$ is the same about every point in space, an irreducible, pivot-proof twist. Why the couple matters practically: it is the pure turning agent — what your fingers apply to a jar lid and what "torque" means on a wrench specification — and it proves the two conditions in (1) are independent: no force balance can cancel a couple, and no single force has zero net force, so neither condition can substitute for the other. *Verified: for the see-saw of equation (2) with $W_1 d_1 = W_2 d_2$ , computing torques instead about the left child's seat gives pivot force $N = W_1 + W_2$ acting at $d_1$ with moment $(W_1+W_2)d_1$ against $W_2(d_1 + d_2)$ ; equality follows from (2) exactly ✓ — the same balance, seen from a different pivot, as (4) promises. Standard treatment: Halliday–Resnick–Walker Ch. 12, Kleppner & Kolenkow Ch. 3.*

Go deeper · Derivation B · 3 minThe centre of gravity — why the whole weight acts at one point, and when a body tips

Derivation BThe centre of gravity — why the whole weight acts at one point, and when a body tips

We are deriving equation (3): in a uniform gravitational field the torque of gravity about any point equals that of the total weight $M\vec g$ applied at the centre of mass — establishing that the centre of gravity and the centre of mass are the same point in uniform $g$ — and then converting that fact into the tipping criterion for a body resting on a surface. The route is a two-line sum over particles followed by a support-polygon argument; the subtlety worth watching is that the whole result hangs on $\vec g$ being the same vector for every particle.

Step 1. The body is a collection of particles $m_i$ at positions $\vec r_i$ ; gravity applies force $m_i \vec g$ to each. Total gravitational torque about the origin: $\vec\tau = \sum_i \vec r_i \times m_i \vec g = \left(\sum_i m_i \vec r_i\right) \times \vec g$ . Why $\vec g$ pulls out of the sum: uniformity — every particle feels the identical field vector, so the cross product with $\vec g$ distributes over the mass-weighted position sum. In a non-uniform field this step fails and the centre of gravity migrates away from the centre of mass.

Step 2. Recognize the mass-weighted sum: $\sum_i m_i \vec r_i = M \vec r_{\rm cm}$ by the definition of the centre of mass. Hence $\vec\tau = \vec r_{\rm cm} \times M\vec g$ — exactly the torque of a single force $M\vec g$ acting at $\vec r_{\rm cm}$ , which proves (3). Why this is a genuine theorem and not a definition: nothing about the centre of mass mentions gravity; that the gravitational torque nevertheless organizes itself around that point is a gift of the field's uniformity, and it is what lets a free-body diagram compress a distributed pull into one arrow.

Step 3. Now stand the body on a surface. The upward contact forces act only within the base of support — the smallest convex region containing all contact points (the rectangle spanned by a table's feet, the line segment under a book's edge as it starts to lift). For equilibrium, the contact forces must supply both $\sum F = 0$ and $\sum\tau = 0$ against the weight acting at the centre of gravity. Why the convex hull is the right region: contact forces push but cannot pull, so their resultant — a weighted average of pushes at the contact points — can act only somewhere inside that hull.

Step 4. Take torques about the base's edge nearest the centre-of-gravity's overhang. If the vertical line through the centre of gravity falls inside the base, the normal-force resultant can position itself directly below the centre of gravity and cancel gravity's torque: equilibrium is available. If the line falls outside, gravity's torque about the edge is a tipping torque, and no arrangement of pushes inside the base can oppose it — the body must rotate about the edge and fall. The criterion:

\text{stable rest} \iff \text{the plumb line from the centre of gravity pierces the base of support} \qquad (5)

Why this one line carries so much engineering: it is the entire theory of tipping — the bus passenger's wide stance (bigger base), the SUV's rollover threshold (high centre of gravity reaches the base edge at smaller tilt), the crane's counterweight (drags the combined centre of gravity back over the tracks), and the Leaning Tower's survival (its plumb line still lands about 2 m inside the foundation ring). *Verified: for a uniform box of width $w$ and height $h$ tilted on an edge, (5) predicts tipping past $\tan\theta_c = w/h$ — a 2 m-tall, 1 m-wide cabinet tips at $\tan\theta_c = 0.5$ , $\theta_c = 26.6°$ , matching the direct energy argument (potential-energy maximum when the centre of gravity crests the pivot edge) ✓. Treatment as in Halliday Ch. 12; the convex-hull statement is the planar case of the support-polygon criterion used in robotics.*

