Friction and Inclined Planes

Rest a book on a table and tip the table gently. For a while the book just sits there — something is quietly holding it against the pull of gravity. Tip it a little more and, at some critical steepness, the book suddenly lets go and slides. What holds it, what lets go, and how fast it then races down are the whole story of this page.

You have already learned to resolve a force into components and you know that friction opposes sliding. Here we put the two together with A-level rigour: we model friction with a single number, the coefficient of friction, and we analyse the classic set-piece of mechanics — a block on an inclined plane. By the end you will be able to say, for any slope and any block, whether it stays or slides, and if it slides, exactly what acceleration it has.

Modelling friction: the coefficient \mu

Friction is really billions of tiny contact points snagging on each other — hopelessly complicated in detail. The triumph of the A-level model is to hide all of that behind one experimental fact: the harder two surfaces are pressed together, the more they grip. "How hard they are pressed" is measured by the normal contact force N — the push the surface gives at right angles to itself. The grip is proportional to it, and the constant of proportionality is the coefficient of friction \mu, a pure number fixed by the pair of surfaces (rubber on tarmac \approx 0.8, steel on ice \approx 0.03).

The subtlety is that friction is not always \mu N. While an object is still at rest, static friction is a self-adjusting force: it grows to exactly cancel whatever is trying to push the object along, but only up to a ceiling. Push harder than that ceiling and the object breaks away and begins to slide.

F_{\text{static}} \le \mu_s N, \qquad F_{\max} = \mu_s N.

Once it is sliding, kinetic (dynamic) friction takes over — and, conveniently, it is simply F_k = \mu_k N, a steady drag against the motion (usually a touch smaller than the static ceiling, which is why a stuck drawer jerks free then glides). At A-level we most often use a single \mu for both and write the model as a limit.

Resolving weight on a slope

Now put a block of mass m on a plane inclined at angle \theta to the horizontal. Its weight mg still points straight down — gravity does not care about the slope. But for a slope the clever axes are not horizontal and vertical; they are along the slope and perpendicular to it. Resolve the weight into those two directions.

The angle between the weight (vertical) and the perpendicular-to-slope direction is itself \theta (rotate both by \theta from vertical and the vertical–horizontal pair). So the weight splits into a part pressing into the slope and a part pulling the block down the slope:

\underbrace{mg\cos\theta}_{\text{perpendicular, into the slope}}, \qquad \underbrace{mg\sin\theta}_{\text{along the slope, down it}}.

Perpendicular to the slope the block does not accelerate (it neither sinks into the plane nor lifts off), so those forces balance: the normal contact force must support the perpendicular part of the weight,

N = mg\cos\theta.

This is the single most important — and most forgotten — line on the page. On a slope the normal force is not mg; it is mg\cos\theta, which shrinks as the slope steepens. And since the maximum friction is \mu N, the grip available is \mu m g\cos\theta — it too falls away as the slope gets steeper, just as the down-slope pull mg\sin\theta is growing. That tug-of-war decides everything.

These are the traps that sink most inclined-plane answers:

See it move: the tug-of-war on a slope

Below is a block resting on a ramp. Drag the slope angle and the coefficient of friction. The weight mg (down) is drawn together with its two dashed components — mg\sin\theta down the slope and mg\cos\theta into it — alongside the normal force N = mg\cos\theta and the friction force. Tilt the ramp gently and friction holds the block; keep tilting and, the moment mg\sin\theta beats the maximum grip \mu N, the block breaks free and the figure reports its acceleration a = g(\sin\theta - \mu\cos\theta).

Stay or slide? The slipping condition

Line up the two forces acting along the slope. Pulling the block down: mg\sin\theta. Available to hold it: friction, up to a maximum F_{\max} = \mu N = \mu m g\cos\theta. The block stays at rest exactly while the down-slope pull does not exceed the grip on offer:

mg\sin\theta \le \mu\, mg\cos\theta.

