AP Physics C: Mechanics • Force and Translational Dynamics

Kinetic and Static Friction

Understanding the contact forces that resist motion and enable traction in mechanical systems.

A motion gauge with symbols for mass and movement.
Inside this lesson 9 sections 5 questions 8 figures

SECTION 1

Historical Context & Motivation

Friction is so ubiquitous that humans have grappled with it since the earliest attempts at engineering. Ancient Egyptians used sledges lubricated with water to move massive stones, and Roman engineers empirically discovered that polished surfaces reduced resistance. Yet a systematic, quantitative understanding of friction took centuries to develop—emerging from careful experiments that revealed surprisingly simple laws governing a remarkably complex phenomenon.

1493 • Leonardo da Vinci's Experiments

Da Vinci conducted some of the earliest systematic friction experiments, discovering that friction is proportional to load and independent of apparent contact area—results he never published.

1699 • Amontons' Laws

Guillaume Amontons rediscovered da Vinci's results and published two empirical laws: friction is proportional to the normal force, and it is independent of the apparent area of contact.

1785 • Coulomb's Friction Model

Charles-Augustin de Coulomb extended Amontons' work, distinguishing between static and kinetic friction and showing that kinetic friction is approximately independent of sliding speed.

1950s–present • Tribology & Microscopic Models

Modern tribology explains friction via asperity interactions, adhesion theory, and deformation at the microscopic level, revealing why the macroscopic Coulomb model works as well as it does.

The central question that the study of friction addresses is deceptively simple: why do objects resist being set into motion, and why does less force seem necessary to keep them moving than to start them? The Amontons–Coulomb model provides a remarkably effective macroscopic framework that AP Physics C leverages to analyze everything from blocks on inclined planes to braking automobiles.

SECTION 2

Core Principles & Definitions

Friction arises whenever two surfaces are in contact and there is either a tendency toward or actual relative motion between them. The classical model captures this behavior through two distinct regimes—static friction and kinetic friction—each characterized by a dimensionless coefficient multiplied by the normal force. These coefficients encapsulate surface material properties, roughness, and microscopic adhesion into a single empirical number.

1

Static Friction (fₛ)

Opposes the tendency of motion between surfaces at rest relative to each other. It is self-adjusting, matching the applied force up to a maximum value fₛ,max = μₛN.
2

Kinetic Friction (fₖ)

Acts on surfaces that are sliding relative to each other. It has a fixed magnitude fₖ = μₖN and always opposes the direction of relative sliding.
3

Normal Force (N)

The perpendicular contact force between surfaces. Friction depends directly on N—not on the weight alone, since applied forces and inclines modify N.
4

μₛ > μₖ (Typically)

The coefficient of static friction typically exceeds that of kinetic friction, meaning it takes more force to initiate sliding than to maintain it once begun.

SECTION 3

Visual Explanation: Forces on a Surface

Free-Body Diagram: Block on a Horizontal SurfacemNmgF_appliedf (friction)Friction opposes the direction of applied force (or tendency of motion)
A block of mass m rests on a horizontal surface. The normal force N (blue) acts upward, weight mg (red) acts downward, the applied force (gold) pushes right, and friction f (green) opposes the applied force.

The diagram above illustrates the four forces that govern the motion of a block on a flat surface. When the block is stationary, static friction adjusts its magnitude to exactly balance the horizontal component of the applied force, maintaining equilibrium. As the applied force increases, static friction increases proportionally until it reaches the threshold fₛ,max = μₛN. Beyond that point, the block begins to slide and kinetic friction—constant in magnitude—takes over. Notice that on a horizontal surface with no vertical applied force component, the normal force equals the object's weight, but this equality breaks down on inclines or when forces have vertical components.

SECTION 4

Mathematical Framework

The Amontons–Coulomb friction model is expressed through two compact relationships. Understanding when each applies—and how to determine the normal force in non-trivial geometries—is the core mathematical skill required for AP Physics C problems involving friction.

STATIC FRICTION (INEQUALITY)

fs≤μsNfₛ ≤ μₛ N

fₛ = magnitude of static friction force; μₛ = coefficient of static friction (dimensionless); N = normal force. The inequality reflects the self-adjusting nature of static friction: it equals μₛN only at the threshold of impending motion.

