Question

A highway curve of radius 833 m is designed for traffic moving at a speed of 82 km/h to be able to traverse the curve without the aid of friction. What is the correct banking angle?

Answer 1

Answer:

$\theta \approx 3.649^{\textdegree}$

Explanation:

Let model an car moving in the highway curve by using Newton's Laws, whose equations of equilibrium is (x is the axis in the radial direction and y' is the axis perpendicular to the highway ground:

$\Sigma F_{x} = N\cdot \sin \theta = m\cdot \frac{v^{2}}{R_{curve}}$

$\Sigma F_{y} = N - m\cdot g \cdot \cos \theta = 0$

The following expression is derived after manipulating both equations of equilibrium algebraically and trigonometrically:

$m\cdot g \cdot \sin \theta\cdot \cos \theta = m \cdot \frac{v^{2}}{R_{curve}}$

$g \cdot \sin \theta\cdot \cos \theta = \frac{v^{2}}{R_{curve}}$

$\frac{1}{2}\cdot g \cdot \sin 2 \theta = \frac{v^{2}}{R_{curve}}$

$\sin 2\theta = \frac{2\cdot v^{2}}{g\cdot R_{curve}}$

$\theta = \frac{1}{2}\cdot \sin^{-1} \left(\frac{2\cdot v^{2}}{g\cdot R_{curve}} \right)$

The banking angle is:

$\theta = \frac{1}{2}\cdot \sin^{-1}\left[\frac{2\cdot[(82\,\frac{km}{h} )\cdot (\frac{1000\,m}{1\,km} )\cdot (\frac{1\,h}{3600\,s} )]^{2}}{(9.807\,\frac{m}{s^{2}} )\cdot (833\,m)} \right]$

$\theta \approx 3.649^{\textdegree}$

Answer 2

Answer:

3.6391° ( degrees )

Explanation:

The banking angle is the angle at which a vehicle is inclined about its longitudinal axis with respect to its horizontal axis when navigating a curve while banking of the road is the raising of the outer side of the road higher than its inner parts using a certain angle.

to calculate banking angle Ф

Ф = tan⁻¹ ( $\frac{V^{2} }{R*g }$ )  (equation 1)

v = speed = 82 km/h

R = radius = 833 m

g = force of gravity = 9.80

first convert V from Km/h to m/s

hence 82 km/h = 22.78 m/s

back to equation 1

Ф = tan⁻¹ ( $\frac{22.78^{2} }{833*9.80}$ )

   = tan⁻¹ ( $\frac{518.9284}{8163.4}$ ) = tan⁻¹ 0.0636

   = 3.6391⁰ ( degrees)