Friction — Physics Module 2

§ 01

The Nature of Friction

Friction is a contact force that acts parallel to a surface, opposing the relative motion (or tendency of motion) between two objects in contact.

At the microscopic scale, all surfaces — however smooth they appear — are covered in asperities: microscopic bumps, ridges, and valleys. When two surfaces are pressed together, these asperities interlock and form microscopic cold-weld junctions. Breaking these junctions requires force; this is friction.

Additionally, at close contact, van der Waals forces (weak electrostatic adhesion between molecules) contribute, particularly for very smooth or very clean surfaces. This is why precision-ground optical surfaces can stick together without any adhesive — a phenomenon called wringing.

🔬 Microscopic origin Real contact area between surfaces is far smaller than the apparent (geometric) contact area — sometimes less than 0.01% of it. Friction depends on the real contact area, which scales with the normal force, not the apparent area. This is why friction is independent of the size of the object's footprint.

What friction is not

Friction is not caused by surface roughness alone. A polished surface can have higher friction than a rough one because polishing increases real contact area. The classic friction model is a macroscopic approximation that holds well for most engineering applications but breaks down at very smooth surfaces, very high pressures, or extremely slow speeds.

⚠ Limits of the model The standard friction model ($f = \mu N$) assumes: (1) $\mu$ is independent of speed, (2) $\mu$ is independent of contact area, (3) $\mu$ is constant for a given surface pair. All three assumptions fail in extreme conditions. For tyres, bearings, or precision manufacturing, more complex tribological models are needed.

§ 02

The Friction Model

Coulomb's friction laws (1781) provide the macroscopic model used in almost all classical mechanics problems.

Static friction

Prevents motion

$f_s \leq \mu_s N$

Adjusts to match the applied force, up to a maximum of $\mu_s N$. The object does not move as long as this condition is satisfied.

Kinetic friction

Opposes sliding

$f_k = \mu_k N$

Once sliding begins, friction becomes constant at $\mu_k N$, regardless of speed (within the basic model). Always $\mu_k < \mu_s$.

Rolling resistance

Opposes rolling

$f_r = C_{rr} N$

Much smaller than sliding friction. Arises from deformation at the contact patch. $C_{rr} \approx 0.01$ for car tyres on tarmac.

The full friction equation

Friction force — general statement

Static regime (not moving)

$$f_s \leq \mu_s N$$

$f_s$ adjusts automatically to whatever is needed to prevent motion, up to the maximum.

Kinetic regime (sliding)

$$f_k = \mu_k N$$

Exact equality — kinetic friction is constant once motion begins. Direction is always opposite to velocity.

Why $\mu_k < \mu_s$?

When surfaces are in static contact, asperity junctions have time to form and grow — adhesion increases with contact time. Once sliding begins, junctions shear and re-form rapidly, but fewer bonds exist at any instant. The force needed to maintain sliding is therefore less than the force needed to initiate it.

This is why a heavy box is hardest to get moving, and once it's moving, a slightly smaller force keeps it going. This hysteresis is exploited in door stoppers, anti-skid systems, and walking biomechanics.

Symbol reference

Symbol	Quantity	SI Unit	Notes
$f_s$	Static friction force	N	Variable; $0 \leq f_s \leq \mu_s N$
$f_k$	Kinetic friction force	N	Constant once sliding; $f_k = \mu_k N$
$\mu_s$	Static friction coefficient	dimensionless	Always $> \mu_k$ for the same pair
$\mu_k$	Kinetic friction coefficient	dimensionless	Approximately constant with speed
$N$	Normal force	N	Perpendicular to surface; not always $= mg$

Friction force vs Applied force — phase diagram

§ 03

Static vs Kinetic — In Detail

The transition from static to kinetic friction is one of the most practically important phenomena in mechanics. Understanding it prevents costly engineering errors.