KINETIC FRICTION

fk=μkNfₖ = μₖ N

fₖ = magnitude of kinetic friction force; μₖ = coefficient of kinetic friction. This is an equality: once sliding occurs, kinetic friction has a fixed magnitude regardless of speed (in the Coulomb model).

NORMAL FORCE ON AN INCLINE

N=mgcosθN = mg cos θ

On an inclined plane of angle θ with no additional applied forces, the normal force equals the component of gravity perpendicular to the surface. The friction force then becomes f = μ mg cos θ, while the gravitational pull along the plane is mg sin θ.

CRITICAL ANGLE FOR IMPENDING SLIP

tanθc=μstan θ_c = μₛ

Setting mg sin θ = μₛ mg cos θ and solving yields the critical angle θ_c at which a block on an incline is on the verge of sliding. This is a clean derivation frequently tested on AP exams.

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Common Exam Pitfall

SECTION 5

Friction on Inclined Planes

Block on an Inclined Plane with FrictionSurfacemmgNfmg sin θθAlong the Plane (x′):ΣF = mg sin θ − f = maPerpendicular (y′):N − mg cos θ = 0
A tilted coordinate system aligned with the incline simplifies the analysis. The component of gravity mg sin θ drives the block down the plane, while friction f opposes the motion (or tendency of motion). Perpendicular to the surface, N = mg cos θ.

The inclined-plane scenario is one of the most commonly tested friction configurations on the AP Physics C exam. By choosing a coordinate system with one axis parallel to the incline and the other perpendicular, you decompose gravity into two components: mg sin θ along the plane and mg cos θ into the surface. The perpendicular equilibrium immediately yields N = mg cos θ (assuming no other perpendicular forces), so friction becomes f = μ mg cos θ. The net force along the plane is then mg sin θ − μ mg cos θ = ma for a block sliding down, or you reverse the friction direction for a block being pushed up.

A particularly elegant result emerges at the critical angle θ_c, where static friction reaches its maximum and the block is on the verge of sliding. Setting mg sin θ_c = μₛ mg cos θ_c gives tan θ_c = μₛ. This relationship means you can experimentally measure μₛ by slowly tilting a surface until the block just begins to slide, then measuring the angle—a classic laboratory technique.

SECTION 6

Worked Example: Acceleration on an Incline

A 5.0 kg block is placed on a 30° incline. The coefficient of kinetic friction between the block and the surface is μₖ = 0.20. The block is sliding down the incline. Find the acceleration of the block down the incline.

Worked example

  1. 1

    Step 1 — Draw the Free-Body Diagram & Choose Coordinates

    Orient the x-axis along the incline (positive down the slope) and the y-axis perpendicular to the surface (positive away from the surface). The forces are: weight mg downward, normal force N perpendicular to the surface, and kinetic friction fₖ up the incline (opposing the sliding direction).
  2. 2

    Step 2 — Apply Newton's Second Law Perpendicular to the Incline

    In the y-direction there is no acceleration, so ΣFyF_y = N − mg cos θ = 0, giving N = mg cos θ = (5.0)(9.8) cos 30° = (49)(0.866).
    N = 42.4 N
  3. 3

    Step 3 — Calculate the Kinetic Friction Force

    With the block sliding, fₖ = μₖ N = (0.20)(42.4 N).
    fₖ = 8.49 N
  4. 4

    Step 4 — Apply Newton's Second Law Along the Incline

    ΣFxF_x = mg sin θ − fₖ = ma. Substituting: (5.0)(9.8) sin 30° − 8.49 = (5.0) a, so 24.5 − 8.49 = 5.0 a, which gives 16.01 = 5.0 a.
    a = 3.2 m/s² down the incline
  5. 5

    Step 5 — Verify & Interpret

    Without friction the acceleration would be g sin 30° = 4.9 m/s². Friction reduces this to 3.2 m/s², a decrease of about 35%. The result is positive (down the slope), confirming our assumed direction was correct.

