Engineering a highway curve.If a car goes through a curve too fast, the car tends to slide out of the curve. For a banked curve with friction, a frictional force acts on a fast car to oppose the tendency to slide out of the curve; the force is directed down the bank (in the direction water would drain). Consider a circular curve of radius R 200 mand bank angle u, where the coefficient of static friction between tires and pavement is. A car (without negative lift) is driven around the curve as shown in Fig. 6-11. (a) Find an expression for the car speed ${\mathit{V}}_{\mathbf{m}\mathbf{a}\mathbf{x}}$that puts the car on the verge of sliding out.

(b) On the same graph, plot ${\mathit{V}}_{\mathit{m}\mathit{a}\mathit{x}}$versus angle u for the range $\mathbf{0}\mathbf{°}\mathbf{}\mathbf{to}\mathbf{}\mathbf{50}\mathbf{°}$, first for${\mathit{\mu }}_{\mathbf{s}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{60}$(dry pavement) and then for${\mathit{\mu }}_{\mathbf{s}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{050}$(wet or icy pavement). In kilometers per hour, evaluate${\mathit{V}}_{\mathbf{max}}$for a bank angle of$\mathit{\theta }\mathbf{=}\mathbf{10}\mathbf{°}$and for

(c)${\mathit{\mu }}_{\mathbf{s}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{60}$and

(d)${\mathit{\mu }}_{\mathbf{s}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{050}$. (Now you can see why accidents occur in highway curves when icy conditions are not obvious to drivers, who tend to drive at normal speeds.)

• The radius of the circular curve, R = 200 m.
• The bank angle is$\theta$.
• The coefficient of friction is ${\mu }_s$.

The problem deals with Newton’s second law of motion, which states that the acceleration of an object is related to the net force acting upon the object and the mass of the object.

Use Newton’s second law of motion with zero acceleration to find the expression of$V_{max}$in terms of bank angle and frictional coefficient.

Formula:

F = ma

The angle of$\overrightarrow{F}$(measured from the vertical axis):

$\varphi =\theta +{\theta }_s$

Where,

$\tan {\theta }_s={\mu }_s,\mathrm{and},\theta$is the bank angle.

The vector sum of $\overrightarrow{F}$ and the vertically downward pull (mg) of gravity must be equal to the (horizontal) centripetal force $(m{v}^2/R)$:

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

Substitute the expression of $\varphi$in the above expression and simplify:

$$\begin{gathered} V_{max}^2=Rg\tan \left(\theta +{\tan }^{-1}{\mu }_s\right) \\ {V}_{max}=\sqrt{Rg\times \frac{\tan \left(\theta \right)+{\mu }_s}{1-{\mu }_s\tan \left(\theta \right)}} \end{gathered}$$

Thus, the expression for v_max is $V_{max}=\sqrt{Rg\times \frac{\tan \left(\theta \right)+{\mu }_s}{1-{\mu }_s\tan \left(\theta \right)}}$.

The graph is shown below (with $\theta$in radian):

Thus the graph is

For${\mu }_s=0.60$( the upper curve ), when$\theta =10°=0.175$radian, the max velocity can be calculated as:

$$\begin{gathered} V_{max}=\sqrt{200\times 9.8\times \frac{\tan (10°)+0.60}{1-0.60\times \tan (10°)}} \\ {V}_{max}=41.3\mathrm{m}/\mathrm{s} \\ {V}_{max}=149\mathrm{km}/\mathrm{h} \end{gathered}$$

Thus, the value of max velocity is 149km/h.

For${\mu }_s=0.050$( the lower curve ), when$\theta =10°=0.175$radian, the max velocity can be calculated as:

$$\begin{gathered} V_{max}=\sqrt{200\times 9.8\times \frac{\tan (10°)+0.50}{1-0.50\times \tan (10°)}} \\ {V}_{max}=21.2\mathrm{m}/\mathrm{s} \\ {V}_{max}=76.2\mathrm{km}/\mathrm{h} \end{gathered}$$

Thus, the value of max velocity is 76,2 km/h.