Car's maximum acceleration on a road is proportional to what?

[haruspex]

It resists the roll.

I worry that the various contributors are now causing destructive interference with slightly different ways of expressing things.
It resists sliding, but neither would I say it "allows" the roll. It encourages rolling rather than sliding.

 

[rudransh verma]

The roll causes the strain that causes the static friction.

If there is static friction where is the force that generates it? Please answer this one question.

 

[bob012345]

It is the OP that has complicated it by worrying about the wheels separately. See post #15.

Please see post#14.

 

[bob012345]

I worry that the various contributors are now causing destructive interference with slightly different ways of expressing things.
It resists sliding, but neither would I say it "allows" the roll. It encourages rolling rather than sliding.

I agree and I'd be willing to work privately with the OP to resolve the issues if you like. Or would you rather?

 

[jack action]

I may have missed something before but I am again asking if it’s static friction between the road and tyre how is the tyre/car accelerating?

Looking at the following figure:

• Ignore the green arrows;
• The wheel torque $T_a$ is shown in blue (torque direction). That is the applied torque coming from the motor;
• The reaction force to this torque is the Friction force $F_f$. It is pushing the wheel forward;
• The blue force at the axle (center of the wheel) is equal and opposite to the friction force $F_f$. The distance between the two blue forces (wheel radius $r$) times the force's magnitude represents a couple that opposes the wheel torque;
• $F_f = \frac{T_a}{r}$. The maximum possible value is equal to the static friction force;
• If you would do the free body diagram for the car's body, that axle force would show as equal and opposite, thus pushing the car's body forward - so with the same direction as $F_f$.

The lesson in that is that the direction of the torque dictates the direction of the friction force at the tire contact patch and that friction force is the "force applied" that you talk about and pushes the entire car, coming directly from the torque applied at the wheel.

 

[bob012345]

Looking at the following figure:

View attachment 296562​

• Ignore the green arrows;
• The wheel torque $T_a$ is shown in blue (torque direction). That is the applied torque coming from the motor;
• The reaction force to this torque is the Friction force $F_f$. It is pushing the wheel forward;
• The blue force at the axle (center of the wheel) is equal and opposite to the friction force $F_f$. The distance between the two blue forces (wheel radius $r$) times the force's magnitude represents a couple that opposes the wheel torque;
• $F_f = \frac{T_a}{r}$. The maximum possible value is equal to the static friction force;
• If you would do the free body diagram for the car's body, that axle force would show as equal and opposite, thus pushing the car's body forward - so with the same direction as $F_f$.

The lesson in that is that the direction of the torque dictates the direction of the friction force at the tire contact patch and that friction force is the "force applied" that you talk about and pushes the entire car, coming directly from the torque applied at the wheel.

Is the blue force at the axle on the wheel or not?

 

[jack action]

Is the blue force at the axle on the wheel or not?

Yes. It is the force at the axle bearing.

 

[Steve4Physics]

@rudransh verma, maybe this helps...

In this problem the internal forces and torques between parts of the car don’t matter - they are in Newton 3rd law pairs and cancel overall.
_____________

Q: You sit in your 4-wheel drive car, on perfectly smooth ice, and try to drive off. What happens and why? [Click fuzzy area to check your answer.]

A: The wheels turn but the car doesn’t move. That’s because there is zero friction (no ‘grip’) between road and tyres. There is no horizontal applied force ($F_a=0$) on the car to cause acceleration.
_____________

Q: You do the same on a rough road. What happens and why? [Click fuzzy area to check your answer.]

A: The wheels turn, the tyres exert a backwards frictional force on the ground. Because of Newton’s 3d law, the ground exerts a forwards frictional force on the tyres. This is the force that makes the car accelerate.

In this situation, the applied force ($F_a$) which causes the car’s acceleration is just another name for the frictional force ($f$) of the ground on the tyres. They are not two different forces.

All things being equal, the frictional force on each of the tyre is ¼ of the total. (You might want to consider if this is still true for a 2-wheel drive car.

 

[bob012345]

Yes. It is the force at the axle bearing.

Then how can the wheel accelerate if the two blue forces are equal and opposite and both act on the wheel?

