Question

A curve of radius 69 m is banked for a design speed of 80 km/h . if the coefficient of static friction is 0.33 (wet pavement) at what range of speeds can a car safely make the curve?

Answer 1

The aviator's normally solid work has missed the mark here. "a design speed of 80 km/h" (= 22.2 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.
gsinΘ = v²cosΘ / r
sinΘ / cosΘ = tanΘ = v² / gr = (22.2m/s)² / (9.8m/s² * 69m) = 0.7288
Θ = 36.1º

normal acceleration a_n = gcosΘ + v²sinΘ/r
and friction acceleration a_f = µ*a_n

At the maximum speed, friction and gravity point downslope, so
gsinΘ + µ(gcosΘ + v²sinΘ/r) = v²cosΘ/r
Plugging in knowns (and dropping units for ease),
9.8sin36.1 + 0.33(9.8cos36.1 + v²sin36.1/69) = v²cos36.1/69
5.77 + 2.61 + 0.0028v² = 0.0117v²
0.0089v² = 8.38
v = 30.7 m/s = 110.5 km/h

At the minimum speed, friction points upslope, so
gsinΘ - µ(gcosΘ + v²sinΘ/r) = v²cosΘ/r
Plugging in knowns (and dropping units for ease),
9.8sin36.1 - 0.33(9.8cos36.1 + v²sin36.1/69) = v²cos36.1/69
5.77 – 2.61 - 0.0028v² = 0.0117v²
0.0145v² = 3.16
v² = 217.93
v = 14.8 m/s = 53.3 km/h