Question

A car moving at 12.5m/s rounds a 140m radius banked turn designed for 16.67m/s. What coefficient of friction is needed to keep the car on the road?

I find the slope of the roads curve = 11.6 degrees, but I can't seem to get the correct coefficient of friction.

Answer 1

Without friction, you calculate the slope,
\[\theta=\arctan\left(\frac{v^2}{Rg}\right)=11.45º\] Then, we use the Newton's laws,

Y axis: \[N-\frac{mv^2}{R}\sin\theta-mg\cos\theta=0\]
X axis: \[-F_r-\frac{mv^2}{R}\cos\theta+mg\sin\theta=0\]
So,
\[\mu=\frac{g\tan\theta-v^2/R}{\tan\theta\cdot v^2/R+g}=0.0867\]