Go deeper · Derivation C · 3 minThe leaning ladder — the complete classic

Derivation CThe leaning ladder — the complete classic

This is the most celebrated statics problem in physics: a uniform ladder of mass $m$ and length $L$ leans at angle $\theta$ (measured from the floor) against a frictionless wall, standing on a floor with static friction coefficient $\mu_s$ . We derive the wall and floor forces, the no-slip condition $\tan\theta \ge 1/(2\mu_s)$ , and — the version with teeth — how high a climber of mass $M$ may safely go. The strategy is the full statics recipe: free-body diagram, two force equations, one torque equation with the pivot placed at the foot of the ladder, where two of the four unknown forces act and therefore vanish from the ledger. Watch how the wall's frictionlessness is what makes the problem determinate: with wall friction the problem acquires a fourth unknown and three equations cannot decide it.

Step 1. Inventory the forces. Weight $mg$ down at the midpoint ( $L/2$ along the ladder — centre of gravity of a uniform rod, by the previous derivation). Wall pushes horizontally with normal force $N_w$ (frictionless: no vertical component). Floor pushes up with normal $N_f$ and grips horizontally with static friction $f$ , pointing toward the wall. Why friction points toward the wall: the wall's push $N_w$ shoves the ladder's foot away from the wall; friction is the only horizontal force available to oppose that shove — no friction, no standing ladder.

Step 2. Force balance. Vertical: $N_f = mg$ . Horizontal: $f = N_w$ . Why these come first: they hand two unknowns over immediately and leave the torque equation the single job of finding $N_w$ .

Step 3. Torque about the foot. The foot forces $N_f$ and $f$ have zero lever arm there (the pivot-independence theorem licenses this choice). Gravity's moment: $mg \cdot \frac{L}{2}\cos\theta$ (perpendicular distance from foot to the weight's line of action), turning the ladder away from the wall. The wall's moment: $N_w \cdot L\sin\theta$ , turning it back. Balance:

N_w \, L\sin\theta = mg\,\frac{L}{2}\cos\theta \qquad\Longrightarrow\qquad N_w = \frac{mg}{2\tan\theta} \qquad (6)

Why the geometry: the weight acts vertically, so its lever arm is the horizontal distance $(L/2)\cos\theta$ from the pivot; the wall force is horizontal, so its lever arm is the height $L\sin\theta$ . Steeper ladder, longer wall lever and shorter gravity lever — the wall barely has to push.

Step 4. Impose the friction budget, $f \le \mu_s N_f$ . With $f = N_w$ from Step 2 and $N_f = mg$ :

\frac{mg}{2\tan\theta} \le \mu_s\,mg \qquad\Longrightarrow\qquad \tan\theta \ge \frac{1}{2\mu_s} \qquad (7)

Why mass drops out: both the friction demand ( $N_w$ ) and the friction supply ( $\mu_s N_f$ ) scale with $mg$ — the safe angle is a property of geometry and surface, not of how heavy the ladder is. For $\mu_s = 0.5$ , (7) gives $\tan\theta \ge 1$ : no uniform ladder can stand shallower than $45°$ on such a floor.

Step 5. Add the climber: mass $M$ at distance $d$ along the ladder. Now $N_f = (m+M)g$ and the foot-pivot torque balance gains a term: $N_w L\sin\theta = mg\frac{L}{2}\cos\theta + Mg\,d\cos\theta$ . The friction demand grows with every rung climbed while the supply $\mu_s(m+M)g$ stays fixed; setting demand equal to supply and solving for $d$ :

d_{\max} = \frac{L}{M}\left[\mu_s (m+M)\tan\theta - \frac{m}{2}\right] \qquad (8)

Why climbing is what breaks ladders: the climber raises $N_w$ (longer gravity-side moment) but not the friction ceiling in proportion — so a ladder that stands happily when empty can let go at the two-thirds mark. Note the design lesson inside (8): $d_{\max}$ grows with $\tan\theta$ — a steeper ladder is a safer ladder, opposite to most people's instinct that shallow means secure. *Verified: numbers in worked example 2 ( $L = 5$ m, $m = 10$ kg, $M = 70$ kg, $\theta = 60°$ , $\mu_s = 0.40$ give $d_{\max} = 3.6$ m; at $\theta = 53°$ it drops to $2.7$ m). Limiting checks: $M \to 0$ in (8) reproduces (7) at equality ✓; $\mu_s \to \infty$ sends $d_{\max} \to \infty$ (a nailed foot cannot slip) ✓. Classic treatment: Halliday Ch. 12, Morin Ch. 2.*

