AP® Physics 1: Algebra-Based review sheet from Aim for Five (aimforfive.com/physics/units/2/2-7)
Unit 2 · Topic 2.7
2.7 Kinetic and Static Friction
Friction is the force a surface exerts parallel to itself. Kinetic friction acts while surfaces slide and has a fixed size, μ_k F_N. Static friction acts while they don't slide and adjusts its size and direction to prevent slipping, up to a maximum of μ_s F_N.
Key terms
- kinetic friction
- static friction
- coefficient of friction (μ)
- normal force
- maximum static friction
Kinetic friction
Kinetic friction acts when two surfaces slide across each other. On each surface it points opposite to that surface's motion relative to the other one.
Its size is F_f,k = μ_k F_N, where F_N is the normal force between the surfaces and μ_k (the coefficient of kinetic friction) is a number with no units that depends on the two materials. Rubber on dry concrete has a large μ; ice on ice has a small one.
In this model, friction doesn't depend on how much surface area is touching or on how fast the surfaces slide. A brick slides with the same friction whether it's lying flat or on its side.
Static friction
Static friction acts when two surfaces touch but don't slide relative to each other. It takes whatever size and direction is needed to stop slipping, up to a maximum: F_f,s ≤ μ_s F_N. The equation sheet writes this as |F_f| ≤ |μF_N|.
Push gently on a heavy box and static friction matches your push exactly. Push harder and it matches again. Only when your push exceeds μ_s F_N does the box start to slide, and then kinetic friction takes over.
For the same pair of surfaces, μ_s is usually larger than μ_k. That's why it takes more force to start a box moving than to keep it moving.
| Your horizontal push | Box's motion | Friction force |
|---|---|---|
| less than μ_s F_N | stays at rest | static, equal to your push |
| just above μ_s F_N | starts to slide | drops to kinetic, μ_k F_N |
| larger still | slides and speeds up | kinetic, still μ_k F_N |
The normal force isn't always mg
Friction depends on the normal force, so find F_N carefully from a free-body diagram. On an incline, F_N = mg cos θ. Pulling up at an angle reduces F_N; pushing down at an angle increases it. In an accelerating elevator, F_N changes too.
Static friction can point forward
Friction opposes slipping, not motion. When you walk, your foot pushes backward on the ground and static friction pushes you forward. A car accelerates because static friction on its tires points forward. A box riding on a speeding-up truck bed is pushed forward by static friction from the bed.
On the verge of slipping
Raise a ramp until a block just starts to slide. At that moment static friction is at its maximum and balances the downhill pull: mg sin θ = μ_s mg cos θ, so μ_s = tan θ. This gives a simple lab method for measuring μ_s.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Does the box move? (classic trap)
A 20 kg box sits on a level floor with μ_s = 0.50 and μ_k = 0.30. Use g = 9.8 m/s². Find the friction force and the acceleration when you push horizontally with (a) 60 N and (b) 120 N.
Show the solutionHide the solution
- Step 1: Normal force: F_N = mg = (20)(9.8) = 196 N. Maximum static friction: μ_s F_N = (0.50)(196) = 98 N.
- Step 2: (a) 60 N is less than 98 N, so the box stays put. Static friction is exactly 60 N, not 98 N. That's the trap: μ_s F_N is only the maximum. Acceleration = 0.
- Step 3: (b) 120 N is more than 98 N, so the box slides. Kinetic friction = μ_k F_N = (0.30)(196) = 58.8 N.
- Step 4: a = (120 − 58.8) ÷ 20 ≈ 3.06 m/s².
Answer: (a) friction 60 N, a = 0; (b) friction ≈ 59 N, a ≈ 3.1 m/s²
- Example 2Calculator allowed
Block about to slip on a ramp
A block on a ramp starts to slide when the ramp is tilted to 25°. Derive an expression for μ_s and find its value.
Show the solutionHide the solution
- Step 1: Just before slipping, the block is in equilibrium with static friction at its maximum, pointing up the slope.
- Step 2: Perpendicular to the slope: F_N = mg cos θ.
- Step 3: Along the slope: mg sin θ = μ_s F_N = μ_s mg cos θ.
- Step 4: Cancel mg: μ_s = sin θ / cos θ = tan θ = tan 25° ≈ 0.47.
Answer: μ_s = tan θ ≈ 0.47
- Example 3Calculator allowed
Angled pull on a sled
A 10 kg sled is pulled across level snow with a 40 N force at 30° above horizontal. μ_k = 0.20. Use g = 9.8 m/s². Find the sled's acceleration.
Show the solutionHide the solution
- Step 1: Vertical (no vertical acceleration): F_N + 40 sin 30° − mg = 0, so F_N = 98 − 20 = 78 N.
- Step 2: Kinetic friction: F_f = μ_k F_N = (0.20)(78) = 15.6 N.
- Step 3: Horizontal: 40 cos 30° − 15.6 = ma, so 34.6 − 15.6 = 10a.
- Step 4: a ≈ 1.9 m/s². Using F_N = mg would give a wrong friction force of 19.6 N.
Answer: a ≈ 1.9 m/s²
Common mistakes
- Always setting static friction equal to μ_s F_N. That's only its maximum; usually it's just enough to prevent slipping.
- Using F_N = mg on inclines or with angled forces. Find the normal force from the perpendicular forces each time.
- Thinking friction always points backward against motion. Static friction can point forward, as with walking and car tires.
- Thinking a larger contact area means more friction. In this model, area doesn't matter.
On the exam
- Graph questions are common: sketch friction versus applied force. It rises along a straight line with slope 1 (friction = push), peaks at μ_s F_N, then drops to the flat value μ_k F_N.
- In lab questions, you might design a way to measure μ. Tilting a ramp until slipping (μ_s = tan θ) or pulling at constant velocity with a force sensor (μ_k = F/F_N) are both good plans.
Connected topics
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Check yourself
4 questions on 2.7 Kinetic and Static Friction. Pick an answer to see if you got it, and why.
A 10 kg box rests on a level floor. The coefficient of static friction is 0.50 and the coefficient of kinetic friction is 0.30. A student pushes horizontally on the box with a 20 N force. What is the friction force on the box? Use g = 10 m/s².
A 10 kg crate is pulled across a level floor at constant speed by a 40 N horizontal force. What is the coefficient of kinetic friction? Use g = 10 m/s².
A 4.0 kg block sits at rest on a ramp tilted 30° above the horizontal. The coefficient of static friction is 0.80. What is the size of the friction force on the block? Use g = 10 m/s².
A loaded truck and an empty truck of the same model skid to a stop with locked wheels from the same speed on the same road. The coefficient of kinetic friction is the same for both. How do their stopping distances compare?
0 of 4 answered