A banked circular highway curve is designed for traffic moving at 57 km/h. The radius of the curve is 216 m. Traffic is moving along the highway at 42 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.)

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Fn sinθ = mv^2/r

Fn cosθ = mg
so tanθ = v^2/rg
Now you have θ go back and add friction to your FBD. Find mu required for extra friction not handled by the normal force

To solve this problem, we need to use the concept of centripetal force and the forces involved in circular motion.

1. The first step is to find the speed of the cars in meters per second. We can convert the given speed of 42 km/h to meters per second by dividing it by 3.6 (1 km/h = 1000 m / 3600 s).
Speed in meters per second = 42 km/h ÷ 3.6 = 11.67 m/s (rounded to two decimal places).

2. Next, we need to calculate the centripetal force acting on a car moving in a circular path. The centripetal force required to keep the car on the curved road is given by the equation:
Centripetal force = (mass of the car) x (velocity of the car)^2 / (radius of the curve)

3. However, in order for a car to move in a curved path without sliding off, the frictional force between the tires and the road must provide the necessary centripetal force. The maximum frictional force is given by:
Maximum Frictional Force = (coefficient of friction) x (normal force)

4. To find the normal force, we use the equation:
Normal force = (mass of the car) x (acceleration due to gravity)

5. Equating the centripetal force and the maximum frictional force, we can solve for the coefficient of friction needed using the following equation:
(coefficient of friction) = (mass of the car) x (velocity of the car)^2 / [(radius of the curve) x (mass of the car) x (acceleration due to gravity)]

Now, let's calculate the minimum coefficient of friction:

Mass of the car is not given, but we can cancel it out, so it won't be needed for our calculation.

(coefficient of friction) = (velocity of the car)^2 / [(radius of the curve) x (acceleration due to gravity)]

Substituting the values:
(coefficient of friction) = (11.67 m/s)^2 / [(216 m) x (9.8 m/s^2)]

Calculating the expression on the right side:
(coefficient of friction) = 0.15218

Therefore, the minimum coefficient of friction between the tires and road that will allow the cars to take the turn without sliding off the road is approximately 0.152 (rounded to three decimal places).