91影视

Consider the given data as below.

The radius of curve, R =150 m

Speed of the car is,

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

As known, that,

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

Consider that the car is 鈥渙n the verge of sliding out鈥� 鈥� meaning that the force of static friction is acting 鈥渄own the bank鈥� (or 鈥渄ownhill鈥� from the point of view of an ant on the banked curve) with maximum possible magnitude. First, consider the vector sum $\stackrel{\mathbf{鈫�}}{\mathbf{F}}$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 $\stackrel{\mathbf{鈫�}}{\mathbf{F}}$is at an angle (measured from the vertical axis) $\mathbf{蠒}\mathbf{=}\mathbf{胃}\mathbf{+}{\mathbf{胃}}_{\mathbf{s}}$, where ${\mathbf{迟补苍胃}}_{\mathbf{s}}\mathbf{=}{\mathbf{渭}}_{\mathbf{s}}$(compare with Eq. 6- 13), and $\mathit{胃}$is the bank angle.

Now, the vector sum of $\stackrel{鈫�}{F}$ 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:

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

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

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

$v_{max}=\sqrt{\frac{Rg\left(\tan 胃+{渭}_s\right)}{1-{渭}_s\tan 胃}}$ 鈥�.. (1)

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

If you do not want the cars to 鈥渄epend鈥� on the static friction to keep from sliding out (that is, if you want the component 鈥渄own the back鈥� of gravity to be sufficient), then you can set ${渭}_s=0$in the equation (1), and you obtain

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

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

Substitute the known values in the above equation.

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

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

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$胃=0掳$) then the above expression becomes

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

Solving this for the coefficient of static friction, you get

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

Substitute known values in the above equation.

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

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