Question

A curve of radius 40 m is banked for a design speed of 60 km/h. If the coefficient of static friction is 0.30 (wet pavement), at what range of speeds can a car safely make the curve?
minimum

maximum

Answer 1

here,

radius , r = 40 m

speed , v = 60 km/h

v = 16.67 m/s

us = 0.3

The aviator's normally solid work has missed the mark here. "a design speed of 60 km/h" (= 16.67 m/s) means that there is no friction force on the ramp at this speed. This means that the downslope component of gravity is equal to the upslope component of centripetal acceleration.

g * sin(theta) = v^2 * cos(theta) / r

sin(theta) / cos(theta) = tan(theta) = v^2 / gr = (16.67 m/s)^2 / (9.8m/s^2 * 40 m) = 0.71

(theta) = 35.3 degree

normal acceleration a_n = gcos(theta) + v^2sin(theta)/r

and friction acceleration a_f = u * a_n

At the maximum speed, friction and gravity point downslope, so
gsin(theta) + u * (gcos(theta) + v^2 * sin(theta)/r) = v^2 * cos(theta)/r

Plugging in knowns (and dropping units for ease),

9.8 * sin(35.3) + 0.3 * (9.8 * cos(35.3) + v^2 * sin(35.3) /40) = v^2 * cos(35.3)/40

v = 22.4 m/s

the maximum speed is 22.4 m/s

At the minimum speed, friction points upslope,

so
g * sin(theta) - u * (gcos(theta) + v^2 * sin(theta)/r) = v^2 * cos(theta)/r

Plugging in knowns (and dropping units for ease),

9.8 * sin(35.3) - 0.3 * (9.8 * cos(35.3) + v^2 * sin(35.3)/40) = v^2 * cos(35.3)/40

v = 11.5 m/s

the minimum speed is 11.5 m/s