The static friction regime

Static friction is a reactive force — it does not have a fixed value. It takes whatever value is needed to prevent motion, up to its maximum $f_{s,\text{max}} = \mu_s N$.

Consider a block on a surface. If you apply a horizontal force $F = 3\ \text{N}$ and $f_{s,\text{max}} = 10\ \text{N}$, then $f_s = 3\ \text{N}$ (exactly opposing the applied force). The net force is zero and the block doesn't move. The friction force is not 10 N — it is exactly 3 N.

✗ Critical error to avoid Static friction does not always equal $\mu_s N$. That is only its maximum value. In most static problems, $f_s < \mu_s N$. Many students apply $f_s = \mu_s N$ universally and get incorrect answers. Use the maximum only when you know the object is on the verge of sliding.

Free body diagrams — static and kinetic

Static — sub-maximum applied force

Kinetic — sliding with friction

Onset of motion — the critical condition

Motion begins when the applied force just exceeds $f_{s,\text{max}}$. At that exact threshold:

Condition for impending motion

$$F_\text{applied} = f_{s,\text{max}} = \mu_s N$$

The instant after this condition is met, friction drops to $f_k = \mu_k N < f_{s,\text{max}}$. The net force suddenly becomes nonzero and the object accelerates.

Worked Example 1 — Is the block moving?

Static friction

A 10 kg block sits on a surface with $\mu_s = 0.4$, $\mu_k = 0.3$. A horizontal force of 30 N is applied. Does the block move? If so, find its acceleration.

m = 10 kg μ_s = 0.4 μ_k = 0.3 F = 30 N

Normal force: $N = mg = (10)(9.8) = 98\ \text{N}$
Maximum static friction: $f_{s,\text{max}} = \mu_s N = (0.4)(98) = 39.2\ \text{N}$
Since $F = 30\ \text{N} < f_{s,\text{max}} = 39.2\ \text{N}$, the block does not move.
Actual static friction: $f_s = 30\ \text{N}$ (exactly opposing the applied force).
Net force = 0, acceleration = 0.

Block remains stationary. f_s = 30 N, a = 0

Worked Example 2 — Block starts sliding

Kinetic friction

Same block as above, but now F = 50 N. Find the acceleration.

m = 10 kg μ_k = 0.3 F = 50 N N = 98 N

$F = 50\ \text{N} > f_{s,\text{max}} = 39.2\ \text{N}$, so the block slides.
Kinetic friction: $f_k = \mu_k N = (0.3)(98) = 29.4\ \text{N}$
Net force: $F_\text{net} = F - f_k = 50 - 29.4 = 20.6\ \text{N}$
Acceleration: $a = F_\text{net}/m = 20.6/10 = 2.06\ \text{m s}^{-2}$

a = 2.06 m/s² in direction of applied force

§ 04

Friction on an Incline

Inclined plane problems are the most common application of friction in classical mechanics. The key is correctly resolving the normal force.

On a slope at angle $\theta$, the normal force is not equal to $mg$. You must resolve gravity into components perpendicular and parallel to the slope:

Forces on an incline

Normal force (⊥ to slope)

$$N = mg\cos\theta$$

This is always less than $mg$ for $\theta > 0$.

Gravity component (∥ to slope)

$$mg\sin\theta$$

This drives the object down the slope.

Friction (up the slope if sliding down)

$$f_k = \mu_k mg\cos\theta$$

Net acceleration down the slope

$F_\text{net} = mg\sin\theta - f_k$

Net force along slope = driving force − opposing friction

$F_\text{net} = mg\sin\theta - \mu_k mg\cos\theta$

Substituting $f_k = \mu_k N = \mu_k mg\cos\theta$

$F_\text{net} = mg(\sin\theta - \mu_k\cos\theta)$

Factor out $mg$

$$\boxed{a = g(\sin\theta - \mu_k\cos\theta)}$$

Divide by $m$ — the mass cancels entirely. Acceleration on a slope with friction is mass-independent.

Critical angle — when does the block slide?

