PHYSICS · LESSON 05 OF 9
Friction and inclined surfaces
Friction is the contact force along a surface. Static friction adjusts up to a limit; kinetic friction has a fixed size, μₖN, and opposes sliding.
What this lesson explains
Friction lets you walk, lets car tyres grip and brakes stop, and keeps loads from sliding off conveyors. It also wastes energy in machines. Inclined planes appear in ramps, conveyors, roads and chutes; resolving forces along and perpendicular to a slope is a skill used throughout mechanics.
Before you begin
Free-body diagrams and ΣF = ma
Draw only forces acting on the object; resolve along chosen axes; apply ΣFx = max, ΣFy = may (see Newton laws and force diagrams).
Resolving on a slope
The angle between the weight and the perpendicular to the slope equals the slope angle θ. So the components are mg sin θ (along) and mg cos θ (perpendicular). Check with θ = 0: nothing along, all perpendicular.
The idea, made visible
When two surfaces are pressed together, the contact force has a part perpendicular to the surface (the normal force N) and a part along it (friction). Friction opposes relative sliding — or, if the surfaces are not sliding, the tendency to slide.
Static friction acts when there is no sliding. It is not a fixed number: it takes whatever value (from 0 up to a maximum μsN) is needed to prevent slipping. Push a heavy box gently and it does not move: static friction exactly matches your push. Push harder than μsN and the box starts to slide.
Kinetic friction acts once sliding occurs: fk = μkN, directed opposite to the velocity of sliding. Usually μk < μs, which is why it is harder to start something sliding than to keep it sliding. The coefficients depend on the pair of materials, not (to a good approximation) on the contact area or the speed.
On an incline of angle θ, choose axes along the slope and perpendicular to it. Then only the weight needs resolving: mg sin θ acts down the slope and mg cos θ acts into the slope. With no other perpendicular forces, N = mg cos θ, not mg.
A block rests on an incline without sliding as long as mg sin θ ≤ μsmg cos θ, that is tan θ ≤ μs. The angle at which it just starts to slide satisfies tan θ = μs; notice that the mass cancels.
| surfaces | μs | μk |
|---|---|---|
| rubber on dry concrete | 1.0 | 0.8 |
| steel on steel (dry) | 0.74 | 0.57 |
| wood on wood | 0.25–0.5 | 0.2 |
| rubber on wet concrete | 0.3 | 0.25 |
| ice on ice | 0.1 | 0.03 |
Treat tabulated coefficients as rough guides. In problems, use the values given.
Key terms
- Static friction fs
- Friction when surfaces do not slide: 0 ≤ fs ≤ μsN, with whatever direction and size prevents slipping.
- Kinetic friction fk
- Friction during sliding: fk = μkN, opposite to the direction of sliding.
- Coefficient of friction μ
- A dimensionless number describing a pair of surfaces; μs for static, μk for kinetic.
- Angle of repose
- The steepest incline angle on which an object stays at rest: tan θ = μs.
The formulas and what they mean
| Symbol | Meaning | Unit |
|---|---|---|
| μs, μk | static and kinetic friction coefficients | no unit |
| N | normal force | N (newton) |
| fs, fk | static and kinetic friction forces | N |
| θ | incline angle above horizontal | ° |
Static friction
fs ≤ μsN
Conditions and limits: An inequality: use fs = μsN only at the point of slipping. Otherwise find fs from ΣF = 0.
Kinetic friction
fk = μkN
Conditions and limits: Surfaces sliding. Direction opposite to the relative velocity. Approximately independent of speed and contact area.
Incline components
along the slope: mg sin θ; perpendicular: mg cos θ
Conditions and limits: θ measured from the horizontal.
Sliding down with friction
a = g(sin θ − μk cos θ)
Conditions and limits: Block sliding down, no other forces. If the result is negative, the block decelerates (and eventually stops).
Onset of sliding
tan θmax = μs
Conditions and limits: Block at rest on an incline with no other forces along the slope.
Why it works: Acceleration of a block sliding down a rough incline
Axes: x down the slope (the direction of motion), y perpendicular to it.
y: N − mg cos θ = 0 ⇒ N = mg cos θ.
No acceleration perpendicular to the slope.
Friction: fk = μkN = μkmg cos θ, pointing up the slope.
Opposes the sliding.
x: mg sin θ − μkmg cos θ = ma.
Down-slope component of weight minus friction.
a = g(sin θ − μk cos θ).
The mass cancels.
With θ = 30° and μₖ = 0.20: a = 9.8(0.500 − 0.173) = 3.20 m/s². Heavy and light blocks of the same material slide down with the same acceleration.
A first worked example
Friction problems
- Draw the FBD. Decide whether the surfaces slide (kinetic) or not (static), and in which direction motion occurs or tends to occur. Friction points opposite.
- Find N from the perpendicular equation — do not assume N = mg.
