Question

A turn of radius $20\text{m}$ is banked for the vehicle of mass $200\text{kg}$ going at a speed of $10\text{m/s}$. The minimum coefficient of static friction required, if the vehicle moves with a speed of $15\text{m/s}$ is given by $\frac{n}{8}$. Then, $n$ is
[Take $g=10{\text{m/s}}^2$]

Answer 1

The turn is banked for a speed of $10\text{m/s}$. As speed of the vehicle is $15\text{m/s}$ i.e. greater than banked speed, therefore vehicle will have a tendency to slip outwards. To prevent it, static friction will act inwards on the banked surface.


On applying equilibrium condition along vertical direction,
$N\cos \theta -mg-f\sin \theta =0$
$\Rightarrow N\cos \theta -f\sin \theta =mg...\left(1\right)$
On applying equation of dynamics towards the centre of horizontal circular path:
$N\sin \theta +f\cos \theta =\frac{m{v}^2}{r}...\left(2\right)$
On mulitplying (1) with $-\text{sin}\theta$ and (2) with $\text{cos}\theta$ and adding them gives,
$-N\cos \theta \sin \theta +f{\sin }^2\theta +N\sin \theta \cos \theta +f{\cos }^2\theta =-mg\sin \theta +\frac{m{v}^2}{r}\cos \theta$
$\Rightarrow f({\sin }^2\theta +{\cos }^2\theta )=\frac{m{v}^2}{r}\text{cos}\theta -mg\sin \theta$
$f=\frac{m{v}^2}{r}\text{cos}\theta -mg\sin \theta ...\left(3\right)$
Since $\text{tan}\theta =\frac{v^2}{rg}$
$\left[\text{Condition for banking of road}\right]$
$\Rightarrow \text{tan}\theta =\frac{{10}^2}{20\times 10}$
$\Rightarrow \text{tan}\theta =\frac{1}{2}=\frac{t}{2t}$


From triangle shown in figure,
$\text{cos}\theta =\frac{2}{\sqrt{5}}$
$\text{sin}\theta =\frac{1}{\sqrt{5}}$
Putting the values in Eq.$\left(3\right)$,
$\Rightarrow f=\frac{200\times {15}^2}{20}\times \frac{2}{\sqrt{5}}-200\times 10\times \frac{1}{\sqrt{5}}$
$\because v=15\text{m/s}$
$\Rightarrow f=500\sqrt{5}$
Since static friction will act at its limiting value to prevent slipping of vehicle
$\Rightarrow \mu N>500\sqrt{5}$
$\Rightarrow \mu \times mg\cos \theta >500\sqrt{5}$
$\Rightarrow \mu >\frac{500\sqrt{5}}{200\times 10\times \frac{2}{\sqrt{5}}}$
$\Rightarrow \mu >\frac{2500}{4000}=\frac{5}{8}$
Thus, minimum required coefficient of static friction $\mu =\frac{5}{8}$
On comparing with $\mu =\frac{n}{8}$, we get $n=5$