Finding maximum acceleration with COF of Ms

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Homework Help Overview

The problem involves determining the maximum acceleration of a car based on the coefficient of static friction between the tires and the ground, which is given as 0.80. Participants express confusion regarding the application of the coefficient of friction and the role of normal force in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between static friction and normal force, questioning how to derive maximum acceleration without specific mass values. There is an exploration of the implications of rolling motion and the assumption of a horizontal surface.

Discussion Status

Some participants have provided guidance on how to approach the problem by suggesting to treat mass as a variable and focus on the forces involved. Others are still seeking clarification on the concepts of rolling and static friction.

Contextual Notes

There is an acknowledgment of missing information, particularly the mass of the car, which leads to uncertainty in calculating the normal force. The discussion reflects a lack of consensus on the initial steps to take in solving the problem.

koolj
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This problem seems simple enough but this coefficient of friction stuff easily confuses me..

Homework Statement


What is the maximum acceleration a car can undergo if the coefficient of static friction between the tires and ground is .80?


Homework Equations


FFs = MsFn ?
a = f/m ?


The Attempt at a Solution



Im not really sure where to start with this, I have little idea how I am suppose to find the maximum acceleration with just the coefficient of friction given and nothing else. I assume I might need to switch some things around to find it?
 
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koolj said:
This problem seems simple enough but this coefficient of friction stuff easily confuses me..

Homework Statement


What is the maximum acceleration a car can undergo if the coefficient of static friction between the tires and ground is .80?

I'm assuming that you've talked about rolling in your course, at least enough to say that the portions of the tires in contact with the pavement are instantaneously stationary. This is why it is static friction that is involved in rolling, even though the vehicle is in motion.

I'll also point out that the question is assuming that the road is horizontal, which would lead to obtaining the maximum acceleration of the car. In that case, what is the normal force acting on each tire from the ground and what would the maximum static frictional force on each tire be? (For convenience, assume that the car's weight is evenly distributed on all four tires.)

Consider Newton's Third Law in dealing with the tires and the ground. Which way does the friction on each surface point? How does friction make the motion of the car possible?
 
Well that's the thing, I can't find normal force because nothing other than the question was given. No masses at all, just the CoF. I believe that they would cancel out anyway and that's why they didnt list mass, but I am still stuck in how I am suppose to solve this. We didn't really go over rolling, but i know what you mean when you say the tires surfaces are instantly stationary.
 
Just call the mass "m" and continue. You won't need it. Find the maximum force of friction, then apply Newton's 2nd law to find the corresponding maximum acceleration.
 
ahhhh, wait I think I get it now. Thanks for the help, that's all I needed for now =)
 
could someone please clarify this? I am studying this on my own (no teacher) and we haven't really gone over rolling i have the exact same problem but have no idea how to start Thanks.
 
alphadog0309 said:
could someone please clarify this? I am studying this on my own (no teacher) and we haven't really gone over rolling i have the exact same problem but have no idea how to start Thanks.
Follow my advice in post #4. If you still have problems, post your attempt and we can take a look at what you're doing.
 
ok so here's how i went about this:
Ffr [tex]\leq[/tex] uk x Fn

Fn = mg
Fn = 9.8m

Ffr [tex]\leq[/tex] .80 x 9.8m

m is of no regard in this problem therefore

Ffr [tex]\leq[/tex] 7.84 m/s2
 
alphadog0309 said:
ok so here's how i went about this:
Ffr [tex]\leq[/tex] uk x Fn

Fn = mg
Fn = 9.8m

Ffr [tex]\leq[/tex] .80 x 9.8m
Good. Since you want the maximum acceleration, choose the maximum value of static friction:
Ffr = μN = μmg.

m is of no regard in this problem therefore

Ffr [tex]\leq[/tex] 7.84 m/s2
Let's clean up your reasoning by applying Newton's 2nd law:
ΣF = ma
Ffr = ma
μmg = ma

so: a = μg
 
  • #10
thanks i just checked in the answer book and that is the answer
thanks a lot!
 

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