Friction Forces

Every step you take is a small act of faith in friction. Your foot pushes backward on the pavement, the pavement grips back, and you move forward — on wet ice, where the grip fails, walking becomes a controlled stagger. The same force holds nails in wood, lets a violin bow sing through stick-slip, and decides whether a car rounds a curve or slides off it. Yet friction is the odd one out among the forces of mechanics: it has no clean fundamental law. It emerges from countless microscopic contact points — asperities — welding, deforming, and shearing; everything we use in practice is an empirical model built on two dimensionless numbers. That model, the Coulomb–Amontons law, hides one genuinely subtle idea. Static friction is not a formula but an inequality: a self-adjusting constraint force that matches whatever push you apply, up to a ceiling $\mu_s N$ reached only at the verge of slipping. Once sliding starts the rules change — friction snaps to a fixed magnitude $\mu_k N$ , opposing the relative motion and typically weaker than the static maximum. That drop is the lurch of a stuck crate breaking free, and the reason anti-lock brakes keep wheels rolling. Master that one distinction — inequality when stuck, equality when sliding — and the classic problems (inclines, stacked blocks, cornering cars, braking distances) reduce to bookkeeping. Get it wrong and every answer that follows is wrong too.

Key takeaways

Friction is a contact force tangent to two surfaces that opposes (or would oppose) relative sliding; it arises from microscopic adhesion at asperities, summarized by the Coulomb–Amontons laws.
Static friction is a constraint force: it takes whatever value prevents sliding, up to the ceiling $f_s^{\max} = \mu_s N$ — an inequality, not an equality.
Once sliding occurs, kinetic friction has the definite magnitude $f_k = \mu_k N$ , opposite the relative velocity, with $\mu_k < \mu_s$ for most material pairs.
Friction is nearly independent of apparent contact area and sliding speed — the true contact area at asperities scales with the normal force, not the visible footprint.
Kinetic friction dissipates energy as heat at rate $P = f_k v_{\mathrm{rel}}$ ; static friction on a rolling wheel does zero work, because the contact point is instantaneously at rest.
Key idea: static friction is an inequality $|f_s| \le \mu_s N$ — it takes whatever value enforces no slipping; kinetic friction is an equality $f_k = \mu_k N$ opposite the relative velocity. Decide which regime you are in before writing any equation.

The core insight

Static friction is an inequality $|f_s|\le\mu_s N$ — it takes whatever value is needed (up to the maximum) to enforce no slipping; kinetic friction is an equality $f_k=\mu_k N$ directed opposite the relative velocity. Decide which regime you are in before writing any equation.

Where this lives today

Friction is estimated to consume around a fifth of the world's energy output — in engines, bearings, gears, and brakes — which is why tribology, the science of rubbing surfaces, is a serious industrial discipline. Its modern face runs from atomic-force microscopes dragging single-asperity tips across crystals (where friction can vanish entirely in the superlubric regime) to earthquake physics, where a fault sticking and slipping under tectonic load is the planet-scale crate that lurches free: stress builds while static friction holds, then releases catastrophically at the ceiling. Rate-and-state friction laws — Coulomb's model refined so $\mu$ depends weakly on sliding history — are the standard language of seismic modeling. In your car, the static-versus-kinetic distinction is enforced by software. Anti-lock braking and traction control exist for one reason: $\mu_s > \mu_k$ , so a rolling contact grips better than a sliding one. Wheel-speed sensors detect incipient lock-up and pulse the brake pressure tens of times per second, holding each tire at the verge of slipping — riding the top of the static budget without spending past it. Every hard stop you walk away from is inequality (1), policed in real time.

The physical picture

Push a heavy crate, gently at first. Nothing happens — but not because friction is stronger than you. Friction is matching your push exactly, newton for newton, the way a wall matches your lean. Push harder and it matches harder, up to a ceiling. There the crate breaks free with a lurch, and suddenly it is easier to keep moving than it was to start. That single experiment contains the whole theory: a self-adjusting static regime, a breakaway threshold $\mu_s N$ , and a weaker, constant kinetic regime $\mu_k N$ beyond it.

