Inclined Plane and Friction

Earth pulls the block down the slope and friction holds it back. Turn the angle α and the roughness μ — and watch the moment friction gives in and the block starts to slide. The diagram shows everything: the forces, their components, the threshold angle and the numbers.

The idea in 30 seconds

Earth pulls down — the slope “splits” that force

Gravity mg points straight down. On a slope it is convenient to split it into two parts: mg·sinα pulls the block down along the slope, and mg·cosα presses it into the slope. The steeper the slope, the bigger the first part and the smaller the second.

Friction answers exactly as hard as needed

While the block stays, static friction equals the pull: F_fr = mg·sinα — like pulling a wardrobe that does not budge. But friction has a ceiling μN = μ·mg·cosα. The rougher the surface and the harder the body is pressed, the higher the ceiling.

The threshold: tan α = μ — and mass does not matter

The block slides as soon as mg·sinα exceeds μ·mg·cosα, that is, when tan α > μ. The mass m is in both parts and cancels: a light and a heavy block start at the same angle and move with the same acceleration a = g(sinα − μ·cosα).

Try it yourself

Choose the “Wood” surface (μ = 0.40) and slowly drag the top of the slope up. The block starts to move at α ≈ 21.8° — that is α_crit = arctan μ. Check it against the yellow dashed line.
Press “m ×2”: the forces N, mg·sinα and F_fr double, but the threshold and the acceleration stay the same. The mass cancelled out!
Choose “Ice” (μ = 0.05): the slope is nearly flat and yet the block already slides. Choose “Rubber” (μ = 0.80): the block moves only at α > 38.7°.
Give a push v₀ at a small angle, α < α_crit: the block slides a little and stops — sliding friction brakes it. At α just below α_crit it brakes so gently that it seems to move uniformly.

Where you meet it

Black ice: ice has a small μ (≈ 0.05–0.1), so a car or a pedestrian slips on a slope of just 3–6°, while on dry asphalt (μ ≈ 0.7) you can stand on a slope of up to 35°.
Angle of repose: dry sand cannot lie steeper than α = arctan μ. That is why piles of sand, gravel and grain always form a cone with the same base angle.
Skiers wax their skis to reduce μ and go faster, while drivers fit studded winter tyres to increase μ and not slip.

Graphs: what changes and how

The two upper graphs show how the forces and the acceleration depend on the angle α. The two lower ones are drawn live during the descent, and the grey dashed line is the formula from the start: the simulator’s points fall exactly on it.

Forces and angle: where the pull overtakes the friction ceiling

Acceleration and angle: zero up to α_crit, then it grows

Speed over time v(t)

Distance over time s(t)

α slope angle 30°

Slope surface

μ friction coefficient 0,30

m mass of the block 2,0 kg

v₀ push down the slope 0,0 m/s

Try it in one click

Playback speed

Show

components of mg (dashed) and the angle α at the block

threshold α_crit — the yellow “ghost” slope

distance ruler (0…4 m)

Lab: find μ of an unknown surface

For simplicity the static and sliding friction coefficients are taken equal, as in school problems. Mass affects neither the threshold α_crit nor the acceleration.

KEY FORMULAS

Splitting gravity into components

$$mg\sin\alpha,\quad mg\cos\alpha$$

Gravity on a slope: mg·sinα pulls along the slope downwards, mg·cosα presses the body into the slope

▶ Detailed Explanation

Gravity mg points straight down and is awkward to use on a slope as it is — the slope works along and across itself. So it is split into two perpendicular components: along the slope mg·sinα (the one that pulls the body down) and across the slope mg·cosα (the one that presses the body into the slope, and from which the normal force N arises). The angle between mg and the component mg·cosα equals the slope angle α — it is marked at the block on the diagram. The two components add up to mg again: the dashed arrows and the grey arrow form a rectangle. Example: m = 2 kg, α = 30°, g = 9.8 m/s² → mg = 19.6 N, mg·sin30° = 9.8 N, mg·cos30° ≈ 17.0 N. Extreme cases: at α = 0 (a floor) all of gravity presses, sin 0 = 0; at α = 90° (a wall) all of gravity pulls along, cos 90° = 0.