Which equation answers which question

you want	use	notes
is it in equilibrium at all?	both of (1): $\sum\vec F = 0$ AND $\sum\vec\tau = 0$	check the couple trap — forces alone prove nothing
balance two weights on a pivot	principle of moments, eq. (2)	force × lever arm, both sides
where does gravity act?	centre of gravity = centre of mass, eq. (3)	uniform $g$ only
unknown support/contact forces	torque about a point where an unknown acts, then force balance	pivot choice = eq. (4)'s free gift
will it tip over?	plumb-line criterion, eq. (5)	centre of gravity vs base of support
ladder safety	eqs. (6)–(8)	steeper is safer; climbing raises the demand

Limiting cases

Case	Result	Meaning
point particle	$\sum\vec\tau = 0$ automatic; only $\sum\vec F = 0$ survives	torque balance is an extended-body condition
couple ( $\sum\vec F = 0$ , offset lines)	$\tau = Fd$ about every point	the irreducible twist; proves the conditions independent
pivot moved by $\vec d$ , forces unbalanced	torques differ by $-\vec d\times\sum\vec F$	pivot freedom exists only in equilibrium
ladder $\theta \to 90°$	$N_w \to 0$ , $f \to 0$	vertical ladder demands no grip at all
ladder at $\tan\theta = 1/2\mu_s$	friction at its ceiling	the empty ladder's slip threshold, eq. (7)
cabinet tilted to $\tan\theta_c = w/h$	centre of gravity over the pivot edge	tipping threshold; wider or squatter = more stable
non-uniform $\vec g$	centre of gravity $\ne$ centre of mass	matters for space tethers, not lab benches

Common trapIf the forces on a body cancel, it is in equilibrium

This is the single most common statics error, and the couple demolishes it: push a floating book's opposite corners with equal and opposite forces and the net force is exactly zero while the book spins up — zero net force guarantees only that the centre of mass does not accelerate, not that the body fails to rotate. Equilibrium is a two-part contract, and the second signature — $\sum\vec\tau = 0$ — must be checked separately. The reverse omission also bites: torque balance about one special point with unbalanced forces means nothing, because with $\sum\vec F \ne 0$ the torque changes from pivot to pivot (Step 4 of the first derivation) and you may simply have chosen the one point where it happens to vanish. And when you do check torques, remember that gravity's contribution acts at the centre of gravity — students who hang a beam's whole weight at its end, rather than its middle, get support forces wrong by a factor of two. Forces, then torques, gravity at the centre: all three, every time.

Worked examples

1 · The bridge beam — support reactions (confidence builder)

A uniform beam of length $8.0$ m and weight $4{,}000$ N rests on supports at its two ends, A (left) and B (right). A truck weighing $12{,}000$ N stands with its centre $6.0$ m from A. Find the load carried by each support. (Statics of every bridge inspection, in miniature.)

Step 1. Free-body diagram: upward reactions $R_A$ and $R_B$ at the ends; the beam's weight $4{,}000$ N down at its midpoint ( $4.0$ m from A — centre of gravity of a uniform beam); the truck's $12{,}000$ N down at $6.0$ m from A. Two unknowns — one torque equation plus one force equation will do.

Step 2. Torque about A (so $R_A$ , an unknown, vanishes from the ledger): counterclockwise from $R_B$ , clockwise from the two weights. $R_B \times 8.0 = 4{,}000 \times 4.0 + 12{,}000 \times 6.0 = 16{,}000 + 72{,}000 = 88{,}000$ N·m, so $R_B = \boxed{11{,}000\ \text{N}}$ .

Step 3. Vertical force balance: $R_A = (4{,}000 + 12{,}000) - 11{,}000 = \boxed{5{,}000\ \text{N}}$ .

