Friction and Inclined Planes

Beginner

Shove a box across the floor and it soon stops. Friction is the force that resists one surface sliding over another. It acts along the surfaces, against the sliding, or against the slide an object is about to make.

Push a heavy sofa gently and it does not move: the floor pushes back just as hard, through static friction. Push harder and harder, and at some point static friction reaches its limit and the sofa starts to slide. Once it moves, kinetic friction takes over; it is usually a little smaller, which is why starting a sofa is harder than keeping it going.

Quick check: you push a box so that it slides at a steady speed. How does your push compare with the friction? Answer: they are equal, because a steady speed means no net force.

How strong is friction? It grows with how hard the surfaces are pressed together, a force called the normal force $N$, and depends on the materials, through a number called the coefficient of friction $\mu$. For sliding, $f=\mu N$. A 10 kg box (weight about 98 N) sliding across a floor with $\mu =0.3$ feels about 29 N of friction; stack a second box on top and the friction doubles.

Quick check: rubber on a dry road has $\mu$ near 0.8, a steel skate on ice near 0.02. Which needs the longer distance to stop? Answer: the skate.

On a slope, part of the weight pulls the object downhill. In the animation a block slides down a 30° slope and speeds up. The weight arrow points straight down, the support arrow (the normal force) points out of the surface, and the small friction arrow points up the slope. Friction holds the block back, but on a slope this steep it cannot stop it.

Intermediate

Normal force and the two friction laws

Whether a tyre grips the road or a crate stays put on a ramp depends on the force between surfaces in contact. When two surfaces touch, that force has two parts. The normal force $N$ is perpendicular to the surface and only as large as needed to stop the object sinking in. Friction $f$ acts along the surface and opposes sliding. Both are measured in newtons. How much friction a given normal force can produce is set by the coefficient of friction $\mu$, a pure number that depends on the two materials.

While the surfaces do not slip, friction is static: it takes whatever size and direction prevent slipping, up to a ceiling:
$f_s\le {\mu }_{s}N$
A box on a level floor that nobody pushes feels no friction at all. Push it with 50 N and friction answers with 50 N; only a push above the ceiling breaks it free.

Once the surfaces slide, friction is kinetic. It points against the sliding and has a definite size:
$f_k={\mu }_{k}N$
Usually ${\mu }_k$ is a little below ${\mu }_s$, which is why a sofa is harder to start than to keep moving. Typical values are 0.8 for rubber on a dry road, 0.3 for wood on wood and 0.02 for a skate on ice, and neither law depends much on the contact area or the speed.

Splitting the weight on a slope

On an incline at angle $\theta$, take one axis along the slope, positive downhill, and one perpendicular to it. Then $N$ and $f$ lie along the axes, and only the weight $mg$ must be split. Tilting the surface by $\theta$ tilts its perpendicular by the same angle, so the weight makes the angle $\theta$ with it. The part along the slope pulls the block downhill; the part across presses it into the surface and is balanced by the normal force:

• $F_{\parallel }=mg\sin \theta$
• $N=mg\cos \theta$

Check the extremes: on level ground, $\theta =0$, nothing pulls along the surface; on a vertical wall, $\theta =90^{\circ }$, nothing presses on it. On the animation's 30° slope, $\sin 30^{\circ }=0.5$ and $\cos 30^{\circ }\approx 0.866$: half the weight pulls downhill, and $N$ is 87% of the weight, not all of it.

Newton's second law along the slope

For a block sliding down, kinetic friction ${\mu }_{k}mg\cos \theta$ points up the slope. Newton's second law along the slope, $ma=mg\sin \theta -{\mu }_{k}mg\cos \theta$, has the mass in every term, so it cancels:
$a=g(\sin \theta -{\mu }_k\cos \theta )$
A heavy and a light block of the same material therefore slide down side by side. For a block sliding up, friction points down the slope as well, and the block slows at $g(\sin \theta +{\mu }_k\cos \theta )$.

