Question

The friction coefficient between a road and the tyre of a vehicle is $\frac{4}{3}$ . Find the maximum incline the road may have 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 5 m.

Answer 1

Given $S=5m,\mu =\frac{4}{3},g=10m/s^{2}u=36km/h=10m/sv=0a=\frac{v^2-u^2}{2s}=\frac{0-{10}^2}{2\times 5}=10m/s^2$
From the freebody diagrams
$R-mg\cos \theta =0\Rightarrow R=mg\cos \theta$
Again $ma+mg\sin \theta -\mu R=0\Rightarrow ma+mg\sin \theta -\mu mg\cos \theta =0\Rightarrow a+g\sin \theta -\mu g\cos \theta =0\Rightarrow 10+10sin\theta -\left(\frac{4}{3}\right)\times 10\cos \theta =0\Rightarrow 30+30\sin \theta -40\cos \theta =0\Rightarrow 3+3\sin -4\cos \theta =0\Rightarrow 4\cos \theta -3\sin \theta =3\Rightarrow 4\sqrt{(1-si{n}^2\theta )}=3+3\sin \theta$
On squaring, we get, $16(1-{\sin }^2\theta )$
$\Rightarrow \sin \theta =\frac{18+\sqrt{{18}^2-4\left(25\right)(-7)}}{2\times 25}=\frac{-18+32}{50}=\frac{14}{50}=0.28\Rightarrow \theta ={\sin }^{-1}\left(0.28\right)={16}^{\circ }$
Hence maximum inclination,
$\theta ={16}^{\circ }$