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)rArrtheta=tan^(-1) ((1)/(2))`
Now as speed is decreased force of friction 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 =300 sqrt5N` ltbr gt(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 =500 sqrt5 N` .