A resting block starts to slide once the pull $mg\sin \theta$ beats the most static friction can give, ${\mu }_{s}mg\cos \theta$. Dividing both by $mg\cos \theta$ leaves
$\tan \theta >{\mu }_s$
The mass drops out again. The limiting angle, where $\tan {\theta }_0={\mu }_s$, is the angle of repose.

Worked example: the block in the animation

The block slides from rest down a 30° slope with ${\mu }_k=0.2$. No mass is needed, but take $m=2\mathrm{kg}$ to see the forces in newtons. The weight is $mg=2\cdot 9.81=19.62\mathrm{N}$, so
$F_{\parallel }=19.62\cdot 0.5=9.81\mathrm{N},N=19.62\cdot 0.866\approx 16.99\mathrm{N}$
$f_k=0.2\cdot 16.99\approx 3.40\mathrm{N},a=\frac{9.81-3.40}{2}\approx 3.21m/{\mathrm{s}}^2$
The formula gives the same without the mass: $a=9.81(0.5-0.2\cdot 0.866)\approx 3.21m/{\mathrm{s}}^2$, the value printed under it in the animation, against $g\sin 30^{\circ }\approx 4.91m/{\mathrm{s}}^2$ without friction.

With $a$ constant, straight-line kinematics applies. After 1 s the block moves at 3.21 m/s and has slid $\frac{1}{2}\cdot 3.21\cdot 1^2\approx 1.60\mathrm{m}$; after 2 s, it moves at 6.41 m/s and has slid 6.41 m. These are the grey outlines. The slide is 7.8 m long and lasts about 2.2 s, and the readouts end near $v=7.1m/s$ and $s=7.8\mathrm{m}$, as $v^2=2as$ predicts. Without friction the block would leave the same slope at 8.7 m/s.

Worked example: a crate on a loading ramp

A 30 kg crate rests on a ramp tilted at 20°, with ${\mu }_s=0.5$ and ${\mu }_k=0.4$. Does it stay put, and what push along the ramp moves it up at a steady speed?

Its weight is $30\cdot 9.81=294.3\mathrm{N}$, so $F_{\parallel }=294.3\cdot \sin 20^{\circ }\approx 100.7\mathrm{N}$ and $N=294.3\cdot \cos 20^{\circ }\approx 276.6\mathrm{N}$. Since $\tan 20^{\circ }\approx 0.36$ is below 0.5, the crate stays put, and static friction equals the downhill pull: 100.7 N up the ramp, not the ceiling ${\mu }_{s}N\approx 138.3\mathrm{N}$.

At a steady speed up the ramp the net force is zero. Friction now opposes the upward sliding, so it points down the ramp, alongside the downhill pull:
$F=mg(\sin \theta +{\mu }_k\cos \theta )=100.7+0.4\cdot 276.6\approx 211.3\mathrm{N}$
Lifting the crate straight up would take its full 294 N; a frictionless ramp would cut that to 100.7 N, and friction raises it to 211 N.

Reading the force diagram

Every force arrow in the animation uses the same scale, so the diagram can be read as numbers. The green $mg$ arrow is the diagonal of a rectangle whose sides are its two thin green components: the component down the slope is exactly half of $mg$, the one into the slope 87% of it, and the blue $N$ arrow cancels the latter. The small gold $f$ arrow, a fifth of $N$, cancels about a third of the downhill pull, leaving $0.5mg-0.173mg\approx 0.327mg$ as the net force.

Added together, $N$ and $f$ give the slope's whole force on the block. A resting block needs it to point straight up and cancel the weight; for the sliding block friction falls short, and the leftover points downhill. The readouts show the consequence: $v$ gains 3.21 m/s every second, while $s$ grows by 1.60 m in the first second and 4.81 m in the next.