The block starts to slide when the gravitational component down the slope exceeds maximum static friction:

Critical angle for sliding

$$mg\sin\theta_c = \mu_s mg\cos\theta_c \implies \tan\theta_c = \mu_s \implies \theta_c = \arctan(\mu_s)$$

The critical angle depends only on $\mu_s$. This gives a neat experimental method: tilt a surface until the object just starts to slide, measure the angle, and compute $\mu_s = \tan\theta_c$.

Interactive Angle Explorer

Live · F = ma

Normal force N

— N

mg sin θ (down slope)

— N

Max static friction

— N

Status

—

Acceleration

— m/s²

Critical angle θ_c

— °

Angle θ30°

Mass m5 kg

Static μ_s0.45

Worked Example 3 — Block sliding down a slope

Incline + friction

A 6 kg block slides down a 40° slope. $\mu_k = 0.25$. Find the acceleration.

m = 6 kg θ = 40° μ_k = 0.25

Normal force: $N = mg\cos40° = (6)(9.8)(0.766) = 45.0\ \text{N}$
Kinetic friction: $f_k = \mu_k N = (0.25)(45.0) = 11.25\ \text{N}$ (acts up the slope)
Driving force down slope: $mg\sin40° = (6)(9.8)(0.643) = 37.8\ \text{N}$
Net force: $F_\text{net} = 37.8 - 11.25 = 26.55\ \text{N}$
$a = F_\text{net}/m = 26.55/6 = 4.43\ \text{m s}^{-2}$
Verify using the formula: $a = g(\sin40° - 0.25\cos40°) = 9.8(0.643 - 0.191) = 9.8 \times 0.452 = 4.43\ \text{m s}^{-2}$ ✓

a = 4.43 m/s² down the slope

Worked Example 4 — Pushing up a slope

Incline, friction reversal

The same block is now pushed up the 40° slope at constant velocity by an applied force parallel to the slope. Find the required force.

m = 6 kg θ = 40° μ_k = 0.25 constant velocity → a = 0

Block moves up → friction acts down the slope (opposing motion).
For constant velocity: $F_\text{net} = 0 \Rightarrow F = mg\sin\theta + f_k$
$f_k = \mu_k mg\cos\theta = (0.25)(6)(9.8)(0.766) = 11.25\ \text{N}$ (now pointing down slope)
$F = mg\sin\theta + f_k = 37.8 + 11.25 = 49.1\ \text{N}$
Compare to pushing down at constant velocity: $F = mg\sin\theta - f_k = 37.8 - 11.25 = 26.6\ \text{N}$

F = 49.1 N up the slope (vs 26.6 N down at same speed)

§ 05

Coefficient of Friction — Reference Table

$\mu$ depends entirely on the surface pair. These values are approximate — real-world values vary with temperature, surface condition, lubrication, and speed.

Surface Pair	μ_s (static)	μ_k (kinetic)
Rubber on dry concrete	0.7 – 0.8	0.5 – 0.7
Steel on steel (dry)	0.74	0.57
Wood on wood	0.25 – 0.5	0.2 – 0.4
Steel on steel (lubricated)	0.15	0.08
Rubber on wet concrete	0.4 – 0.5	0.2 – 0.35
Teflon on steel	0.04	0.04
Ice on ice	0.05 – 0.15	0.03 – 0.05
Synovial joints (human knee)	0.001 – 0.003	0.001 – 0.003

🦴 Biological engineering Human synovial joints — lubricated by synovial fluid — have friction coefficients lower than almost any engineered surface, including Teflon. The fluid acts as a hydrostatic bearing at high loads and as a boundary lubricant at low loads. This dual mechanism is why joints last decades under enormous repetitive loads.

§ 06

Rolling Resistance & Fluid Drag

Beyond sliding friction, two other resistive forces appear frequently: rolling resistance (for wheels) and fluid drag (for motion through air or liquid).

