Kinetic Friction Force

Equivalent forms

F_k = \mu_k m g

f_k = \mu_k F_N

A surprisingly simple law for a microscopically chaotic phenomenon.

Symbol	Name	SI unit	Dimension	Range
$F_k$	Kinetic friction forceoutput Force opposing sliding motion	N	MLT^-2	0 – 500
$mu_k$	Coefficient of kinetic friction Dimensionless surface property		1	0.01 – 1.5
$N$	Normal force Force perpendicular to surface	N	MLT^-2	1 – 1000

Unit systems

Symbol	SI	CGS	Imperial
$F_k$	N	dyn	lbf
$mu_k$	—	—	—
$N$	N	dyn	lbf

Where it holds

Classical Amontons–Coulomb friction for dry contact; breaks down for lubricated, very high-speed, or microscopic contacts.

Dimensional analysis

F_k \to [MLT^{-2}]

\mu _k\cdot N \to [1]\cdot [MLT^{-2}] = [MLT^{-2}]

Both sides newtons.

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Discovery

Guillaume Amontons / Charles-Augustin de Coulomb · 1699

Amontons rediscovered Leonardo da Vinci's friction laws in 1699; Coulomb refined them in 1785, distinguishing static from kinetic friction.

Try this

Push a 20 kg crate across a concrete floor — how much force does friction steal back?

Compute the kinetic friction force on a sliding crate of mass 20 kg with μ_k = 0.35 on a flat floor.

Research status: stable

Real-world applications

Brake system design
Tire-road contact modeling
Machine wear and lubrication engineering
Sport (curling, skating, climbing)

Common misconceptions

Friction depends on contact area — it doesn't, to first order
Friction depends strongly on sliding speed — only weakly, for dry contact
$\mu _k$ > $\mu _s$ — actually $\mu _s \ge \mu _k$ for almost all material pairs

Experimental verification

Verified by inclined-plane and tow-cart experiments since Coulomb (1785); modern tribology measures

\mu _k

to four decimal places.

Derivation

Empirical law: experiments show that the tangential force needed to keep two solid surfaces sliding is approximately proportional to the normal load and nearly independent of contact area and sliding speed.

Defining the proportionality constant

as \mu _k

gives

F_k = \mu _k N

.

Limiting cases

\mu _k \to 0

⟶

F_k \to 0

Perfectly slick surfaces (idealization).

N \to 0

⟶

F_k \to 0

Object floats free; no contact means no friction.

v \to 0

⟶ Switches to static regimeOnce motion stops,

\mu _s

applies; usually

\mu _s

>

\mu _k

.

What if…

What if I double the normal force?

Friction doubles — linear in N.

What if surfaces are wet?

Water acts as a lubricant; $\mu _k$ can drop by 5– $10\times$ .

What if the object stops?

You must overcome the larger static friction $F_s_\max = \mu _s N$ to restart motion.

1

Crate on concrete

Given ·

mu k:: 0.35
N:: 196

Find · F_k

Steps

Identify $\mu _k = 0.35$ (rubber-on-concrete typical)
$N = mg = 20 \times 9.8 = 196\,\mathrm{N}$
$F_k = \mu _k N = 68.6\,\mathrm{N}$

Result ·

F_k = 0.35 \times 196 = 68.6\,\mathrm{N}

2

Hockey puck on ice

Given ·

mu k:: 0.05
N:: 1.6

Find · F_k

Steps

$\mu _k \approx 0.05$ for puck on ice
$N = mg = 0.16 \times 9.8 \approx 1.6\,\mathrm{N}$
$F_k \approx 0.08\,\mathrm{N}$ — barely slows the puck

Result ·

F_k = 0.05 \times 1.6 = 0.08\,\mathrm{N}

Maximum Static Friction→Newton's Second Law→Normal Force on an Incline→