I've attempted to solve this problem but I'm not really sure if I'm doing it right. I've looked up other threads containing the same question, but they just don't have the answers I'm looking for. Thank you in advance to anyone who helps.(adsbygoogle = window.adsbygoogle || []).push({});

1. The problem statement, all variables and given/known data

A race-car driver is driving her car at a record-breaking speed of 225 km/h (62.5 m/s). The first turn on the course is banked at 15 degrees, and the car's mass is 1450 kg.

a) Calculate the radius of curvature for this turn.

b) Calculate the centripetal acceleration of the car.

c) If the car maintains a circular track around the curve (does not move up or down the bank), what is the magnitude of the force of static friction?

d) What is the coefficient of static friction necessary to ensure the safety of this turn?

2. Relevant equations

tan 15 degrees = v^2/rg

ac = v^2/r

mv^2/r = Fs (unsure)

Ff = uFN

FN = mgcos15 (unsure)

3. The attempt at a solution

a) r = v^2/gtantheta

= (62.5^2)/(9.8tan15)

= 1487.584027

= 1.49 x 10^3 m

b) ac = v^2/r

= (62.5^2)/(1.49 x 10^3)

= 2.625902086

= 2.63 m/s^2

c) mv^2/r = Fs

(1450)(62.5^2)/(1.48 x 10^3) = Fs

3807.558025 = Fs

Fs = 3.81 x 10^3 N

d) Ff/FN = u

3.81 x 10^3/ mgcos15 = u

3.81 x 10^3/ (1450)(9.8cos15) = u

0.277 = u

I'm pretty sure parts a and b are correct, but part c is where it gets confusing. Also, I'm getting a little confused in what is really the normal force for a banked curve. Is it mgcostheta as with an inclined plane, or mg/costheta?

Thanks a lot!

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# Homework Help: Centripetal Force and Banked Curves

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