How High is the Coefficient of Static Friction for a Drag Car's Tires?

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Homework Help Overview

The discussion revolves around determining the coefficient of static friction for a drag car's tires, given its acceleration over a quarter mile distance. The context involves physics concepts related to motion and friction.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between acceleration, distance, and friction, with some attempting to calculate the coefficient of static friction based on provided data. Questions arise regarding the necessity of additional information, such as the mass of the car, to solve the problem accurately.

Discussion Status

The discussion is active, with participants offering various calculations and corrections. Some express uncertainty about the equations used, while others clarify the principles of motion involved. There is no explicit consensus on the correct approach or final answer.

Contextual Notes

Participants note that the original problem statement contained a correction regarding the distance, which may affect the calculations. The need for additional information, such as the mass of the car, is also highlighted as a potential constraint in solving the problem.

liljediboi
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a drag car's tires have some of the highest friction coefficients, so the book says.

it takes 6 s to go a quarter mile (about 402.5m) with constant acceleration, no skidding

what is the coefficient for the static friction?

A CORRECTION HAS BEEN MADE, IT IS 402.5m NOT 602.5
 
Last edited:
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is it 1.14?
 
You need more information to solve this problem, maybe the mass of the car. Atleast I think this problem is unsolvable based on the information you gave us.
 
First, we need to know the amount of acceleration that makes the car travels 602.5 m in 6 s (starting from zero velocity, I presume). Using d = 0.5 at^2, we get a = 33.47 m/s/s.

If the car has mas M, then the force that makes the car accelerates as much as 33.47 m/s/s is Ma. But that Ma is the friction force (static) between the tires and the road (Since no other force could cause it). Since Ffriction = uN where N is the normal force and u is the coefficient of static friction, and since N equals to Mg where g is gravity, then

Ffriction= uN
Ma = u Mg
a = ug
u = a/g
u = 33.47/9.8
u = 3.41

Correct me if I'm wrong...
 
umm.. the equation is d=at, not d=at^2 that's for falling objects. isint it?
 
It is true that d = 0.5*at^2 is the equation for falling objects with zero initial velocity. More generally, it is the equation of motion for objects with constant acceleration and zero initial velocity. Falling objects happens to undergo constant acceleration (the acceleration of gravity), and so does your constantly accelerated car!

it takes 6 s to go a quarter mile (about 402.5m) with constant acceleration, no skidding

the equation d = vt (not d = at) is for objects moving with a constant velocity (zero acceleration). Your car's velocity is not constant, it starts from zero and increases constantly (constant acceleration).

Anything else that still puzzles you?

Isn't it wonderful that a falling apple and an accelerating car have something in common :) ?
 

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