Banking of roads — derivation: Visual walkthrough

We are hunting for one relationship: how steeply must a curved road tilt so that a car going speed $v$ around a circle of radius $r$ stays on its path — even with no grip at all?

Step 1 — The problem: a turn needs a sideways pull

WHAT. A car drives around a flat circular track. Watch its velocity arrow: the direction keeps changing even if the speed stays the same.

WHY. Anything moving in a circle is constantly being turned. Turning means the velocity vector rotates, and a rotating velocity is an acceleration — pointed straight at the centre of the circle. This is the centripetal ("centre-seeking") acceleration. To cause it, some real force must push the car inward. See Uniform Circular Motion for where this comes from.

PICTURE. The car (top view) with its velocity arrow tangent to the circle, and a red arrow pointing to the centre — the force we must somehow supply.

Step 2 — On flat ground, only friction can do it (and that's the danger)

WHAT. On a flat road the only horizontal force available is friction between tyre and road. So friction alone must equal $\frac{m v ^{2}}{r}$ .

WHY. Weight points straight down, the road's push (normal force) points straight up — both are vertical. Neither has a sideways part. The one force with a horizontal component is Friction. If the road is wet, icy, or oily, friction shrinks and the car slides outward off the curve.

PICTURE. Side view of a car on flat ground: weight down, normal up (both vertical, cancelling), and a lone horizontal friction arrow straining to point inward.

Step 3 — Tilt the road: the normal force learns to lean

WHAT. Raise the outer edge of the road by an angle $θ$ . The road surface is now a slope. The normal force — the push the road gives, always perpendicular to the surface — no longer points straight up. It tilts inward by that same angle $θ$ .

WHY perpendicular? A surface can only push, never pull sideways, along an ideal frictionless contact — so the push is at right angles to the surface. Tilt the surface by $θ$ , and the perpendicular tilts by $θ$ too. Now the normal force has grown a horizontal, inward-pointing component — exactly the direction we needed in Step 1.

PICTURE. The banked road in cross-section. The normal force $N$ (blue) leans off vertical by the banking angle $θ$ (yellow), while weight $m g$ (red) still hangs straight down.

Step 4 — Split $N$ into a vertical part and a horizontal part

WHAT. We break the tilted arrow $N$ into two arrows at right angles: one straight up, one straight sideways (inward).

WHY these two directions? Because the motion itself splits cleanly along them. Vertically the car does not accelerate — it neither sinks into the road nor lifts off, so vertical forces must balance. Horizontally the car does accelerate toward the centre. Choosing "up" and "inward" as our axes lets us write one honest equation per direction. (Same trick as the Inclined plane.)

PICTURE. A right triangle built on $N$ : the hypotenuse is $N$ itself, the vertical leg is $N cos θ$ , the horizontal leg is $N sin θ$ .

Step 5 — Write the two balance equations (frictionless case)

WHAT. Assume for now perfectly slippery ice — friction $f = 0$ . Write one equation up–down and one equation in–out.

WHY. With no friction, only $N$ and gravity act. This gives the cleanest possible result — the "design" case the road is built for.

Vertical — nothing accelerates up or down: $\underbrace{N\cos\theta}_{\text{up-part of the road's push}} \;=\; \underbrace{mg}_{\text{weight pulling down}} \tag{1}$ The upward slice of $N$ must exactly cancel gravity, or the car would sink or fly.

Horizontal — this is the centripetal force: $\underbrace{N\sin\theta}_{\text{inward slice of the road's push}} \;=\; \underbrace{\frac{mv^2}{r}}_{\text{inward force the turn demands}} \tag{2}$ The inward slice of $N$ is the entire centripetal force — no friction helping.

PICTURE. The two triangle legs colour-matched to their jobs: the vertical leg battling $m g$ , the horizontal leg supplying $\frac{m v ^{2}}{r}$ .

Step 6 — Divide the equations: the magic cancellation

WHAT. Divide equation (2) by equation (1), left side over left, right side over right.

