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Friction

Static and kinetic friction forces

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Friction

Introduction to Friction

Friction is a force that opposes motion or attempted motion between surfaces in contact.

Key Characteristics

  • Always parallel to the surface
  • Always opposes motion (or attempted motion)
  • Caused by microscopic interactions between surfaces
  • Converted to thermal energy (heat)
  • Can be helpful (walking, driving) or harmful (wear, energy loss)

Types of Friction

Static Friction (fsf_s)

Static friction opposes the tendency to move when surfaces are not sliding.

fs≤μsNf_s \leq \mu_s N

Where:

  • fsf_s = static friction force
  • μs\mu_s = coefficient of static friction (no units)
  • NN = normal force
  • ≤\leq means "less than or equal to"

Key points:

  • Static friction adjusts to prevent motion (up to a maximum)
  • Maximum value: fs,max=μsNf_{s,max} = \mu_s N
  • When applied force < fs,maxf_{s,max}: object doesn't move
  • When applied force = fs,maxf_{s,max}: object is on verge of moving
  • When applied force > fs,maxf_{s,max}: object starts to slide

Kinetic Friction (fkf_k)

Kinetic friction opposes motion when surfaces are sliding.

fk=μkNf_k = \mu_k N

Where:

  • fkf_k = kinetic friction force
  • μk\mu_k = coefficient of kinetic friction (no units)
  • NN = normal force

Key points:

  • Kinetic friction has a constant value (for given surfaces)
  • Always less than maximum static friction: fk<fs,maxf_k < f_{s,max}
  • Therefore: μk<μs\mu_k < \mu_s (always!)
  • Direction: opposite to velocity

Coefficients of Friction

Properties of μ\mu (mu)

  • Dimensionless (no units)
  • Depends on surfaces in contact (materials, roughness)
  • Ranges from 0 (frictionless) to >1 (very rough)
  • Typical values: 0.10.1 to 1.51.5
  • μs>μk\mu_s > \mu_k (harder to start moving than to keep moving)

Common Values

Surfacesμs\mu_sμk\mu_k
Rubber on dry concrete1.00.8
Wood on wood0.50.3
Steel on steel0.70.6
Ice on ice0.10.03
Teflon on Teflon0.040.04

Direction of Friction

Static friction: Opposes the direction the object would move if there were no friction

Kinetic friction: Opposes the direction of motion (opposite to velocity)

On a horizontal surface:

  • Object pushed to the right → friction points left
  • Object moving left → friction points right

On an incline:

  • Object tendency to slide down → friction points up the incline
  • Object sliding down → friction points up the incline
  • Object being pulled up → friction points down the incline

Relationship to Normal Force

f∝Nf \propto N

Important: Friction depends on normal force, NOT on:

  • Surface area (wider tire doesn't reduce friction)
  • Weight directly (only through NN)
  • Speed (in simple models)

On horizontal surface: N=mgN = mg, so fs,max=μsmgf_{s,max} = \mu_s mg

On incline: N=mgcos⁡θN = mg\cos\theta, so fs,max=μsmgcos⁡θf_{s,max} = \mu_s mg\cos\theta

Static Friction: Adjust or Maximum?

Static friction can be any value from 00 to μsN\mu_s N:

Case 1: Object at Rest (Equilibrium)

Applied force Fapp=20F_{app} = 20 N, fs,max=50f_{s,max} = 50 N

  • Static friction adjusts: fs=20f_s = 20 N
  • Object doesn't move: ∑F=0\sum F = 0

Case 2: On the Verge of Moving

Applied force Fapp=50F_{app} = 50 N, fs,max=50f_{s,max} = 50 N

  • Static friction at maximum: fs=50f_s = 50 N
  • Object about to slide: ∑F=0\sum F = 0 (barely)

Case 3: Sliding

Applied force Fapp=60F_{app} = 60 N, fs,max=50f_{s,max} = 50 N

  • Static friction can't hold: object slides!
  • Now kinetic friction applies: fk=μkN<50f_k = \mu_k N < 50 N
  • Object accelerates: ∑F=Fapp−fk>0\sum F = F_{app} - f_k > 0

Motion on a Horizontal Surface

Free Body Diagram:

  • Weight W=mgW = mg (down)
  • Normal force NN (up)
  • Applied force FappF_{app} (horizontal)
  • Friction ff (opposite to motion/tendency)

Equations:

Vertical: N−mg=0N - mg = 0, so N=mgN = mg

Horizontal:

  • If not moving: Fapp−fs=0F_{app} - f_s = 0, so fs=Fappf_s = F_{app} (up to μsN\mu_s N)
  • If moving: Fapp−fk=maF_{app} - f_k = ma, where fk=μkNf_k = \mu_k N

