A 950-kg race car can drive around an unbanked turn at a maximum speed of 41 m/s

Question

A 950-kg race car can drive around an unbanked turn at a maximum speed of 41 m/s without slipping. The turn has a radius of 180 m. Air flowing over the car's wing exerts a downward-pointing force (called the downforce) of 12000 N on the car. (a) What is the coefficient of static friction between the track and the car's tires? (b) What would be the maximum speed if no downforce acted on the car?

Explanation / Answer

a) Centripital force = friction ==> m * v^2/r = friction friction = coefficient * Normal force, and Normal force = gravity + airflow = 830kg * 9.8 m/s^2 + 1100 = 8134 + 1100 = 9234 N Now set up the following equation for the coefficient (µ) centripital = friction ==> (830)*(58)^2/160 = 9234µ... solve for µ µ = 1.890 b) use the above value for µ, solving for v 830v^2/160 = 8134 * 1.890 v^2 = 2963.52 v = 54.4 m/s

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