Engineering Applying Physics to Vehicle Dynamics: Do Wheel Number Matter?

[Saalz]

Homework Statement

Vehicle acceleration from wheel torque

Relevant Equations

Ftan = Torque / r
F = m * a

I was modeling the dynamics of a vehicle for a project, and started doubting about the way of applying physics in this particular case.

The thing is, I know the torque in the wheels from the torque the electric motor I designed do provide, multiplied by the gearbox ratio. I also know the radius of the wheels, and so, the tangential force in the surface of the tyre.

I'm under the case the wheels do not slip and therefore, the velocity at the circumference of the wheel its the same as the velocity of the car.

Thus, I do apply F = m * a, and isolate acceleration in that equation to calculate it as the ratio between the friction force on the road equal to the tangential force in the wheels, and the mass of the vehicle.

My question following that is, do wheel number matter? As the force is calculated from a single wheel, do I have to multiply it by the number of wheels the vehicle has as each one generates it's own friction force in the road (and, not even of the same magnitude sometimes, as in turns, the differential makes each wheel rotate at different speeds)?

Also, the mass used in the equation is the total mass of the vehicle and each wheel holds a percentage of the total mass of the vehicle. Does that even matter if I use the force in a single wheel?

Or I'm just overthinking and it's as simple and physically correct as using the friction force generated by one wheel and the total mass of the vehicle to know the linear acceleration it has? In that case, why?

Thanks for the replies.

 

[Lnewqban]

Welcome!
You should also consider the angular inertia of all the wheels.

 

[Saalz]

Welcome!
You should also consider the angular inertia of all the wheels.

What would be the procedure then?

 

[erobz]

What would be the procedure then?

If there is no-slip between the tire and the road there is a relation between the angular acceleration of the wheel and the linear acceleration of the vehicle that must be held.

Then you get a system of equations using the following laws (one for a driven wheel, one for a non-driven wheel, and one for the entire vehicle):

$$ \sum \tau = I \alpha $$

$$ \sum F = m a $$

The last equation contains elements of the other two which you will use to solve for $a$.

 

[Lnewqban]

What would be the procedure then?

For all rotating parts of the car, the engine must use energy for increasing both, the linear velocity and the angular one.
The heavier and bigger the tires, the higher their moment of inertia and the slower they can be accelerated.

We can always establish a parallel between linear and rotational movements.
Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html

 

[erobz]

For all rotating parts of the car, the engine must use energy for increasing both, the linear velocity and the angular one.
The heavier and bigger the tires, the higher their moment of inertia and the slower they can be accelerated.

We can always establish a parallel between linear and rotational movements.
Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html

This made me realize that we have to include a friction force acting on the non-driven wheels too (to the OP think about its direction in comparison to the static friction acting on the driven wheels). Even though they are not driven, they too have angular acceleration, and the force of static friction acting on those wheels is responsible for it. I edited my post to show that.

 

[jack action]

If you look at the top view of your vehicle, you will have longitudinal and lateral forces acting on each wheel. All of these forces should be taken into account to evaluate not only the resulting motion of the vehicle but also the weight transfer on each wheel that will in turn influence the longitudinal and lateral forces on these wheels.

That being said, simplification can be made by combining forces from different wheels together - for example, wheels on the same axle when examining the side view of the vehicle. For simplicity and good approximations, combining all wheels into a single wheel might be acceptable - for example, assuming the rolling friction acts on a single wheel with the entire weight of the vehicle on it.

Don't forget that if a single source drives more than one wheel, its power is split between all of them. The ratio will depend on the mechanical design of the drivetrain.

 

[erobz]

Does static friction force do work in this case? It seems like the frictional forces must be doing work, but it often seems like this is "explained away" in some way I still don't fully understand.

If I'm doing a force balance to solve the OP's problem, there are frictional forces acting as external forces on the vehicle. The rear frictional forces assisting acceleration and the front tires frictional forces opposing it. These forces must be doing the work...they are the only external forces (in the direction of motion). But I so often hear about static frictional forces not doing work because the point of contact between the tire and the road is instantaneously at rest w.r.t the road. Am I still confused, because I feel a bit unsure of my understanding?[EDIT]
I believe I'm confusing the case where the wheel rolls without slipping at constant velocity. That must be the context in which static friction does no work. I feel like I've been over this a lot, and it still escapes me initially.

But then again...in both cases ( accelerating/not accelerating) the point of contact is instantaneously at rest w.r.t. the ground. So, I still am unsure what the actual resolution is! It all seems like a bit of a circular argument. It feels like whether or not the static frictional force does work or not is whether or not the body is accelerating, and whether or not the body is accelerating is whether or not the static frictional forces are doing work.

 

[jack action]

The rear frictional forces assisting acceleration and the front tires frictional forces opposing it.

To be clear, it is the rolling resistance that opposes the motion, not friction.

