Question

The friction coefficient between a road and the tyre of a vehicle is $4/3$. 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 $36km/hr$ is stopped within $5m$.

Answer 1

Given that,

$s=5m$

$\mu =4/3$

$g=10m/s^2$

$u=36km/hr=36x5/18=10m/s$

$V=0m/s$

Now, from equation of motion

$v^2-u^2=2as$

$a=\frac{v^2-u^2}{2s}$

$a=\frac{0-{10}^2}{2\times 5}$

$a=-10m/s^2$

Now,

Normal force $N=mg\cos \theta$

Frictional force

$F_f=\mu N$

$F_f=\frac{4}{3}\times mg\cos \theta$

Net force along the incline against the motion

$=\frac{4}{3}mg\cos \theta -mg\sin \theta$

Now, newton’s second law

$\frac{4}{3}mg\cos \theta -mg\sin \theta =ma$

$4mg\cos \theta -3mg\sin \theta =30$

$40\cos \theta -30\sin \theta =30$

$3\sin \theta -4\cos \theta +3=0$

$3(1+\sin \theta )=4\cos \theta$

$\frac{1+\sin \theta }{\cos \theta }=\frac{4}{3}$

$\frac{(\cos \frac{\theta }{2}+\sin \frac{\theta }{2})}{(\cos \frac{\theta }{2}-\sin \frac{\theta }{2})}=\frac{4}{3}$

$\frac{(1+\tan \frac{\theta }{2})}{(1-\tan \frac{\theta }{2})}=\frac{4}{3}$

$3+3\tan \frac{\theta }{2}=4-4\tan \frac{\theta }{2}$

$7\tan \frac{\theta }{2}=1$

$\tan \frac{\theta }{2}=\frac{1}{7}$

$\frac{\theta }{2}=8.13$

$\theta =16.2$

$\theta ={16}^0$

Hence, the maximum incline is ${16}^0$