Figure

The breakaway experiment — friction force versus applied push

Why does the ceiling track the normal force but ignore the size of the contact face? Zoom in. No surface is flat: at the micron scale both faces are mountain ranges, touching only where peak meets peak. The true contact area is a scattering of tiny junctions, often a ten-thousandth of the visible footprint. Press harder and the peaks flatten, so the true area — and the number of load-bearing junctions that must be sheared for sliding — grows in proportion to $N$ . Rest a brick on its broad face instead of its narrow one and you get more junctions, each carrying less load; the total resistance is unchanged. Area independence is asperity bookkeeping.

Figure

Why friction tracks load, not area — the asperity picture

Two habits follow. First, treat static friction like a rope's tension: a reaction, set by what equilibrium demands, never by a multiplication — $\mu_s N$ is only its budget, spent in full at breakaway. Second, find the direction of relative sliding (or the sliding that would happen) first, then place friction against it. Kinetic friction opposes $\mathbf{v}_{\mathrm{rel}}$ , not absolute velocity — the subtlety that decides conveyor-belt, stacked-block, and rolling-wheel problems.

The friction laws

Everything here flows from two statements — one inequality, one equality. Start with the stuck regime. A crate at rest under a sideways push is in equilibrium, so friction must exactly oppose the push; the law only says how much the interface can supply before it fails:

|f_s|\le \mu_s N \qquad (1)

where: $f_s$ — static friction force (self-adjusting); $\mu_s$ — coefficient of static friction (dimensionless); $N$ — normal force pressing the surfaces together. Read the $\le$ literally: equation (1) never tells you what $f_s$ is — Newton's second law does that. Sliding begins only when equilibrium would demand more than $\mu_s N$ . And $\mu_s$ is a pure number — a ratio of two forces on the same contact — which is why a protractor alone can measure it. The moment sliding starts, friction stops negotiating. The junctions are sheared continuously, and experiment says the resistance settles to a fixed value, slightly below the static ceiling and nearly independent of sliding speed:

f_k=\mu_k N \qquad (2)

where: $f_k$ — kinetic friction magnitude during sliding; $\mu_k$ — coefficient of kinetic friction, with $\mu_k \lesssim \mu_s$ for most dry material pairs. Because $\mu_k < \mu_s$ , the force needed to keep something moving is less than the force needed to start it — the breakaway lurch, and the reason a skidding tire grips worse than a rolling one. Magnitude is half the law; direction is the other half. Friction acts on each surface against its motion relative to the other, which the vector form makes explicit:

\mathbf{f}_k=-\mu_k N\,\hat{\mathbf{v}}_{\mathrm{rel}} \qquad (3)

where: $\hat{\mathbf{v}}_{\mathrm{rel}}$ — unit vector along the velocity of the surface relative to the other. Friction comes from junctions between the two surfaces, so it resists their relative motion: if they move together there is no sliding and only static friction may act. A package dropped on a moving conveyor belt feels friction forward — it slides backward relative to the belt, so the belt drags it up to speed. Sliding friction is mechanics' canonical energy sink. Because the force always opposes the relative slip, its net work on the pair of surfaces is negative whenever any sliding occurs, and the lost energy shows up as heat at the interface:

P_{\mathrm{diss}}=\mathbf{f}_k\cdot\mathbf{v}_{\mathrm{rel}}=-\mu_k N\,v_{\mathrm{rel}} \qquad (4)

where: $P_{\mathrm{diss}}$ — rate of mechanical energy loss to heat (always $\le 0$ ); $v_{\mathrm{rel}}$ — relative sliding speed at the contact. Two readings matter. Friction cannot be written as $-\nabla U$ — no potential stores what it takes, so the mechanical subsystem genuinely loses energy. And when $v_{\mathrm{rel}} = 0$ , as at the contact patch of a rolling wheel, dissipation is zero: static friction can accelerate a whole car without doing any work on it. The engine, not the road, pays the energy bill. Which law when? The decision tree fits in one table:

