A car of mass  $$m$$ moves on a banked road having radius  $$' r '$$ and banking angle  $$\theta.$$ To avoid slipping from the banked road, the maximum permissible speed of the car is  $$v_0.$$  The coefficient of friction $$\mu$$ between the wheels of the car and the banked road is:

consider forces on the car on a banked road

forces acting:

• weight = mg (vertical downward)
• normal reaction = N (perpendicular to surface)
• friction = f (along surface)

for maximum speed, the car tends to slip up the incline, so friction acts down the slope

resolve forces:

along horizontal (towards center → provides centripetal force)

N sinθ + f cosθ = m v₀² / r

along vertical (no vertical acceleration)

N cosθ − f sinθ = mg

at limiting condition:

f = μN

substitute f = μN

horizontal:

N sinθ + μN cosθ = m v₀² / r

N (sinθ + μ cosθ) = m v₀² / r …(1)

vertical:

N cosθ − μN sinθ = mg

N (cosθ − μ sinθ) = mg …(2)

divide (1) by (2):

(sinθ + μ cosθ) / (cosθ − μ sinθ) = (v₀² / rg)

cross multiply:

sinθ + μ cosθ = (v₀² / rg)(cosθ − μ sinθ)

expand:

sinθ + μ cosθ = (v₀² / rg)cosθ − (v₀² / rg)μ sinθ

collect μ terms:

μ cosθ + (v₀² / rg)μ sinθ = (v₀² / rg)cosθ − sinθ

factor μ:

μ [cosθ + (v₀² / rg) sinθ] = (v₀² / rg)cosθ − sinθ

final expression:

μ = [ (v₀² / rg)cosθ − sinθ ] / [ cosθ + (v₀² / rg) sinθ ]

$$ \mu=\frac{v_0^2-rg\tan\theta}{rg+v_0^2\tan\theta} $$

this is the required coefficient of friction for maximum speed.