A 1600 kg car rounds a curve of 61 m banked at an angle of 18 ° . If the car is traveling at 97 km/h , will a friction force be required? If so, in what direction? How much friction force will be required?

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

the components of a vector are the parts of the vectors that are to each other.

b perpendicular