A car is negotiating a curved road of radius R. The road is banked at an angle $\mathrm{\theta }$. The coefficient of friction between the types of the car and the road is ${\mathrm{\mu }}_{\mathrm{s}}$. The maximum safe velocity on this road is 

In the case of vertical equilibrium

$\begin{array}{r}\mathrm{Ncos}\mathrm{\theta }=\mathrm{mg}+{\mathrm{f}}_1\sin \mathrm{\theta }\\ \end{array}$

$\mathrm{mg}=\mathrm{Ncos}\mathrm{\theta }-{\mathrm{f}}_1\sin \mathrm{\theta }.........\left(1\right)$

In the case of horizontal equilibrium,

$\mathrm{Nsin}\mathrm{\theta }+{\mathrm{f}}_1\cos \mathrm{\theta }=\frac{{\mathrm{mv}}^2}{\mathrm{R}}\dots \dots \dots \left(2\right)$

On dividing equation (1) and (2), we get,

$\frac{{\mathrm{v}}^2}{\mathrm{Rg}}=\frac{\sin \mathrm{\theta }+{\mathrm{\mu }}_{\mathrm{s}}\cos \mathrm{\theta }}{\cos \mathrm{\theta }-{\mathrm{\mu }}_{\mathrm{s}}\sin \mathrm{\theta }}\dots \dots .\left({\mathrm{f}}_1\propto {\mathrm{\mu }}_{\mathrm{s}}\right)$

On dividing both the numerator and denominator by $\cos \theta$,

$\mathrm{v}=\sqrt{\mathrm{Rg}\left(\frac{\tan \mathrm{\theta }+{\mathrm{\mu }}_{\mathrm{s}}}{1-{\mathrm{\mu }}_{\mathrm{s}}\tan \mathrm{\theta }}\right)}$