Uniform circular motion problem

Question

A banked circular highway curve is designed for traffic moving at 60km/h. The radius of the curve is 200m. Traffic is moving along the highway at 40km/h on a rainy day. What is the minimum coefficient of friction between tires and road that will allow cars to make the turn w/o sliding off the road?(Assume the cars do not have negative lift).

Please explain. i dont know where to start

Answer 1

You're needing a little bit more info in your problem.  The way I did it with what you provided was this:  WHICH IS INCORRECT

  F = ma(x)

ma(x) = u(x)mg     NOTE: u = mu

   a(x) = u(x)g

   u(x) = 1

This value for mu doesn't make sense.  Like the other person (above) answered, you're missing some vital info.

Answer 2

you need the angle at which the road is banked to answer this.

Basically, the banking allows the road to provide a normal force, contributing to the overall centripetal force necessary to round the corner, this lessens the reliance on the frictional force that the road provides.

You need to draw a diagram showing the normal force acting on the tyres, and the frictional force, both as components of the centripetal force.

F(centripetal) = mv^2/R

F(normal) = mgcos(theta)

F(friction) = mgsin(theta) =

theta is banking angle

solution is found by

F(centripetal) = F(normal) + F(friction)

therefore

F(friction) = F(centrip)-F(normal)

This will give you friction force which you can use the relation

F(fr) = mu(k) x F(N) where mu(k) coefficient of kinetic friction, F(N) normal force of tyres on road

anyway you dont have banking angle, and you need it.