Question

A circular curve of highway is designed for traffic moving at 60 km/h. Assume the traffic consists of cars without negative lift. (a) If the radius of the curve is 150 m, what is the correct angle of banking of the road? (b) If the curve were not banked, what would be the minimum coefficient of friction between tires and road that would keep traffic from skidding out of the turn when traveling at60 km/h?

Short answer

(a) The correct angle of banking of the road is $\theta =11°$.

(b) The minimum coefficient of friction between tires and road is ${\mu }_s=0.19$.

Step-by-step solution

Given data:

Consider the given data as below.

The radius of curve, R =150 m

Speed of the car is,

$$\begin{gathered} v_{max}=60\mathrm{km}/\mathrm{h} \\ =60\mathrm{km}/\mathrm{h}\times \left(\frac{1\mathrm{m}/\mathrm{s}}{3.6\mathrm{km}/\mathrm{g}}\right) \\ =16.7\mathrm{m}/\mathrm{s} \end{gathered}$$

As known, that,

Acceleration due to gravity,$g=9.8\mathrm{m}/{\mathrm{s}}^2$

Understanding the concept:

Consider that the car is “on the verge of sliding out” – meaning that the force of static friction is acting “down the bank” (or “downhill” from the point of view of an ant on the banked curve) with maximum possible magnitude. First, consider the vector sum $\overrightarrow{\mathbf{F}}$of the (maximum) static friction force and the normal force.

Due to the fact that they are perpendicular and their magnitudes are simply proportional (Eq. 6-1), you can find $\overrightarrow{\mathbf{F}}$is at an angle (measured from the vertical axis) $\mathbf{\varphi }\mathbf{=}\mathbf{\theta }\mathbf{+}{\mathbf{\theta }}_{\mathbf{s}}$, where ${\mathbf{tan\theta }}_{\mathbf{s}}\mathbf{=}{\mathbf{\mu }}_{\mathbf{s}}$(compare with Eq. 6- 13), and $\mathit{\theta }$is the bank angle.

Now, the vector sum of $\overrightarrow{F}$ and the vertically downward pull (mg)of gravity must be equal to the (horizontal) centripetal force $\left(\frac{m{v}^2}{R}\right)$, which leads to a surprisingly simple relationship:

$$\begin{gathered} \tan \varphi =\frac{m{v}^2/R}{mg} \\ =\frac{v^2}{Rg} \end{gathered}$$

Writing this as an expression for the maximum speed, you have

role="math" localid="1661151842310" $v_{max}=\sqrt{Rg\tan \left(\theta +{\tan }^{-1}{\mu }_s\right)}$

$v_{max}=\sqrt{\frac{Rg\left(\tan \theta +{\mu }_s\right)}{1-{\mu }_s\tan \theta }}$ ….. (1)

Calculate the correct angle of banking of the road if the radius of the curve is 150m:

Note that the given speed is (in SI units) roughly 17 m/s .

If you do not want the cars to “depend” on the static friction to keep from sliding out (that is, if you want the component “down the back” of gravity to be sufficient), then you can set ${\mu }_s=0$in the equation (1), and you obtain

$$\begin{gathered} v_{max}=\sqrt{\frac{Rg\left(\tan \theta +0\right)}{1-\left(0\right)\tan \theta }} \\ =\sqrt{\frac{Rg\left(\tan \theta +0\right)}{1}} \\ =\sqrt{Rg\left(\tan \theta +0\right)} \end{gathered}$$

$$\begin{gathered} v_{max}^2=Rg\left(\tan \theta \right) \\ \theta ={\tan }^{-1}\left(\frac{v_{max}^2}{Rg}\right) \end{gathered}$$

Substitute the known values in the above equation.

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{\left(16.7\mathrm{m}/{\mathrm{s}}^2\right)}{\left(150\mathrm{m}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}\right) \\ ={\tan }^{-1}\left(\frac{278.89m^2/s^2}{1470m^2/s^2}\right) \\ ={\tan }^{-1}\left(0.189\right) \\ =10.7° \\ \theta \approx 11° \end{gathered}$$

Hence, the correct angle of banking of the road is $11°$.

The minimum coefficient of friction between tires and road:

Calculate the minimum coefficient of friction between tires and road that would keep traffic from skidding out of the turn when traveling at if the curve were not banked

If, however, the curve is not banked (so$\theta =0°$) then the above expression becomes

$$\begin{gathered} v=\sqrt{Rg\tan \left({\tan }^{-1}{\mu }_s\right)} \\ =\sqrt{Rg{\mu }_s} \end{gathered}$$

Solving this for the coefficient of static friction, you get

$\begin{array}{l}{v}^2=Rg{\mu }_s\\ {\mu }_s=\frac{v^2}{Rg}\end{array}$

Substitute known values in the above equation.

$$\begin{gathered} {\mathrm{\mu }}_{\mathrm{s}}=\frac{{\left(16.7\mathrm{m}/\mathrm{s}\right)}^2}{\left(150\mathrm{m}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)} \\ =\frac{\left(278.89{\mathrm{m}}^2/{\mathrm{s}}^2\right)}{\left(1470{\mathrm{m}}^2/{\mathrm{s}}^2\right)} \\ =0.19 \end{gathered}$$

Hence, the minimum coefficient of friction between tires and road is 0.19.