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

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

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: $f_s \leq \mu_s N$ (adjusts to prevent motion)
   • Kinetic: $f_k = \mu_k N$ (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: $F_{min} = f_{s,max} = \mu_s N$

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.

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.

Find: Tension $T$ in the rope

Free Body Diagram:

• Weight: $W = mg = 200$ N (down)
• Normal: $N$ (up)
• Tension: $T$ at angle $30°$
  • 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!