Friction

Static and kinetic friction forces

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 ( $f_s$ )

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

$f_s \leq \mu_s N$

Where:

$f_s$ = static friction force
$\mu_s$ = coefficient of static friction (no units)
$N$ = normal force
$\leq$ means "less than or equal to"

Key points:

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

Kinetic Friction ( $f_k$ )

Kinetic friction opposes motion when surfaces are sliding.

$f_k = \mu_k N$

Where:

$f_k$ = kinetic friction force
$\mu_k$ = coefficient of kinetic friction (no units)
$N$ = normal force

Key points:

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

Coefficients of Friction

Properties of $\mu$ (mu)

Dimensionless (no units)
Depends on surfaces in contact (materials, roughness)
Ranges from 0 (frictionless) to >1 (very rough)
Typical values: $0.1$ to $1.5$
$\mu_s > \mu_k$ (harder to start moving than to keep moving)

Common Values

| Surfaces | $\mu_s$ | $\mu_k$ | |----------|---------|---------| | Rubber on dry concrete | 1.0 | 0.8 | | Wood on wood | 0.5 | 0.3 | | Steel on steel | 0.7 | 0.6 | | Ice on ice | 0.1 | 0.03 | | Teflon on Teflon | 0.04 | 0.04 |

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

$f \propto N$

Important: Friction depends on normal force, NOT on:

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

On horizontal surface: $N = mg$ , so $f_{s,max} = \mu_s mg$

On incline: $N = mg\cos\theta$ , so $f_{s,max} = \mu_s mg\cos\theta$

Static Friction: Adjust or Maximum?

Static friction can be any value from $0$ to $\mu_s N$ :

Case 1: Object at Rest (Equilibrium)

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

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

Case 2: On the Verge of Moving

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

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

Case 3: Sliding

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

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

Motion on a Horizontal Surface

Free Body Diagram:

Weight $W = mg$ (down)
Normal force $N$ (up)
Applied force $F_{app}$ (horizontal)
Friction $f$ (opposite to motion/tendency)

Equations:

Vertical: $N - mg = 0$ , so $N = mg$

Horizontal:

If not moving: $F_{app} - f_s = 0$ , so $f_s = F_{app}$ (up to $\mu_s N$ )
If moving: $F_{app} - f_k = ma$ , where $f_k = \mu_k N$

Problem-Solving Strategy

Draw free body diagram
Identify whether static or kinetic friction applies
- At rest or no sliding → static
- Sliding → kinetic
Find normal force (from vertical equilibrium)
Calculate friction force
- Static: $f_s \leq \mu_s N$ (adjusts to prevent motion)
- Kinetic: $f_k = \mu_k N$ (constant)
Apply Newton's Second Law in direction of motion
Solve for unknowns

Common Scenarios

Starting Motion

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

Constant Velocity

Object moving at constant speed:

$a = 0$ , so $\sum F = 0$
Applied force = kinetic friction
$F_{app} = f_k = \mu_k N$

Accelerating Motion

Object speeding up or slowing down:

$a \neq 0$
$\sum F = ma$
$F_{app} - f_k = ma$ (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 1medium

❓ 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

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

2Problem 2easy

❓ Question:

A $10$ kg box rests on a horizontal surface. The coefficient of static friction is $\mu_s = 0.4$ . What is the maximum static friction force?

💡 Show Solution

Given:

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

Find: Maximum static friction $f_{s,max}$

Step 1: Find normal force

Draw free body diagram (vertical direction):

Weight down: $W = mg$
Normal force up: $N$

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

Step 2: Calculate maximum static friction

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

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.

3Problem 3medium

❓ Question:

💡 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

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

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)

5Problem 5hard

❓ Question:

💡 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)

6Problem 6medium

❓ Question:

A $5$ kg block on a horizontal surface experiences a horizontal applied force of $30$ N. The coefficients of friction are $\mu_s = 0.5$ and $\mu_k = 0.3$ . Does the block move? If so, what is its acceleration? (Use $g = 10$ m/s²)

💡 Show Solution

Given:

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

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

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

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

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

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

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

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

Choose right as positive direction: $\sum F_x = ma$ $F_{app} - f_k = ma$ $30 - 15 = 5a$ $15 = 5a$ $a = 3 \text{ m/s}^2$

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.

7Problem 7hard

❓ Question:

A $20$ kg crate is pulled across a horizontal floor at constant velocity by a rope making an angle of $30°$ above the horizontal. The coefficient of kinetic friction is $\mu_k = 0.25$ . Find the tension in the rope. (Use $g = 10$ m/s², $\cos 30° = 0.866$ , $\sin 30° = 0.5$ )

💡 Show Solution

Given:

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

Find: Tension $T$ in the rope

Free Body Diagram:

Weight: $W = mg = 200$ N (down)
Normal: $N$ (up)
Tension: $T$ $T$ at angle $30°$ $30°$
- Horizontal component: $T_x = T\cos 30°$
- Vertical component: $T_y = T\sin 30°$
Kinetic friction: $f_k = \mu_k N$ (left, opposing motion)

Step 1: Vertical direction (no vertical acceleration)

$\sum F_y = 0$ $N + T\sin 30° - mg = 0$ $N + 0.5T - 200 = 0$ $N = 200 - 0.5T$

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

$\sum F_x = 0$ $T\cos 30° - f_k = 0$ $T\cos 30° = f_k$

Step 3: Express friction in terms of normal force

$f_k = \mu_k N = 0.25N$

Substituting $N$ from Step 1: $f_k = 0.25(200 - 0.5T)$ $f_k = 50 - 0.125T$

Step 4: Substitute into horizontal equation

$T\cos 30° = 50 - 0.125T$ $0.866T = 50 - 0.125T$ $0.866T + 0.125T = 50$ $0.991T = 50$ $T = \frac{50}{0.991} \approx 50.5 \text{ N}$

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

Check:

$N = 200 - 0.5(50.5) = 174.75$ N
$f_k = 0.25(174.75) = 43.7$ N
$T\cos 30° = 50.5(0.866) = 43.7$ N ✓

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

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Friction

Friction

Introduction to Friction

Key Characteristics

Types of Friction

Static Friction (fsf_sfs​)

Kinetic Friction (fkf_kfk​)

Coefficients of Friction

Properties of μ\muμ (mu)

Common Values

Direction of Friction

On a horizontal surface:

On an incline:

Relationship to Normal Force

Static Friction: Adjust or Maximum?

Case 1: Object at Rest (Equilibrium)

Case 2: On the Verge of Moving

Case 3: Sliding

Motion on a Horizontal Surface

Free Body Diagram:

Equations:

Problem-Solving Strategy

Common Scenarios

Starting Motion

Constant Velocity

Accelerating Motion

Air Resistance (Brief)

📚 Practice Problems

1Problem 1medium

❓ Question:

2Problem 2easy

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3Problem 3medium

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4Problem 4hard

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5Problem 5hard

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6Problem 6medium

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7Problem 7hard

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Static Friction ( $f_s$ )

Kinetic Friction ( $f_k$ )

Properties of $\mu$ (mu)