Common mistakes

• The normal force is not always the weight: on the 30° slope the 2 kg block has $N\approx 16.99\mathrm{N}$, not 19.62 N.
• Static friction is not ${\mu }_{s}N$ by default: the crate resting on the 20° ramp feels 100.7 N, not the 138.3 N maximum, and an unpushed box on level ground feels none.
• Friction does not always point up the slope. It opposes the sliding, or the slide about to happen: for the crate pushed up the ramp it points down, so the push needs 211.3 N.
• Heavier blocks do not slide faster: the mass cancels, so a 20 kg block of the same material also accelerates at 3.21 m/s² on the animation's slope.
• Adding friction to the downhill pull for a block sliding down gives $9.81(0.5+0.2\cdot 0.866)\approx 6.60m/{\mathrm{s}}^2$, which is how fast a block sliding up slows. Check sine against cosine with $\theta =0$: on level ground nothing pulls along the surface.

Where you meet it

A rolling tyre grips through static friction, because its contact patch does not slide on the road; anti-lock brakes keep the wheels turning so that the weaker kinetic friction never takes over. Wheelchair ramps are kept to a gradient of about 1 in 12, under 5°, to keep the push needed small. Heaps of sand or gravel stand at their angle of repose, about 30–40°, and most snow avalanches start on slopes of 30–45°. Next, Work, Energy and Power shows where the energy taken by friction goes, Circular Motion shows friction holding a car on a bend, and Simple Machines: Levers, Pulleys and Gears treats the ramp, the wedge and the screw as machines.

Advanced

The smooth slope

Start without friction, because the smooth slope fixes the two pieces of the weight that every later equation uses. A block of mass $m$ lies on a plane tilted at the angle $\theta$ to the horizontal. Its weight $m\vec{g}$ points straight down. Split it into a part along the slope and a part across it, along the normal (the line at right angles to the surface). The weight makes the angle $\theta$ with the normal, so the part along the slope is $mg\sin \theta$ and the part pressing into the surface is $mg\cos \theta$. The block neither sinks into the slope nor lifts off it, so the slope pushes back with the normal force $N=mg\cos \theta$.

The part along the slope also follows from energy alone. Push a weight $W$ slowly up a frictionless slope of length $L$ and height $h$. The push $F$ does the work $FL$, and all of it becomes potential energy $Wh$, so $F=Wh/L=W\sin \theta$.

Simon Stevin reached the same result in 1586 with a thought experiment: a closed loop of beads draped over a triangular wedge and hanging freely below it. The loop never starts to circulate by itself, since that would be perpetual motion, and its hanging part is symmetric and balances itself. So the many beads on the long, gentle face must pull along their face as hard as the few beads on the short, steep face. The number of beads on a face is proportional to its length $L$, so the pull of one bead is proportional to $1/L$. Both faces rise through the same height $h$, so that pull is proportional to $h/L$, which is $\sin \theta$.

Two kinds of friction

Now roughen the surfaces. Friction is the part of the contact force that lies along the surface, and it comes in two kinds that obey different rules. Telling them apart decides every problem below.

When the surfaces slide over each other, kinetic friction is a definite force. Its size is ${\mu }_{k}N$, where the kinetic coefficient ${\mu }_k$ is a pure number for the pair of materials, and to a first approximation it does not depend on how fast they slide. It points against the velocity $\vec{v}$ of one surface relative to the other. Dividing $\vec{v}$ by its length $|\vec{v}|$ gives a unit vector along the motion, so
${\vec{f}}_k=-{\mu }_{k}N\frac{\vec{v}}{|\vec{v}|}$

When the surfaces do not slide, static friction is a constraint force, like the normal force. Its size and its direction in the contact plane are whatever keeps the surfaces from slipping. The only law is a ceiling, set by the static coefficient ${\mu }_s$:
$|{\vec{f}}_s|\le {\mu }_{s}N$

For most pairs ${\mu }_s>{\mu }_k$. Friction is therefore not a function of velocity: at $v=0$ it can take any value up to ${\mu }_{s}N$, and the moment sliding starts it jumps to ${\mu }_{k}N$.

