A turn of radius $20 \mathrm{m}$ is banked for the vehicle of mass $200 \mathrm{kg}$ going at a speed of $10 \mathrm{m}/\mathrm{s}$. Find the direction and magnitude of frictional force acting on a vehicle if it moves with a speed $5 \mathrm{m}/\mathrm{s}$. Assume that friction is sufficient to prevent slipping. $\left(g=10 \mathrm{m}/{\mathrm{s}}^2\right)$

 The turn is banked for speed $v=10 \mathrm{m}/\mathrm{s}$

Therefore,

$\tan \theta =\frac{v^2}{rg}=\frac{(10{)}^2}{\left(20\right)\left(10\right)}=\frac{1}{2}$

Now, as the speed is decreased, force of friction f acts upwards.

Using the equations.

$\Sigma F_x=\frac{m{v}^2}{r}$

and

$\Sigma F_y=0,\text{ we get }$

$N\sin \theta -f\cos \theta =\frac{m{v}^2}{r}$

$N\cos \theta +f\sin \theta =mg$

Substituting, $\theta ={\tan }^{-1}\left(\frac{1}{2}\right),v=5 \mathrm{m}/\mathrm{s},m=200 \mathrm{kg}$ and $r=20 \mathrm{m}$, in the above equations, we get

$f=300\sqrt{5} \mathrm{N}$                                                                               (Upwards)