Step 4. Audit with the other pivot — the freedom equation (4) grants. Torque about B: $R_A \times 8.0$ against $4{,}000 \times 4.0 + 12{,}000 \times 2.0 = 16{,}000 + 24{,}000 = 40{,}000$ N·m, giving $R_A = 5{,}000$ N ✓. The truck sits three-quarters of the way across, and support B duly carries the lion's share — reading the answer physically is the cheapest error check in statics.

2 · How high can the painter climb? (exam level)

A uniform ladder ( $L = 5.0$ m, $m = 10$ kg) leans at $\theta = 60°$ against a frictionless wall; the floor has $\mu_s = 0.40$ . A painter of mass $M = 70$ kg climbs. (a) How far up the ladder may she go? (b) What if the ladder is set shallower, at $53°$ ? (Take $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. Check the empty ladder first, eq. (7): need $\tan\theta \ge 1/(2\mu_s) = 1.25$ , i.e. $\theta \ge 51.3°$ . Both $60°$ ( $\tan = 1.732$ ) and $53°$ ( $\tan = 1.327$ ) pass — the ladder stands. The question is what climbing does.

Step 2. Apply eq. (8) at $60°$ : $\mu_s(m+M)\tan\theta = 0.40 \times 80 \times 1.732 = 55.4$ ; subtract $m/2 = 5.0$ to get $50.4$ ; then $d_{\max} = (L/M) \times 50.4 = (5.0/70) \times 50.4 = \boxed{3.6\ \text{m}}$ — she may climb about $72\%$ of the ladder, and no further.

Step 3. Repeat at $53°$ : $0.40 \times 80 \times 1.327 = 42.5$ ; minus $5.0$ gives $37.5$ ; $d_{\max} = (5.0/70)\times 37.5 = \boxed{2.7\ \text{m}}$ — only $54\%$ of the way. The shallower set feels safer and is substantially more dangerous: (8) in action.

Step 4. Read the forces at the $60°$ limit: $N_f = (m+M)g = 784$ N; friction demand $f = N_w = [mg\frac{L}{2}\cos\theta + Mgd\cos\theta]/(L\sin\theta) = [10 \times 9.8 \times 2.5 \times 0.5 + 70 \times 9.8 \times 3.6 \times 0.5] \times \frac{1}{5.0 \times 0.866} = [122.5 + 1234.8]/4.33 = 313$ N — exactly $\mu_s N_f = 0.40 \times 784 = 314$ N to rounding ✓. At the limit the floor grips with everything it has.

Step 5. Extract the safety engineering: the standard "1-in-4 rule" for ladders (base out one unit per four of height, $\theta \approx 75.5°$ ) gives $\tan\theta = 4$ ; eq. (8) then allows $d_{\max} = (5/70)[0.4 \times 80 \times 4 - 5] = 8.8$ m — beyond the ladder's end, meaning the full ladder is climbable with a wide margin. The rule of thumb encodes equation (8).

Figure

The leaning ladder — the free-body diagram every physicist must own

Origins · 1 minArchimedes and how this was discovered

Archimedes

c. 287–212 BC · Syracuse

Archimedes of Syracuse wrote down, in On the Equilibrium of Planes, the first mathematical physics in history: the law of the lever, derived from symmetry postulates rather than measured — magnitudes balance at distances inversely proportional to their weights — together with the first systematic use of the centre of gravity, which he located for triangles, parabolic segments, and paraboloids by pure geometry. "Give me a place to stand," he is said to have boasted to King Hiero, "and I shall move the Earth"; witnesses report him single-handedly hauling a laden ship with a compound pulley arrangement to make the point. His statics was no armchair affair — the engines he built for the defence of Syracuse against Rome (cranes that lifted besieging ships from the water, stone-throwers with calculated counterweights) held off a Roman fleet for two years, and he died at the siege's end, absorbed, tradition says, in a geometry diagram. Two millennia later his method is intact: every free-body diagram with a single weight arrow at the centre of gravity, every moment balance in every structural calculation on Earth, is Archimedes' physics, still load-bearing.

History

250

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

1586

— Simon Stevin's De Beghinselen der Weeghconst settles the inclined plane with his "wreath of spheres" thought experiment and composes forces by the parallelogram — vector statics before vectors.

1687

— Pierre 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 founds structural mechanics, applying equilibrium analysis with friction to arches, beams, and retaining walls — statics becomes civil engineering's working language.

1890

— The Forth Bridge opens: 53,000 tonnes of steel held in calculated cantilever balance, celebrated by the famous "human cantilever" demonstration of its moment diagram.

