A banked circular highway curve is designed for traffic moving at $\mathbf{60}\mathbf{}\mathbf{km}\mathbf{/}\mathbf{h}$. The radius of the curve is$\mathbf{200}\mathbf{}\mathbf{m}$. Traffic is moving along the highway at$\mathbf{40}\mathbf{}\mathbf{km}\mathbf{/}\mathbf{h}$on a rainy day. What is the minimum coefficient of friction between tires and road that will allow cars to take the turn without sliding off the road? (Assume the cars do not have negative lift.)

• Optimum speed,$\mathrm{v}=60\text{km/hr}$.
• The radius of the curve,$\mathrm{r}=200\text{\hspace{0.17em}m}$.

• Speed of traffic on a rainy day, $\mathrm{v}=40\text{\hspace{0.17em}km/hr}$

The problem deals with Newton's second law of motion, which states that the acceleration of an object is dependent upon the net force acting upon the object and the mass of the object. Also, it deals with the centripetal force. This is the force. It is a force that makes a body follow a curved path.

$\mathrm{\theta }={\tan }^{-1}\left(\frac{{\mathrm{v}}^2}{\mathrm{gR}}\right)$

where,

$\begin{array}{c}\mathrm{v}=60(1000/3600)\\ =17\text{\hspace{0.17em}m/s}\end{array}$

$\mathrm{R}=200\text{\hspace{0.17em}m}$

Substitute the values, and we get,

$\mathrm{\theta }={\text{tan}}^{-1}\left(\frac{{17}^2}{10\times 200}\right)$

$\mathrm{\theta }=8{.1}^{\text{o}}$

Now we consider a vehicle taking this banked curve at,

$\begin{array}{c}{\mathrm{v}}^{\text{'}}=40(1000/3600)\\ =11\text{\hspace{0.17em}m/s}\end{array}$

Its (horizontal) acceleration is ${\mathrm{a}}^{\text{'}}={\mathrm{v}}^{\text{'}2}/\mathrm{R}$, which has components parallel to the incline and perpendicular to it:

$\begin{array}{c}{\mathrm{a}}_{\parallel }={\mathrm{a}}^{\text{'}}\text{cos}\mathrm{\theta }\\ =\frac{\mathrm{v}{\text{'}}^2\text{cos}\mathrm{\theta }}{\mathrm{R}}\end{array}$

$\begin{array}{c}{\mathrm{a}}_{\perp }={\mathrm{a}}^{\text{'}}\text{sin}\mathrm{\theta }\\ =\frac{\mathrm{v}{\text{'}}^2\text{sin}\mathrm{\theta }}{\mathrm{R}}\end{array}$

These enter Newton's second law as follows (choosing downhill as the $+\mathrm{x}$direction and away-from-incline as $+\mathrm{y}$):

$\mathrm{mg}\text{sin}\mathrm{\theta }-{\mathrm{f}}_{\mathrm{s}}={\mathrm{ma}}_{\parallel }$

${\mathrm{F}}_{\mathrm{N}}-\mathrm{mg}\text{cos}\mathrm{\theta }={\mathrm{ma}}_{\perp }$

$\frac{{\mathrm{f}}_{\mathrm{s}}}{{\mathrm{F}}_{\mathrm{N}}}=\frac{\mathrm{mg}\text{sin}\mathrm{\theta }-{\mathrm{mv}}^{\text{'}2}\text{cos}\mathrm{\theta }/\mathrm{R}}{\mathrm{mg}\text{cos}\mathrm{\theta }+{\mathrm{mv}}^{\text{'}2}\text{sin}\mathrm{\theta }/\mathrm{R}}$

We cancel the mass and plug in the values,

$\begin{array}{c}\frac{{\mathrm{f}}_{\mathrm{s}}}{{\mathrm{F}}_{\mathrm{N}}}=\frac{\left(9.8\right)\text{sin}{8.1}^{\circ }-{11}^2\text{cos}{8.1}^{\circ }/200}{\left(9.8\right)\text{cos}{8.1}^{\circ }+{11}^2\text{sin}{8.1}^{\circ }/200}\\ \frac{{\mathrm{f}}_{\mathrm{s}}}{{\mathrm{F}}_{\mathrm{N}}}=0.078\\ {\mathrm{\mu }}_{\mathrm{s}}=0.078\end{array}$

Thus, the minimum coefficient of friction between tires and road that will allow cars to take the turn without sliding off the road is$0.078$ .