Applying Physics to Vehicle Dynamics: Do Wheel Number Matter?

[erobz]

@erobz , what is in contradiction (in your mind) with this result?

This is what is confusing.

$$ F_{net} = 2(f_r - f_f) \implies \delta W = \int F_{net} \, dx $$

Which should be equivalent to what I've shown in #24. Yet the static frictional forces apparently "do no work" on the vehicle.

Don't even bother though, I can tell people are already getting upset.

 

[jack action]

The torque ##T## in post #24 is a net torque. It should at least be detailed as:
$$T= T_{in} - (F_r + F_d)R$$
Where ##T_{in}## is the input torque, ##F_r## is the rolling resistance from all wheels, and ##F_d## is the aerodynamic drag on the body.

By doing so, you have assumed the whole car as a single block and modified the mass ##m## definition to an equivalent mass ##m_e## including the wheel rotating inertia. Where:
$$m_e = m\left(1+\frac{4I}{mR^2}\right)$$
and thus reducing everything to:
$$F_{net} = m_e a$$
as I explained in post #30.

 

[jack action]

This is what is confusing.

$$ F_{net} = 2(f_r - f_f) \implies \delta W = \int F_{net} \, dx $$

Which should be equivalent to what I've shown in #24. Yet the static frictional forces apparently "do no work" on the vehicle.

Yes, it does. The static frictional forces move with the vehicle. A force times a displacement is work.

The static point at which it applies might be at rest but the force constantly moves to the next point that will be at rest.

 

[Lnewqban]

This is what is confusing.

$$ F_{net} = 2(f_r - f_f) \implies \delta W = \int F_{net} \, dx $$

Which should be equivalent to what I've shown in #24. Yet the static frictional forces apparently "do no work" on the vehicle.

Don't even bother though, I can tell people are already getting upset.

No reason to get upset; it is important that you arrive to an understanding on this simple, but confusing subject.
I have the whole morning for you.

I see zero contradiction on your previous posts.
Perhaps, we should look at it as a case of cause-effect.

The pneumatic effect (expanding hot gasses inside the cylinders) are the cause of the rotation of the wheels.
Instantaneous static friction at the contact patch of the tires is the effect or consequence.

Then, the combination of energy-providing-torque and resistance-providing friction is the following cause, and its effect is the forward movement of the axle and rest of the attached car (reaction to all the above).

 

[erobz]

Yes, it does. The static frictional forces move with the vehicle. A force times a displacement is work.

The static point at which it applies might be at rest but the force constantly moves to the next point that will be at rest.

Yes it does... as in...the frictional force does work on the vehicle? Because that's what it seems like you are saying. And I think it does too.

There is a net external force on the car provided by the difference in the rear ground-tire friction and the front. This does not say that the ground does work on the car because the contact point is static. The contact point is static because the car is not a rigid body and is continuously deformed by the engine (it takes energy to make the rear wheels turn.

It seems like they are saying the work is done by "dead dinosaurs", not by the frictional force.

 

[erobz]

No reason to get upset; it is important that you arrive to an understanding on this simple, but confusing subject.
I have the whole morning for you.

I see zero contradiction on your previous posts.
Perhaps, we should look at it as a case of cause-effect.

The pneumatic effect (expanding hot gasses inside the cylinders) are the cause of the rotation of the wheels.
Instantaneous static friction at the contact patch of the tires is the effect or consequence.

Then, the combination of energy-providing-torque and resistance-providing friction is the following cause, and its effect is the forward movement of the axle and rest of the attached car (reaction to all the above).

I'm fine with all of that. Is this all just a semantical argument based on the scope of the model? @jack action seems to be saying that the static frictional forces do work on the vehicle. I think they do too.

So does Author Giancolli: Its right in the text of the problem. The friction forces ##F_1## and ##F_2## decelerate the car.

The force moves with the vehicle and hence has a displacement. The static friction force does work to slow the vehicle.

@hutchphd say its "dead dinosaurs" that do the work, but not the friction. I can see a perspective where that is true too. The dinos act through the engine which provides the torque! Should people really be getting frustrated with me over this?

