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

[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.

 

[Steve4Physics]

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

The Post #85 diagram shows rolling resistance as a simple horizontal force acting at the base of the wheel, vertically below the wheel's centre. This is an inaccurate simplfication and (as has already been noted) gives a torque which would tend to accelerate the wheel!

A better representation is this: https://www.lockhaven.edu/~dsimanek/scenario/rollres3.gif

For this diagram and accompanying explanation, see the section entitled “Rolling Resistance” about halfway down this link: https://www.lockhaven.edu/~dsimanek/scenario/rolling.htm

 

[jack action]

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 $\alpha= \large \frac{a}{R}$.View attachment 296601

The diagram is not a true free body diagram (FBD).

It's a simplified schematic version to explain the relationship between force and torque directions.

What is important to understand is that a rolling wheel can reverse the friction force direction by reversing the wheel torque, even if the rolling direction stays the same.

In an accurate FBD, yes, there would be wheel accelerations to consider (both linear and angular). BUT THIS IS IRRELEVANT TO THE PROBLEM AT HAND. It would only confuse the OP to make an FBD for each wheel and the car's body. Just imagine the wheel is massless and the car's body is heavier by the equivalent of the wheels' masses.

Rolling resistance should also be ignored for this discussion. IT IS IRRELEVANT TO THE PROBLEM AT HAND and can only confuse the OP.

I can't believe we have 91 posts in this discussion.

 

[kuruman]

Here are 3 FBDs related to the one-wheel skateboard mentioned in post #21. To make things simple, only horizontal forces are shown.

$f_{\text{S}} = ~$ force of static friction exerted by the road on the wheel.
$f_{\text{WB}}=~$ contact force exerted by the wheel on the board at the axle.
$f_{\text{BW}}=~$ contact force exerted by the board on the wheel at the axle.

To @rudransh verma:
The sum of the board and wheel FBDs is the FBD of the one-wheel skateboard. The board, the wheel and the combined board + wheel have a common acceleration $a$. What is the net force in each case that provides this acceleration?

A picture is worth 10³ words.

 

[sysprog]

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

If you have access to PF, you presumably also have access to some great physics-friendly drawing tools available on the net, e.g. GeoGebra $-$ https://geogebra.org/m/NvzumAMV, LaTeX/TIKZ $-$ https://texample.net/tikz/examples/all/, Inkscape $-$ https://inkscape.org/gallery/?tags=physics.

 

[hutchphd]

A picture is worth 10³ words.

All preamble aside, what if one wishes to use the term "rolling friction"? Personally I would be inclined not to use it ever. Is there a reason to put it in the elementary curriculum (as yet another frictional force??) Just askin'