# Banked Curve , w/ friction

• blueray4
In summary, if the coefficient of friction between tires and road is 2, the turn is banked at 45 degrees.

## Homework Statement

A banked circular highway curve is designed for traffic moving at 60 km/h. The radius of the curve is 230 m. Traffic is moving along the highway at 35 km/h on a rainy day. What is the minimum coefficient of friction between tires and road that will allow cars to negotiate the turn without sliding off the road?

NO IDEA.

## The Attempt at a Solution

You have any thoughts on how to solve it or even what eqns might be useful. This is la lot like an incline-block problem, with centripetal force thrown in.

maybe find the angle of the curve? i honestly have no clue...

Lets see if I can help. First problem is it suggests the turn is banked but gives no angle of the banking. So maybe the problem designer hinted at what it might be. He says safely negotiated at 60km/hr. Ok, so we have more data. with 230 m radius, we could calculate, the bank of the turn if we knew the coefficient of friction, or vice versa. From there we could figure the wet coefficient and be done.

This is where the problem gets nasty and assumption laden. Is there any more data, besides that given?

no , nothing else

hells bells. All i can add at this time is to call the angle of the bank theta.

We know that sin(theta)*g=V^2/R -mu(dry)*cos(theta).

and that when wet, sin(theta)*g=V'^2/R'-mu(wet)*cos(theta)

Two eqns, 3 unknowns. Anyone else out there with another approach?

Looks good if you assume a wet tire will grab equally well as a dry one, try telling that to a race car driver!

It seems assumed this is not the case by the way the question was worded. If your stuck for time, I would try to express the coefficient of friction as a ratio between wet and dry coefficients of friction. Then the banking angle no longer matters and it becomes a simple ratio.