On an incline, the weight of a block splits into W·sinθ down the slope and W·cosθ into it. Friction resists sliding up to a limit of μs·N, where N is the normal force. If the force trying to slide the block is less than that limit, it stays put; at the limit, motion is impending; once it slides, friction drops to μk·N.
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Dry (Coulomb) friction
Friction acts along the contact surface, opposite to the motion or the tendency to move. Its size is whatever equilibrium needs, from zero up to a maximum F_max = μs·N. Only at impending motion can you write F = μs·N; before that, friction is an unknown found from the equilibrium equations.
Once the surfaces slide, friction is F = μk·N, with μk a little smaller than μs for most material pairs. Coefficients depend on the materials, finish, contamination and lubrication, so textbook values are only typical; use measured values for real work.
Setting up an incline problem
Tilt your axes: x along the slope and y perpendicular to it. Then the normal force and friction each have one component, and only the weight (and any horizontal push) needs resolving. Draw the free body diagram with friction pointing opposite the direction the block would move.
Check first whether the block slides on its own: it holds if tanθ ≤ μs.
Decide which way motion is impending. To hold a block that would slide down, friction points up the slope. To push it up, friction points down the slope.
If a force is not parallel to the slope, it changes N. Write ΣFy = 0 for N before using μN.
Angle of repose
Tilt a plane until a block just starts to slide. At that angle φs, the block’s weight component down the slope equals the maximum friction, which gives tanφs = μs. This tilt test is a quick way to measure the static coefficient, and it is why bulk materials pile at a characteristic slope.
Key formulas
Formulas used on this page, with the textbook each is cited from
Formula
Expression
Where
Cited from
Weight components on an incline
W∥ = W·sinθ, W⊥ = W·cosθ, W = m·g
W = weight (N), m = mass (kg), g = 9.81 m/s², θ = slope angle from horizontal (degrees)
Hibbeler, Engineering Mechanics: Statics
Static friction limit
F ≤ μs·N (F = μs·N at impending motion)
F = friction force (N), μs = coefficient of static friction (dimensionless), N = normal force (N)
Beer & Johnston, Vector Mechanics for Engineers: Statics
Kinetic friction
F = μk·N
μk = coefficient of kinetic friction (dimensionless), surfaces sliding
Beer & Johnston, Vector Mechanics for Engineers: Statics
Angle of repose (angle of static friction)
tan φs = μs
φs = slope angle at which a block on its own just begins to slide (degrees)
Meriam & Kraige, Engineering Mechanics: Statics
Worked examples
Example: Holding and pushing a crate on a ramp
Given
A 40 kg crate rests on a ramp inclined at 25°.
Assume μs = 0.35 and μk = 0.28 for this crate and ramp surface (illustrative values; measure them for a real ramp).
Any applied force P acts parallel to the ramp. g = 9.81 m/s².
Find
Whether the crate slides on its own.
The force P needed to hold it at rest.
The force P needed to push it up the ramp at constant speed.
Solution
Compare the slope with the friction coefficient:tan 25° = 0.466 > μs = 0.35 → the crate slides down unless held
Weight and its components:W = 40 × 9.81 = 392.4 N; W·sin25° = 165.8 N (down the slope), W·cos25° = 355.6 N
P is parallel to the slope, so it does not change N:ΣFy = 0: N = 355.6 N
To hold it, motion down the slope is impending, so friction acts up the slope at its maximum:P + μs·N − W·sinθ = 0 → P = 165.8 − 0.35(355.6) = 165.8 − 124.5 = 41.4 N
Pushing up at constant speed is equilibrium too (no acceleration), with kinetic friction down the slope:P − μk·N − W·sinθ = 0 → P = 165.8 + 0.28(355.6) = 165.8 + 99.6 = 265.4 N
Answer: The crate slides on its own because tan 25° exceeds μs. About 41 N up the slope holds it; about 265 N keeps it moving up at constant speed. Pushing up costs far more because weight and friction then act the same way.
Example: Range of a horizontal force that keeps a block still
Given
A 25 kg block sits on a 20° incline. Assume μs = 0.30 (illustrative).
A horizontal force P pushes the block toward the slope. g = 9.81 m/s².
Find
The smallest and largest P for which the block stays at rest.
Solution
Weight, and the slope check:W = 25 × 9.81 = 245.25 N; tan 20° = 0.364 > 0.30, so P > 0 is needed
Resolve P into the slope axes. Its normal component pushes the block into the slope and raises N:N = W·cosθ + P·sinθ
Largest P: motion up the slope is impending, friction acts down the slope. Substitute N and solve:P·cosθ − W·sinθ − μs(W·cosθ + P·sinθ) = 0 → P = W(sinθ + μs·cosθ) / (cosθ − μs·sinθ)
Numbers for the largest P:P_max = 245.25 × (0.342 + 0.282) / (0.940 − 0.103) = 245.25 × 0.624 / 0.837 = 182.8 N
Smallest P: motion down the slope is impending, friction acts up the slope:P_min = W(sinθ − μs·cosθ) / (cosθ + μs·sinθ) = 245.25 × 0.0601 / 1.042 = 14.1 N
Answer: The block stays at rest for 14.1 N ≤ P ≤ 182.8 N. Below the range it slides down; above it, it slides up. Because a horizontal push also raises N, the answer is not simply W·sinθ ± μs·W·cosθ.
Common mistakes
Writing F = μs·N when motion is not impending. μs·N is the maximum. Until the block is about to slip, find F from equilibrium and check that F ≤ μs·N.
Using N = W·cosθ when another force has a normal component. Write ΣF = 0 perpendicular to the slope with every force included, then use that N.
Pointing friction the wrong way. Friction opposes the motion that would happen without it. Holding a block on a steep ramp and pushing it up need opposite friction directions.
Using μs for a sliding body. Once it slides, use μk. “Constant speed” still means equilibrium, but with kinetic friction.
Treating textbook coefficients as design values. They vary widely with finish, moisture, oil and wear. Measure them, or design with a margin for the full plausible range.
In industry
Frequently asked questions
Why is static friction larger than kinetic friction?
Surfaces at rest settle into closer contact at their microscopic high points. Once sliding, there is less time for those contacts to form, so the resisting force drops.
Does friction depend on contact area?
In the Coulomb model, no. It depends on the normal force and the coefficient. The model works well for hard, dry surfaces, less well for soft or lubricated ones.
Does mass matter if the block slides on its own?
No. Whether it slides depends only on tanθ versus μs, because both the driving force and the friction limit are proportional to the weight.
How does this relate to screw threads?
A screw thread is an inclined plane wrapped around a cylinder. The same equations, with the lead angle as the slope, give the torque to raise or lower a load on a screw and explain why many threads are self-locking.