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 15 m/s. Assume that friction is sufficient to prevent slipping. $(g=10m/s^2)$

Answer 1

In the second case force of friction f will act downwards.
Using $\Sigma F_x=\frac{m{v}^2}{r}$
and $\Sigma F_y=0,$ we get
$N\sin \theta +f\cos \theta =\frac{m{v}^2}{r}$ ...(iii)
$N\cos \theta -f\sin \theta =mg$ ...(iv)
Substituting $\theta ={\tan }^{-1}\left(\frac{1}{2}\right),v=15m/s,m=200kg$
and r=20 m in the above equations, we get
$f=500\sqrt{5}N$ (downwards)