Question

A banked circular highway curve is designed for traffic moving at 60 km/h. The radius of the curve is 200 m. Traffic is moving along the highway at 40 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

Answer 1

The minimum coefficient of friction which allow cars to take the turn without sliding off the road is 0.617

• What is friction force:

Friction is the force resisting the relative motion of the surface sliding against each other.

here, given that

m = mass of car

The velocity of the cars, v= 60 km/h = 17 m/s

The radius of the curve, R =200 m.

The velocity of the cars on the highway curve, v' = 40 km/h = 11 m/s

In order to allow the cars to take the turn without sliding off the road, the Centripetal Force must be equal to the frictional force between the tires  of the car and road.

Centripetal Force = Frictional Force

• centripetal force = mv²/r

• friction force = μR

mv²/r = μR

here

R = Normal Reaction = Weight of Car = mg

Therefore,

mv²/r = μmg

μ = v²/gr

μ = (11)² / (9.8)(200)

μ = 0.0617

hence

The minimum coefficient of friction which allow cars to take the turn without sliding off the road is 0.617

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