derive an expression for banking of road with the help of diagram?

For the vehicle to go round the curved track at a reasonable speed without skidding, the greater centripetal force is managed for it by raising the outer edge of the track a little above the inner edge. It is called banking of circular tracks.

Consider a vehicle of weight Mg, moving round a curved path of radius r, with a speed v, on a road banked through angleθ.

The vehicle is under the action of the following forces:

• The weight Mg acting vertically downwards

• The reaction R of the ground to the vehicle, acting along the normal to the banked road OA in the upward direction

The vertical component R cos θ of the normal reaction R will balance the weight of the vehicle and the horizontal component R sin θ will provide the necessary centripetal force to the vehicle. Thus,

R cosθ = Mg …(i)

$R \sin \theta = \frac{M{v}^2}{r}$   ..........(ii)

On dividing equation (ii) by equation (i), we get

$$\begin{gathered} \frac{R \sin \theta }{R \cos \theta }= \frac{M{v}^2/r}{Mg} \\ \tan \theta = \frac{v^2}{rg} \end{gathered}$$

As the vehicle moves along the circular banked road OA, the force of friction between the road and the tyres of the vehicle, F = μR, acts in the direction AO.