Finding maximum acceleration with COF of Ms

AI Thread Summary
The discussion revolves around calculating the maximum acceleration of a car based on the coefficient of static friction, which is given as 0.80. Participants clarify that the normal force can be represented as the weight of the car, allowing the maximum frictional force to be calculated. By applying Newton's second law, they establish that the maximum acceleration can be derived from the equation a = μg, where μ is the coefficient of friction and g is the acceleration due to gravity. The final conclusion confirms that the maximum acceleration is 7.84 m/s², affirming the importance of understanding static friction in this context. This problem highlights the relationship between friction, normal force, and acceleration in vehicle dynamics.
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 accleration 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 accleration 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 \leq uk x Fn

Fn = mg
Fn = 9.8m

Ffr \leq .80 x 9.8m

m is of no regard in this problem therefore

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

Fn = mg
Fn = 9.8m

Ffr \leq .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 \leq 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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