Normal force

$$N = mg\cos\alpha$$

The normal force on a slope equals the pressing component of gravity

▶ Detailed Explanation

The slope does not let the body fall through, so its reaction N points perpendicular to the surface. Along the normal the body does not accelerate (it neither sinks in nor lifts off), so the forces along that axis balance: N = mg·cosα. As the angle grows, cosα decreases: at α = 60° N = 0.5·mg, and on a vertical wall (α = 90°) N = 0 — the body does not touch the support. Example: m = 2 kg, α = 30° → N = 19.6·0.866 ≈ 17.0 N. This force sets the friction ceiling: F_fr ≤ μN. So on a steep slope friction is weaker than on a gentle one, although μ of the surface is the same.

Friction force: it adapts, but not above its ceiling

$$F_{\text{fr}} \le \mu N,\qquad F_{\text{fr}}^{\max} = \mu mg\cos\alpha$$

Static friction adapts to the pull but cannot exceed the limit μN

▶ Detailed Explanation

While the body stays, static friction is exactly as large as needed to hold it: F_fr = mg·sinα. It is not a “constant” force — it grows together with the pull. But it has a ceiling F_fr,max = μN = μ·mg·cosα: the rougher the surface (larger μ) and the harder the body is pressed (larger N), the higher the ceiling. When the body slides, sliding friction equals this ceiling: F_fr = μN. Example: m = 2 kg, α = 30°, μ = 0.3 → mg·sinα = 9.8 N, the ceiling μN = 0.3·17.0 ≈ 5.1 N. The pull is larger than the ceiling — the body slides and friction is 5.1 N. With μ = 0.8 the ceiling is 13.6 N, above 9.8 N, and the body stays with friction of 9.8 N.

Sliding condition and the critical angle

$$\tan\alpha > \mu \iff \alpha > \alpha_{\text{crit}} = \arctan\mu$$

The body slides when tan α exceeds μ; the critical angle depends only on the friction coefficient

▶ Detailed Explanation

The body starts to slide when the pull exceeds the friction ceiling: mg·sinα > μ·mg·cosα. Divide both sides by mg·cosα and get tan α > μ. The mass and g cancelled, so the critical angle depends only on the surface: α_crit = arctan μ. Examples: ice, μ = 0.05 → α_crit ≈ 2.9°; wood on wood, μ = 0.4 → α_crit ≈ 21.8°; rubber on dry asphalt, μ = 0.8 → α_crit ≈ 38.7°. This is also a way to measure μ: raise the angle slowly until the body moves — then μ = tan α_crit. At tan α = μ exactly the body is “on the edge”: it stays, but any extra angle will break it loose; if it is already moving, at tan α = μ it moves at constant speed.

Acceleration while sliding — no mass in it

$$a = g(\sin\alpha - \mu\cos\alpha)$$

Acceleration along the slope: the mass cancels and does not affect a

▶ Detailed Explanation

Newton’s second law along the slope: ma = mg·sinα − μN = mg·sinα − μ·mg·cosα. Cancelling m gives a = g(sinα − μ·cosα). That is why a 2 kg block and a 10 kg block slide down the same way. At μ = 0 (perfectly smooth) a = g·sinα, and on a vertical wall (α = 90°) a = g — free fall. Example: α = 30°, μ = 0.3 → a = 9.8·(0.5 − 0.3·0.866) ≈ 2.35 m/s². If the bracket is negative, a < 0: the body is already moving but friction is stronger than the pull — it slows down (and at the stop sliding friction gives way to static friction, a becomes 0). The expression makes sense only for a sliding body: a body at rest has a = 0.

Descent time and the speed at the end

$$t = \sqrt{\frac{2L}{a}},\qquad v = \sqrt{2aL}$$

From rest with constant acceleration: the time and the speed after the distance L

▶ Detailed Explanation

The acceleration is constant, so from rest s = at²/2 and v = at. Expressing t from the first formula (t = √(2L/a) at s = L), we get v = at = √(2aL). These are the formulas the simulator uses for its prediction (the “Measurements” tab) — and the points on the s(t) and v(t) graphs fall exactly on the grey dashed line. Example: L = 4 m, a = 2.35 m/s² → t = √(8/2.35) ≈ 1.84 s, v = √(2·2.35·4) ≈ 4.34 m/s. With an initial speed v₀ the time comes from at²/2 + v₀t = L, and the speed is v = √(v₀² + 2aL). Mass is not in these formulas: a heavy block reaches the finish at the same time as a light one.

