Sliding friction and the normal force
You will be able to: Calculate kinetic friction from the actual normal force and choose its direction from relative sliding.
Why does pulling upward change the friction on a box?
A 2 kg box slides right on a level floor. With μₖ = 0.2 and no other vertical force, the 20 N normal force gives a 4 N friction force leftward.
A useful starting point: Gravity inside a sphere and a shell →
Words and symbols before equations
- Kinetic friction fₖ
- Contact force opposing relative sliding between surfaces.
- Coefficient μₖ
- Dimensionless material-dependent parameter in this simplified model.
- Normal force N
- Perpendicular contact force, not automatically mg.
What this picture assumes
Instantaneous rightward sliding relative to a fixed floor; no vertical acceleration. Normal force remains positive throughout the controls. This is not a time simulation through stopping.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- N = 15 N; kinetic friction 3 N left; net 3 N; aₓ = 1.5 m/s² while sliding right.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
First determine N from perpendicular forces and acceleration. Then use fₖ = μₖN. On a level floor with no vertical acceleration or other vertical force, N = mg. An upward pull reduces N, provided contact remains.
Friction opposes sliding relative to the other surface. A box sliding slower than a moving conveyor can feel friction forward relative to the ground. Direction must be determined from the two surfaces, not from a remembered “always backward” rule.
The ideal dry-friction model treats fₖ as approximately independent of apparent contact area and speed. Real surfaces can deviate. When sliding stops, switch to a static-friction analysis; do not let a stopping calculation automatically continue into reverse motion with the same friction direction.
A worked example, step by step
A 2 kg box slides right with μₖ = 0.2. A pull has components 6 N right and 5 N up. Find N, friction and horizontal acceleration.
- Vertical acceleration is zero, so N + 5 − 20 = 0 and N = 15 N.
- Kinetic friction magnitude is 0.2(15) = 3 N, leftward.
- Horizontal net force is 6 − 3 = 3 N.
- aₓ = 3/2 = 1.5 m/s² right while the stated sliding continues.
Using μₖmg without checking the normal force can give the wrong friction.
If a box stops sliding, is f automatically still μₖN?
Compare with an explanation
No. Re-evaluate with static friction and the new tendency to slip.
Predict. Change one thing. Explain.
Keep the rightward sliding state and horizontal pull fixed. Increase the upward pull. Predict the normal force, friction and horizontal acceleration.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
N = 15 N; kinetic friction 3 N left; net 3 N; aₓ = 1.5 m/s² while sliding right.
Instantaneous rightward sliding relative to a fixed floor; no vertical acceleration. Normal force remains positive throughout the controls. This is not a time simulation through stopping.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant force, system boundary, acceleration or calculus 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 slides right. It is pulled 10 N right and 10 N up with μₖ = 0.2. Find normal force, friction, net horizontal force and acceleration.
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Compare with the answer and four-point rubric
- 1 point: N = 40 − 10 = 30 N.
- 1 point: fₖ = 0.2(30) = 6 N left.
- 1 point: Net horizontal force = 10 − 6 = 4 N right.
- 1 point: a = 4/4 = 1 m/s² right.
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 1What determines kinetic friction direction?
Relative sliding direction of the two contacting surfaces.
RECALL 2What are the units of μₖ?
None; it is dimensionless.
RECALL 3Why check contact?
A surface cannot pull inward with a negative normal force in this model.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Sliding friction and the normal force
- fₖ = μₖN during relative sliding.
- Find N from perpendicular force balance or acceleration.
- Friction opposes relative surface motion.
Remember: Using μₖmg without checking the normal force can give the wrong friction.
Conditions: Instantaneous rightward sliding relative to a fixed floor; no vertical acceleration. Normal force remains positive throughout the controls. This is not a time simulation through stopping.
Refresh Kid · AP Physics C: Mechanics Unit 2 (official Unit 2) · Objectives 2.7.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 2.7, objectives 2.7.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026 alongside the Fall 2026 clarifications. This is Mechanics Unit 2: Force and Translational Dynamics. The unit covers Topics 2.1–2.10. Calculus is introduced where it is needed for continuous mass and velocity-dependent forces. Shell theorem is applied without requiring a proof; spring combinations are purely series or purely parallel. 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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