Classical-Mechanics · Unit 8 · Video 2 · Interactive Practice

Two Faces of Friction: Kinetic, Static, and the Instant of Slip

IKey Formulas

Formula	Name	What it says
$f_k = \mu_k N$	Kinetic friction	Fixed; blind to area and speed
$0 \le f_s \le (f_s)_{\max}$	Static friction range	Adjusts to oppose the push
$(f_s)_{\max} = \mu_s N$	Maximum static friction	The slip threshold
$\mu_s > \mu_k$	Coefficient ordering	Stick, then slip

Key Insight: Kinetic friction is a single value; static friction is a whole range from $0$ up to $\mu_s N$ . Because $\mu_s > \mu_k$ , breaking a resting object loose takes more force than keeping it sliding.

IIKinetic Friction ∝ Normal Force

Kinetic friction rises in direct proportion to the normal force pressing the two surfaces together.

Normal force N 8

Neither contact area nor sliding speed appears in $f_k = \mu_k N$ : a block laid flat and the same block stood on end feel identical kinetic friction.

IIIStatic Friction Adjusts and Reverses

Static friction grows and flips direction to oppose the applied force — until that force exceeds its ceiling.

Applied force F 3

IVThe Instant of Slip

How large a push does it take to break a resting block loose and start it sliding?

Your guess: slip force 3

💡 This gap between starting and sliding is stick-slip — the catch of chalk on a blackboard and the voice of a bowed violin string.

VQuiz Questions

Problem 1 · Kinetic Friction from the Law

Given: A crate slides across a level floor with $\mu_k = 0.30$ and normal force $N = 200\text{ N}$ — find the kinetic friction $f_k$ .

✅ Correct!

f_k = \mu_k N = 0.30 \times 200 = 60\text{ N}

❌ That's the normal force. Multiply it by the coefficient:

f_k = \mu_k N = 0.30 \times 200

❌ Check the operation. The law multiplies, it doesn't divide:

f_k = \mu_k N

, not

N / \mu_k

❌ Not quite. Apply

f_k = \mu_k N

: multiply

0.30

200\text{ N}

Show solution

Kinetic friction follows da Vinci's law:

f_k = \mu_k N = 0.30 \times 200 = 60\text{ N}

The coefficient $\mu_k$ is a pure number, so multiplying it by the normal force gives a force in newtons.

Problem 2 · How Much Static Friction?

Given: A book pressing down with $N = 40\text{ N}$ rests on a table, $\mu_s = 0.5$ , $\mu_k = 0.3$ . You push horizontally with $12\text{ N}$ and it does not move — find the static friction force $f_s$ .

✅ Correct! The book is in equilibrium, so static friction exactly cancels the push:

f_s = 12\text{ N}

— well below the ceiling

(f_s)_{\max} = 20\text{ N}

❌ That's the maximum, not the actual value.

(f_s)_{\max} = \mu_s N = 20\text{ N}

is the most it could supply; since the

12\text{ N}

push is smaller, static friction only rises to

12\text{ N}

❌ Static friction isn't zero here. With no push it would be zero, but it grows to match a push — here it supplies

12\text{ N}

to hold the book still.

❌ Not quite. While the book stays put, static friction equals the applied push.

Show solution

First, the ceiling:

(f_s)_{\max} = \mu_s N = 0.5 \times 40 = 20\text{ N}

The $12\text{ N}$ push is below this maximum, so the book does not move.

Equilibrium condition: with no acceleration, static friction balances the push exactly:

f_s = 12\text{ N}

Static friction takes whatever value it needs (up to $20\text{ N}$ ) — it is not automatically $\mu_s N$ .

Problem 3 · Push Hard Enough to Slide

Given: A $100\text{ N}$ box on the floor with $\mu_s = 0.5$ and $\mu_k = 0.35$ . You push horizontally with $60\text{ N}$ .

Does the box start to slide?

What is the friction force once it is sliding?

✅ Correct! The

60\text{ N}

push beats the

50\text{ N}

static ceiling, so the box slides and friction settles at the kinetic value

f_k = \mu_k N = 35\text{ N}

❌ Check the ceiling. The maximum static friction is

(f_s)_{\max} = \mu_s N = 0.5 \times 100 = 50\text{ N}

; a

60\text{ N}

push exceeds it, so the box slides.

❌ Not the static maximum. Once sliding, friction drops to the kinetic value

f_k = \mu_k N = 0.35 \times 100 = 35\text{ N}

Show solution

Step 1 — Find the static ceiling:

(f_s)_{\max} = \mu_s N = 0.5 \times 100 = 50\text{ N}

The $60\text{ N}$ push exceeds $50\text{ N}$ , so the box breaks loose and slides.

Step 2 — Kinetic friction while sliding:

f_k = \mu_k N = 0.35 \times 100 = 35\text{ N}

Friction jumps down from the $50\text{ N}$ static maximum to the $35\text{ N}$ kinetic value — the $\mu_s > \mu_k$ drop.

Problem 4 · Same Mass, Different Footprint

Given: Two identical bricks (same mass, same material) are dragged across the same floor at constant speed. Brick A lies flat (large contact area); brick B stands on end (small contact area). Compare their kinetic friction forces $f_A$ and $f_B$ .

✅ Correct! Same weight → same normal force → same

f_k = \mu_k N

. Contact area and speed never enter the law.

❌ Area doesn't add friction. da Vinci's second law

f_k = \mu_k N

has no area term, so doubling the footprint leaves the friction unchanged.

❌ Higher pressure, but same total friction. The smaller footprint presses harder per unit area, yet

f_k = \mu_k N

depends only on

N

— the same for both bricks.

❌ Speed doesn't matter. Coulomb found kinetic friction is independent of speed for ordinary sliding, so both bricks feel the same

f_k

Show solution

Both bricks have the same mass, so each presses on the floor with the same normal force $N$ . Kinetic friction is

f_k = \mu_k N,

which contains no area term and no speed term. Brick A's wide footprint spreads the weight over more area (lower pressure); brick B's narrow footprint concentrates it (higher pressure) — but the product $\mu_k N$ is identical.

Therefore $f_A = f_B$ .

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