Answer:
Step-by-step explanation:
With no friction, the normal force of slope on car will provide the necessary centripetal force in the horizontal and support the weight of the car in the vertical
tanθ = mv²/R / mg = v²/Rg
97 kph = 26.94 m/s
tanθ = 26.94²/(61(9.8)) = 1.214458...
θ = 50.5°
As this is much steeper than the 18° bank angle, a friction force is required in the down slope direction. or toward the center of curvature.
The ideal speed for an 18° slope is
v = √(Rgtanθ) = √(61(9.8)tan18) = 13.93 m/s
The 18° slope at ideal speed makes a centripetal force of
Fc = 1600(13.93²/61) = 5,094.7
The required centripetal force at 97 kph is
Fh = 1600(26.94²) / 61 = 19,042.7 N
Tire friction must supply 19042.7 - 5094.7 = 13,948 N horizontally
Therefore friction parallel to the slope must be
13,948 / cos18 = 14,665 N = 15 kN