An automobile with all wheel drive has a mass of 1000 kg (this weight is evenly distributed amongst the 4 wheels). The coefficient of static friction with the road is 0.80.
The car starts from rest and the road is horizontal.
What is the maximum acceleration (a.max) this car can attain without spinning it’s wheels?
First person to give me equation/answer will receive maddd karmaaaa y00! 
edit: I got 7.84 meters/second^2 but I took physics 2 years ago…yeah that can’t be right can it?
And your answer is incorrect. You gave a velocity, you are looking for acceleration.
^good point, however this question is more elementary than you are interpreting it. I think it’s just examining the relationship between normal force (exerted opposite of gravity’s pulling force) and coefficient of static friction…
this may be a hint, im not sure
I meant 7.84 m/s^2
corrected, but probably not right. Can you figure this out?
hmmm i too took physics last yr but unfortunately i got a D. sorry i cant help
are we talking an ideal situation where weight transfer, tire size, differential compensation, and road conditions do not apply?
" coefficient of static friction with the road is 0.80"
What about the coefficient of drag? or are you not going that deep. i look at this later
also need to know the contact patch area, since mu is a unitless number. Wider tires, more accereration.
Unfortunately I sold my physics book… but the answer you gave is very believeable.
9.8m/s^2 is the acceleration do to gravity, so if you are accelerating at 7.84m/s^2 you are feeling a little less then 1G of acceleration.
It is very possible to accelerate a car at 1G for an extended period of time… the top fuel drag cars accelerate prolly over 2-3G’s ( Im just guessing )
1320ft / 4.4 seconds = 300 ft/s ( 1ft = .3048m ) thats 91.44 m/s ( Velocity )
91.44 m/s / 4.4 secs = 20.78 m/s^2 ----> 20.78 / 9.8 = 2.12 G of acceleration.
Assuming the 1/4 mile time is 4.4 seconds.
Just to check my answer… if they travel 91.44 m/s *60 = 5486.4 m/min *60 = 329184 m/hr / 1000 = 329.184 km/hr … this of course assume they maintain their velocity from very start to very finish, this is why the Km/hr seems low.
This all assumes constants… Constant accel, constant velocity… etc etc… which we know is not true… hence the slow final speed for a top fuel drag car.
mhill, fastandcheap, kevin are all looking too deep. this problem is for a physics 151 lab, which I forgot to take with the class
Ninja - good insite. the figure 320.184 km/hour is the average speed of a dragcar if you are taking distance traveled and dividing by time elapsed. if my answer is plausible, I’ll just stick with it. A 1 credit hour lab isn’t worth the hassel anyways now that I think about it.
thanks.
Should we start a homework help sub-forum? What’s with all the homework help threads lately? 
^Yes, yes we should.
i would like some homework help on my java labs from u programming gurus 
I got the same answer as you did.
7.84 m/s^2
I don’t know if that would be right though, since thats just the minimum force to get the wheels moving. I dunno if you could you apply more force, and keep the wheels from spinning.
Assumptions:
4-wheel drive
0 weight transfer due to acceleration
0 drag force
Contact patch size doesn’t matter, as coefficient of static friction is a function of contact patch size, tire compositon, tire temp, road temp, etc etc etc…
On with the fun!
M = 1000 kg
Mtire=1000/4 = 250 kg (not necessary to know since we’re about to multiply total frictional force by 4 since there are 4 tires)
Us = 0.8
Fftire = mass x gravity x Us = 250kg x 9.81m/s^2 x 0.8 = 1962 N
Fftotal = 4 x 1962N = 7848N
Frictional force = accelerative force
7848N = Ma = 1000kg (a)
a = 7.848m/s^2
But let’s not forget significant digits…
The correct answer is 7.8 m/s^2
Just to add some fun, let’s figure out the possible 1/4 mile time assuming that these tires are the only limiting factor:
1/4 mile = 1320 feet = 402.336 meters
s = vi x t + 1/2 x a x (tt)
402.336 = 0 + 1/2 x 7.8 x (tt)
402.336=3.9 x (t*t)
103.163=t^2
t=10.1569 seconds
Damn, I will avoid doing real engineering at all costs today…
:tup:
That is the maximum force the tire’s can apply to the road before the tires slip, not before the car begins to roll. If the tires aren’t slipping then the tire/road interface is always static. Theoretically, 7.841 m/s^2 would cause the tires to slip and you would need a coefficient of kinetic friction to continue the problem.
holy fuck i was actually right…
Cool, thanks for the explaination. I was just reading ahead, as we just starting doing static friction, and decided what the hell, why not try it.
One quick question though, if the wheels are actually moving, why wouldn’t that be considered kinetic friction?
just spend 20 hours a week in labs like the rest of us had to do :tup:
Nope. If the tire is slipping against the road then it would be kinetic friction. When a tire is rolling, at any given time the tire/road interface is not slipping so it is continually static friction. Think about driving in snow. You know how there’s a fairly well defined point between when you have traction and when you’re sliding? That’s because you change from static friction to kinetic friction. Or think about how proper braking is not to lock up the tires. You can stop faster with the tires rolling than with the tires locked up because coefficient of static friction is always greater than coefficient of kinetic friction.
While we’re on the subject, that’s also what makes teflon such a cool lubricant. A teflon/teflon interface has an equal static and kinetic frictional coefficient. There’s no stick/slip.