Also, there is not necessarily an acceleration.

But I so often hear about static frictional forces not doing work because the point of contact between the tire and the road is instantaneously at rest w.r.t the road.

Although the tire-road contact point is at rest, the axle is not. The geometry of the wheel transfers the friction force to the axle that is not at rest, thus doing work on the vehicle.

It feels like whether or not the static frictional force does work or not is whether or not the body is accelerating,

Acceleration is not a criterion to do work. Work is a force that is displaced.

and whether or not the body is accelerating is whether or not the static frictional forces are doing work.

Work is not a criterion to cause acceleration. Acceleration is present if there is a resultant force acting on a body. A body at rest can have an acceleration. Otherwise, its speed would never increase.

 

[erobz]

To be clear, it is the rolling resistance that opposes the motion, not friction.

Also, there is not necessarily an acceleration.

Front wheels: There must be a torque that is causing the angular acceleration of the wheel. Which implies there is a force acting at the point of contact.

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

Rear Wheel:

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

Total Vehicle:

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

Although the tire-road contact point is at rest, the axle is not. The geometry of the wheel transfers the friction force to the axle that is not at rest, thus doing work on the vehicle.

The axles in the model above would seem to be incapable of doing work by having the friction force transferred to them. They are non-dimensional in this idealization?

Is this an incorrect model, a failed idealization? I don't want to be leading the OP astray, because this is what I had in mind.

 

[Lnewqban]

If a concentrated moment is acting on the free end of a rod, which is articulated or hinged at the opposite end, that rod will rotate about that hinge.
Now, think of many rods forming the rays of a circunference, as well as many hinges that instantaneously form (contact point or patch of asphalt and rubber) in rapid succesion moving forward.

The top end of each ray will happily rotate around its pivot, only that the road constrains that rotation.
The next ray-pivot pair appears, and so on.
Rather than a rotation of each open end, we obtain a forward translation of the many free ends, now named the driving axle (concentrated moment or driving torque) of the car.

 

[jack action]

Front wheels: There must be a torque that is causing the angular acceleration of the wheel. Which implies there is a force acting at the point of contact.

There is a torque. The torque is caused by the misalignment between the normal load and the normal reaction force at the tire-road contact. That misalignment is caused by the deformation of the wheel and/or ground. That rolling resistance torque can be defined as a perpendicular force acting at the tire-road contact patch when you assume both normal forces are aligned.

The axles in the model above would seem to be incapable of doing work by having the friction force transferred to them. They are non-dimensional in this idealization?

I'm not sure I understand. If you remove the wheel (no matter its size), the friction force transfers directly to the vehicle's axle, thus pushing the vehicle. Do a free body diagram (FBD) for the wheel and another one for the vehicle without the wheel. With the wheel FBD, the wheel center has the same velocity as the axle, thus the wheel also moves forward as it turns.

 

[erobz]

There is a torque. The torque is caused by the misalignment between the normal load and the normal reaction force at the tire-road contact. That misalignment is caused by the deformation of the wheel and/or ground. That rolling resistance torque can be defined as a perpendicular force acting at the tire-road contact patch when you assume both normal forces are aligned.

I'm not talking about rolling resistance though. Does there have to be rolling resistance in the simplest physical model for the front wheels to turn? The static friction force is not capable of causing the rotation?

 

[erobz]

Can someone just show the appropriate set of equations that describes the motion of the car in the diagram? I feel that conceptually it will be easier to examine a relevant chunk of the EOM.

 

[erobz]

If a concentrated moment is acting on the free end of a rod, which is articulated or hinged at the opposite end, that rod will rotate about that hinge.
Now, think of many rods forming the rays of a circunference, as well as many hinges that instantaneously form (contact point or patch of asphalt and rubber) in rapid succesion moving forward.

The top end of each ray will happily rotate around its pivot, only that the road constrains that rotation.
The next ray-pivot pair appears, and so on.
Rather than a rotation of each open end, we obtain a forward translation of the many free ends, now named the driving axle (concentrated moment or driving torque) of the car.

View attachment 314520

Sorry, the massage is not being received here with regard to whether the force of static friction does work. The car is accelerating. There are only forces of friction acting externally on the car as a whole that I can see? If there was no friction the car cannot accelerate.

 

[hutchphd]

There are only forces of friction acting externally on the car as a whole that I can see?

There need to be forces on the motor to keep it from counter-rotating (like a clown car). The car would not be very useful if, when you stepped on the gas, the motor rotated. It is the motor that does the work on the wheel

The axles in the model above would seem to be incapable of doing work by having the friction force transferred to them. They are non-dimensional in this idealization?

In your simplified model, there is no engine. In the simplest model just think of the engine as supplying "negative rolling friction" to the rear wheels. In fact it is supplying a torque which is countered by lessening weight on the front wheels. The work is done by the engine. Energy comes from liquified dinosaaurs. There is a net force on the car from the road and yet the road does no work.

