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jehan4141
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Air rushing over the wings of high-performance race cars generates unwanted horizontal air resistance but also causes a vertical downforce, which helps the cars hug the track more securely. The coefficient of static friction between the track and the tires of a 690-kg race car is 0.87. What is the magnitude of the maximum acceleration at which the car can speed up without its tires slipping when a 4060-N downforce and an 1190 N horizontal air resistance force act on it.
My work:
I get the general concept of the problem, but my confusion is about 1/2 way into the problem.
The free-body diagram would be like so:
The car is moving in the positive direction, +x.
Frictional force and air resistant point in the negative direction, -x.
Normal force points up.
Weight and downward force point down.GIVEN
coefficient of friction, u = 0.87
m = 690 kg
w = 690(9.8) = 6762 N
Downforce, Fd = 4060 N
Air resistance, Fa = 1190 N
Normal Force = Fn = 6762 + 4060
Normal Force = Fn = 10, 822 N
Force of friction = Ff = (Fn)(u)
Ff = (10822)(0.87)
Ff = 9415.14 N
We know that frictional force is in the -x direction.
Net Fx = -Fa + (-Ff) = ma
Net Fx = -1190 + (-9415.14) = (690)a *************
a = (-10605.14) / 690
a = -15.369768 m/s^2
My teacher said that I messed up at the part with the stars. He says that there shouldn't be a negative sign infrom of the Ff, force of air resistance.
Isn't frictional force a vector that points in the -x direction? Could somebody please explain this to me please? Does the frictional force of a tire point in the positive direction? And if yes, is this always true for tires? and lastly, are there other instances besides tires where this is the case? Thank you in advance ! :]
My work:
I get the general concept of the problem, but my confusion is about 1/2 way into the problem.
The free-body diagram would be like so:
The car is moving in the positive direction, +x.
Frictional force and air resistant point in the negative direction, -x.
Normal force points up.
Weight and downward force point down.GIVEN
coefficient of friction, u = 0.87
m = 690 kg
w = 690(9.8) = 6762 N
Downforce, Fd = 4060 N
Air resistance, Fa = 1190 N
Normal Force = Fn = 6762 + 4060
Normal Force = Fn = 10, 822 N
Force of friction = Ff = (Fn)(u)
Ff = (10822)(0.87)
Ff = 9415.14 N
We know that frictional force is in the -x direction.
Net Fx = -Fa + (-Ff) = ma
Net Fx = -1190 + (-9415.14) = (690)a *************
a = (-10605.14) / 690
a = -15.369768 m/s^2
My teacher said that I messed up at the part with the stars. He says that there shouldn't be a negative sign infrom of the Ff, force of air resistance.
Isn't frictional force a vector that points in the -x direction? Could somebody please explain this to me please? Does the frictional force of a tire point in the positive direction? And if yes, is this always true for tires? and lastly, are there other instances besides tires where this is the case? Thank you in advance ! :]
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