Answer:
[tex]\theta \approx 3.649^{\textdegree}[/tex]
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:
[tex]\Sigma F_{x} = N\cdot \sin \theta = m\cdot \frac{v^{2}}{R_{curve}}[/tex]
[tex]\Sigma F_{y} = N - m\cdot g \cdot \cos \theta = 0[/tex]
The following expression is derived after manipulating both equations of equilibrium algebraically and trigonometrically:
[tex]m\cdot g \cdot \sin \theta\cdot \cos \theta = m \cdot \frac{v^{2}}{R_{curve}}[/tex]
[tex]g \cdot \sin \theta\cdot \cos \theta = \frac{v^{2}}{R_{curve}}[/tex]
[tex]\frac{1}{2}\cdot g \cdot \sin 2 \theta = \frac{v^{2}}{R_{curve}}[/tex]
[tex]\sin 2\theta = \frac{2\cdot v^{2}}{g\cdot R_{curve}}[/tex]
[tex]\theta = \frac{1}{2}\cdot \sin^{-1} \left(\frac{2\cdot v^{2}}{g\cdot R_{curve}} \right)[/tex]
The banking angle is:
[tex]\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][/tex]
[tex]\theta \approx 3.649^{\textdegree}[/tex]
