Friction: Static, Kinetic & the Inclined Plane

You’ll be able to

Distinguish static friction as an inequality from kinetic friction as an equality
Resolve forces on an incline into components along and perpendicular to the surface
Determine the angle at which sliding begins from the coefficient of static friction

Static friction is an inequality

This is the distinction that costs the most points. Kinetic friction has a fixed magnitude once sliding occurs: f_k = μ_k N. Static friction is whatever it needs to be to prevent sliding, up to a maximum: f_s ≤ μ_s N. A block at rest on a surface with a 10 N push applied and a maximum static friction of 30 N experiences 10 N of friction, not 30. Writing f_s = μ_s N for a stationary object that is not on the verge of slipping is simply false.

The two friction laws

f_s ≤ μ_s N f_k = μ_k N with generally μ_s > μ_k

Both are empirical approximations, not fundamental laws. Neither depends on contact area, and the normal force N is rarely equal to mg except on a level surface with no vertical applied force.

Choosing axes on an incline

On an incline, tilt your coordinate system so x runs along the surface and y perpendicular to it. Then the normal force and the friction force each lie along one axis, and only gravity needs resolving: the component along the incline is mg sin θ and the component into the surface is mg cos θ. With no acceleration perpendicular to the surface, N = mg cos θ — which is smaller than mg, and shrinks as the incline steepens.

The angle of repose

Tilt the incline until the block is just about to slide. At that instant static friction is at its maximum, so mg sin θ = μ_s mg cos θ, and the mass cancels entirely: tan θ_max = μ_s. This gives a direct experimental method for measuring μ_s — tilt until it slips and take the tangent — and it explains why the result does not depend on how heavy the block is, a fact students frequently find counterintuitive and examiners frequently test.

Worked example

A block slides down a 30° incline with μ_k = 0.20. Find its acceleration. Then find the angle at which a block with μ_s = 0.60 would just begin to slide.

1.Along the incline: mg sin θ − μ_k N = ma, with N = mg cos θ.
2.Substituting: a = g(sin θ − μ_k cos θ), and the mass cancels.
3.a = 9.8(sin 30° − 0.20 cos 30°) = 9.8(0.500 − 0.173) = 9.8(0.327) ≈ 3.2 m/s².
4.For impending motion, tan θ_max = μ_s = 0.60, so θ_max = arctan(0.60) ≈ 31°.

Answer: a ≈ 3.2 m/s² down the incline; sliding begins at θ ≈ 31° for μ_s = 0.60

Checkpoint

A 5.0 kg crate sits at rest on level ground with μ_s = 0.40. A horizontal push of 12 N is applied and the crate does not move. The friction force on the crate is —

On the exam

N = mg only on a horizontal surface with no vertical component of applied force. On an incline N = mg cos θ; with a push at an angle, N changes; in an accelerating elevator, N changes. Deriving N from the perpendicular equilibrium condition every time costs ten seconds and prevents a whole family of errors.

Checkpoint

The angle at which a block begins to slide down an incline depends on —

Answer the 2 checkpoints as you read.