Problem-Solving Strategy

  1. Draw free body diagram
  2. Identify whether static or kinetic friction applies
    • At rest or no sliding → static
    • Sliding → kinetic
  3. Find normal force (from vertical equilibrium)
  4. Calculate friction force
    • Static: fs≤μsNf_s \leq \mu_s N (adjusts to prevent motion)
    • Kinetic: fk=μkNf_k = \mu_k N (constant)
  5. Apply Newton's Second Law in direction of motion
  6. Solve for unknowns

Common Scenarios

Starting Motion

Find minimum force to start an object moving: Fmin=fs,max=μsNF_{min} = f_{s,max} = \mu_s N

Constant Velocity

Object moving at constant speed:

  • a=0a = 0, so ∑F=0\sum F = 0
  • Applied force = kinetic friction
  • Fapp=fk=μkNF_{app} = f_k = \mu_k N

Accelerating Motion

Object speeding up or slowing down:

  • a≠0a \neq 0
  • ∑F=ma\sum F = ma
  • Fapp−fk=maF_{app} - f_k = ma (if moving to the right)

Air Resistance (Brief)

Air resistance (drag) is a type of friction with air:

  • Depends on speed (higher speed → more drag)
  • Depends on surface area and shape
  • Negligible at low speeds
  • At terminal velocity: drag force = weight (no acceleration)

We'll usually neglect air resistance in AP Physics 1 unless stated.

📚 Practice Problems

1Problem 1easy

❓ Question:

A 1010 kg box rests on a horizontal surface. The coefficient of static friction is μs=0.4\mu_s = 0.4. What is the maximum static friction force?

💡 Show Solution

Given:

  • Mass: m=10m = 10 kg
  • Coefficient of static friction: μs=0.4\mu_s = 0.4
  • On horizontal surface
  • g=10g = 10 m/s²

Find: Maximum static friction fs,maxf_{s,max}

Step 1: Find normal force

Draw free body diagram (vertical direction):

  • Weight down: W=mgW = mg
  • Normal force up: NN

Since no vertical acceleration: N−mg=0N - mg = 0 N=mg=(10)(10)=100 NN = mg = (10)(10) = 100 \text{ N}

Step 2: Calculate maximum static friction

fs,max=μsNf_{s,max} = \mu_s N fs,max=(0.4)(100)f_{s,max} = (0.4)(100) fs,max=40 Nf_{s,max} = 40 \text{ N}

Answer: The maximum static friction force is 40 N.

Interpretation: The box will remain at rest as long as the horizontal applied force is less than or equal to 40 N. If the applied force exceeds 40 N, the box will start to slide.

2Problem 2medium

❓ Question:

A 10 kg box rests on a horizontal floor with μₛ = 0.50 and μₖ = 0.30. (a) What is the maximum static friction force? (b) What minimum force is needed to start the box moving? (c) Once moving, what force is needed to keep it moving at constant velocity?

💡 Show Solution

Solution:

Given: m = 10 kg, μₛ = 0.50, μₖ = 0.30, g = 10 m/s²

(a) Maximum static friction: Normal force: N = mg = 10 × 10 = 100 N f_s,max = μₛN = 0.50 × 100 = 50 N

(b) Force to start moving: To overcome static friction: F_min = f_s,max = 50 N

(c) Force for constant velocity: At constant velocity, a = 0, so F_applied = f_k f_k = μₖN = 0.30 × 100 = 30 N

Note: Less force is needed to keep it moving (30 N) than to start it (50 N) because μₖ < μₛ.

3Problem 3medium

❓ Question:

A 55 kg block on a horizontal surface experiences a horizontal applied force of 3030 N. The coefficients of friction are μs=0.5\mu_s = 0.5 and μk=0.3\mu_k = 0.3. Does the block move? If so, what is its acceleration? (Use g=10g = 10 m/s²)

💡 Show Solution

Given:

  • Mass: m=5m = 5 kg
  • Applied force: Fapp=30F_{app} = 30 N (horizontal)
  • μs=0.5\mu_s = 0.5, μk=0.3\mu_k = 0.3
  • g=10g = 10 m/s²

Find: Does it move? If yes, find acceleration.

Step 1: Find normal force N=mg=(5)(10)=50 NN = mg = (5)(10) = 50 \text{ N}

Step 2: Find maximum static friction fs,max=μsN=(0.5)(50)=25 Nf_{s,max} = \mu_s N = (0.5)(50) = 25 \text{ N}

Step 3: Compare applied force to max static friction Fapp=30 N>fs,max=25 NF_{app} = 30 \text{ N} > f_{s,max} = 25 \text{ N}

Since applied force exceeds maximum static friction, the block will slide!

