A turn of radius $20 \mathrm{m}$ is banked for the vehicles going at a speed of $36 \mathrm{km}/\mathrm{h}$. If the coefficient of static friction between the road and the tyre is 0.4. What are the possible speeds of a vehicle, so that it neither slips down nor skids up ? (Take, $\mathrm{g}=9.8 \mathrm{m}/{\mathrm{s}}^2$ )

Banking angle, $\mathrm{\theta }={\tan }^{-1}\left(\frac{{\mathrm{v}}^2}{\mathrm{Rg}}\right)$

$36 \mathrm{km}/\mathrm{h}=10 \mathrm{m}/\mathrm{s}$

$\therefore \mathrm{\theta }={\tan }^{-1}\left(\frac{100}{20\times 9.8}\right)=27°$

Angle of repose,

${\mathrm{\theta }}_{\mathrm{r}}={\tan }^{-1}\left(\mathrm{\mu }\right)={\tan }^{-1}(0.4)=21.8°$

Since $\mathrm{\theta }>{\mathrm{\theta }}_{\mathrm{r}}$, vehicle cannot remain in the given position with v=0 . At rest, it will slide down. To find minimum speed, so that vehicle does not slip down, maximum friction will act up the plane.

To find maximum speed, so that the vehicle does not skid up, maximum friction will act down the plane. Minimum Speed

Equation of motion are,

$\mathrm{Ncos\theta }+\mathrm{\mu Nsin\theta }=\mathrm{mg}$

$\mathrm{Nsin\theta }-\mathrm{\mu Ncos\theta }=\frac{\mathrm{m}}{\mathrm{R}}{\mathrm{v}}_{\min }^2$

Solving these two equations, we get

$v_{\text{min }}=4.2 \mathrm{m}/\mathrm{s}$

Maximum Speed

Equations of motion are,

$\mathrm{Ncos\theta }-\mathrm{\mu Nsin\theta }=\mathrm{mg}$N 

$\mathrm{sin\theta }+\mathrm{\mu Ncos\theta }=\frac{\mathrm{m}}{\mathrm{R}}{\mathrm{v}}_{\max }^2$

Solving these two equations, we have

$v_{\text{max }}=15 \mathrm{m}/\mathrm{s}$