 

[Lnewqban]

I'm fine with all of that. Is this all just a semantical argument based on the scope of the model? @jack action seems to be saying that the static frictional forces do work on the vehicle. I think they do too. So does Author Giancolli:
The force moves with the vehicle and hence has a displacement. The static friction force does work.

Respectfully, I believe that the idealized friction force between wheel and surface does as much work as the reactive force at the pivot of a see-saw or at the hinges of a door: zero.

If you grab the top edge of one door of your kitchen cabinet, halfway between the hinges line and the vertical edge, and exert a twisting force (zero translation), the door will rotate about the hinges, dragging your twisting hand along a circular trajectory.

Now, think of the hinge's line like the contact patch of our tire, the halfway top of the door as the location of our axle, and your twisting effort, like the energy inducing the door's swing.

Applying calculus:
As the number of hinged doors (and/or horizontal length of the door) tend to infinite, the translation trajectory of your hand tends to linear (just like for the actual axle).

IMHO, the only work that the friction force can do in this case, is to heat up the rubber of the wheels, when slipping, skipping and deforming carcass of the tires exist.

 

[erobz]

Respectfully, I believe that the idealized friction force between wheel and surface does as much work as the reactive force at the pivot of a see-saw or at the hinges of a door: zero.

You're certainly not going to offend me.

The Author of my Physics Textbook ( implies ) that the work done to stop the vehicle acts through these friction forces. That has to be the interpretation when we write (as they did) that:

The Friction Forces ##F_1## and ##F_2## decelerate the car, so Newtons Second Law gives:

$$ F_1 + F_2 = Ma$$

etc...

That is the same logic I applied to develop post #10 which is completely consistent with and independent of post #24 (work/energy).

Something is clearly not agreed upon...even among experts if it is to be the case that the static friction can do no work.

 

[jack action]

This:
$$F_{in}-F_{out}=ma$$
Could be rewritten as
$$\int (F_{in}-F_{out})dx=\int madx$$
$$W_{in}-W_{out} = \frac{1}{2}m\Delta(v^2)$$
The ##W_{in}## is done by "dead dinosaurs" (through a friction force) but the ##W_{out}## is done by the resistive forces (rolling resistance and aerodynamic drag). All forces move at velocity ##v## with the vehicle.

If the ##W_{in}## is the same as the ##W_{out}##, then we have no resulting force and the vehicle has a constant speed.

If the ##W_{in}## is greater than the ##W_{out}##, then we have a resulting force that will accelerate the vehicle.

If the ##W_{out}## is greater than the ##W_{in}##, then we have a resulting force that will decelerate the vehicle.

even among experts if it is to be the case that the static friction can do no work.

Stop saying that. It is not the case here. This is not a block on an incline, held by static friction. This is a wheel on an incline held by static friction. The wheel obviously moves. The combination of gravity and static friction creates a couple that gives ##T=mgR\sin \theta## and that does work on the wheel, even without the presence of "dead dinosaurs".

 

[hutchphd]

The wheel obviously moves. The combination of gravity and static friction creates a couple that gives T=mgRsin⁡θ and that does work on the wheel, even without the presence of "dead dinosaurs

Just to be clear, the work is done because the wheel is turning. When the wheel is stationary (and not skidding), no work is done. The nature of the ordinary "static friction" is the same in either case. When the wheel rolls there is additionally what we call "rolling resistance" which might be a brake drum or the mechanism of flexing descriibed. All such mechanisms require deformation of the object (a bearing rotating or tire dimpling being a continuous deformation)

 

[Lnewqban]

...
The Author of my Physics Textbook ( implies ) that the work done to stop the vehicle acts through these friction forces. That has to be the interpretation when we write (as they did) that:

The Friction Forces ##F_1## and ##F_2## decelerate the car, so Newtons Second Law gives:

$$ F_1 + F_2 = Ma$$

etc...

In that case, the book statement seems correct to me, the kinetic energy of the vehicle, being now the cause, is doing all the work.
That work is transformed into thermal energy of braking pads-discs, being the result.