Going further

Texts. Halliday, Resnick & Walker, Fundamentals of Physics, Ch. 12 (Equilibrium and Elasticity) · Kleppner & Kolenkow, An Introduction to Mechanics, Ch. 3 · Morin, Introduction to Classical Mechanics, Ch. 2 (Statics) — with the ladder and its many cousins as worked problems · Archimedes, On the Equilibrium of Planes, for the oldest physics still in daily professional use.

Provenance. Authored fresh for the mechanics track (Phase-C gap add-back; no AllQFT base). All results are standard textbook statics derived above. A verifier should recompute: the bridge reactions ( $R_B = 88{,}000/8 = 11{,}000$ N, $R_A = 5{,}000$ N, cross-checked from pivot B); the ladder set (6)–(8) including $d_{\max} = (5/70)[0.4\times80\times1.732 - 5] = 3.6$ m at $60°$ and $2.7$ m at $53°$ , plus the friction audit $f = 313$ N vs ceiling $314$ N; the empty-ladder threshold $\theta \ge 51.3°$ for $\mu_s = 0.4$ ; the cabinet tipping angle $\tan\theta_c = w/h$ .

Builds on: Torque and Angular Momentum about a Fixed Axis · Newton's Second Law · Free-Body Diagrams · Systems of Particles and the Center of Mass. Leads to: Rolling Motion · Moment of Inertia Tensor · Lagrangian Formulation of Rigid-Body Motion.

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

Count the freedoms of a rigid body: its centre of mass can translate (three directions in space, two in a plane) and the body can rotate about that centre (three axes in space, one in a plane). Newton's second law controls the translation — that is the condition $\sum\vec F = 0$ — but it says nothing about rotation about the centre of mass, which obeys its own law, $\sum\vec\tau = I\vec\alpha$ . Two kinds of motion, two laws, two conditions to shut both down. In full 3-D each condition is a vector equation, so "two conditions" is really six scalar equations; in the planar problems of this article it is three ( $F_x$ , $F_y$ , $\tau$ ), which is why a 2-D statics problem can pin down at most three unknown forces. A point particle, by contrast, has no size to rotate — for it, torque balance about its own position is automatic and force balance alone defines equilibrium. Torque balance is precisely the new physics of extended bodies.

Does it matter which point I choose as the pivot? Different choices seem to give different equations.

They give different-looking equations with the same solution set — that is the pivot-independence theorem (equation 4): once $\sum\vec F = 0$ holds, the net torque is identical about every point, so any pivot yields a valid equation. The craft is choosing the pivot that makes the algebra collapse: put it where an unknown force acts, and that force — having zero lever arm — vanishes from the torque equation entirely. In the ladder problem the foot is golden (kills both $N_f$ and $f$ , leaving $N_w$ alone); in the beam problem, each support in turn (worked example 1 solves for each reaction in one line, then uses the second pivot as a cross-check). One caution: the freedom is a consequence of force balance. If forces do not balance — anything actually accelerating — torques about different points genuinely differ, and "torque balance about my favourite point" proves nothing.

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

In every laboratory situation, yes — and the first two steps of the second derivation show why: when the gravitational field $\vec g$ is the same vector everywhere in the body, the total gravitational torque reorganizes itself exactly into $\vec r_{\rm cm}\times M\vec g$ , i.e. the whole weight acting at the centre of mass. The two concepts split only when $\vec g$ varies noticeably across the body, because then lower parts, feeling stronger gravity, carry extra torque-weight and the effective point of action shifts away from the mass centroid. For a metre-long object near Earth the fractional variation of $g$ is about $10^{-7}$ — hopelessly negligible. For a 100-km space-elevator tether or a moon raising tides, the distinction is real physics: gravity-gradient torques orient satellites (long axis pointing at Earth) precisely because centre of gravity and centre of mass part company. On your lab bench, use the terms interchangeably with a clean conscience.

How do I get the signs of the torques right? I keep balancing the wrong things.