 

[rudransh verma]

The lesson in that is that the direction of the torque dictates the direction of the friction force at the tire contact patch and that friction force is the "force applied" that you talk about and pushes the entire car, coming directly from the torque applied at the wheel.

Edit: So $f_r-Torque=ma$ is wrong or the eqn $f_r-F_a=ma$. Torque is the one which generates static friction and this friction accelerates the car forward.

 

[jack action]

Then how can the wheel accelerate if the two blue forces are equal and opposite and both act on the wheel?

Because the car's body is not fixed and the wheel will accelerate with it.

 

[bob012345]

So $f_r-Torque=ma$ is wrong or the eqn $f_r-F_a=ma$. Torque is the force which generates static friction and this friction accelerates the car forward.

No, torque is not a force, it is a force multiplied by a distance. The units are Nm not N.

 

[bob012345]

Because the car's body is not fixed and the wheel will accelerate with it.

My point is that if you draw a free-body diagram of the wheel alone, which is perfectly valid, it better have a net force on it if it is undergoing acceleration. You cannot say the net force on the wheel is zero but the car carries it along with it. Every part of the car has a net force on it if it accelerates with the car. Even the driver.

 

[jack action]

So $f_r-Torque=ma$ is wrong or the eqn $f_r-F_a=ma$. Torque is the force which generates static friction and this friction accelerates the car forward.

No,
$$F_a = ma$$
And
$$F_a = \frac{T_a}{r}$$
And also
$$F_{a\ max} = \mu mg$$
So (extra info not needed for your problem)
$$T_{a\ max} = F_{a\ max} r$$
$$T_{a\ max} = \mu mg r$$
So the maximum torque you can apply at the wheel depends on the friction coefficient.

But what is important is how the torque applied is transformed into a force applied at the tire contact patch, and what direction that friction force will take.

And therefore, for your problem (getting the maximum acceleration):
$$F_{a\ max} = ma$$
$$\mu mg= ma$$

 

[jack action]

it better have a net force on it if it is undergoing acceleration.

Not if it's massless.

The point of the drawing was just to show how the direction of the friction force can be in the same direction as the body's motion with the introduction of a wheel.

 

[jbriggs444]

If there is static friction where is the force that generates it? Please answer this one question.

Static friction is a force. Forces do not need to be generated by other forces.

Though certainly the torque of the axle sets up the condition under which the force of static friction can continuously exist.

 

[jack action]

I just think it is a misleading diagram in the context of this thread.

If the OP's question would be exactly the same, except for the car that started at an initial velocity until it stops (thus braking), the answer would be exactly the same.

But the free body diagram - representing the car as a block - would show the friction force in the other direction, i.e opposite of motion, i.e. the "expected way" of a block sliding on the ground (even though there is no sliding involved).

How can you explain the change of friction force direction without explaining how a wheel works and how the direction of the applied torque at the wheel plays a major role in this?

 

[bob012345]

If the OP's question would be exactly the same, except for the car that started at an initial velocity until it stops (thus braking), the answer would be exactly the same.

But the free body diagram - representing the car as a block - would show the friction force in the other direction, i.e opposite of motion, i.e. the "expected way" of a block sliding on the ground (even though there is no sliding involved).

How can you explain the change of friction force direction without explaining how a wheel works and how the direction of the applied torque at the wheel plays a major role in this?

Good point. Frankly, I've entered a bit of a brain fog regarding the free-body diagrams that I have to work out on this problem beyond the basic idea of the static friction being the driving force.

 

[bob012345]

Good point. Frankly, I've entered a bit of a brain fog regarding the free-body diagrams that I have to work out on this problem beyond the basic idea of the static friction being the driving force.

Ok, I resolved my brain fog issue over the free body diagram of the wheel alone. What bugged me in the diagram is that it looked like the blue Axle force to the right was exactly equal to the static friction force to the left which could not be true under acceleration. The total forces left and right are slightly different and total to a net force of the exact amount necessary to accelerate the wheel of mass $m$ by acceleration $a$. Likewise the driving torque at the axle is greater than the counter torque due to all other torques by exactly the amount consistent with an angular acceleration $\alpha= \large \frac{a}{R}$.

