Asked by Tiffany
Air rushing over the wings of high-performance race cars generates unwanted horizontal air resistance but also causes a vertical downforce, which helps cars hug the track more securely. The coefficient of static friction between the track and the tires of a 552-kg car is 0.967. What is the magnitude of the maximum acceleration at which the car can speed up without its tires slipping when a 3150-N downforce and an 1300-N horizontal air resistance force act on it?
Answers
Answered by
Isuck
We need the horizontal acceleration so,
ax=Fx/m
Since Fx, or the horizontal forces acting on the car is composed of the air resistance and the force of static friction,
ax=(fs-1300N)/m
Looking for fs(static friction):
fs=us*N
N=mg+3150N
=(552kg)(9.8)+3150N
=8559.6N
fs=(0.967)(8559.6N)
=8277.13N
Solving for ax:
ax=(8277.13-1300N)/553
ax=12.54m/s2
ax=Fx/m
Since Fx, or the horizontal forces acting on the car is composed of the air resistance and the force of static friction,
ax=(fs-1300N)/m
Looking for fs(static friction):
fs=us*N
N=mg+3150N
=(552kg)(9.8)+3150N
=8559.6N
fs=(0.967)(8559.6N)
=8277.13N
Solving for ax:
ax=(8277.13-1300N)/553
ax=12.54m/s2
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