WHY divide? Two unknowns clutter our equations: the normal force $N$ and the mass $m$ . Neither is something we get to choose when designing a road. Dividing makes both vanish at once — $N$ cancels on the left, $m$ cancels on the right — leaving only the quantities a road engineer actually controls: angle, speed, radius, gravity.

$\frac{N s i n θ}{N c o s θ} = \frac{m v ^{2} / r}{m g}$

The $N$ 's cancel; the $m$ 's cancel. And $\frac{s i n θ}{c o s θ}$ is by definition $tan θ$ — the "steepness ratio" of our triangle, how much sideways per unit up.

Step 7 — Friction returns: friction points DOWN when you go too fast

WHAT. Now allow friction. If the car goes faster than the design speed, it tends to slide up-and-outward. Friction opposes that tendency, so it acts down the slope, adding its own inward help — allowing a higher top speed $v_{ma x}$ .

WHY down the slope? Friction always fights the direction the surface would slip. Too fast ⇒ car wants to climb outward ⇒ surface would slip outward-up ⇒ friction grips downward-inward. It takes the biggest value it can, $f = μ N$ , where $μ$ = coefficient of Friction.

Vertical (down-slope friction has a downward slice too): $N\cos\theta = mg + \underbrace{\mu N\sin\theta}_{\text{down-part of friction}} \tag{3}$

Horizontal (both $N$ and friction now push inward): $N\sin\theta + \underbrace{\mu N\cos\theta}_{\text{inward-part of friction}} = \frac{mv_{max}^2}{r} \tag{4}$

Group (3) as $N (cos θ - μ sin θ) = m g$ , then divide (4) by it — again $N$ and $m$ vanish:

PICTURE. Fast car high on the bank: $N$ leaning in, friction arrow pointing down the slope, both feeding the inward total.

Step 8 — Go too slow: friction flips UP

WHAT. If the car creeps slower than the design speed, gravity's pull down the slope wins and the car tends to slide inward-and-down. Friction flips to act up the slope, holding it out — setting a lowest safe speed $v_{min}$ .

WHY. Same rule: friction opposes the slip direction. Now the slip is inward-down, so friction grips outward-up. Mathematically we just flip the sign of $μ$ in the previous result:

PICTURE. Slow car low on the bank: friction arrow now pointing up the slope, propping the car against gravity's inward slide.

Recall Check the two edge cases yourself

What is $v_{ma x}$ when the road is flat, $θ = 0$ ? ::: $r g \cdot μ = μ r g$ — the ordinary flat-road friction limit. The formula stays consistent. What happens to $v_{ma x}$ as $1 - μ tan θ \to 0$ (i.e. $tan θ \to 1/ μ$ )? ::: The denominator vanishes, $v_{ma x} \to \infty$ — a bank so steep friction can hold any speed.

The one-picture summary

Every force, every component, both friction directions, and the resulting formulas — compressed into a single cross-section.

Recall Feynman retelling — the walkthrough in plain words

A turning car is being yanked inward every instant, or it flies off the road (Step 1). On flat ground only the tyres' grip can yank it, and grip fails on ice (Step 2). So we tilt the road like the inside of a bowl. Now the ground's push, which is always straight out of the surface, leans toward the centre and does the yanking for us (Step 3). We slice that tilted push into an up-part and a sideways-part (Step 4): the up-part holds the car's weight, the sideways-part is the inward yank (Step 5). Dividing those two facts wipes out the mass and the size of the push, leaving a pure relation between tilt, speed, and turn: $tan θ = v^{2} / r g$ (Step 6). Add real friction and you get wiggle room — go fast and friction slides down to help (Step 7); go slow and it flips up to hold you (Step 8). That's a whole road-engineering rulebook from one right triangle.

Connections

Centripetal force — the inward force Steps 1–2 demand.
Uniform Circular Motion — where $v^{2} / r$ comes from.
Friction — the $μ$ in Steps 7–8, the safe-speed range.
Inclined plane — same resolve-into-components method.
Conical pendulum — twin geometry, same $tan θ = v^{2} / r g$ .
Newton's Second Law — applied per axis throughout.