 

[jbriggs444]

View attachment 296601

If the green arrows are intended to denote rolling resistance then the nature of that effect could be more accurately denoted by a shift in the upward red arrow (normal force) so that it is no longer directly under the axle. The contact patch experiences more force as it compresses into the ground than as it rebounds from it. So the center of force is forward of the center of the contact patch. That is one source of rolling resistance.

Another source of rolling resistance is frictional torque at the axle.

Either way, when superimposed on everything else, the net effect is more directly equivalent to a retarding torque rather than a retarding force.

 

[bob012345]

If the green arrows are intended to denote rolling resistance then the nature of that effect could be more accurately denoted by a shift in the upward red arrow (normal force) so that it is no longer directly under the axle. The contact patch experiences more force as it compresses into the ground than as it rebounds from it. So the center of force is forward of the center of the contact patch. That is one source of rolling resistance.

Another source of rolling resistance is frictional torque at the axle.

Either way, when superimposed on everything else, the net effect is more directly equivalent to a retarding torque rather than a retarding force.

No doubt the actual forces and torques are complex but with a no-slip condition whatever the retarding torques are there must be a co-existing retarding force since $a$ and $\alpha$ are consistent unless the wheels are either slipping or sliding.

 

[jbriggs444]

No doubt the actual forces and torques are complex but with a no-slip condition whatever the retarding torques are there must be a co-existing retarding force since $a$ and $\alpha$ are consistent.

But a retarding force pair that involves a forward torque on the wheel? That's just wrong.

 

[bob012345]

But a retarding force pair that involves a forward torque on the wheel? That's just wrong.

Not sure what you mean. Please define force pair and forward torque.

 

[jbriggs444]

Not sure what you mean.

Look at your two green arrows. You have one at the axle acting leftward and one at the contact patch acting rightward. That is a counter-clockwise torque acting to make the wheel rotate more rapidly.

The true effect is a clockwise torque acting to make the wheel rotate more slowly.

 

[bob012345]

Look at your two green arrows. You have one at the axle acting leftward and one at the contact patch acting rightward. That is a counter-clockwise torque acting to make the wheel rotate more rapidly.

The true effect is a clockwise torque acting to make the wheel rotate more slowly.

I did not make that diagram nor originally post it but I reposted it to argue I thought it was wrong. Here is another free-body diagram I found of a wheel slowing down by rolling resistance. It looks like it would have a torque that would speed the wheel up but we know it does not so there is something about the way friction works at the point of contact that I'm just not getting. It is easy to see how $F_{rr}$ will act to slow down the wheel but I don't see how the torque $RF_{rr}$ acts in the correct direction. I wonder if it has something to do with the point of contact actually being the center of rotation? And where is the static friction in this diagram? The brain fog continues...
https://www.physicsforums.com/attachments/296631
https://archive.thepocketlab.com/educators/lesson/rolling-resistance-physics-lab

 

[hutchphd]

If the wheel is already rolling at speed then $$F_{rr}=0 ~~exactly$$ Remember $$F_{rr} \leq \mu N $$ and the angular acceleration is 0

 

[jbriggs444]

If the wheel is already rolling at speed then $$F_{rr}=0 ~~exactly$$ Remember $$F_{rr} \leq \mu N $$ and the angular acceleration is 0

If the wheel is rolling at speed then there is a forward torque from the axle and a rearward torque from rolling resistance. Rolling resistance is a torque, not a force.

 

[bob012345]

If the wheel is already rolling at speed then $$F_{rr}=0 ~~exactly$$ Remember $$F_{rr} \leq \mu N $$ and the angular acceleration is 0

The diagram I showed assumes the wheel is slowing down. I think the diagram is again wrong like the other one. There appears to be a lot of incorrect free-body diagrams available.

Wheels rolling at some starting speed will slow down due to rolling friction. That much is certain.

If the wheel is rolling at speed then there is a forward torque from the axle and a rearward torque from rolling resistance. Rolling resistance is a torque, not a force.

It must be both if the wheel is not slipping. What I think resolves this is that the rolling resistance is a distributed force acting over a region of contact (a patch) in such a way that it opposes the forward motion and create a torque which slows the wheel.

 

[hutchphd]

Is there a conventional way to show it on free body diagram? Seems a confusing adjunct to the pedagogy

 

[bob012345]

It would be nice if there was a simple drawing tool here at PF.