How Does Friction Affect Acceleration in High-Performance Race Cars?

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Homework Help Overview

The discussion revolves around the effects of friction on acceleration in high-performance race cars, specifically examining the forces acting on a race car including downforce, air resistance, and friction. The original poster attempts to analyze the forces involved in the car's acceleration and the role of static friction.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between the direction of frictional force and the motion of the car, questioning the signs used in the free-body diagram and the implications of static friction acting in different directions.

Discussion Status

Some participants are clarifying the direction of forces involved, particularly the static friction force, and how it relates to the car's acceleration. There is acknowledgment of differing interpretations regarding the signs of forces, and some guidance has been offered to reconsider the assumptions made in the calculations.

Contextual Notes

There is a mention of confusion regarding the signs of forces in the context of a free-body diagram, as well as a reference to external sources that may influence understanding. The discussion reflects a mix of interpretations and attempts to reconcile different viewpoints on the physics involved.

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 ! :]
 
Last edited:
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hey jehan4141

If the car is moving in +x then friction due to downward forces and air drag should act in -x

if this wasn't true and air drag should be in +x so we can design a car for which air drag>downward force friction and then that car will move on its own ... which is not possible

so air drag should be in -X IMO
ask your teacher again in case he told you this by mistake
 
thank you :) i will ask him again :D
 
In order for the car to accelerate in the +x direction, its tires must push back against the road surface, in the -x direction. The force that prevents the tire from slipping is the static friction. So the static friction must act in the +x direction (thus opposing the tire's backwards push in the road surface).
 
gneill said:
In order for the car to accelerate in the +x direction, its tires must push back against the road surface, in the -x direction. The force that prevents the tire from slipping is the static friction. So the static friction must act in the +x direction (thus opposing the tire's backwards push in the road surface).

OH MY GOD !

I forgot that tires are rolling ... Damn it that was a dumb mistake ...
______________

jehan, I am really sorry for the wrong answer ... :(

Thank you gneill for correcting me.
 

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