# Consider the four wheeled vehicle travelling round a track banked at an angle $\theta$ to the horizontal, as shown. Let the conditions be such that when the vehicle is moving with a speed of $v$ there is no sideways force. The forces acting are then weight acting at the centre of gravity $G$ of the vehicle, and the normal reaction $R_1$ and $R_2$ at the wheels. Which of the following values of $\theta$ satisfies the condition for no sideways force?

$\begin{array}{1 1} \tan^{-1} v^2/gr \\ \cot^{-1} v^2/gr \\ \sin^{-1} v^2/gr \\\cos^{-1} v^2/gr \end{array}$

Answer :$\tan^{-1} v^2/gr$
By Newton's second law the force Fproducing this acceleration on a mass m is $F= ma= \large\frac{mv^2}{r}$
Taking moments about G gives $R_1 \large\frac{l}{2} $$= R_2 \large\frac{l}{2} Therefore R_1=R_2-the reaction forces are the same, we will call them both R from now on. Vertically there is no motion so no acceleration so resolving forces in the vertical direction gives 2R \cos \theta =mg ----(i) Resolving forces horizontally gives \large\frac{mv^2}{r}$$=2R \sin \theta$ -----(ii)
Dividing (ii) by (i) gives
$\tan \theta = \large\frac{\sin \theta}{\cos \theta} =\frac{mv^2}{mgr}$
$\tan \theta =\large\frac{v^2}{gr}$
Therefore $\theta= \tan ^{-1} v^2/gr$
edited Aug 25, 2014 by meena.p