Chapter 5, Problem 23P

To determine

ToDetermine:

Range of speeds so thatthe car safely makes the curve.

Answer to Problem 23P

Solution:

The range of speed of the vehicle so that it neither skids nor slips.

  $v_{\max }=\text{114}\text{.3 km/h}$ and $v_{\min }=\text{59}\text{.11 km/h}$ .

Explanation of Solution

Given:

Banked for design speed, $v=85\text{km/h}$

The radius of curvature, $r=78\text{m}$

The coefficient of static friction, $\mu =0.30$

Formula used:

Centripetal acceleration $\left(a_R\right)=\frac{v^2}{r}$

Where

V is speed of the vehicle

R is Radius of curve

Newton’s second law

  ${\displaystyle \sum F=ma}$

Banking of road:

  $\tan \theta =\frac{v^2}{Rg}$

Where,

v :design speed

r :radius of the curve

g :acceleration due to gravity

$\theta$:Banking angle

Calculation:

  

N is the normal reaction force.

  $\begin{array}{l}N\text{cos}\theta =mg\\ N\text{sin}\theta =\frac{m{v}^2}{r}\end{array}$

  $\tan \theta =\frac{v^2}{rg}$

  $\begin{array}{l}r=78\text{m}\\ g=9.80{\text{m/s}}^{\text{2}}\\ v=\frac{\text{85}\text{km}}{\text{1}\text{h}}\\ \frac{85\times 1000\text{m}}{60\times 60\text{s}}\\ =23.61\text{m/s}\end{array}$

  $\begin{array}{l}\tan \theta =\frac{{23.61}^2}{78\times 9.80}\\ =0.73\\ \theta =\arctan \left(0.73\right)\\ =36.10\end{array}$

If a car istraveling at the maximum speed, then fiction force acts inward and equal to ${\mu }_s\text{N}$

Resolving the vertical and horizontal direction, apply Newton's second law

  $\begin{array}{l}{\displaystyle \sum F_y=o}\\ N\cos \theta =mg+{\mu }_{s}N\sin \theta \\ N(\cos \theta -{\mu }_s\sin \theta )=mg\\ N=\frac{mg}{\left(\cos \theta -{\mu }_s\sin \theta \right)}\cdots \cdots \cdots \left(1\right)\end{array}$

Apply Newton’s second law in the horizontal direction

  $\begin{array}{l}N\sin \theta +{\mu }_{s}N\cos \theta =m{a}_R\\ N\left(\sin \theta +{\mu }_s\cos \theta \right)=\frac{m{v}^2}{r}\end{array}$

Putting the value of Ninabove equation

  $\begin{array}{l}\frac{mg}{\left(\cos \theta -{\mu }_s\sin \theta \right)}\left(\sin \theta +{\mu }_s\cos \theta \right)=\frac{m{v}_{\max }{}^2}{r}\\ v_{\max }{}^2=\frac{gR\left(\sin \theta +{\mu }_s\cos \theta \right)}{\left(\cos \theta -{\mu }_s\sin \theta \right)}\end{array}$

Divide left side by $\text{cos}\theta$ from numerator and denominator.

  $\begin{array}{l}{v}^2=\frac{gR\left(\tan \theta +{\mu }_s\right)}{\left(1-{\mu }_s\tan \theta \right)}\\ v_{\max }=\sqrt{\frac{gR\left(\tan \theta +{\mu }_s\right)}{\left(1-{\mu }_s\tan \theta \right)}}\end{array}$

  $\begin{array}{l}{v}_{\max }=\sqrt{\frac{9.80\times 78\times \left(0.73+0.30\right)}{\left(1-0.30\times 0.73\right)}}\\ =31.75\text{m/s}\\ =\frac{31.75}{1000}\times 60\times 60\text{km/h}\\ \text{=114}\text{.3}\text{km/h}\end{array}$

For $v_{\min }$ , vehicle start sliding down for friction force start acting along the plane in a downward direction, so friction force start acting in an upward direction similar to previous, and we can calculate

  $v_{\min }=\sqrt{\frac{gR\left(\tan \theta -{\mu }_{{}_s}\right)}{\left(1-{\mu }_s\tan \theta \right)}}$

  $\begin{array}{l}\sqrt{\frac{9.8\times 78\times \left(0.73-0.30\right)}{\left(1+0.30\times 0.73\right)}}\\ =16.42\text{m/s}\\ =\frac{16.42\times 60\times 60}{1000}\text{km/h}\\ \text{=59}\text{.11km/h}\end{array}$

Conclusion: Hence, the minimum speed of the vehicle for safety is from $59.11\text{km/h to 114}\text{.30}\text{km/h}$ .