Rolling resistance

A rolling wheel experiences less friction than a sliding one — this is why wheels exist. Rolling resistance arises from the deformation of the tyre (or wheel) at the contact patch: the leading edge of the contact zone is slightly compressed, creating an asymmetric normal force distribution that applies a net retarding torque.

Rolling resistance

$$f_r = C_{rr} \cdot N$$

$C_{rr}$ is the rolling resistance coefficient. Typical values: bicycle tyre on tarmac ≈ 0.004, car tyre on tarmac ≈ 0.01–0.02, steel wheel on steel rail ≈ 0.001. Much smaller than $\mu_k$.

Fluid drag

When an object moves through air or water, fluid is displaced and a resistive force acts opposite to velocity. At low speeds (viscous regime), drag is proportional to velocity. At high speeds (inertial regime), drag is proportional to $v^2$.

Drag force models

Low speed — Stokes' Law

$$F_D = 6\pi\eta r v$$

$\eta$ = dynamic viscosity, $r$ = sphere radius. Valid for small, slow objects (dust particles, cells).

High speed — quadratic drag

$$F_D = \tfrac{1}{2}\rho v^2 C_D A$$

$\rho$ = fluid density, $C_D$ = drag coefficient, $A$ = cross-sectional area. Used for cars, aircraft, falling objects.

Terminal velocity

A falling object accelerates under gravity but its drag increases with speed. Eventually, drag equals weight and acceleration becomes zero — this is terminal velocity:

$mg = \tfrac{1}{2}\rho v_t^2 C_D A$

At terminal velocity, net force = 0, so drag = weight

$$v_t = \sqrt{\frac{2mg}{\rho C_D A}}$$

Solve for $v_t$. Terminal velocity increases with mass and decreases with drag area — why skydivers deploy parachutes.

🪂 Terminal velocity examples Skydiver (spread-eagle): $v_t \approx 55\ \text{m s}^{-1}$ (200 km/h). Skydiver (head-down): $v_t \approx 85\ \text{m s}^{-1}$. Raindrop (2mm): $v_t \approx 6\ \text{m s}^{-1}$. Ant: $v_t \approx 1\ \text{m s}^{-1}$ — ants survive falls from any height because their terminal velocity is low enough to be survivable.

Worked Example 5 — Terminal velocity

Quadratic drag

A 80 kg skydiver has $C_D = 1.0$, cross-sectional area $A = 0.7\ \text{m}^2$. Air density $\rho = 1.2\ \text{kg m}^{-3}$. Find terminal velocity.

m = 80 kg C_D = 1.0 A = 0.7 m² ρ = 1.2 kg/m³

$v_t = \sqrt{\dfrac{2mg}{\rho C_D A}} = \sqrt{\dfrac{2(80)(9.8)}{(1.2)(1.0)(0.7)}}$
Numerator: $2 \times 80 \times 9.8 = 1568$
Denominator: $1.2 \times 1.0 \times 0.7 = 0.84$
$v_t = \sqrt{1568/0.84} = \sqrt{1866.7} \approx 43.2\ \text{m s}^{-1}$

Terminal velocity ≈ 43 m/s ≈ 155 km/h

§ 07

Interactive Simulation

All forces computed from Newton's laws each frame. Explore static/kinetic transition, incline dynamics, and drag effects in real time.

Friction Lab

v — t

F_net — t

Friction &Surface Forces

The Nature of Friction

What friction is not

The Friction Model

The full friction equation

Why $\mu_k < \mu_s$?

Symbol reference

Static vs Kinetic — In Detail

The static friction regime

Free body diagrams — static and kinetic

Onset of motion — the critical condition

Friction on an Incline

Net acceleration down the slope

Critical angle — when does the block slide?

Coefficient of Friction — Reference Table

Rolling Resistance & Fluid Drag

Rolling resistance

Fluid drag

Terminal velocity

Interactive Simulation

Friction &
Surface Forces