Static and Kinetic Friction

Core idea

Friction depends on contact and relative motion

Friction is an interaction between a pair of contacting surfaces. The coefficients belong to that pair, not to one object by itself. Decide first whether the surfaces remain stuck or slide relative to one another, then choose the model.

Static friction prevents attempted relative motion and adjusts from zero up to a maximum: 0 ≤ f_s ≤ μ_sF_N. Kinetic friction acts after sliding begins with the model f_k = μ_kF_N. In both cases, friction points opposite the relative sliding or attempted sliding at the contact.

Guided lesson path

Start with a horizontal block, then turn the surface

1. Identify the surface pair. Moving a block from wood to ice can change both friction coefficients because the contacting materials changed.
2. Find the normal force. A 4 kg block on a level surface with no other vertical force has F_N = mg = 39.2 N.
3. Check for attempted sliding. With no horizontal force, actual static friction is zero. It is not automatically at its maximum.
4. Calculate the static limit. For μ_s = 0.50, f_s,max = μ_sF_N = 19.6 N.
5. Apply an 8 N push. Because 8 N is below the limit, static friction is 8 N left and the block remains at rest.
6. State the static-friction rule. The coefficient gives an upper limit. The actual value is whatever is needed to prevent relative motion, from zero through that limit.
7. Exceed the limit. A 24 N push is too large for 19.6 N of static friction, so the block begins sliding right.
8. Switch to kinetic friction. With μ_k = 0.25, sliding friction is 9.8 N left. The 24 N push leaves 14.2 N net force and a_x = 3.55 m/s² right.
9. Compare starting and sliding. The maximum static friction is 19.6 N, while kinetic friction is 9.8 N. Starting the slide requires more force than maintaining it.
10. Keep a sliding block at constant speed. A block already moving right at 2 m/s needs a 9.8 N applied force right to balance 9.8 N kinetic friction left.
11. Press down on the block. A 12 N downward force raises the normal force from 39.2 N to 51.2 N, so kinetic friction becomes 12.8 N.
12. Move to a rough incline. On a 20° ramp, gravity attempts to slide a 4 kg block downhill, so static friction points up the ramp when it prevents that attempted motion.
13. Resolve the normal force. Only the perpendicular weight component presses into the ramp: F_N = mg cos20° ≈ 36.84 N, less than the 39.2 N weight.
14. Find actual static friction. The downhill component is mg sin20° ≈ 13.41 N. Because this is below the 18.42 N static limit, actual friction is 13.41 N up the ramp and acceleration is zero.
15. Derive the critical angle. At the slipping threshold, mg sinθ_c = μ_smg cosθ_c, so tanθ_c = μ_s. For μ_s = 0.50, θ_c ≈ 26.6°.
16. Test mass independence. Mass and g cancel from the critical-angle equation. Doubling mass changes force magnitudes but not the angle at which slipping begins.
17. Push a sliding crate up a ramp. For a 4 kg crate sliding down a 45° ramp, an 8 N up-ramp push and 6.93 N kinetic friction still leave 12.79 N down-ramp net force, or about 3.20 m/s² down.
18. Use the complete strategy. Choose the object, find the normal force, classify sticking or sliding, set the friction direction from relative motion, and then apply Newton’s Second Law along the chosen axis.

Worked examples

Separate the limit from the actual force

For the level 4 kg block:

F_N = mg = (4.0 kg)(9.8 m/s²) = 39.2 N
f_s,max = μ_sF_N = (0.50)(39.2 N) = 19.6 N

With a 24 N push right after sliding begins:

f_k = (0.25)(39.2 N) = 9.8 N
ΣF_x = 24 N − 9.8 N = 14.2 N; a_x = 14.2/4.0 = 3.55 m/s²

With a 12 N downward press while sliding:

F_N = 39.2 N + 12 N = 51.2 N; f_k = (0.25)(51.2 N) = 12.8 N

Friction on an incline

Use ramp components before choosing friction

On a ramp, resolve weight into mg sinθ down the surface and mg cosθ into the surface. The normal force is usually F_N = mg cosθ when no other perpendicular force acts. The required static friction is the amount needed to balance the parallel forces, not automatically μ_sF_N.

After sliding begins on the 35° ramp, the downhill net force is mg sin35° − μ_kmg cos35°. Static and kinetic friction are different models, so do not keep using the static limit after the block has slipped.

Normal force

Friction changes when the contact force changes

Friction depends on F_N, not directly on the block’s speed. A downward applied force increases F_N and therefore increases kinetic friction. An upward force can reduce contact force and may eventually lift the block off the surface, at which point the contact-friction model no longer applies.

On an incline, the normal force is set by the perpendicular force balance. This is why a ramp’s normal force is smaller than mg and why changing the ramp angle changes both the friction limit and the sliding acceleration.

For teachers

Make students classify the contact first

Begin with the level block and ask students to predict the actual friction for 0 N and 8 N before showing the maximum. Then exceed the 19.6 N limit with a 24 N push and have them switch explicitly to μ_kF_N.

Use the downward-press trial to separate speed from normal force, then move to the 20° incline. Require students to calculate the required static friction and compare it with its maximum before deciding whether the block moves. Finish with θ_c = tan⁻¹(μ_s) and a sliding crate with an applied up-ramp force.

Continue with Free-Body Diagrams for contact-force models, Inclined Planes for ramp components, and the Critical Angle and Friction experiment for the full simulation workflow.