A banked circular highway curve is designed for traffic moving at 55 km/h. The radius of the curve is 220 m. Traffic is moving along the highway at 38 km/h on a rainy day. What is the minimum coefficient of friction between tires and road that will allow cars to take the turn without sliding off the road? (Assume the cars do not have negative lift.)

Question

I know this question has already been posted here before BUT I'm still confused as to the way it was solved.

The way I solved it was by determining the banking angle first which I found to be 6.2 degrees. Next, I solved for the normal force from the vertical components and I got:

FNcos A + mu*FNsin A = mg
FN = mg / (cos A + mu*sin A)

And I basically just substituted that into:

FNsin A - mu*FNcos A = m*v^2/R

I solved this and got an answer of 0.0567, which is the same answer that was obtained in the previous question. So I'm just wondering how is it that you can obtain this same answer by substituting mg for FN instead of what I did?

Answer 1

I agree with the two equations you have written and that substitution of FN into the second equation is the way to solve for mu. I have not gone through the numbers but assume the 0.0567 figure is correct for the minimum mu(static)

Perhaps the answer to your question has soimething to do with the mu*sinA term being very small compared to 1. The effect of the upward component of the friction force is quite small compared to Mg

Answer 2

So does that mean you can just use friction force without resolving it into its components, and so what you use in you equation would be:

M V^2/R = M g sin A - M g cosA*mu

(This is what you did in the previous example)

So did you basically just ignore the vertical component of friction because it is extremely tiny and assume that FN = mg?