An 860-kg race car can drive around an unbanked turn at amaximum speed of 62 m/s

Question

An 860-kg race car can drive around an unbanked turn at amaximum speed of 62 m/s without slipping. The turn has aradius of curvature of155 m. Air flowing over the car's wing exertsa downward-pointing force (called the downforce)of 3000 N on the car.
(a) What is the coefficient of static friction between the trackand the car's tires?

(b) What would be the maximum speed if no downforce acted on thecar?



(a) What is the coefficient of static friction between the trackand the car's tires?

(b) What would be the maximum speed if no downforce acted on thecar?



(a) What is the coefficient of static friction between the trackand the car's tires?

(b) What would be the maximum speed if no downforce acted on thecar?

Explanation / Answer

F = m v2 / R      centripetal force required to keep car in circular path F = (m g + 3000)     frictionalforce due to normal force of car on track = m v2 / [R * (m g +3000)]       combiningequations = 860 * 622 / [155 * (860 * 9.8 +3000)] = 1.86 Without the downward force of 3000 N : m v2 / R = m g v = ( g R) = (1.86 * 9.8 * 155) = 53.2m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---