Situation	Equation to use	What friction equals
Surfaces stuck, not on the verge	$\|f_s\| \le \mu_s N$ , value from $\Sigma F = ma$	whatever equilibrium/motion demands
Impending slip (verge, breakaway, "maximum" problems)	$f_s = \mu_s N$	the static ceiling, exactly
Surfaces sliding	$f_k = \mu_k N$ , direction $-\hat{\mathbf{v}}_{\mathrm{rel}}$	fixed magnitude, opposing relative slip
Rolling without slipping	static law (1), contact point at rest	as demanded, up to $\mu_s N$ ; zero work

Derivation AThe angle of repose

Tilt a board with a block on it, slowly. The block sits still, sits still — then at one sharp angle it lets go, and that critical angle is what we now compute: equation (5), $\tan\theta_c=\mu_s$ , which turns a protractor into an instrument that measures a friction coefficient. The strategy is the cleanest static-friction argument there is — resolve the equilibrium along and across the incline, then push $\theta$ to the verge where the demand $mg\sin\theta$ first equals the budget $\mu_s mg\cos\theta$ . We take this route because it uses inequality (1) exactly once, at the single angle where it becomes an equality; before the verge, $f_s=\mu_s N$ is simply false. Watch both trends at once: tilting raises the demand and shrinks the budget, which is why the crossing is sharp. And note what cancels — the mass drops out entirely, so a coin and an anvil let go at the same angle.

Step 1. Resolve along and across the incline. With the block at rest on a slope at angle $\theta$ , the two balances read $N = mg\cos\theta$ and $f_s = mg\sin\theta$ . Why: at rest both components of Newton's second law vanish separately — the normal force carries the perpendicular part of the weight, friction the tangential part.

Step 2. Push to the verge. As $\theta$ grows, the demand $mg\sin\theta$ rises while the budget $\mu_s mg\cos\theta$ falls. At the critical angle they meet: $mg\sin\theta_c = \mu_s\,mg\cos\theta_c$ . Why: slipping begins exactly when required friction first equals the maximum available — the only moment inequality (1) becomes an equality.

Step 3. Cancel and read off the law.

\tan\theta_c=\mu_s \qquad (5)

where: $\theta_c$ — critical incline angle at which a resting block just begins to slide. Mass and $g$ have cancelled: a coin and an anvil let go at the same angle. This is why sand piles and scree slopes relax to a material-specific cone — the angle of repose is a friction coefficient wearing geometric clothing. Verified: When $\mu_s\to0$ the critical angle goes to zero (anything slides on a frictionless slope); the same balance with $\mu_k$ gives the angle of steady, unaccelerated sliding.

Figure

Block on an incline — the tug-of-war that defines the angle of repose

Go deeper · Derivation B · 1 minStopping distance of a sliding block

Derivation BStopping distance of a sliding block

A block slides on level ground at speed $v_0$ and kinetic friction alone brings it to rest — we want the distance it travels, equation (7): $d=v_0^2/(2\mu_k g)$ . This is the skid-mark calculation accident investigators run in reverse: measure the mark, infer the speed. The route has three moves, each locking down one quantity — the vertical balance fixes $N$ and hence the friction magnitude via (2), the horizontal Second Law converts that into a constant deceleration, and the kinematic identity $v^2=v_0^2+2ad$ finishes without ever needing the time. We choose $v^2$ -kinematics precisely because the question asks for distance, not duration. Watch two features of the answer: the mass cancels (a truck and a hockey puck stop in the same distance), and the dependence on speed is quadratic — doubling $v_0$ quadruples $d$ , which is the whole science behind following-distance rules.

Step 1. Vertical balance. $N = mg$ . Why: on level ground with no vertical acceleration, the normal force must equal the weight — friction's magnitude is now fixed by (2).

Step 2. Horizontal dynamics. $ma = -f_k = -\mu_k N = -\mu_k mg$ , so

a=-\mu_k g \qquad (6)

where: $a$ — acceleration of a block sliding on level ground. Why: kinetic friction is the only horizontal force, and it opposes the motion. The mass cancels — a truck and a hockey puck on the same surface decelerate identically.