Step 4: Find kinetic friction (now that it's moving) fk=μkN=(0.3)(50)=15 Nf_k = \mu_k N = (0.3)(50) = 15 \text{ N}

Step 5: Apply Newton's Second Law (horizontal)

Choose right as positive direction: ∑Fx=ma\sum F_x = ma Fapp−fk=maF_{app} - f_k = ma 30−15=5a30 - 15 = 5a 15=5a15 = 5a a=3 m/s2a = 3 \text{ m/s}^2

Answers:

  • Yes, the block moves (applied force > max static friction)
  • Acceleration: 3 m/s² to the right

Key insight: Once sliding starts, kinetic friction (15 N) is less than max static friction (25 N), so there's a net force causing acceleration.

4Problem 4hard

❓ Question:

A 5.0 kg block on a horizontal surface is pulled by a 40 N force at 30° above the horizontal. The coefficient of kinetic friction is μₖ = 0.25. (a) Find the normal force. (b) Find the friction force. (c) Find the block's acceleration.

💡 Show Solution

Solution:

Given: m = 5.0 kg, F = 40 N at 30°, μₖ = 0.25, g = 10 m/s²

Force components: F_x = F cos 30° = 40(0.866) = 34.6 N F_y = F sin 30° = 40(0.50) = 20 N (upward)

(a) Normal force: Vertical equilibrium (a_y = 0): N + F_y = mg N = mg - F_y = 5.0(10) - 20 = 30 N

(b) Friction force: f_k = μₖN = 0.25 × 30 = 7.5 N (opposing motion)

(c) Acceleration: Horizontal direction: F_net,x = F_x - f_k = ma 34.6 - 7.5 = 5.0a 27.1 = 5.0a a = 5.4 m/s²

5Problem 5hard

❓ Question:

A 2020 kg crate is pulled across a horizontal floor at constant velocity by a rope making an angle of 30°30° above the horizontal. The coefficient of kinetic friction is μk=0.25\mu_k = 0.25. Find the tension in the rope. (Use g=10g = 10 m/s², cos⁡30°=0.866\cos 30° = 0.866, sin⁡30°=0.5\sin 30° = 0.5)

💡 Show Solution

Given:

  • Mass: m=20m = 20 kg
  • Angle of rope: θ=30°\theta = 30°
  • Constant velocity (so a=0a = 0)
  • μk=0.25\mu_k = 0.25
  • g=10g = 10 m/s²

Find: Tension TT in the rope

Free Body Diagram:

  • Weight: W=mg=200W = mg = 200 N (down)
  • Normal: NN (up)
  • Tension: TT at angle 30°30°
    • Horizontal component: Tx=Tcos⁡30°T_x = T\cos 30°
    • Vertical component: Ty=Tsin⁡30°T_y = T\sin 30°
  • Kinetic friction: fk=μkNf_k = \mu_k N (left, opposing motion)

Step 1: Vertical direction (no vertical acceleration)

∑Fy=0\sum F_y = 0 N+Tsin⁡30°−mg=0N + T\sin 30° - mg = 0 N+0.5T−200=0N + 0.5T - 200 = 0 N=200−0.5TN = 200 - 0.5T

Step 2: Horizontal direction (constant velocity, so a=0a = 0)

∑Fx=0\sum F_x = 0 Tcos⁡30°−fk=0T\cos 30° - f_k = 0 Tcos⁡30°=fkT\cos 30° = f_k

Step 3: Express friction in terms of normal force

fk=μkN=0.25Nf_k = \mu_k N = 0.25N

Substituting NN from Step 1: fk=0.25(200−0.5T)f_k = 0.25(200 - 0.5T) fk=50−0.125Tf_k = 50 - 0.125T

Step 4: Substitute into horizontal equation

Tcos⁡30°=50−0.125TT\cos 30° = 50 - 0.125T 0.866T=50−0.125T0.866T = 50 - 0.125T 0.866T+0.125T=500.866T + 0.125T = 50 0.991T=500.991T = 50 T=500.991≈50.5 NT = \frac{50}{0.991} \approx 50.5 \text{ N}

Answer: The tension in the rope is approximately 50.5 N.

Check:

  • N=200−0.5(50.5)=174.75N = 200 - 0.5(50.5) = 174.75 N
  • fk=0.25(174.75)=43.7f_k = 0.25(174.75) = 43.7 N
  • Tcos⁡30°=50.5(0.866)=43.7T\cos 30° = 50.5(0.866) = 43.7 N ✓

Key insight: The upward component of tension reduces the normal force, which reduces friction, which reduces the horizontal force needed!

Explain using:

📌 Related Topics in Dynamics

❓ Frequently Asked Questions

What is Friction?▾
Static and kinetic friction forces
How can I study Friction effectively?▾
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 5 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
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Friction is part of the AP Physics 1 course on Study Mondo, specifically in the Dynamics section. You can explore the full course for more related topics and practice resources.
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Yes, this page includes 5 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.