What is the minimum required value of the coefficient of static friction between

Question

What is the minimum required value of the coefficient of static friction between the surface of the road and the tires so that the car can exit the highway safely without sliding at a constant speed of 41.3 km/h?

The radius of curvature of a highway exit is r = 47.5 m. The surface of the exit road is horizontal, not banked. (See figure.) What is the minimum required value of the coefficient of static friction between the surface of the road and the tires so that the car can exit the highway safely without sliding at a constant speed of 41.3 km/h?

Explanation / Answer

If we look at the horizontal forces acting on the car, the only force is friction of the tires which will allow the car to "accelerate" centripetally towards the center of the circle. So in the x direction, we can say:

Fx = max. The only x force if friction, and the x acceleration is circular ( centripetal, v2/r )

Ff = mv2/r

Now, the frictional force can be found using the relationship Ff = Fn. The normal force, Fn, is simply equal to the weight, mg, on a flat surface with no other y forces present, so plug it all in:

mg = mv2/r

Notice the mass cancels out, it is not needed, all car with the same tires will behave identically.

g = v2/r

Lastly, since we are using g=9.8 m/s2, we need to convert  km/h to m/s,

( 41.3 km/hour )( 1 hour/3600 sec )( 1000 m/1 km ) = 11.47 m/s

( 9.8 ) = ( 11.47 m/s )2 / 47.5 m

= .283

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