On the slope, measure $x$ down the slope, so that $\ddot{x}$ is the acceleration down it. A block sliding down feels friction up the slope, and Newton's second law along the slope, divided by $m$, gives
$\ddot{x}=g(\sin \theta -{\mu }_k\cos \theta )$
A block sliding up feels friction down the slope, adding to gravity:
$\ddot{x}=g(\sin \theta +{\mu }_k\cos \theta )$
This $\ddot{x}$ is positive while the velocity $\dot{x}$ is negative, so it slows the block. A block at rest needs the static friction $mg\sin \theta$ up the slope, which stays under the ceiling ${\mu }_{s}mg\cos \theta$ exactly when $\tan \theta \le {\mu }_s$; then the block stays at rest.

One consequence is easy to miss. If $\tan \theta <{\mu }_k$, the first equation gives a negative $\ddot{x}$, so a block launched downhill slows and stops. It is then held by static friction: friction stops motion, but it never pushes the block back up.

Static friction adjusts to the pull

Legend

$f$	Friction along the slope, positive up the slope.	$\mathrm{N}$
$F$	Pull along the slope, positive up the slope; a negative value pulls down the slope.	$\mathrm{N}$
${\mu }_{s}N$	Largest possible static friction: the dashed ceiling the line may not cross.	$\mathrm{N}$
${\mu }_{k}N$	Size of kinetic friction once the block slides.	$\mathrm{N}$
$mg\sin \theta$	Weight component along the slope: the pull at which friction is zero.	$\mathrm{N}$
$F_1$	Lowest pull at which the block stays at rest, $mg(\sin \theta -{\mu }_s\cos \theta )$; negative here.	$\mathrm{N}$
$F_2$	Highest pull at which the block stays at rest, $mg(\sin \theta +{\mu }_s\cos \theta )$.	$\mathrm{N}$
$f_s$	Static friction: whatever balances the other forces, $mg\sin \theta -F$.	$\mathrm{N}$
$f_k$	Kinetic friction, always against the sliding: $-{\mu }_{k}N$ here, where the block slides up the slope.	$\mathrm{N}$

Under the static law, the size and direction of static friction are set by the other forces, not by a formula of their own. Figure 1 follows the friction $f$ on a block that rests by itself on a gentle slope, with $\tan \theta <{\mu }_s$, while a pull $F$ along the slope is increased slowly. Both $F$ and $f$ are counted positive up the slope.

While the block stays at rest, $\ddot{x}=0$, so the forces along the slope balance: the up-slope forces $f+F$ equal the down-slope component $mg\sin \theta$. The friction the block needs is therefore
$f=mg\sin \theta -F$
This is the descending line $f_s$ in Figure 1. Read it from left to right. At $F=0$ friction carries the whole component $mg\sin \theta$ and points up the slope. It shrinks as the pull takes over the job, is zero where the line crosses the axis at $F=mg\sin \theta$ (the green dot), and beyond that point it is negative: friction now points down the slope and holds the block back against the pull.

The line is allowed only between the dashed levels $+{\mu }_{s}N$ and $-{\mu }_{s}N$. Setting $f=+{\mu }_{s}N$ and $f=-{\mu }_{s}N$, with $N=mg\cos \theta$, gives the two ends of the allowed stretch:
$F_1=mg(\sin \theta -{\mu }_s\cos \theta )$
$F_2=mg(\sin \theta +{\mu }_s\cos \theta )$
The block stays at rest for every pull from $F_1$ to $F_2$, the shaded range. Here $F_1$ is negative: even a pull down the slope is resisted, until its size exceeds $|F_1|$. On a steeper slope, with $\tan \theta >{\mu }_s$, $F_1$ would be positive: the block needs at least that much pull up the slope to be held at all.

Outside the shaded range the block slides, and friction steps down from ${\mu }_{s}N$ to ${\mu }_{k}N$, to the open circles. For pulls below $F_1$ the block slides down, so friction $+{\mu }_{k}N$ points up the slope. For pulls above $F_2$ it slides up, so friction $f_k=-{\mu }_{k}N$ points down the slope. The rule for problems follows: decide which way the block moves, or is about to move, before writing the direction of friction into the equations.

