The friction coefficient between a road and the tyre of a vehicle is 4/3. Find the maximum incline the road may thave so that once hard brakes are applied and the wheel starts skidding, the vehicle going down at a speed of 36 km/hr is stopped within 5m.

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`