The mass m and gravity g cancel from both sides — a beautiful result: whether a block slides does not depend on how heavy it is. Divide by \cos\theta and it collapses to a statement about the angle alone:

\tan\theta \le \mu \quad\Longrightarrow\quad \text{stays};\qquad \tan\theta > \mu \quad\Longrightarrow\quad \text{slides}.

The steepest angle at which the block is still on the point of resting — where \tan\theta = \mu — is called the angle of repose (or angle of friction). Below it, everything sticks; above it, everything of any weight slides. That is why \mu can be measured with nothing more than a tilting board and a protractor: tilt until the object just slips, read off \theta, and \mu = \tan\theta.

If it slides: the acceleration down the slope

Once \tan\theta > \mu the block is moving, so friction is now kinetic and equal to \mu N = \mu m g\cos\theta, directed up the slope (against the motion). Apply Newton's second law along the slope, taking down-slope as positive:

\underbrace{mg\sin\theta}_{\text{down}} - \underbrace{\mu m g\cos\theta}_{\text{friction, up}} = ma.

The mass cancels once more, leaving the acceleration in its standard form:

a = g\left(\sin\theta - \mu\cos\theta\right).

Two sanity checks fall straight out. Set \mu = 0 (a frictionless slope) and you recover a = g\sin\theta — the pure resolved-gravity result. And set \tan\theta = \mu (the angle of repose) and the bracket is zero, so a = 0: the block is exactly on the verge, neither held firmly nor accelerating. Everything joins up.

Pour dry sand, sugar or gravel into a heap and it always builds to a cone with the same steepness, however much you pour. That steepness is precisely the angle of repose: a grain on the surface sits still only while \tan\theta \le \mu, so the pile grows until its sides reach \theta = \arctan\mu and then any extra grain simply tumbles down. For dry sand that angle is about 34^\circ; it is why a well-designed hourglass keeps such steady time, and why grain silos and mine tips are engineered around it.

Snow-covered mountainsides play the same game with lethal stakes. A slab of snow clings to the slope while the friction holding it exceeds the down-slope pull. Fresh snowfall, a warming thaw that lubricates the layer beneath (dropping the effective \mu), or a skier's added load can tip mg\sin\theta past \mu N — and the whole slab lets go at once. Most avalanches release on slopes of 30^\circ–45^\circ: gentler and the snow never slips, steeper and it never builds up enough to be dangerous.

Worked examples

Example 1 — will it slip? A crate rests on a ramp inclined at 20^\circ. The coefficient of friction between crate and ramp is \mu = 0.5. Does it slide? Compare \tan\theta with \mu:

\tan 20^\circ \approx 0.364 \; < \; 0.5 = \mu,

so mg\sin\theta < \mu mg\cos\theta — the grip wins and the crate stays put. (Note we never needed the mass.) The actual friction holding it is only mg\sin 20^\circ, comfortably below the maximum \mu mg\cos 20^\circ on offer.

Example 2 — find the acceleration. A 4\ \text{kg} block is on a slope where \sin\theta = 0.6 and \cos\theta = 0.8 (a 3-4-5 slope), with \mu = 0.25. Take g = 10\ \text{m/s}^2. First the normal force and the two along-slope forces:

N = mg\cos\theta = 4\times 10\times 0.8 = 32\ \text{N}, mg\sin\theta = 4\times 10\times 0.6 = 24\ \text{N}, \qquad \mu N = 0.25\times 32 = 8\ \text{N}.

Since 24 > 8 it slides. Newton's second law along the slope gives

a = g(\sin\theta - \mu\cos\theta) = 10(0.6 - 0.25\times 0.8) = 10(0.6 - 0.2) = 4\ \text{m/s}^2.

Example 3 — the angle of repose. On a tilting board a coin just begins to slip when the board reaches 31^\circ. What is the coefficient of friction? At the point of slipping \tan\theta = \mu, so

\mu = \tan 31^\circ \approx 0.60.

No masses, no forces — just an angle. This is exactly how \mu is measured in the laboratory.