Laws of Motion · Lesson 12 of 14

Banked Roads and Safe-Turn Conditions

“Resolve normal and friction forces on a banked curve and derive the friction-free and limiting-speed relations.”

Learning Objectives

• Derive the design speed from normal-force components. • Determine friction direction above and below the design speed. • Derive the maximum-speed formula with static friction. • Recognise the assumptions and conditions on the formula.

Banking gives the normal force an inward component

On a banked road the outer edge is higher than the inner edge. The normal force is tilted from vertical toward the curve’s centre, so it can supply horizontal inward acceleration without requiring all of it from tyre friction. Use vertical and horizontal inward axes. The car’s height is constant during the ideal steady turn, so vertical acceleration is zero while radial acceleration is v²/R.

The friction-free design speed

No lateral friction neededLaTeX

N cos θ = m g, N sin θ = \frac{m v _{0}^{2}}{R}

θ is the road banking angle from horizontal. The normal is tilted by the same angle from vertical.

Design speedLaTeX

tan θ = \frac{v _{0}^{2}}{R g} ⟹ v_{0} = R g tan θ

At this speed, normal-force components alone support weight and supply inward acceleration in the ideal model.

Divide the inward equation by the vertical equation to eliminate N and mass. A higher bank angle or a larger turn radius allows a larger design speed. A frictionless bank permits this particular speed, not an arbitrary speed range. The expression is a model value, rather than a practical driving instruction for changing road conditions.

Which way does friction act?

Above the design speed, the car would tend to slip outward and up the bank. Static friction then acts down the slope, with an inward horizontal component and a downward vertical component. Below the design speed, the car tends to slide down the bank, so friction points up the slope. Draw the actual tendency before assigning friction direction. At v = v₀ the required lateral friction is zero.

Forces near the high-speed banked-road limit— Cross-section: the road rises outward to the right. At high speed, normal points upward and inward (left), while friction down the bank points inward and downward. The inward axis is left in this sketch. Arrow lengths are schematic unless stated otherwise.

Derive the high-speed limit

At impending upward/outward slip, friction points down the bank and has magnitude f = μ_sN. Its vertical component is downward f sinθ; its radial component is inward f cosθ. Vertical balance and inward acceleration therefore give two equations. Substituting the limiting friction equality and dividing them eliminates N and m.

Force components with down-bank frictionLaTeX

N cos θ - f sin θ = m g, N sin θ + f cos θ = \frac{m v ^{2}}{R}

These directions describe the high-speed case. At the maximum no-slip speed, put f = μ_sN.

Maximum no-slip speed in the modelLaTeX

v_{m a x} = R g \frac{tan θ + μ _{s}}{1 - μ _{s} tan θ}

This finite upper-bound formula assumes 1 − μ_s tanθ > 0, steady turning, maintained contact, and the stated high-speed friction direction.

To see the algebra, vertical balance gives N(cosθ − μ_s sinθ) = mg. Substitute into the inward equation, divide numerator and denominator by cosθ, and obtain v²/(Rg) = (tanθ + μ_s)/(1 − μ_s tanθ). With μ_s = 0, it reduces to the design-speed result. With θ = 0, it reduces to the level-road limit √(μ_sRg). These limiting cases check the signs.

Low speed and the possibility of parking

At low speed, friction points up the bank, so its vertical component is upward and its inward component is outward. At impending downhill slip, equations become N cosθ + μ_sN sinθ = mg and N sinθ − μ_sN cosθ = mv²/R. The corresponding nonnegative lower speed bound is useful when the bank is steep relative to the available friction.

Lower no-slip speed, when neededLaTeX

v_{m i n}^{2} = R g \frac{tan θ - μ _{s}}{1 + μ _{s} tan θ} if tan θ > μ_{s}

If tanθ ≤ μ_s, static friction can hold a parked car in this simplified model and the lower bound is zero. This extension explains the friction-direction reversal.

A 300 m radius track banked at 15°

Problem
Take μ_s = 0.20 and g = 9.8 m/s². Find friction-free design speed and maximum no-slip speed.

1.The bank factor is tan15° = 0.267949.
2.v₀ = √(300 × 9.8 × 0.267949) = 28.067 m/s, about 28.1 m/s.
3.v_max = √[2940(0.267949 + 0.20)/(1 − 0.20 × 0.267949)] = 38.127 m/s, about 38.1 m/s.
4.The denominator is positive, as required. The larger speed needs down-bank static friction, while the design speed needs no lateral friction in the model.
5.For completeness, because tan15° exceeds 0.20, the simplified lower bound is about 13.77 m/s. A parked vehicle could not be held by that friction coefficient alone on this bank.

The normal force is not simply mg on a bank

Even at design speed, N = mg/cosθ, which exceeds mg because only the vertical component supports weight. At the high-speed friction limit, N = mg/(cosθ − μ_s sinθ). Force components, not whole force magnitudes, belong in the vertical balance. This is another reason to determine the normal force from a diagram rather than setting it equal to weight by habit.

Quick check

At a banked road’s friction-free design speed, what supplies the inward force?

Key Takeaways

• Banking tilts the normal force inward. • Design speed follows from dividing inward and vertical component equations. • Friction points down the bank above design speed and up the bank below it. • The maximum-speed formula has assumptions and a denominator condition. • A banked-road normal force generally differs from mg.

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Forces in Circular Motion

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Changing Forces, Motion Graphs and Multi-Stage Problems