Step 3. Kinematics with constant deceleration. Apply $v^2 = v_0^2 + 2ad$ with final speed zero: $0 = v_0^2 - 2\mu_k g\,d$ , giving

d=\frac{v_0^2}{2\mu_k g} \qquad (7)

where: $d$ — distance to stop from initial sliding speed $v_0$ . Why: constant force means constant deceleration. Doubling the speed quadruples the stopping distance — the nonlinear rule behind following-distance guidance. *Verified: Energy check: friction work $=f_k d=\mu_k mg\cdot v_0^2/(2\mu_k g)=\tfrac12 m v_0^2$ , exactly the initial kinetic energy. Consistent.*

Go deeper · Derivation C · 2 minThe two-block slip threshold

Derivation CThe two-block slip threshold

A block of mass $m$ rests on a larger block of mass $M$ on a frictionless floor; the static coefficient between the blocks is $\mu_s$ , and we pull the bottom block with force $F$ . For small $F$ they move as one; for large $F$ the top block gets left behind — the goal is the exact dividing line, $F_{\max}=\mu_s g(M+m)$ . The strategy is the canonical one for every "does it slip?" problem: assume no slipping, compute the common acceleration of the pair, then ask how much friction the top block needs to keep up and cap that demand at the interface's ceiling $\mu_s N$ . This order matters — friction here is a follower, computed from the Second Law, not from $\mu N$ ; setting $f_s=\mu_s N$ from the start begs the question. The two classic wrong turns are both about $N$ : the normal force between the blocks is the top block's weight $mg$ alone (not the system weight), and the frictionless floor means $F$ is the only external horizontal force on the pair.

Step 1. Assume no slipping and find the common acceleration. Treat the pair as one body: $a = F/(M+m)$ . Why: while the contact holds, the two blocks share one acceleration, and the frictionless floor means $F$ is the only external horizontal force on the system.

Step 2. Ask what the top block needs. The only horizontal force on the top block is static friction from below, so it must supply $f_s = ma = mF/(M+m)$ . Why: the top block has no push of its own — friction is its sole means of keeping up, computed from the second law, not from $\mu N$ .

Step 3. Cap the demand. The interface supplies at most $\mu_s N = \mu_s mg$ , so no-slip requires $mF/(M+m) \le \mu_s mg$ . Why: the normal force between the blocks is just the top block's weight, and inequality (1) is the interface's contract.

Step 4. Solve for the threshold.

F_{\max}=\mu_s\,g\,(M+m)

Why: the top-block mass cancels, leaving the total. Beyond $F_{\max}$ the contact slides, the law switches to $\mu_k$ , and the top block lags with its own smaller acceleration $\mu_k g$ . *Verified: Limit checks: as $m\to0$ , $F_{\max}\to\mu_s gM$ (only the bottom block matters); the top-block mass cancels from the per-block requirement, leaving the system mass — dimensionally a force, as required.*

Go deeper · Derivation D · 1 minThe equilibrium window on a steep incline

Derivation DThe equilibrium window on a steep incline

Now a block on an incline too steep to hold unaided ( $\tan\theta > \mu_s$ ), supported by a force $P$ applied up the slope — and the result we are after is not a single value of $P$ but a whole window, $mg(\sin\theta-\mu_s\cos\theta)\le P\le mg(\sin\theta+\mu_s\cos\theta)$ . The reason is the defining quirk of static friction: it can point either way along the surface — uphill when the block is about to slide down, downhill when $P$ is about to shove it up. So the strategy is to work both verges separately: fix the normal force once (it never changes, since $P$ acts along the slope), then write the on-the-verge balance twice, once with friction at full ceiling pointing up-slope and once pointing down-slope. The wrong turn to avoid is assuming friction's direction in advance and solving a single equation — that finds one edge of the window and silently discards the other. Between the edges friction simply sits below its ceiling, adjusting continuously, which is inequality (1) doing its everyday work.

Step 1. Fix the normal force. $N = mg\cos\theta$ . Why: $P$ is along the slope, so it never enters the perpendicular balance; the budget $\mu_s N$ is the same at both edges of the window.

Step 2. Find the smallest $P$ . At the lower edge the block is about to slide down, so friction points fully up-slope: $P_{\min} + \mu_s N = mg\sin\theta$ , giving $P_{\min} = mg(\sin\theta - \mu_s\cos\theta)$ . Why: friction and $P$ are allies here, jointly resisting gravity; the verge uses the full ceiling.