The tire's contact patches (and their respective friction forces) are, again, nothing but four non-energy-producing links in the chain through which that energy flows.
In real life, the patches consume some of that kinetic energy and use it for rubber deformation and heat.

Would you mind commenting about the above posted video (Post #34)?

 

[erobz]

Would you mind commenting about the above posted video (Post #34)?

Its neat, but I'm not understanding it.

 

[erobz]

In that case, the book statement seems correct to me, the kinetic energy of the vehicle, being now the cause, is doing all the work.
That work is transformed into thermal energy of braking pads-discs, being the result.

But they are the net force on the object in the very definition of work, that is allowing the work from the brakes to be converted into a change in the vehicle's translational kinetic energy. No force there...no change. No change...no work. Why should it matter where the energy is stored, whether that be in initial kinetic energy or dead dinos. The path to "ground" is the crucial part for work to be done on the vehicle. That connection is what is converting that stored energy to heat, or kinetic energy. To me, from its definition it seems like work is the conversion of energy, not the bucket from which it is being pulled? Can we do work without a bucket, No. But the bucket can't do work by itself either?

 

[hutchphd]

No force there...no change. No change...no work.

But notice that all forces in that chain do not do work. There is always a particular place where the mechanical energy is converted to heat. For friction it is often (always?) a motion interface. For a sliding (or squirming) tire it is at the ground/tire interface. For a brake it is at the pad/disc interface.
Similarly for a heat engine there is a localized place where the "heat" energy is converted into mechanical energy. Carnot tells us that even when clever we can't get it all.
It does matter that we can correctly identify where, in the chain of energy, this happens.

 

[Lnewqban]

... The path to "ground" is the crucial part for work to be done on the vehicle. That connection is what is converting that stored energy to heat, or kinetic energy...

To me, there is no path of energy to ground.
The road is just a surface to which the wheels anchor in a continuous fluid manner.
The direction of the flow of mechanical energy matters regarding how the wheels gain or lose impulse.

Accelerating car: Energy flows from the engine, bounces on the road surface, and continues on to the mass of the car via axle and suspension.

Braking car: Energy flows from the mass of the car, bounces on the road surface, and continues on to the brakes, which fixed parts are anchored to the chassis

Without grip on that bouncing surface (lifted wheels or slippery mud), the mechanical energy still flows from engine to wheels, or from spinning wheels to brakes.
The only difference is that the car does not change its velocity respect to the ground.

 

[erobz]

All these things make sense. My only issue is why it appears as though friction does work in the definition. If have 10$ and I'm in a box and a pass it to Lenewqban , and Lenewqban ( the only one seen holding the money) passes it to hutchphd. Did I give hutchphd $10, or did Lenewqban? Lenewqban is the friction force here.

 

[Lnewqban]

Al these things make sense. My only issue is why it appears as though friction does work in the definition.

Starting off a torque, what does work should be a pair of forces (couple).
Both forces induce movement of the points that are free to move about the point that is unable to move (pivot, instantaneous contact patch).

Please, see:
https://en.wikipedia.org/wiki/Couple_(mechanics)

The driving entity could be a shaft output torque (as represented), or a tangential force applied on the top of the wheel (as when you have to push a heavy car and you push with your hands located on the top of one tire and use mechanical advantage of 2 to do it).

 

[erobz]

I want this to be over as much as anyone else. So I'm going to be as concise as possible about what I find confusing still:

$$\int F_{net} \, dx = \int 2 \left( f_r - f_f \right) dx = 0 $$

Will anyone explain this?

 

[hutchphd]

Simple. Its wrong. Are we done? Is it copied from somewhere (perhaps I miss a nuance) ?

 

[erobz]

Simple. Its wrong. Are we done? Is it copied from somewhere (perhaps I miss a nuance) ?

No, it's not copied. It is the sum of forces in post#10. What is wrong about it? Static friction does zero work. Certainly that implies what is above?

 

[jack action]

What is wrong about it? Static friction does zero work. Certainly that implies what is above?

No, it implies the work done by one force is equal and opposite to the work done by the other force, thus the acceleration is zero, thus the velocity is constant. It's like having a container with two holes, filled with water. If water gets in one hole as much as the water gets out the other one, the water volume in the container stays the same.

