Check out some similar questions!

srob106 · Dec 1, 2010, 06:16 PM

harum · Dec 1, 2010, 09:18 PM

It depends on the slope. What is the slope?

Unknown008 · Dec 1, 2010, 11:05 PM

Could you re-submit your question by typing it in the 'Answer Box' please? When you typed your question the first time, it was in the title area and the latter cannot support all the question. Thank you.

ebaines · Dec 2, 2010, 08:34 AM

+1 on Harum's note that it depends on the slope. But it also depends on the percentage of the car's weight that is supported by tires that are driven by the engine (an all-wheel drive car will have a different answer than a rear- or front-wheel drive car). And if the car does not have all wheel drive then it also depends on the height of the car's center of gravity above the ground (as the car accelerates more of its weight is shifted to the rear tires).

harum · Dec 2, 2010, 11:20 AM

Friction F(total) = F1 + F2 + F3 + F4 = k*(N1 + N2 + N3 + N4) = k*m*g*cos(alpha)

ebaines · Dec 2, 2010, 12:27 PM

Originally Posted by harum

Friction F(total) = F1 + F2 + F3 + F4 = k*(N1 + N2 + N3 + N4) = k*m*g*cos(alpha)

Perhaps. For acceleration this must be modified if all four wheels aren't driven by the engine, to reflect an increase in normal force at the driving wheels under acceleration (for rear-wheel drive cars) or a decrease (front-wheel drive), because the center of mass of the car is above the ground.

Unknown008 · Dec 2, 2010, 12:33 PM

Well... I think that this is getting far, while I'm almost certain the actual question is something along the lines of using

$P = Fv$

and $F = \mu R$

or along those lines, only.

harum · Dec 2, 2010, 01:16 PM

Normal support reactions most definitely vary from wheel to wheel. But their sum, to which the total friction is proportional, does not depend on the weight distribution. Wouldn't it be the only thing that matters, because the friction force is additive?

ebaines · Dec 2, 2010, 01:32 PM

Originally Posted by harum

Normal support reactions most definitely vary from wheel to wheel. But their sum, to which the total friction is proportional, does not depend on the weight distribution. Wouldn't it be the only thing that matters, because the friction force is additive?

Yes, for an all-wheel drive car. But for a rear wheel drive car the normal force on the rear wheel depends on both the front-to-rear weight distribution and the car's acceleration. So first you need to know how much weight is carried by the drive wheels when standing still - the higher the percentage the better (this is one reason why dragsters and Porsches have the engine in the back - the added weight over the drive wheels allows them to accelerate faster). Then this normal force on the drive wheels is increased for rear wheel drive cars under acceleration by an amount equal to m*a*h/L, where h is the height of the center of gravity and L is the length of the wheel base length. Similarly, the normal force on the front wheels is decreased by the same amount. For an all-wheel drive car the effect cancels out. But for a rear wheel drive car this means that you get more traction on the wheels that are being driven, and for a front wheel drive car you get less, if acceleration is greater than 0. If we let N1 and N2 be the normal force for the rear wheels when standing still, then for a rear wheel drive car on level ground your formula becomes:

$ k(N_1 + N_2 + ma \frac h L) = ma $

which can be rearranged to solve for a:

$ a = \frac {k(N_1+N_2)}{m(1-\frac {kh} L)} $

and for a front wheel drive car:

$ k( N_3 + N_4 - ma \frac h L) = ma \\ a = \frac {k(N_3+N_4)}{m(1+\frac {kh} L)} $

Conversely - under braking the deceleration of the car puts more downward force on the front wheels than the back. This is why the majority of braking is performed by the front brakes, not the back.

harum · Dec 2, 2010, 02:27 PM

We have to agree that the maximum acceleration of the car is achieved at the very start, when v=0m/s and when the wheels do not slip. This is because the static friction is always larger than the dynamic friction. Whatever happens afterwards is irrelevant. This is one thing.

Second, you are solving a different problem. What I meant was that the slope was relevant. What you are saying is that the car's make and model is relevant. Yes, if we measure the maximum acceleration of our car and then re-measure it again after rebuilding the car by moving its engine to the other end and by raising its center of mass above the ground and after the tire change, the results will be different. They will. No argument here. But those cars would be called different cars in my book. What would your book say?

One is most probably assumed here to think about accelerations of a car with fixed geometry/anatomy and not some shape-changing transformer. Are we comparing different cars or there is only one car?

ebaines · Dec 2, 2010, 02:56 PM

Originally Posted by harum

We have to agree that the maximum acceleration of the car is achieved at the very start, when v=0m/s and when the wheels do not slip. This is because the static friction is always larger than the dynamic friction. Whatever happens afterwards is irrelevant. This is one thing.

I don't agree, but it really doesn't matter, since as both you and I have asumed so far the max force between tire and pavement is equal to kN. Anything greater than that leads to the wheel spinning - and I do agree that once that happens the force (and resulting acceleratoin) is less. But there's nothing that says that the wheels have to slip, and thus the acceleration can be constant as long as k is constant. But even if you're right that the max acceleration is at v = 0, so what?

Originally Posted by harum

Second, you are solving a different problem. What I meant was that the slope was relevant. What you are saying is that the car's make and model is relevant. Yes, if we measure the maximum acceleration of our car and then re-measure it again after rebuilding the car by moving its engine to the other end and by raising its center of mass above the ground and after the tire change, the results will be different. They will. No argument here. But those cars would be called different cars in my book. What would your book say?

My book says you're missing the point.

Originally Posted by harum

One is most probably assumed here to think about accelerations of a car with fixed geometry/anatomy and not some shape-changing transformer. Are we comparing different cars or there is only one car?

One car - all the OP need do is define it and we can calculate its max acceleration. And to your point - he also needs to tell us if it's on a slope or not.

Look, in reality the OP is probably a high school kid taking intro physics and the problem is defined so that the answer is simply a = kg. But since he didn't finish typing out the problem statement, we may never know!

harum · Dec 2, 2010, 03:08 PM

Originally Posted by ebaines

the answer is simply a = kg.

My point is that the only condition that may or may not be missing in this unfinished problem is the slope. Everything else is pretty much given. Adding the slope would not make the problem harder, but would change the answer. All the assumptions about the wheelbase and the model sound like a bit of a stretch and would turn this problem into a totally different animal. You would then have to know what wheels are driving, how the weight is distributed or all kinds of geometry of the car -- i.e. many more things might be missing which, if they really are, would change the problem.