# Solving acceleration of a car with torques

1. Apr 4, 2016

### whatdoido

Hi!

I cannot get rid of a mass and having some confusion about this whole problem.

1. The problem statement, all variables and given/known data

A car's wheelbase is 300 cm and its center of gravity horizontal distance from rear axle is 120 cm and its distance from the ground is 75 cm. Wheels and ground's coefficient of friction is 0,50. Specify maximum acceleration on a flat road if the car is rear wheel drive. With what solution this acceleration could be improved?

$x=120 cm=1,2 m$
$\mu=0,50$
$a=300 cm=3,0 m$
$b=75 cm=0,75 m$

2. Relevant equations

$M=J\alpha$
$M=Fr$
$J=\frac{1}{2}mr^2$
$\sum M=0$

3. The attempt at a solution

I drew an illustration about the problem:

A is the center of gravity. I understand that the wheel creates a torque $F_1b$ when $F_\mu\mu=F_1$. $F_3a$ is the torque that keeps the car's front down in relation to wheel. As far as I can see, I can try solving the acceleration with torques $F_1b$ and $F_3a$ in relation to the wheel(s) but obviously $F_2$ is the force that keeps car's rear down and cannot be ignored. I can solve car's proportional masses in relation to A with torques $F_1x$ and $F_3(a-x)$ but I don't know how that helps. So first step would be to know in what relation I should try solving torques. Also how to solve acceleration without knowing wheel's mass.

2. Apr 4, 2016

### CWatters

The force that accelerates the car is the friction force Fμ. That depends on the force on the rear wheels F2. The weight of the car (F1) is distributed between F2 and F3 (eg F3 reduces and F2 increases if the car accelerates). Under what condition is F2 a maximum and what's it's value?

Edit: Sorry I think I might be leading you astray here. Typically F2 is a max when the car is just about pulling a wheelie and so F3 = zero. However after some thought I'm not sure you can assume this us true for this problem.

Last edited: Apr 4, 2016
3. Apr 4, 2016

### CWatters

What your diagram doesn't show is the rearward acting force, call it Fi = ma acting through the centre of gravity due to the cars acceleration.

It appears you are also assuming the wheel radius is 0.75m which need not be the case.

Last edited: Apr 4, 2016
4. Apr 4, 2016

### haruspex

Your diagram is unnecessarily complicated. Treat the car as a rigid body. When the car is on the point of flipping, only three forces act on it. What are they? Take torques about the car's mass centre.

5. Apr 6, 2016

### whatdoido

Yes I assumed that because I thought I could not solve it without that assumption. But true, I should not assume something like that..

So acceleration creates a force in point A? Then it would point upwards?

6. Apr 6, 2016

### whatdoido

$F_1, F_2$ and $F_3$

I would mark $F_3$ as zero when $F_1 = F_2$ around tipping point, but CWatters is unsure I can assume $F_3$ as zero.

7. Apr 6, 2016

### haruspex

There are two separate constraints that may limit the acceleration. One involves tipping, the other involves the coefficient of static friction. To answer the problem completely, you have to analyse each separately and see which one imposes the lower limit.
In my post #4, I was concerned with the tipping constraint. CWatters' point is that F3 need to be zero if the other other constraint is the limiting one. But let's go through the tipping first, so we can take F3 as zero.
There are still three forces that act on the car as a whole. In your diagram, I assume the circle is a rear wheel. I don't know what F1 represents, but it does not appear to be an external force acting on the car. F2 seems to be a force on the rear wheel from the car, so again that is an internal force. So, having discounted F1, F2 and F3, what three external forces act on the car as a whole?

8. Apr 8, 2016

### whatdoido

Ok, I understand the goal of this problem a bit more clearly. $F_1$ represents the force created by friction, but let's leave it out. If I would list three external forces acting on the car, then $F_\mu$ the friction force, applied force of engine and grounds normal force would be them. So engine creates a torque to wheel and act on rear wheels and normal force acts on all four wheels. The friction force should be considered on rear wheels only? Since that's where the acceleration happens

EDIT: Yes, circle is the rear wheel

9. Apr 8, 2016

### haruspex

The engine is part of the car, so that is not an external force. For the third force, consider why there is a normal force on the wheels.
As for that normal force, is it the same on all four wheels?

10. May 21, 2016

### whatdoido

Because of gravity? So gravity is external force?

I would think that normal force is higher on the rear wheels since it is closer to the center of gravity

11. May 22, 2016

### haruspex

Yes. If the car is on the point of tipping, can you be more specific about the distribution of the normal forces?

12. Jun 1, 2016

### whatdoido

At point B it would be 0 meaning that on rear wheels the normal force $N=mg$. $m$ is the weight of the car

13. Jun 1, 2016

Right.