Static friction and sliding friction
You will be able to: Decide whether friction adjusts to prevent slipping or follows the kinetic model.
Why can a small push fail to move a box?
Push a box gently with 2 N and it stays still. Push with 4 N and it may still stay still. Friction can match both pushes; it does not always have one fixed magnitude.
A useful starting point: Free-body diagrams →
Words and symbols before equations
- Static friction f_s
- Contact force when surfaces are not slipping relative to each other.
- Kinetic friction f_k
- Contact force while surfaces slide relative to each other.
- Coefficient μ
- Dimensionless number describing the surface pair in this model.
- Threshold
- Largest static friction available before slipping begins.
What this picture assumes
Every setting is a fresh trial from rest, not a history-dependent sliding animation. m=2 kg, μs=0.5, μk=0.3, g=10 m/s². At exactly 10 N, limiting static equilibrium remains possible.
Connect the picture to the physics
Static friction adjusts in magnitude and direction as needed, up to μ_s N. First calculate the force needed to avoid slipping; then compare it with the maximum. Equality is the limiting case, not the rule for every stationary object.
Once sliding, the ideal dry-friction model gives f_k=μ_k N opposite relative sliding. The normal force is determined by the other forces and geometry; it is not always mg.
Friction opposes relative sliding or its tendency at a contact, not necessarily the object’s velocity relative to the ground. Static friction can propel a walking person forward. This simple model assumes fixed surface coefficients; real surfaces may behave more intricately.
| Feature | Static | Kinetic |
|---|---|---|
| Relative sliding | Absent | Present |
| Magnitude | Adjusts up to μsN | μkN in this model |
| Direction | Prevents slipping tendency | Opposes relative sliding |
A worked example, step by step
A 2 kg box is on a level floor with μ_s=0.5 and μ_k=0.3. Use g=10 m/s². Compare a 6 N horizontal push from rest with a 12 N push from rest.
- Vertical balance gives N=20 N. Maximum static friction is 0.5×20=10 N.
- At 6 N, static friction is 6 N opposite the push: no slipping, zero acceleration.
- At 12 N, static friction cannot hold. Once sliding, friction is 0.3×20=6 N.
- Net force is 12−6=6 N, so acceleration is 6/2=3 m/s² in the push direction.
f_s=μ_s N is only the maximum static friction, not its automatic value.
If a stationary box needs 3 N to balance a push and its limit is 10 N, how much friction acts?
Compare with an explanation
3 N. The 10 N value is a limit, not the actual force required.
Predict. Change one thing. Explain.
Each slider setting starts a fresh trial from rest. Keep m=2 kg, μ_s=0.5 and μ_k=0.3. Increase the push from 6 to 12 N. Locate the threshold; explain the friction change at the start of sliding.
Net horizontal force 0 N; acceleration 0 m/s² (right positive). Force arrows share a scale of 4.75 drawing units/N. No slip in this trial.
Every setting is a fresh trial from rest, not a history-dependent sliding animation. m=2 kg, μs=0.5, μk=0.3, g=10 m/s². At exactly 10 N, limiting static equilibrium remains possible.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use a force or motion relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA 4 kg box has μ_s=0.4 and μ_k=0.25 on a level floor. (a) Find N, (b) find maximum static friction, (c) determine friction for a 10 N push from rest, and (d) find acceleration for a 20 N push once sliding.
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Compare with the answer and four-point rubric
- 1 point: 40 N.
- 1 point: 16 N.
- 1 point: 10 N opposite the push, because 10<16.
- 1 point: Kinetic friction 10 N; a=(20−10)/4=2.5 m/s².
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Is static friction always maximum?
No; it adjusts up to its limit.
RECALL 2What does μ measure?
A dimensionless property of the surface pair in the chosen model.
RECALL 3Can friction point forward?
Yes, such as static friction on a foot during walking.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Static friction and sliding friction
- 0≤|f_s|≤μ_s N; |f_k|=μ_k N while sliding.
- On a level floor with only horizontal other forces and no vertical acceleration, N=mg.
Remember: f_s=μ_s N is only the maximum static friction, not its automatic value.
Conditions: Every setting is a fresh trial from rest, not a history-dependent sliding animation. m=2 kg, μs=0.5, μk=0.3, g=10 m/s². At exactly 10 N, limiting static equilibrium remains possible.
Refresh Kid · Unit 2 · Objectives 2.7.A, 2.7.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 2.7, objectives 2.7.A, 2.7.B. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. Fall-2026 corrections also checked. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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