For a vehicle moving on a banked curved road,using a free body diagram $(FBD)$,obtain the formula for the maximum safe speed $(v_{max})$.

(N/A) For a vehicle of mass $m$ moving on a banked road with angle $\theta$ and radius $R$,the forces acting are the normal force $N$,the gravitational force $mg$,and the maximum static friction $f_s = \mu_s N$.
Resolving forces vertically:
$N \cos \theta = mg + f_s \sin \theta$
$N \cos \theta = mg + \mu_s N \sin \theta$
$N(\cos \theta - \mu_s \sin \theta) = mg$ --- $(1)$
Resolving forces horizontally (providing centripetal force):
$N \sin \theta + f_s \cos \theta = \frac{mv_{max}^2}{R}$
$N \sin \theta + \mu_s N \cos \theta = \frac{mv_{max}^2}{R}$
$N(\sin \theta + \mu_s \cos \theta) = \frac{mv_{max}^2}{R}$ --- $(2)$
Dividing $(2)$ by $(1)$:
$\frac{\sin \theta + \mu_s \cos \theta}{\cos \theta - \mu_s \sin \theta} = \frac{v_{max}^2}{Rg}$
Dividing numerator and denominator by $\cos \theta$:
$\frac{\tan \theta + \mu_s}{1 - \mu_s \tan \theta} = \frac{v_{max}^2}{Rg}$
Thus,$v_{max} = \sqrt{Rg \left( \frac{\mu_s + \tan \theta}{1 - \mu_s \tan \theta} \right)}$.