Step 3. Find the largest $P$ . At the upper edge the block is about to slide up, so friction flips fully down-slope: $P_{\max} = mg\sin\theta + \mu_s N = mg(\sin\theta + \mu_s\cos\theta)$ . Why: friction has changed sides and now resists $P$ — same maximum magnitude, opposite sign.

Step 4. Read the window.

mg(\sin\theta-\mu_s\cos\theta)\le P\le mg(\sin\theta+\mu_s\cos\theta)

Why: between the edges, static friction sits below its ceiling and adjusts continuously — inequality (1) at work, friction sweeping from full-up to full-down as $P$ grows. *Verified: When $\mu_s\to0$ the window collapses to the single point $P=mg\sin\theta$ (frictionless balance). The window width $2\mu_s mg\cos\theta$ is exactly twice the maximum static friction, as expected.*

Friction and circular motion

One more equation belongs in the core set, because it governs every flat curve ever driven. A car rounding an unbanked bend needs a centripetal force $mv^2/R$ , and the only horizontal force available is static friction at the tires (static, because a rolling tire does not slide). The demand cannot exceed the budget: $mv^2/R = f_s \le \mu_s mg$ . Setting friction to its maximum and cancelling the mass gives the speed limit the road itself imposes:

v_{\max}=\sqrt{\mu_s\, g\, R} \qquad (8)

where: $R$ — radius of the circular path; $g$ — gravitational acceleration, $\approx 9.81\,\mathrm{m/s^2}$ . The mass cancels: a loaded truck and a motorcycle share the same limiting speed. The numbers are sobering — for $R = 50\,\mathrm{m}$ , dry asphalt ( $\mu_s \approx 0.8$ ) allows $19.8\,\mathrm{m/s}$ ( $71\,\mathrm{km/h}$ ); ice ( $\mu_s \approx 0.1$ ) cuts that to $7.0\,\mathrm{m/s}$ ( $25\,\mathrm{km/h}$ ). Banking tilts the normal force so it shoulders part of the centripetal demand — the design principle of highway interchanges and velodromes.

Limiting cases

Case	Result	Meaning
Verge of slipping	$f_s=\mu_s N$	Static friction reaches its maximum exactly when sliding is about to begin; this is the only situation where you may set $f_s=\mu_s N$ .
Steady sliding	$f_k=\mu_k N,\quad \mu_k<\mu_s$	During sliding friction is fixed in magnitude and independent of speed; kinetic friction is typically less than the static maximum.
No-slip (rolling) contact	$f_s\le\mu_s N,\ \ v_{\mathrm{contact}}=0$	For a rolling wheel the contact point is instantaneously at rest, so static friction acts and does zero work even while propelling the body.
Frictionless limit $\mu_s\to0$	$\theta_c\to0$ , equilibrium window collapses	Any slope sheds the block; only exact force balance holds anything in place.
High speed / lubrication	$f\to f(v),\ \ f\propto v\ \text{or}\ v^2$	Coulomb's speed-independence breaks down: lubricated or fluid contact gives velocity-dependent drag, not constant kinetic friction.

Worked examples

1 · Will the crate move? (confidence builder)

A $12\ \mathrm{kg}$ crate rests on a floor with $\mu_s = 0.45$ , $\mu_k = 0.30$ . You push horizontally with $40\ \mathrm{N}$ . Does it move, and what is the friction force? You then push with $80\ \mathrm{N}$ — find the acceleration.

Step 1. Find the budget. $N = mg = 12 \times 9.81 = 117.7\ \mathrm{N}$ , so the static ceiling is $f_s^{\max} = \mu_s N = 0.45 \times 117.7 = 53.0\ \mathrm{N}$ .

Step 2. Compare demand with budget. The push demands $40\ \mathrm{N}$ of opposing friction, and $40 < 53.0$ . The crate does not move, and the friction force is exactly $40\ \mathrm{N}$ — not $53\ \mathrm{N}$ ; the ceiling is not reached.

Step 3. Now push with $80\ \mathrm{N}$ . Since $80 > 53.0$ , static friction cannot hold; the crate slides and the law switches to kinetic: $f_k = \mu_k N = 0.30 \times 117.7 = 35.3\ \mathrm{N}$ .