- Kinetic: fk = μkN. Static: first find the friction needed for equilibrium, then compare it with μsN. If the need exceeds μsN, the object slides.
- Apply ΣF = ma along the motion.
- Check: does the answer make sense when μ = 0 or θ = 0?
Kinetic friction on a level floor
Problem. A 20 kg crate is pushed across a floor by a horizontal 80 N force. μk = 0.30. Find the friction force and the acceleration (g = 9.8 m/s2).
N = mg = 20 × 9.8 = 196 N.
Horizontal push: no other vertical forces, so here N = mg.
fk = 0.30 × 196 = 58.8 N, opposite to the motion.
Kinetic friction formula.
a = 80 − 58.8 / 20 = 1.06 m/s2.
Net force divided by mass.
Result: fk = 58.8 N; a ≈ 1.06 m/s2.
What it means: If the push were exactly 58.8 N the crate would slide at constant velocity.
A different case
Does it move? (static friction)
Problem. The same 20 kg crate is at rest; μs = 0.50. A horizontal push of 60 N is applied. Does it move, and what is the friction force?
Maximum static friction: μsN = 0.50 × 196 = 98 N.
The largest friction the floor can supply without slipping.
The push 60 N is less than 98 N, so the crate stays at rest.
Compare the need with the limit.
Equilibrium: fs = 60 N (not 98 N).
Static friction only supplies what is needed.
Result: It does not move; fs = 60 N.
What it means: Writing f = μₛN = 98 N would predict a 38 N net force backwards on a crate at rest — impossible.
More worked cases
Each case below uses a different skill. Every step and result is shown.
Sliding down an incline
Problem. A block slides down a 30° incline with μk = 0.20. Find its acceleration.
a = g(sin θ − μk cos θ).
Derived above; mass cancels.
= 9.8(0.500 − 0.20 × 0.866) = 9.8 × 0.327.
Substitute.
Result: a ≈ 3.20 m/s2 down the slope.
What it means: Without friction it would be g sin 30° = 4.9 m/s².
Angle at which sliding starts
Problem. A box rests on a board whose angle is slowly increased. μs = 0.40. At what angle does it start to slide?
On the point of slipping: mg sin θ = μsmg cos θ.
Down-slope pull equals maximum static friction.
tan θ = μs = 0.40 ⇒ θ = arctan 0.40.
Divide by mg cos θ.
Result: θ ≈ 21.8°.
What it means: This gives a simple experiment to measure μₛ.
Net force with friction
Problem. An object slides on a horizontal surface. N = 60 N, μk = 0.2, and a forward applied force is 30 N. Find the net forward force.
fk = 0.2 × 60 = 12 N backwards.
Opposes sliding.
ΣF = 30 − 12 = 18 N forward.
Forward positive.
Result: 18 N forward.
What it means: Friction reduces the effect of the applied force; it does not depend on the size of the push.
Common misunderstandings
Misunderstanding: Always writing f = μN for static friction.
Correct idea: Static friction is ≤ μₛN. Use the equality only at the point of slipping.
Misunderstanding: Taking N = mg on an incline.
Correct idea: On an incline N = mg cos θ (if no other perpendicular forces).
Misunderstanding: Friction always points backwards.
Correct idea: It opposes relative sliding. A crate on an accelerating truck bed is pushed forward by static friction.
Misunderstanding: Swapping sin and cos on the incline.
Correct idea: Check θ = 0: the along-slope component must vanish, so it is mg sin θ.
Misunderstanding: Friction depends on contact area.
Correct idea: In the simple model it depends only on μ and N.
Going further (optional): Air resistance and terminal speed (check whether your outline includes it)
A body moving through air or water feels a drag force opposite to its velocity, and drag grows with speed. A falling object therefore speeds up less and less: once drag equals the weight, the net force is zero and the speed stops increasing. That constant speed is the terminal speed.
In the simple linear model Fdrag = bv, terminal speed satisfies mg = bvt, so vt = mg/b. For m = 1.0 kg and b = 2.0 N·s/m (with g = 9.8 m/s2), vt = 9.8/2.0 = 4.9 m/s. Many real objects follow a quadratic law (drag ∝ v²) instead; the model must match the situation.
Keep in mind
- fs ≤ μsNStatic friction
- fk = μkNKinetic friction
- along the slope: mg sin θ; perpendicular: mg cos θIncline components
- a = g(sin θ − μk cos θ)Sliding down with friction
- tan θmax = μsOnset of sliding
Scope of this lesson
- Coulomb’s friction model (constant μ) is an approximation. Rolling resistance, lubrication and air drag need other models.
Next: Work energy and power. Work transfers energy. Kinetic and potential energy trade places, and the total is conserved unless friction or other forces add or remove energy. Power is the rate of energy transfer.
Original study text. Sources and credits.