Sliding down a 30° incline; a separate idealized course-specific preset.Sliding down a 30° incline; a separate idealized course-specific preset.

Sliding block • Physics C; exact schematic, not a measured apparatus.Sliding block • Physics C; exact schematic, not a measured apparatus.

SECTION 7

Static vs. Kinetic Friction: Key Comparisons

PropertyStatic FrictionKinetic Friction
When it actsSurfaces at rest relative to each otherSurfaces sliding relative to each other
MagnitudeVariable: 0 ≤ fₛ ≤ μₛNConstant: fₖ = μₖN
DirectionOpposes tendency of relative motionOpposes direction of relative sliding
Coefficient sizeμₛ (typically larger)μₖ (typically smaller)
Depends on speed?No (object is not moving)Approximately no (Coulomb model)
Can do positive work?Yes (e.g., friction accelerates a passenger in a car)No—always negative work on the system, converting kinetic energy to thermal energy

Comparison of static and kinetic friction properties

SECTION 8

Connection to Advanced Theory

The Coulomb friction model is a powerful approximation, but it has well-known limitations. At the microscopic level, friction arises from the interlocking of surface asperities—tiny peaks on each surface that bond through adhesion and must be sheared for sliding to occur. Modern tribology and condensed-matter physics treat friction through statistical mechanics and surface science, revealing phenomena such as velocity-dependent friction, stick-slip dynamics (which cause earthquakes and squeaking brakes), and ultra-low friction in atomically smooth materials.

FeatureCoulomb Model (AP Level)Advanced Tribology
Speed dependencefₖ independent of speedFriction can vary with velocity (e.g., Stribeck curve)
Area dependenceIndependent of apparent contact areaDepends on real contact area (sum of asperity contacts)
Temperature effectsNot consideredCoefficients change with temperature; surface melting can reduce friction
Mathematical frameworkAlgebraic inequalities and equalitiesStatistical mechanics, contact mechanics (Hertz theory), molecular dynamics

Coulomb model versus modern friction theory

For AP Physics C, the Coulomb model is fully sufficient and is the framework used on the exam. However, awareness of its limitations enriches your understanding—particularly in experimental-design FRQs, where you may need to discuss sources of error. The fact that μₖ can vary slightly with speed, that surfaces deform under high loads, or that environmental factors like humidity change friction coefficients are all valid error-analysis points.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A heavy crate sits at rest on a rough floor. You push horizontally with a force of 50 N and the crate does not move. Which statement about the friction force is correct?
  • A. The static friction force is equal to μₛN.
  • B. The static friction force is 50 N, directed opposite your push.
  • C. The kinetic friction force is 50 N, directed opposite your push.
  • D. The friction force is zero because the crate is not moving.
Show answer (B)
B is correct. Since the crate remains stationary, Newton's second law requires the net horizontal force to be zero, so static friction must equal the 50 N applied force in the opposite direction. A is wrong because fₛ = μₛN only at the threshold of impending motion—here the crate is not on the verge of sliding. C is wrong because kinetic friction applies only when the object is sliding, and the crate is at rest. D is wrong because friction is needed to maintain equilibrium against the applied force.

Question 2: BASIC CALCULATION

A 10 kg block rests on a horizontal surface with μₛ = 0.40 and μₖ = 0.30. What is the minimum horizontal force needed to start the block moving?
  • A. 29.4 N
  • B. 39.2 N
  • C. 49.0 N
  • D. 98.0 N
Show answer (B)
B is correct. To initiate motion, the applied force must just exceed fₛ,max = μₛ mg = (0.40)(10)(9.8) = 39.2 N. A (29.4 N) is the kinetic friction force μₖ mg = (0.30)(10)(9.8), which describes friction after the block is already sliding. C (49.0 N) appears to use μ = 0.50, which is not given. D (98.0 N) equals the full weight mg, which would require μₛ = 1.0.

