Posted by D on .
A banked circular highway curve is designed for traffic moving at 57 km/h. The radius of the curve is 211 m. Traffic is moving along the highway at 48 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.

Physics Mechanics 
drwls,
First, convert km/h speeds to m/s.
57 km/h = 15.83 m/s
48 km/h = 13.33 m/s
Next, compute the bank angle of the road, knowing that it is designed for 57 km/h. That means the normal force of the road on the tires is resolvable into a vertical force Mg and a horiziontal force MV^2/R, with no friction force required.
M V^2/R = M g tan A
tan A = V^2/(Rg) = 0.1212
A = 6.9 degrees
Finally, at the lower speed, require that there be a tire friction force Ff sufficient to keep the car from sliding toward the center of the circle.
Ff + Mg sin A = (M V^2/R)cos A
M*g*cosA*Us + M*g sin A = (M V^2/R)cos A
Us = [(V^2/Rg)cos A  g sin A]/(g cosA)
= V^2/(Rg) + tan A
= 0.0859 + 0.1212 = 0.035 
Physics Mechanics 
D,
Thank you so much drwls. You're a life saver. Thanks a lot! :)

Physics Mechanics 
your friend,
jk...i love you