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 (a) `5 m//s`
(b) `15 m//s`
Assume that friction is sufficient to prevent slipping. `(g=10 m//s^(2))`

Explanation

`v=10m//s`
`tan theta=(v^(2))/(rg)=((10)^(2))/((20)(10))=(1)/(2)" "rArr theta=tan^(-1)((1)/(2))`
Now, as speed is decreased, force of friction f acts upwards. Using the equations
`N sin theta -f cos theta=(mv^(2))/(r ), N cos theta+f sin theta=mg`
Substituting `theta=tan^(-1)((1)/(2)), v=5m//s, m=200kg`
and `r=20m`, in the above equations, we get
`f=300sqrt5N`
b) In the second case force of friction f will act downwards
`N sin theta+f cos theta=(mv^(2))/(r), N cos theta-f sin theta=mg`
Substituting `theta=tan^(-1)((1)/(2)),v=15m//s`,
`m=200kg and r=20m,` in the above
equations, we get `f=500sqrt5N`