The only case where there wouldn't be work done is if the vehicle's velocity is zero. Like if no water got in or out with our container example, even though there would be pressure present at the holes.

But as long as it moves, work is being done, even if it's "negative work" (Remember, work is defined as the dot product of two vectors). The rear [driven] wheels add energy (positive work) to the vehicle, and the front wheels remove energy (negative work) from the vehicle. Since they both have the same velocity (from the vehicle), It must then mean that the net force acting on the vehicle is zero.

 

[erobz]

No, it implies the work done by one force is equal and opposite to the work done by the other force, thus the acceleration is zero, thus the velocity is constant.

But acceleration isn't zero.

$$a = \frac{2TR}{mR^2+4I}$$

This is the solution to the system in #10 as well as Work/Energy in #24.

 

[jack action]

I told you in post #37 that the ##T## in that equation represents the net torque, i.e. it might be zero.

 

[erobz]

I told you in post #37 that the ##T## in that equation represents the net torque, i.e. it might be zero.

No, you are evoking forces that are not present in the idealization of #10. You are unwittingly setting up a strawman argument to knock it down. This is a simple physics problem. Fundamentally it's about an exchange of energy. It does not require the extra complexity of drag, rolling resistance, etc.. you are attempting to add in #37. It is a foundational argument. ##F = ma##

it might be zero.

This is not an argument. It might be ##|a|>0##. What is your point? You are flat out telling me the car can't accelerate under the action of constant torque, because it is not easy to explain otherwise.

 

[jack action]

So you think that the torque input from your engine is completely transferred to acceleration, regardless of the other forces acting on the vehicle? If the front end of your vehicle is resting against a wall, do you think that having a rear wheel torque will give your vehicle an acceleration?

Before anything else, you must first admit you are wrong. The proof is that even you can see that your modelization gives results that don't make sense.

There are two ways that you can model your vehicle:

1. You assume the vehicle as a whole block with forces acting on it, and you do a single FBD;
2. You assume the vehicle as three blocks - one body & two axles - and you do a FBD on each of them.

The way you are doing it - one block with two free rotating axles in a single FBD - is not good. It doesn't give results that make sense. It's not used anywhere, ever.

I would tell you to do the FBD correctly and get back to us, but I already did it for you, for each method. All you have to do is accept that these are the right methods to do the work properly and try to understand them.

If you have further questions to help you understand, I'll be glad to answer them.

 

[erobz]

So you think that the torque input from your engine is completely transferred to acceleration, regardless of the other forces acting on the vehicle? If the front end of your vehicle is resting against a wall, do you think that having a rear wheel torque will give your vehicle an acceleration?

Now you are adding a wall. Another external force that is not present in the idealization.

Before anything else, you must first admit you are wrong. The proof is that even you can see that your modelization gives results that don't make sense.

Yes...I've admitted this over and over. It doesn't make sense! Clearly I must be wrong, but where specifically remains to be addressed... Because it works. The formulation works either way. I'm just a little circus monkey spinning the crank on the jack in the box!

The way you are doing it - one block with two free rotating axles in a single FBD - is not good. It doesn't give results that make sense. It's not used anywhere, ever.

I can take any part of the system I'd like to be the system. The laws of physics don't "break". They might be uninteresting for some choices, but they don't break.

I've shown that a widely published author and physicist has done the very thing I have. So don't say "Its never used".

I would tell you to do the FBD correctly and get back to us, but I already did it for you, for each method. All you have to do is accept that these are the right methods to do the work properly and try to understand them.

Your model will yield to the same conclusion as mine under the same assumptions. Yours just initially has extra stuff in it. They are fundamentally the same models.

 

[jack action]

Your equation:
$$a = \frac{2TR}{mR^2+4I}$$
Your drawing:

The force at the rear wheel is ##F= \frac{T}{R}##. So rewriting your equation:
$$a = \frac{2FR^2}{mR^2+4I}$$
Where's the front wheel force in that equation?

How can this equation be good if the force at the front wheel is not present?

If the force at the front wheel is equal and opposite to the force at the rear wheel, shouldn't the acceleration be zero? If it is greater, shouldn't the vehicle decelerate?