The slide in the animation

The animation puts numbers on the sliding equation: a block released from rest on a 30° slope with ${\mu }_k=0.2$. Its acceleration down the slope is
$a=g(\sin \theta -{\mu }_k\cos \theta )=9.81(0.5-0.2\cdot 0.866)\approx 3.21m/{\mathrm{s}}^2$

On the expert graph of the animation, $v(t)$ is therefore a straight solid line of slope 3.21 m/s². Above it runs the dashed line of a frictionless slope, whose slope is $g\sin \theta \approx 4.91m/{\mathrm{s}}^2$. The two lines part by the friction term, $g{\mu }_k\cos \theta \approx 1.70m/s$ every second. The slide lasts 2.2 s, so the block ends at 7.1 m/s instead of 10.8 m/s. The area under the solid line is the distance travelled, $\frac{1}{2}\cdot 2.2\cdot 7.06\approx 7.8\mathrm{m}$: the length of the slide.

Where the energy goes

Legend

$s$	Distance the block slides along the slope.	$\mathrm{m}$
$h$	Height the block drops: $h=s\sin \theta$.	$\mathrm{m}$
$\theta$	Angle of the slope to the horizontal.	$\mathrm{rad}$
$W_g$	Work done by gravity, $mgh=mgs\sin \theta$: the energy released.	$\mathrm{J}$
$Q$	Heat made by friction, $f_{k}s={\mu }_{k}mgs\cos \theta$.	$\mathrm{J}$
$K$	Kinetic energy gained, $\frac{1}{2}m{v}^2=W_g-Q$.	$\mathrm{J}$

The same slide can be followed through energy instead of forces. This shows where the energy released by gravity ends up, and it gives the final speed without needing the time.

In Figure 2 the block starts from rest at the pale position and slides a distance $s$ to the solid one. Read the drop on the left: $h=s\sin \theta$. Gravity does the work $W_g=mgh=mgs\sin \theta$, the green bar, drawn to the same height scale as $h$ so that the two match. Friction has the constant size $f_k={\mu }_{k}N={\mu }_{k}mg\cos \theta$ over the same distance $s$, so it turns the energy $Q=f_{k}s={\mu }_{k}mgs\cos \theta$ into heat: the gold part of the second bar. The rest, the purple part, is the kinetic energy $K=\frac{1}{2}m{v}^2$. This is the work–energy theorem, $K=W_g-Q$:
$\frac{1}{2}m{v}^2=mgs(\sin \theta -{\mu }_k\cos \theta )$

Dividing $Q$ by $W_g$ cancels $m$, $g$ and $s$: friction takes the share $Q/W_g={\mu }_k\cos \theta /\sin \theta ={\mu }_k\cot \theta$ of the released energy. Here that is $0.2\cdot 1.732\approx 0.35$, the gold third of the second bar. For the animation's 7.78 m slide, per kilogram of block, gravity releases $9.81\cdot 0.5\cdot 7.78\approx 38.2\mathrm{J}$; 13.2 J become heat and 24.9 J kinetic energy, so $v=\sqrt{2\cdot 24.9}\approx 7.06m/s$, the speed the velocity graph gave.

The heat grows with the length of the path, not just with where the path starts and ends. That is why no potential energy can be defined for friction.

The friction cone

Legend

$\vec{R}$	Contact force of the slope on the block: $\vec{N}+\vec{f}$.	$\mathrm{N}$
$\vec{N}$	Normal force: the part of $\vec{R}$ at right angles to the slope.	$\mathrm{N}$
$\vec{f}$	Friction: the part of $\vec{R}$ along the slope, here up the slope.	$\mathrm{N}$
$m\vec{g}$	Weight, which $\vec{R}$ must cancel for the block to stay at rest.	$\mathrm{N}$
$\theta$	Angle of the slope, which is also the angle between $\vec{R}$ and the normal.	$\mathrm{rad}$
${\phi }_s$	Half-angle of the static friction cone: $\tan {\phi }_s={\mu }_s$.	$\mathrm{rad}$
${\phi }_k$	Half-angle of the kinetic friction cone: $\tan {\phi }_k={\mu }_k$.	$\mathrm{rad}$

