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.
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
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