Step 4. Second law in the sliding regime: $a = (F - f_k)/m = (80 - 35.3)/12 = \boxed{3.7\ \mathrm{m/s^2}}$ . Notice the discontinuity: just below $53\ \mathrm{N}$ of push the crate sits still; just above, it breaks away and accelerates briskly, because the resistance has dropped from $53$ to $35\ \mathrm{N}$ . That jump is the lurch of every stuck drawer and dragged sofa.

2 · The incline, both regimes (exam level)

A block rests on an adjustable ramp with $\mu_s = 0.60$ , $\mu_k = 0.40$ . Find the angle at which it starts to slide. The ramp is then fixed at $35^\circ$ — find the acceleration once sliding, and the speed after sliding $3.0\ \mathrm{m}$ from rest.

Step 1. Angle of repose from (5): $\tan\theta_c = \mu_s = 0.60$ , so $\theta_c = \arctan(0.60) = \boxed{31.0^\circ}$ . Below this the block sits still with $f_s = mg\sin\theta$ under budget; at $31.0^\circ$ demand meets budget.

Step 2. At $\theta = 35^\circ > \theta_c$ the block slides. Along the slope, the second law with kinetic friction gives $a = g(\sin\theta - \mu_k\cos\theta)$ — mass cancels again.

Step 3. Numbers: $\sin 35^\circ = 0.574$ , $\cos 35^\circ = 0.819$ , so $a = 9.81\,(0.574 - 0.40 \times 0.819) = \boxed{2.4\ \mathrm{m/s^2}}$ .

Step 4. Speed after $3.0\ \mathrm{m}$ from rest: $v = \sqrt{2ad} = \sqrt{2 \times 2.41 \times 3.0} = \boxed{3.8\ \mathrm{m/s}}$ .

Step 5. Regime check: using $\mu_s$ in Step 3 would give $0.78\ \mathrm{m/s^2}$ — a number that means nothing, because $\mu_s$ never governs a sliding contact. Regime first, arithmetic second.

3 · Why ABS brakes win (exam level)

A car brakes from $100\ \mathrm{km/h}$ ( $27.8\ \mathrm{m/s}$ ) on dry asphalt. Rolling (ABS) braking uses static friction, $\mu_s = 0.80$ ; a locked-wheel skid uses kinetic, $\mu_k = 0.60$ . Compare stopping distances.

Step 1. Rolling wheels: the tire patch does not slide, so braking can use the static budget, $a = \mu_s g = 7.85\ \mathrm{m/s^2}$ . From (7): $d = v_0^2/(2\mu_s g) = 27.8^2/(2 \times 7.85) = \boxed{49\ \mathrm{m}}$ .

Step 2. Locked wheels: the tire slides, the law switches to kinetic, $a = \mu_k g = 5.89\ \mathrm{m/s^2}$ , and $d = 27.8^2/(2 \times 5.89) = \boxed{66\ \mathrm{m}}$ .

Step 3. The skidding car needs $17\ \mathrm{m}$ more road — about four car lengths — purely because $\mu_k < \mu_s$ . A locked skid also surrenders steering: sliding friction points opposite $\mathbf{v}_{\mathrm{rel}}$ regardless of where the wheels aim.

Origins · 1 minCharles-Augustin de Coulomb and how this was discovered

Charles-Augustin de Coulomb

1736–1806 · Paris

Coulomb was a military engineer before he was a physicist, and it shows: his masterpiece on friction, the 1785 Théorie des machines simples, grew out of an Académie des Sciences prize problem about ropes, pulleys, and ship-launching ways. Working with full-scale naval hardware rather than laboratory toys, he separated static from kinetic friction and confirmed Amontons' proportionality to load across an enormous range of materials. The friction law $f = \mu N$ still bears his name — as does the unit of electric charge, earned a few years later with his torsion balance. Few careers have stamped one name onto two entirely different chapters of physics.

History

1493

— Leonardo da Vinci records in his notebooks that friction is proportional to load and independent of contact area — unpublished for four centuries.