The two friction laws have a geometric form that turns slipping into a comparison of angles. Instead of treating $N$ and $f$ separately, Figure 3 draws the whole force of the slope on a resting block as one vector, the contact force $\vec{R}=\vec{N}+\vec{f}$. Its part at right angles to the slope is the normal force $\vec{N}$, and its part along the slope is the friction $\vec{f}$; the dashed lines complete the sum. The angle $\phi$ by which $\vec{R}$ leans away from the normal has $\tan \phi =f/N$.

The static law $f\le {\mu }_{s}N$ therefore reads $\tan \phi \le {\mu }_s$: $\vec{R}$ must lie inside the friction cone, of half-angle ${\phi }_s$ with $\tan {\phi }_s={\mu }_s$, the wide shaded region about the normal. Now look at the block at rest. Only two forces act on it, $\vec{R}$ and the weight $m\vec{g}$, so $\vec{R}$ must point straight up. On a slope tilted by $\theta$, a vertical line makes the angle $\theta$ with the normal, the angle marked between $\vec{N}$ and $\vec{R}$. So the block can rest only while $\theta \le {\phi }_s$.

The steepest slope that holds the block, the angle of repose ${\theta }_0$, is therefore the friction angle: $\tan {\theta }_0={\mu }_s$, which is 26.6° for ${\mu }_s=0.5$. This gives a simple way to measure ${\mu }_s$: tilt a board until a block on it slips. The mass does not appear, so a loaded block slips at the same angle.

Once the block slides, $f={\mu }_{k}N$ exactly, so $\vec{R}$ lies on the edge of the narrower kinetic cone, of half-angle ${\phi }_k$ with $\tan {\phi }_k={\mu }_k$, the darker region inside the first. In Figure 3 the slope angle lies between the two: ${\phi }_k<\theta <{\phi }_s$. The vertical $\vec{R}$ falls inside the static cone, so a block at rest stays at rest. It falls outside the kinetic cone, so the contact force on a sliding block cannot be vertical and cannot balance the weight: that block speeds up. On a slope with $\theta <{\phi }_k$ it is the other way round: friction outweighs the pull of gravity along the slope, and a sliding block slows to a stop.

The best angle for a pull

Legend

$\vec{F}$	Pull of the rope; drawn is the shortest pull that balances the other two forces.	$\mathrm{N}$
$m\vec{g}$	Weight of the crate.	$\mathrm{N}$
$\vec{R}$	Force of the floor on the crate: normal force plus friction.	$\mathrm{N}$
$\phi$	Friction angle between $\vec{R}$ and the vertical: $\tan \phi ={\mu }_k$.	$\mathrm{rad}$
$\alpha$	Angle of the rope above the horizontal.	$\mathrm{rad}$

The friction angle also answers a practical question: at what angle should a rope pull to drag a heavy crate with the least force? On the left of Figure 4 a 30 kg crate slides steadily across a level floor with ${\mu }_k=0.4$, pulled by a rope at the angle $\alpha$ above the horizontal.

Steady sliding means zero net force. Three forces act: the pull $\vec{F}$, the weight $m\vec{g}$ and the floor's force $\vec{R}=\vec{N}+\vec{f}$. Their sum vanishes, $\vec{F}+\vec{R}+m\vec{g}=0$, so drawn head to tail they close a triangle, on the right of Figure 4.

Two sides of the triangle are already determined. The weight $m\vec{g}$ is fixed in size and direction. The direction of $\vec{R}$ is fixed too: the crate slides, so $f={\mu }_{k}N$, and $\vec{R}$ leans back from the vertical, against the motion, by the friction angle $\phi$ with $\tan \phi ={\mu }_k$. Only $\vec{F}$ is free: it must start on the line of $\vec{R}$ and end at the tail of $m\vec{g}$. The dashed lines are some of the other pulls that would also balance, each at its own $\alpha$.

