In the section 3.7.3 (Banking of road) we have not included the friction exerted by the road on the car. Suppose the coefficient of static friction between the car tyre and the surface of the road is `mu_s`, calculate the minimum speed with which the car can take safe turn? When the car takes turn in the banked road, the following three forces act on the car. (1) The gravitational force mg acting downwards (2) The normal force N acting perpendicular to the surface of the road (3) The static frictional force f acting on the car along the surface.

Explanation

Given `S=5m, mu=4/3 ,g= 10 m/s^2
v=36 km/h = 10 m/s,
v=0
`a=(v^2-u^2)/(2s)=(0-10^2)/(2xx5)`
=10 m/s^2
From the freebody diagrams
`R-mg costheta=0`
`rarr R=mg cos theta ………i
Again `ma+mg isn theta - mu R =0`
`rarr ma+mg sintheta=mu mg cos theta=0`
`a+g sin theta-mug costheta =0`
`rarr 10+10 sin theta -(4/3)xx10 cos theta=0`
`rarr `30+30 sin theta -40 cos theta =0`
`rarr 3+3 sin -4costheta=0`
`rarr 4 cois theta -3 sin theta =3`
`rarr 4 sqrt(1-sin^2theta)=3+3sin theta`
On squaring, we get, `16(1-sin^2theta)`
`=9+9 sin^2theta+18sintheta`
`rarr sin theta= (18+sqrt(18^2-4(25)(-7))/(2xx25)`
`=(-18+32(/50=14/50=0.58`
Taking +ve sign only
`rarr theta = sin^-1 (0.28)=16^0`
Hence maximum incliN/Ation,
`theta=16^0`