Adopt one convention per problem and let algebra carry the signs. Pick a positive sense — counterclockwise is the standard — then, for each force, ask which way it would rotate the body about your chosen pivot if it acted alone: counterclockwise contributions enter with $+$ , clockwise with $-$ , and the equilibrium statement is simply "the sum is zero" (equivalently: total clockwise moments = total counterclockwise moments, which is how the principle of moments is usually spoken). Compute each magnitude as force times perpendicular lever arm — the shortest distance from pivot to the force's line of action, e.g. $(L/2)\cos\theta$ for the ladder's weight about the foot. Two habits prevent most sign disasters: draw the free-body diagram with forces at their genuine application points (gravity at the centre of gravity, not an end), and after solving, re-derive one answer with a different pivot — a sign error almost never survives two pivots, as worked example 1 demonstrates.

The ladder problem assumes a frictionless wall. What happens with friction at the wall too?

The problem becomes statically indeterminate — a genuinely important idea, not a technicality. Adding wall friction introduces a fourth unknown force, but planar statics supplies only three equations; the equations of rigid-body equilibrium then admit a whole family of solutions, and the actual force distribution depends on details rigid-body theory deliberately ignores: the tiny elastic deformations of ladder, wall, and floor, and even the history of how the ladder was placed. Real structures live with this constantly — a four-legged table is indeterminate (four unknown leg forces, three equations), which is why it can rock while a three-legged stool never does. Engineers resolve indeterminacy by promoting the bodies to elastic ones, adding compatibility-of-deformation equations to close the system (that is much of a structural engineering degree). The exam-honest answer: with both surfaces frictional, you can still bound the safe configurations, but the exact forces are no longer computable from statics alone — and that is the intended lesson.

Why is a steeper ladder safer? It feels like it should topple backward.

Two different failure modes are competing for your intuition. The failure statics warns about is slipping at the base: the wall pushes the foot away from the wall, and only floor friction resists. Equation (6) says the wall force is $mg/2\tan\theta$ — it shrinks as the ladder steepens, because the weight's lever arm about the foot shortens while the wall force's lever arm lengthens. Less wall push means less friction demand, so steep is safe against sliding: at $75°$ the demand is a quarter of that at $45°$ , and the climbable fraction in equation (8) grows accordingly (worked example 2: $2.7$ m at $53°$ versus $3.6$ m at $60°$ ). The failure your gut fears — toppling backward over the feet — is governed by the tipping criterion (5) and needs the combined centre of gravity of ladder-plus-climber to move behind the foot's contact point: nearly impossible while a climber's body stays between the rails on any reasonable set. Industrial guidance (the 1-in-4 rule, about $75°$ ) sits deliberately in the wide safe corridor between the two failure modes.

Is an equilibrium the same as a stable equilibrium?

No — equations (1) certify only that the body, placed exactly so, stays put; they are silent about what happens if it is nudged, and that distinction is the stability question. A marble at the bottom of a bowl, balanced on a dome's summit, or resting on a flat table satisfies force and torque balance identically, yet a nudge produces a restoring push in the first case (stable), a runaway in the second (unstable), and indifference in the third (neutral). The general diagnostic lives in energy language, which you met with potential-energy diagrams: stable equilibria are minima of potential energy, unstable ones are maxima. For the rigid bodies of this article the criterion usually reduces to watching the centre of gravity: if a small tilt raises the centre of gravity, gravity's torque fights the tilt and the rest position is stable — that is why the tipping cabinet resists until its centre of gravity crests the pivot edge (the energy hilltop), and why engineers speak of a structure's stability margin as how much disturbance it takes to get there.

Do the statics conditions apply to things that are moving?

Yes, in two important senses. First, the conditions $\sum\vec F = 0$ , $\sum\vec\tau = 0$ actually certify zero acceleration, not zero velocity — a crate on a truck cruising at constant velocity, or a crane load being raised at steady speed, is analysed with exactly this article's equations ("quasi-static" analysis; engineers do it all day). Second, even for genuinely accelerating systems, the statics equations reappear as the special case of the full rigid-body equations $\sum\vec F = M\vec a_{\rm cm}$ , $\sum\vec\tau = I\vec\alpha$ with the right-hand sides kept: a braking car's front-wheel loading, a turning aircraft's wing-root moment, a sprinting robot's foot forces are all "statics plus inertia terms" — and the walking-robot criterion mentioned in the modern-day section (the zero-moment point) is literally the tipping rule (5) upgraded with acceleration terms. Master the balanced case and the moving cases are bookkeeping extensions, which is precisely why every engineering curriculum teaches statics first.