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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 40 km/hr on a rainy day. What is the minimum coefficient of friction between tired and road that will allow cars to take the turn without sliding off the road? (Assume the cars do not have negative lift; curious to know what is negative lift?, a wiki search on car handling will help.)

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The Coefficient of friction u = 0.078    

Calculate the following?

Radius of curvature r = 200 m

First we need banking angle at v = 60 Km/h = 60(1000 m/3600 s) = 16.66 m/s

At this speed (for which the road is designed)

Coefficient of friction  v^2/rg = sin A

sin A  = (16.66)^2/(200*9.8)

sin A  = 0.1416  

A = 8.141 degrees

For no slipping at the banking angle at  the lower Speed v = 40 Km/h = 40 (1000 m/3600 s) =11.11 m/s  ,

mv^2/r = mg sin A - u mg cos A

0.617 = (9.8*0.1416) - u (0.9899*9.8)

0.617 = 1.387 - u (9.7)

9.78 u = 0.77

∴ Coefficient of friction u = 0.078    

To know more about friction, visit:

https://brainly.com/question/24338873

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