Finding the smallest of these pulls is now a question of geometry. The shortest line from a point to a straight line meets it at a right angle, so the smallest pull is perpendicular to $\vec{R}$. Since $\vec{R}$ is $\phi$ from the vertical, that pull is $\phi$ from the horizontal: the best rope angle is $\alpha =\phi$, and then $F=mg\sin \phi$.

The algebra agrees. The rope lifts part of the load, so the floor presses up with only $N=mg-F\sin \alpha$. Steady sliding needs the horizontal part of the pull to equal the friction, $F\cos \alpha ={\mu }_k(mg-F\sin \alpha )$, which gives
$F=\frac{{\mu }_{k}mg}{\cos \alpha +{\mu }_k\sin \alpha }$
The pull is smallest when the denominator is largest, where its derivative $-\sin \alpha +{\mu }_k\cos \alpha$ is zero, that is at $\tan \alpha ={\mu }_k$, here $\alpha =21.8^{\circ }$. Then $F=109.3\mathrm{N}$, against 117.7 N for a horizontal pull, and the triangle gives the same, $mg\sin \phi =294.3\cdot \sin 21.8^{\circ }\approx 109.3\mathrm{N}$. The friction angle sets the best pulling angle.

Why friction is proportional to the load

The friction laws are empirical rules, not fundamental laws like gravitation. This section explains where they come from. Leonardo da Vinci noticed around 1500 that friction does not depend on the contact area; Guillaume Amontons stated its proportionality to the load in 1699, and Charles-Augustin de Coulomb added in 1781 that kinetic friction barely depends on speed.

The key fact is that even polished surfaces touch only at the tips of their highest bumps. These tips squash until together they carry the load, and a squashed tip bears a pressure about equal to the hardness $H$ of the softer material. The real contact area is therefore about $A_{\mathrm{real}}\approx N/H$. A 2 kg steel block on steel, with $N\approx 20\mathrm{N}$ and $H\approx 2\times 10^9\mathrm{Pa}$, touches over only $10^{-8}{\mathrm{m}}^2$, about a millionth of a 10 cm by 10 cm face. This is why the apparent area does not matter: on a larger face each tip is pressed less, but the total real area is still fixed by $N$.

At the tips the atoms of the two surfaces bond. Sliding must shear these junctions, which takes a force $\tau A_{\mathrm{real}}$, where $\tau$ is the shear strength of the softer material. Putting the two steps together,
$f\approx \tau A_{\mathrm{real}}\approx \frac{\tau }{H}N$
a friction law with $\mu \approx \tau /H$. For metals $\tau /H$ is roughly 1/6, about 0.2. Measured values run from about 0.1 to 1, because junctions grow as they are sheared and hard bumps plough grooves in the softer surface. The work done ends up as vibrations of the atoms in the lattice, that is, as heat. Lifting the slider over the bumps cannot be the cause, since the energy spent climbing a bump would come back as the slider drops down its far side.

Stick–slip and rolling

Two everyday effects follow: one from the gap between the two coefficients, one from avoiding sliding altogether.

Because ${\mu }_s>{\mu }_k$, a slider driven through anything springy tends to move in jerks. It sticks while the pull of the spring builds up to ${\mu }_{s}N$. Then it slips, friction drops to ${\mu }_{k}N$, and the slider lurches forward, slackening the spring, until it sticks again. This stick–slip motion makes doors squeak, keeps a bowed violin string vibrating, and lets faults in the Earth's crust lock for centuries and then slip in earthquakes.

A rolling wheel avoids sliding. Its contact point is momentarily at rest on the ground, so static friction acts there but does no work. Only rolling resistance remains, with coefficients near 0.001 for a train wheel and 0.01 for a car tyre.

Worked problem: the block sent back up

At the foot of the animation's slope the block moves at 7.06 m/s. A spring bumper there sends it back up at the same speed; take ${\mu }_k=0.2$ and ${\mu }_s=0.3$. How far up does it go, and how fast does it return?