1699

— Guillaume Amontons rediscovers and publishes the two laws of dry friction; the Académie initially doubts the strange area-independence.

1785

— Charles-Augustin de Coulomb's prize memoir distinguishes static from kinetic friction and establishes

\mu_k < \mu_s

1950

— Bowden and Tabor's The Friction and Lubrication of Solids explains the laws microscopically: true contact at asperities grows with load, so friction tracks

N

, not area.

2004

— Structural superlubricity is measured between incommensurate graphite surfaces — near-zero friction at the nanoscale, far beyond the Coulomb model.

Going further

References. Morin, Introduction to Classical Mechanics, Ch. 2–3 (friction, constraint forces) · Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., §1.1 (dissipative forces, Rayleigh dissipation) · Landau & Lifshitz, Mechanics (Vol. 1), §25 (dissipative forces) · Taylor, Classical Mechanics, Ch. 2–3 (friction, resistive forces)

Provenance. Coulomb–Amontons friction laws and the static-inequality / kinetic-equality distinction follow Morin Ch. 2–3 and Taylor Ch. 2–3. Dissipative-force treatment (power $P=f_k v_{\mathrm{rel}}$ , non-potential nature) per Goldstein §1.1 and Landau-Lifshitz §25. A verifier should recompute: angle of repose $\tan\theta_c=\mu_s$ ; stopping distance $d=v_0^2/(2\mu_k g)$ via energy ( $f_k d=\tfrac12 mv_0^2$ ); flat-curve $v_{\max}=\sqrt{\mu_s gR}$ (numerical example: $R=50$ m, $\mu_s=0.8$ , $g=9.81$ gives $19.8$ m/s); incline equilibrium window endpoints; two-block threshold $F_{\max}=\mu_s g(M+m)$ . Coefficient values quoted ( $\mu_s\approx0.8$ dry tire-road, $\approx0.1$ icy) are standard order-of-magnitude figures, not precise constants.

Prerequisites. Newton's Second Law · Normal Force · Newton's Third Law · Vectors in Kinematics and Coordinate Systems · Gravitational Force Near Earth's Surface. Leads to. Circular Motion and Centripetal Acceleration · Projectile Motion · One-Dimensional Kinematics.

If friction always opposes motion, how can it push a car — or a walker — forward?

Friction opposes relative slipping between the two surfaces, not motion through space. A driven tire pushes backward on the road at the contact patch; the road's reaction pushes the tire forward — and since a rolling tire does not slide, this is static friction. Walking is the same trade: your foot shoves the pavement backward, the pavement grips your foot forward, and that forward reaction is the only external horizontal force available to accelerate you. The vector law (3) makes the rule precise: kinetic friction points along $-\hat{\mathbf{v}}_{\mathrm{rel}}$ , and static friction points wherever the no-slip constraint demands. "Friction opposes motion" is a slogan that fails exactly where the interesting problems live — driving wheels, conveyor belts, stacked blocks.

How does static friction know exactly how hard I'm pushing?

The same way a table "knows" how heavy the book is, or a rope knows its load: static friction is a constraint force, fixed by Newton's second law, not by a formula of its own. Microscopically, the asperity junctions deform elastically under your push — like millions of tiny springs — and they deform exactly as much as needed to transmit whatever shear force equilibrium requires. Inequality (1) never tells you the value of $f_s$ ; it only states the budget $\mu_s N$ the interface can supply before the junctions shear through. Push with 30 N and friction is 30 N; push with 45 and it is 45 — right up until demand exceeds the ceiling, the junctions fail, and the law switches to $f_k=\mu_k N$ .

If friction is independent of contact area, why do racing cars run wide tires?

Because rubber is where the Coulomb–Amontons model frays. Area-independence is asperity bookkeeping for hard surfaces: fewer junctions each carrying more load, or more junctions each carrying less — the same total either way. Rubber, though, is soft and adhesive, and its friction is not strictly proportional to $N$ : the effective $\mu$ falls as contact pressure rises. A wide tire spreads the load, keeps the pressure low, and so buys a higher effective coefficient. Width also manages heat — a flexing, partially slipping tire dissipates energy furiously, in the spirit of (4), and cooler rubber both grips and survives better. The textbook law is not wrong; it is an idealization for dry, hard contacts, and racing rubber deliberately lives outside its jurisdiction.

