Solving acceleration of a car with torques

[whatdoido]

Hi!

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

1. Homework Statement

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?

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

Homework Equations

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

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.

 

[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.

 

[CWatters]

What your diagram doesn't show is the rearward acting force, call it F_i = 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.

 

[haruspex]

the wheel creates a torque F1b when Fμ=F1. F3a is the torque that keeps the car's front down in relation to wheel.

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.

 

[whatdoido]

What your diagram doesn't show is the rearward acting force, call it F_i = 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.

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?

 

[whatdoido]

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.

$F_1, F_2$ 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.

 

[haruspex]

$F_1, F_2$ 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.

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?

 

[whatdoido]

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?

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

 

[haruspex]

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

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?

 

[whatdoido]

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?

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

 

[haruspex]

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

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

 

[whatdoido]

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

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

 

[haruspex]

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

Right.