A sports car is accelerating up a hill that rises 22.0° above the horizontal. The coefficient of static friction between the wheels and the road is µs = 0.88. It is the static frictional force that propels the car forward.
a) What is the magnitude of the maximum acceleration that the car can have?
b) What is the magnitude of the maximum acceleration if the car is being driven down the hill?
2006-10-20 19:38:35 · 2 answers · asked by Alan l 1 in Education & Reference ➔ Homework Help
The maximum acceleration is equal to the static friction force, minus the weight component opposing friction, divided by the mass of the car. The static friction force is the coeff of friction µs times the normal component of the car's weight. The weight component opposing friction is m*g*sin(A). The normal force is m*g*cos(A), so the max accleration is
[µs*m*g*cos(A)-m*g*sin(A)]/m = µs*g*cos(A)-g*sin(A)
When driven down the hill, the weigh component adds to the fricion force giving µs*g*cos(A)+g*sin(A)
2006-10-20 20:33:58 · answer #1 · answered by gp4rts 7 · 0⤊ 0⤋
a. 44
b.88.7
2006-10-21 02:41:04 · answer #2 · answered by Anonymous · 0⤊ 0⤋