Static friction coefficient question

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Homework Help Overview

The problem involves estimating the coefficient of static friction for drag race tires on an asphalt surface, based on a scenario where a drag racer covers a quarter mile in 6.0 seconds with constant acceleration and no tire slipping.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply kinematic equations to find acceleration and relates forces acting on the car. Some participants question the role of applied force versus friction force in the context of the problem.

Discussion Status

Participants are exploring the relationship between applied force and friction force, with some guidance provided on their equivalence under certain conditions. There is an ongoing discussion about the assumptions made regarding forces acting on the car.

Contextual Notes

There is a mention of confusion regarding the inclusion of applied force in the equations, as well as the need to consider the conditions under which the car accelerates without losing traction.

hamsterpower7
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Homework Statement



from giancoli 5th ed (algebra based)
chapter 4, question 42


Drag race tires in contact with an asphalt surface probably have one of the higher coefficients of static friction in the everyday world. Assuming a constant acceleration and no slipping of tires estimate the coefficient of static friction for a drag racer that covers the quarter mile in 6.0s


Homework Equations





The Attempt at a Solution



I converted quarter mile to m ----> 402.5m

using kinematic formula i found the acceleration a=22.36 m/s^2

\Sigma F = F_A - F_fr

F_A - \mu_s F_N = mass 22.36 m/s^2

and than I got stuck...
the answer key states that I shouldn't have put applied force in the equation
but how does that make sense? the person should have put applied for to have constant acceleration

anyways if there wasn't a applied force I can solve for static friction from the equation above

help me please
 
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Keep in mind that the friction force (coefficient of friction times mass of car times acceleration of gravity) will be equal to the applied force (mass of car times its acceleration) if you assume that the car is accelerating as fast as it can without losing traction.
 
obafgkmrns said:
Keep in mind that the friction force (coefficient of friction times mass of car times acceleration of gravity) will be equal to the applied force (mass of car times its acceleration) if you assume that the car is accelerating as fast as it can without losing traction.


I don't get this, isn't applied force and friction force two separate force?
 
In this case, they are one and the same! The car's tire is exerting a friction force parallel to the ground that accelerates the car. That friction force can be no greater than the tire's normal force times the coefficient of friction.
 

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