Question 3: INTERMEDIATE

A 4.0 kg block slides down a 37° incline at constant velocity. What is the coefficient of kinetic friction between the block and the incline? (Use sin 37° ≈ 0.60, cos 37° ≈ 0.80)
  • A. 0.60
  • B. 0.80
  • C. 0.75
  • D. 0.48
Show answer (C)
C is correct. Constant velocity means zero acceleration, so mg sin θ = fₖ = μₖ mg cos θ. The mass cancels, giving μₖ = sin θ / cos θ = tan θ = 0.60/0.80 = 0.75. A (0.60) confuses μₖ with sin 37°. B (0.80) confuses μₖ with cos 37°. D (0.48) appears to be sin × cos = 0.60 × 0.80, which has no physical basis in this problem.

Question 4: APPLIED

A 3.0 kg block is pushed against a rough vertical wall by a horizontal force F. The coefficients of friction are μₛ = 0.50 and μₖ = 0.35. (a) Draw a free-body diagram for the block. (b) Determine the minimum force F required to keep the block from sliding down the wall. (c) If F is removed, describe the motion of the block and calculate its initial acceleration.
Show answer
(a) FBD: horizontal applied force F into the wall, normal force N from wall outward (both horizontal), weight mg downward, static friction fₛ upward along the wall surface. (b) For the block not to slide, fₛ must equal mg. The normal force equals F (horizontal equilibrium), and at the minimum F, fₛ = μₛN = μₛF = mg, so F = mg/μₛ = (3.0)(9.8)/0.50 = 58.8 N. (c) Once F is removed, the normal force becomes zero, which means friction is also zero. The only force is gravity, so the block falls freely with a = g = 9.8 m/s² downward.

Question 5: CRITICAL THINKING

Two blocks are stacked on a frictionless surface. Block A (mass mAm_A = 2.0 kg) sits on top of Block B (mass mBm_B = 5.0 kg). The coefficient of static friction between A and B is μₛ = 0.40 and the coefficient of kinetic friction is μₖ = 0.30. A horizontal force F is applied to Block B. (a) Determine the maximum force F that can be applied to Block B such that both blocks accelerate together as a single system. (b) If F = 40 N, determine the acceleration of each block. (c) Derive a general expression for the acceleration of Block A as a function of F for F greater than the value found in part (a). (d) Briefly explain, with supporting calculation, how the answer to part (a) would change if the force F were instead applied to Block A.
Show answer
(a) When both blocks move together with acceleration a, Block A's only horizontal force is static friction from B: fₛ = mAm_A × a. The maximum acceleration before slipping is a_max = fₛ,max / mAm_A = μₛ mAm_A g / mAm_A = μₛ g = (0.40)(9.8) = 3.92 m/s². The force on the whole system: F_max = (mAm_A + mBm_B) × a_max = (7.0)(3.92) = 27.4 N. (b) At F = 40 N > 27.4 N, the blocks slide relative to each other. For Block A: kinetic friction is the only horizontal force, fₖ = μₖ mAm_A g = (0.30)(2.0)(9.8) = 5.88 N, so aAa_A = 5.88 / 2.0 = 2.94 m/s². For Block B: net force = F − fₖ = 40 − 5.88 = 34.12 N, so aBa_B = 34.12 / 5.0 = 6.82 m/s². (c) For F > F_max, Block A always experiences kinetic friction fₖ = μₖ mAm_A g, so aAa_A = μₖ g = (0.30)(9.8) = 2.94 m/s², independent of F. (d) If F is applied to Block A instead, the maximum force is significantly smaller. Now friction on Block B (from A) is the only horizontal force accelerating B, and friction on A from B acts backward, opposing A's tendency to slide forward. The normal force between the surfaces is still N = mAm_A g, so fₛ,max = μₛ mAm_A g = (0.40)(2.0)(9.8) = 7.84 N. This maximum friction must accelerate Block B: a_max = fₛ,max / mBm_B = 7.84 / 5.0 = 1.568 m/s². Therefore F_max = (mAm_A + mBm_B) × a_max = (7.0)(1.568) ≈ 11.0 N. The maximum force is much smaller (11.0 N vs. 27.4 N) because the lighter block's weight determines the contact normal force, which limits the friction available to accelerate the heavier bottom block. In the original scenario, friction only needed to accelerate the light top block; now it must accelerate the heavy bottom block with the same maximum friction force.

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