Can a friction coefficient be greater than 1?

Yes — nothing in the physics caps it. $\mu$ is a ratio of two forces, shear resistance over normal load, and no theorem forces junctions to shear more easily than they compress. Drag-racing slicks reach $\mu_s\sim 3$ ; clean metal surfaces in vacuum can cold-weld, giving effectively enormous coefficients. What $\mu_s>1$ means geometrically, via the angle of repose (5), is simply $\theta_c>45^\circ$ : the block clings past a forty-five-degree tilt. The intuition that friction "obviously" cannot exceed the weight comes from everyday pairs like wood on wood, where $\mu\approx0.3$ – $0.5$ — a fact about common materials, not about the law.

If static friction does zero work on a rolling wheel, what actually gives the car its kinetic energy?

Separate force from energy. Static friction at the contact patch is the external horizontal force in $\Sigma F=ma$ — without it the wheels would spin uselessly — so it absolutely causes the acceleration. But the bit of tire touching the road is instantaneously at rest, $v_{\mathrm{rel}}=0$ , so by (4) friction transfers no energy across the contact. The kinetic energy is generated internally: the engine does work through the drivetrain, torquing the wheels against the very constraint friction enforces. Braking mirrors it — with ABS keeping the tires rolling, friction decelerates the car while doing no work, and the car's kinetic energy ends up as heat in the brake discs, not in the road. Force flows through the contact; energy flows through the machinery.

A package dropped on a conveyor belt gets pushed forward by friction — isn't that backwards?

Only if you use the wrong velocity. Kinetic friction opposes the surfaces' relative motion, $\mathbf{f}_k=-\mu_k N\,\hat{\mathbf{v}}_{\mathrm{rel}}$ per equation (3), not motion relative to the floor. At the instant it lands, the package is at rest while the belt streaks forward beneath it, so the package slides backward relative to the belt — and friction on the package therefore points forward, dragging it up to belt speed. During the skid, energy is dissipated at the rate $\mu_k N v_{\mathrm{rel}}$ from (4). Once the speeds match, $v_{\mathrm{rel}}=0$ , sliding ends, and only static friction remains — supplying whatever the motion demands, which on a straight, steady belt is nothing at all.

Shouldn't a heavy truck take much longer to stop than a small car?

Not in the idealized Coulomb world, thanks to a clean cancellation: doubling the mass doubles the normal force, which doubles the available friction $f=\mu mg$ , which exactly compensates the doubled inertia. The deceleration is $a=\mu_k g$ — equation (6) — mass-free, and the skid distance $d=v_0^2/(2\mu_k g)$ from (7) depends only on speed and surface. The same cancellation produces the mass-free angle of repose and the mass-free cornering limit $v_{\max}=\sqrt{\mu_s gR}$ . Real trucks do stop longer, but for reasons outside the model: brakes that fade, tires whose effective $\mu$ drops under heavy load, cargo-stability limits. Within dry Coulomb friction, a hockey puck and a loaded truck skid to rest in the same distance.

Why is it easier to keep something moving than to start it — where does the drop from

\mu_s

\mu_k

come from?

Junctions strengthen with time in contact. While the surfaces sit still, the asperity contact spots creep, flatten, and bond — the interface ages toward higher shear strength. Once sliding starts, each junction exists only fleetingly, sheared before it can fully form, so the average resistance is lower: $\mu_k<\mu_s$ . That drop is the breakaway lurch of Worked Example 1, where resistance falls from 53 N to 35 N the instant the crate moves; it also powers stick-slip oscillation — the violin bow's song, a door hinge's squeak — and, scaled up, the earthquake cycle, where a fault holds under growing tectonic stress and releases catastrophically at its ceiling. Rate-and-state friction laws model exactly this history dependence.

Friction Forces

Where this lives today

The physical picture

The friction laws

Friction and circular motion

Limiting cases

Worked examples

1 · Will the crate move? (confidence builder)

2 · The incline, both regimes (exam level)

3 · Why ABS brakes win (exam level)

Going further

Before this

Where this leads