Q: 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 a) 5 m/s b) 15 m/s Assume that friction is sufficient to prevent slipping (g = 10m/s^{2})

Sol: v = 10 m/s

$\large tan\theta = \frac{v^2}{r g} $

$\large tan\theta = \frac{(10)^2}{20 \times 10 } = \frac{1}{2}$

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

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

$\large N cos\theta + f sin\theta = m g $ ;

Substituting $\large tan\theta = \frac{1}{2} $ , v = 5 m/s, m = 200 kg and r = 20 m,

We get , f = 300√5 N