How large must the coefficient of static friction be between the tires and the road if a car is to round a level curve of radius 125 m at a speed of 95 km/h?

The car moves on a frictional surface around a curve. Then the friction force balances the centripetal force.When the speed of the car is maximum, the car may cause slipping. At this time, the car will experience maximum static frictional value.

The given data can be listed below as,

• The radius of curve is, $r=125\mathrm{m}$.
• The speed of the car is, $v=95\mathrm{km}/\mathrm{h}\left(\frac{1\mathrm{m}/\mathrm{s}}{3.6\mathrm{km}/\mathrm{h}}\right)=26.389\mathrm{m}/\mathrm{s}$
• The acceleration due to gravity is, $g=9.81\mathrm{m}/{\mathrm{s}}^2$.

The diagram of the forces on the car can be represented as,

Here, $F_{\mathrm{N}}$is the normal force, mgis the weight of the car, m is the mass of the car, $F_{\mathrm{fr}}$is the frictional force.

From the above figure, apply the equilibrium equation in the vertical direction. As the car is not accelerating, then the vertical forces balance each other. The equation can be expressed as,

$\begin{array}{c}\sum F_y=0\\ F_{\mathrm{N}}-mg=0\\ F_{\mathrm{N}}=mg\end{array}$

In the above figure, the car is shown as it's emerging out of the page. The circular path centre is represented in the right direction on the page. The frictional force will cause the rotational motion of the car. Then, equate the friction force and the centripetal force of the car.

$\begin{array}{c}{F}_{\mathrm{c}}=F_{\mathrm{fr}}\\ m{a}_{\mathrm{c}}={\mu }_s{F}_{\mathrm{N}}\\ \frac{m{v}^2}{r}={\mu }_{s}mg\\ \frac{v^2}{r}={\mu }_{s}g\end{array}$

Here, $a_{\mathrm{c}}$is the centripetal acceleration of the car, v is the car’s speed, ${\mu }_{\mathrm{s}}$is the coefficient of static friction, $F_{\mathrm{c}}$is the centripetal force.

Substitute the values in the above equation.

$\begin{array}{c}\frac{{\left(26.389\mathrm{m}/\mathrm{s}\right)}^2}{125\mathrm{m}}={\mu }_{\mathrm{s}}\times 9.81\mathrm{m}/{\mathrm{s}}^2\\ {\mu }_{\mathrm{s}}=\frac{{\left(26.389\mathrm{m}/\mathrm{s}\right)}^2}{125\mathrm{m}\times 9.81\mathrm{m}/{\mathrm{s}}^2}\\ {\mu }_{\mathrm{s}}\approx 0.57\end{array}$

Thus, the coefficient of static friction is $0.57$.