Answer:
a) [tex]\tan \theta = \frac{v^{2}}{g\cdot R}[/tex]
Explanation:
a) The Free Body Diagram of the vehicle and reference axis are included in the image attached below. Equations or equilibrium are presented below:
[tex]\Sigma F_{x} = N\cdot \sin \theta = m\cdot \frac{v^{2}}{R}[/tex]
[tex]\Sigma F_{y} = N\cdot \cos \theta - m\cdot g = 0[/tex]
Tangent of the highway can be found by dividing the first expression by the second one:
[tex]\tan \theta = \frac{m\cdot \frac{v^{2}}{R} }{m\cdot g}[/tex]
[tex]\tan \theta = \frac{v^{2}}{g\cdot R}[/tex]
Part b) :
Calculate the angle of banking on the highway
Answer:
a)[tex]tan \theta = \frac{v^{2} }{rg}[/tex]
b) [tex]\theta = 10.97^{0}[/tex]
Explanation:
traffic speed, v = 88 km/ h = 88 * (1000/3600) = 24.44 m/s
Radius of the curve, r = 314 m
a) Write an equation for the tangent of the highway's angle of banking.
The traffic should move both in the vertical and horizontal directions
In the vertical direction, it is moving against gravity, therefore
[tex]\sum F_{y} = 0\\Ncos \theta - mg =0\\[/tex]
[tex]Ncos \theta = mg[/tex]..............(1)
For the horizontal movement,
[tex]\sum F_{y} = 0\\Nsin \theta - ma =0\\[/tex]
[tex]Nsin \theta = ma[/tex]................(2)
To get the tangent equation, divide equation (2) bu equation (1)
[tex]\frac{N sin \theta}{N cos \theta} = \frac{ma}{mg}[/tex]
[tex]tan \theta = \frac{a}{g}[/tex]...........(3)
Centripetal acceleration, [tex]a = \frac{v^{2} }{r} \\[/tex]..........(4)
Substituting (4) into (3)
[tex]tan \theta = \frac{v^{2} }{r} /g\\tan \theta = \frac{v^{2} }{rg}[/tex]
b) Angle of banking on the high way
v = 24.44 /s
r = 314 m
g = 9.81 m/s^2
Substituting those values into the result gotten in part a
[tex]tan \theta = \frac{24.44^{2} }{314*9.81}\\tan \theta = 0.194\\\theta = tan^{-1} 0.194\\[/tex]
[tex]\theta = 10.97^{0}[/tex]