From my post #37:
$$T= T_{in} - (F_r + F_d)R$$
##T_{in}## is what you simplify as ##T##, and ##F_r## (rolling resistance) is the force at the front wheel shown on your model. I admit that I added the drag force ##F_d## just to show you that forces acting on rigid bodies that are connected somehow to the rear axle (i.e. both the vehicle's body and the front axle) does influence the net force, thus the acceleration as well.

 

[erobz]

Your equation:
$$a = \frac{2TR}{mR^2+4I}$$
Your drawing:

View attachment 314627​

The force at the rear wheel is ##F= \frac{T}{R}##. So rewriting your equation:
$$a = \frac{2FR^2}{mR^2+4I}$$
Where's the front wheel force in that equation?

How can this equation be good if the force at the front wheel is not present?

If the force at the front wheel is equal and opposite to the force at the rear wheel, shouldn't the acceleration be zero? If it is greater, shouldn't the vehicle decelerate?

From my post #37:
$$T= T_{in} - (F_r + F_d)R$$
##T_{in}## is what you simplify as ##T##, and ##F_r## (rolling resistance) is the force at the front wheel shown on your model. I admit that I added the drag force ##F_d## just to show you that forces acting on rigid bodies that are connected somehow to the rear axle (i.e. both the vehicle's body and the front axle) does influence the net force, thus the acceleration as well.

You are adding a force that doesn't exist. Where is ##F## in that diagram?

Here are the EOM.

Front wheels:

$$ \circlearrowright^+ \sum \tau = R f_f = I \alpha = I \frac{a}{R} $$

Rear Wheel:

$$ \circlearrowright^+ \sum \tau = T - R f_r = I \alpha = I \frac{a}{R} $$

Total Vehicle:

$$ \rightarrow^+\sum F = 2 \left( f_r - f_f \right) = Ma $$

 

[jack action]

You are adding a force that doesn't exist. Where is ##F## in that diagram?

The force at the rear wheel. Sorry, it is not labeled in your drawing.

 

[erobz]

The force at the rear wheel. Sorry, it is not labeled in your drawing.

Thats is not correct though. ##\sum \tau = I \alpha## must be satisfied on the rear wheel (and the front wheel). Under the no-slip condition this is what follows:

Rear Wheel:

$$ \circlearrowright^+ \sum \tau = T - R f_r = I \alpha = I \frac{a}{R} $$

There is no other ##F## that is causing a torque, that has already not been listed.

And just to be clear, ##f_r## is not rolling friction. It is static friction acting on the rear wheel. The ##r## subscript refers to rear, not rolling. There is no rolling friction in this problem.

 

[jack action]

So why isn't your acceleration set up as:
$$a = \frac{(2T-Rf_f)R}{mR^2+4I}$$
(The only reason I put a ##2## in there is that you assume that ##T## was for each wheel in one of your equations but you didn't in the other. So it is kind of confusing)

Or as I put it in my post #37 (ignoring the drag force ##F_d## I added):
$$a = \frac{F_{net}}{m_e} = \frac{\frac{T}{R}}{m\left(1+ \frac{4I}{mR^2}\right)} = \frac{\frac{T_{in}-F_r R}{R}}{\frac{1}{R^2}(mR^2+ 4I)} = \frac{(T_{in}-F_r R)R}{mR^2+ 4I}$$
Do you see the resemblance?

 

[erobz]

So why isn't your acceleration set up as:
$$a = \frac{(2T-Rf_f)R}{mR^2+4I}$$
(The only reason I put a ##2## in there is that you assume that ##T## was for each wheel in one of your equations but you didn't in the other. So it is kind of confusing)

Because it's not the solution to the system of equations.

There is only ##T## on each rear wheel, not ##2T##.

 

[jack action]

Why are you ignoring ##f_r## in your final solution? (It is the rolling resistance, right?)

 

[erobz]

Why are you ignoring ##f_r## in your final solution? (It is the rolling resistance, right?)

No its the static friction on the rear wheel. I stated that in #65?

 

[jack action]

Where is the rolling resistance?