Braking: the body moves, but friction is stronger

$$a = -g(\mu\cos\alpha - \sin\alpha) < 0,\qquad s = \frac{v_0^2}{2|a|}$$

At tan α < μ and an initial speed v₀ the body slows down and stops after the distance s = v₀²/(2|a|)

▶ Detailed Explanation

If tan α < μ, the pull is smaller than the friction ceiling. A body at rest stays at rest, but if it is pushed, sliding friction μN acts on it, and it is larger than the pull. The net force points against the motion: a = g(sinα − μ·cosα) < 0. The speed decreases as v = v₀ − |a|·t, the body stops after t = v₀/|a| and covers the distance s = v₀²/(2|a|). Example: α = 10°, μ = 0.3, v₀ = 3 m/s → a = 9.8·(0.174 − 0.3·0.985) ≈ −1.19 m/s², t ≈ 2.51 s, s ≈ 3.77 m. If the braking distance is longer than the slope (4 m in the simulator), the block manages to reach the end. After stopping on a gentle slope, sliding friction is replaced by static friction: F_fr = mg·sinα, a = 0.

Energy: where the work of gravity goes

$$mg\,s\sin\alpha = \frac{mv^2}{2} - \frac{mv_0^2}{2} + \mu mg\cos\alpha\cdot s$$

Work of gravity = gain of kinetic energy + heat from friction

▶ Detailed Explanation

Gravity does the work A = mg·sinα·s (the component mg·cosα does no work — it is perpendicular to the displacement). This work goes two ways: it speeds the body up (the gain of kinetic energy ΔEk = mv²/2 − mv₀²/2) and it heats the surface — that is the work of friction, Q = μN·s = μ·mg·cosα·s. In the simulator the three numbers are accumulated independently (the “Measurements” tab) and the balance “work − Ek − Q” is zero. Example: m = 2 kg, α = 30°, μ = 0.3, distance 4 m → A = 39.2 J, Q ≈ 20.4 J, Ek ≈ 18.8 J. Note that more than half went into heat — friction “ate” a large part of the energy. Mass enters all three terms the same way, so the acceleration does not depend on it.

What is going on. A block lies on a slope. Earth pulls it down with the force of gravity mg — always straight down. The slope answers with two forces: the normal force N (perpendicular to the slope — it does not let the block “fall through”) and the friction force F_fr (along the slope — it opposes sliding). To see which one wins, gravity is split into two parts: mg·sinα along the slope (it pulls the block down) and mg·cosα across it (it presses the block into the slope). On the diagram these parts are dashed, and together with mg they form a rectangle.

Friction “adjusts itself” — up to a limit. While the block stays, static friction balances the pull: F_fr = mg·sinα. When the slope gets steeper, friction grows with it — you can see that in the orange arrow and in the “Who is stronger” bars. But static friction has a ceiling: F_fr ≤ μN = μ·mg·cosα. As soon as mg·sinα outgrows this ceiling, friction cannot keep up — the block breaks loose and slides. Now sliding friction acts on it, equal to μN (in school problems the static and sliding coefficients are taken equal).

Threshold and acceleration. The condition for sliding is mg·sinα > μ·mg·cosα, that is tan α > μ. The angle at which this happens is called critical: α_crit = arctan μ (for μ = 0.3 it is 16.7°, for μ = 0.8 it is 38.7°). If the block slides, its acceleration is a = g(sinα − μ·cosα). There is no mass in the formula: it cancelled, so a light block and a heavy one start at the same angle and move the same way. Mass changes only the forces (N, F_fr) and the energy — see the “Measurements” tab.

What else you can see. Give the block a push v₀ down the slope at an angle below α_crit — it slides a while and stops: sliding friction is stronger than the pull, so the acceleration is negative. The braking distance s = v₀²/(2|a|) is computed by the formula and matches the simulation. The yellow “ghost slope” shows the threshold α_crit. In the “Lab” mode the surface is hidden and you must find μ yourself — from the angle at which the block starts to move: μ = tan α_crit. This is how friction coefficients are measured in practice.

How the calculation works. On every stretch the acceleration is constant, so the simulator uses the exact formulas s = v₀t + at²/2 and v = v₀ + at, and it finds the moments of events — stopping and reaching the end of the slope — by solving an equation, not by “the nearest frame”. The numbers on screen do not depend on the frame rate or the playback speed and match the formulas: the descent time from rest t = √(2L/a), the final speed v = √(2aL).