Question

An engineer must design a curved exit ramp for a highway in such a way that a car, exiting at the posted speed limit of 15.64 m/s (35 mi/hr), does not depend on friction to round the curve without skidding. The radius of the curve is 182.0 m. At what angle with respect to the horizontal must the curve be banked (in degrees)?

Answer 1

This is a famous dynamics problem. Everyone who has taken a dynamics course has had this problem.

Let's draw a free body diagram of the car on the banked road. Let positive x be towards the center of the radius of curvature and y be positive upwards from the road surface. Let \(\theta\) be the angle the road makes with the horizontal.

We can balance the forces in the x direction as such. \[\sum F_x = ma \rightarrow F_N \sin(\theta) = m {v^2 \over \rho}\]where \(\rho\) is the radius of curvature of the circle and \(F_N\) is the normal force of the car to the road.
We can balance the forces in the y direction as such. \[\sum F_y = 0 \rightarrow F_N \cos(\theta) - mg = 0\] Solving both equations for \(F_N\) and setting them equal yields \[{m v^2 \over \rho \sin(\theta)} = {mg \over \cos(\theta)}\]Solving for \(\theta\) \[{\sin(\theta) \over \cos(\theta)} = {v^2 \over g \rho} \rightarrow \tan(\theta) = {v^2 \over g \rho}\]

Answer 2

equation along x axis
\[N \sin \theta=M \times (V ^{2}\div R) .....eq1\]
\[N \cos \theta=Mg.....eq2\]
divide equation 1 by 2
\[\tan \theta=V ^{2}/(R \times G)\]
\[\theta =\tan^{-1} (V ^{2}\div(R \times G)\]
\[\theta=7.4 degrees\]