Question

A turn of radius 20 m is banked for the vehicle of mass 200 kg going at a speed of 10 m/s. Find the direction and magnitude of frictional force acting on a vehicle if it moves with a speed 5 m/s.

Answer 1

The turn is banked for speed v=10 m/s
Therefore, $\tan \theta =\frac{v^2}{rg}=\frac{{\left(10\right)}^2}{\left(20\right)\left(10\right)}=\frac{1}{2}$
Now, as the speed is decreased, force of friction f acts upwards.
Using the equations ${\Sigma }_x=\frac{m{v}^2}{r}$
and $\Sigma F_y=0,$ we get
$N\sin \theta -f\cos \theta =\frac{m{v}^2}{r}$
$N\cos \theta +f\sin \theta =mg$
Substituting, $\theta ={\tan }^{-1}\left(\frac{1}{2}\right),v=5m/s,m=200kgandr=20m,$ in the above equations, we get
$f=300\sqrt{5}N$ (outwards)