Static friction coefficient question

AI Thread Summary
The discussion revolves around calculating the coefficient of static friction for drag race tires based on a quarter mile time of 6.0 seconds. The user correctly converts the distance to meters and calculates the acceleration at 22.36 m/s² but becomes confused about the role of applied force in the equations. It is clarified that, under constant acceleration without slipping, the friction force equals the applied force, meaning they are effectively the same in this scenario. The friction force is determined by the coefficient of friction multiplied by the normal force, which is essential for maintaining traction. Understanding this relationship is crucial for solving the problem accurately.
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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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