Going up, gravity and friction both point down the slope, so the block slows at
$g(\sin \theta +{\mu }_k\cos \theta )=9.81(0.5+0.2\cdot 0.866)\approx 6.60m/{\mathrm{s}}^2$
At a steady deceleration the speed falls to zero after $t=v/a=7.06/6.60\approx 1.07\mathrm{s}$, and from $v^2=2ad$ the block has climbed $d=7.06^2/(2\cdot 6.60)\approx 3.78\mathrm{m}$ up the slope.

It cannot stay there: $\tan 30^{\circ }\approx 0.577$ exceeds ${\mu }_s$, so static friction cannot hold it. It slides back at the animation's 3.21 m/s² and reaches the bumper after $\sqrt{2\cdot 3.78/3.21}\approx 1.53\mathrm{s}$, at $\sqrt{2\cdot 3.21\cdot 3.78}\approx 4.9m/s$.

The descent takes longer than the climb because its acceleration is smaller. Both legs cover the same $d$, so $v^2=2ad$ makes the returning speed $\sqrt{3.21/6.60}\approx 0.70$ times the launch speed, and the kinetic energy 0.49 times. The rest became heat over the 7.55 m travelled: friction of $g{\mu }_k\cos \theta \approx 1.70\mathrm{N}$ per kilogram over 7.55 m is 12.8 J per kilogram, which matches the kinetic energy lost.

With down the slope positive, $v(t)$ is two straight lines meeting at $v=0$, with slopes of 6.60 and 3.21 m/s². The velocity is continuous, but the acceleration jumps, because friction reverses with the motion. Each bounce repeats the factor 0.70, and every time in a trip is proportional to its launch speed, so each round trip lasts 0.70 times the one before. The first takes $1.07+1.53=2.60\mathrm{s}$, so the ever shorter trips form a geometric series lasting $2.60/(1-0.697)\approx 8.6\mathrm{s}$ in all.

Limits of validity

Newton's mechanics gives way near the speed of light or on atomic scales. The friction law has no such single boundary: it fails whenever the surfaces stop behaving like an average dry pair. Clean metals in a vacuum, with no oxide layer between them, bond as one piece: $\mu$ rises above 1 and parts can weld cold, a real concern on spacecraft. Tables of $\mu$ for steel on steel really describe oxide films and grease.

Rubber flows into the roughness of the road, so its grip depends on area, speed and temperature, and racing tyres exceed $\mu =1$. Very flat surfaces such as polished glass cling by adhesion even without a load. A skate on ice has ${\mu }_k\approx 0.02$ because a thin liquid-like layer lubricates it. At high sliding speeds the contact tips flash to hundreds of degrees, and $\mu$ drops.

The point model has a limit of its own: a real block can topple instead of sliding. On a slope it tips over the lower edge once $\tan \theta$ exceeds its length along the slope divided by its height, and it slips once $\tan \theta$ exceeds ${\mu }_s$. So a block more than $1/{\mu }_s$ times as tall as it is long, twice as tall for ${\mu }_s=0.5$, tips over before it slips.

Where this leads

In Work, Energy and Power, kinetic friction is the standard non-conservative force, its work $-{\mu }_{k}Ns$ becoming heat. In Circular Motion, static friction holds a car on a flat bend up to $v=\sqrt{{\mu }_{s}gr}$, and banked curves reuse this lesson's components. Rotational Dynamics and Torque explains rolling: a solid ball rolls down the 30° slope at $\frac{5}{7}g\sin \theta \approx 3.50m/{\mathrm{s}}^2$, faster than the sliding block, losing nothing to friction. Static Equilibrium and Levers handles ladders and toppling; Simple Machines: Levers, Pulleys and Gears turns the ramp into the wedge and the screw, which holds its load because its thread angle is below the friction angle; and Drag and Terminal Velocity covers fluid friction, which grows with speed.