 

[jack action]

For the front wheel, the axle pushes on the wheel center. The friction force reaction must be equal to that axle force. Those two misaligned forces cause a torque.

That torque must be equal to another torque otherwise, the wheel would accelerate. (Assuming constant speed here.) What is that resistive torque? It is the one due to the misalignment of the two normal forces acting on the wheel I mentioned in my previous post.

So by pushing on the wheel center, you create a misalignment of the normal forces, and you have to push as hard as required by this resistive torque created, which depends only on the wheel/road deformation and the normal force acting on the wheel.

The force from the axle, pushing the wheel, moves so there is a displacement. Work is required to move a force.

 

[erobz]

Can someone identify $F_{net}$ using work/energy?

$$ \begin{align} \int F_{net} dx &= \frac{1}{2} m v^2 + 4 \frac{1}{2} I \omega^2 \tag*{} \\ &= \frac{1}{2} m v^2 + \frac{1}{2 R^2} 4 I v^2 \tag*{} \\ &= \frac{1}{2} \left( \frac{mR^2 + 4I}{R^2}\right) v^2 \tag*{} \end{align} $$

If we differentiate both side with respect to $x$:

$$ \begin{align} \end{align} $$

$$ \frac{d}{dx} \int F_{net} dx = \frac{d}{dx} \left( \frac{1}{2} \left( \frac{mR^2 + 4I}{R^2}\right) v^2 \right) \tag*{} $$

$$ \implies F_{net} = \left( \frac{1}{2} \left( \frac{mR^2 + 4I}{R^2}\right) \right) 2 v \frac{dv}{dx} \tag*{} $$

$$ \implies F_{net} = \left( \frac{mR^2 + 4I}{R^2}\right) a \tag*{} $$

$$ \implies a = \frac{F_{net} R^2}{mR^2 + 4I} \tag*{} $$

So where is $F_{net}$ on the diagram?

 

[erobz]

There need to be forces on the motor to keep it from counter-rotating (like a clown car). The car would not be very useful if, when you stepped on the gas, the motor rotated. It is the motor that does the work on the wheel

In your simplified model, there is no engine. In the simplest model just think of the engine as supplying "negative rolling friction" to the rear wheels. In fact it is supplying a torque which is countered by lessening weight on the front wheels. The work is done by the engine. Energy comes from liquified dinosaaurs. There is a net force on the car from the road and yet the road does no work.

Isn't the engine in my model represented by the torque acting on the wheels?

 

[Lnewqban]

Sorry, the massage is not being received here with regard to whether the force of static friction does work. The car is accelerating. There are only forces of friction acting externally on the car as a whole that I can see? If there was no friction the car cannot accelerate.

That is correct, both extremes are zero static friction (wheels on ice) and maximum value of adherence.
The car does not move forward in the first case, even if the driving wheels where spinning, as you well know.

 

[hutchphd]

It is the one due to the misalignment of the two normal forces acting on the wheel I mentioned in my previous post.

Yes flexion of the tire requires work to be done. But this is not "work done by the friction force" in the usual sense of that term. This is a squirm issue. The simplest model is a train

 

[hutchphd]

The train will not move without friction force.
The train will not move without gravity force.
The train will not move without engine.

Only the engine does work.

 

[erobz]

The train will not move without friction force.
The train will not move without gravity force.
The train will not move without engine.

Only the engine does work.

But the engine is modeled as the applied torque $T$. Are the equations in #10 incorrect? #18? What are the proper EOM?

I'm afraid if someone doesn't set the record straight soon with a reasonable set of equations, they are going to call me a "crank" and shut this kid's thread down for being a disruptor.

 

[erobz]

Picking up with Work/Energy replacing the LHS by the work done by a constant torque $2T$ ( $T$ is applied to each wheel).

$$ \frac{d}{dx} \int 2T \, d \theta = \frac{d}{dx} \left( \frac{1}{2} \left( \frac{mR^2 + 4I}{R^2}\right) v^2 \right) \tag*{} $$

$$ \frac{d}{dx}\frac{2T}{R} \int \, dx = \frac{d}{dx} \left( \frac{1}{2} \left( \frac{mR^2 + 4I}{R^2}\right) v^2 \right) \tag*{} $$

$$ \implies \frac{2T}{R} = \left( \frac{1}{2} \left( \frac{mR^2 + 4I}{R^2}\right) \right) 2 v \frac{dv}{dx} \tag*{} $$

$$ \implies \frac{2T}{R} = \left( \frac{mR^2 + 4I}{R^2}\right) a \tag*{} $$

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

This and the system of equations in #10 using Newtons Laws yield equivalent results.

 

[jack action]

Yes flexion of the tire requires work to be done. But this is not "work done by the friction force" in the usual sense of that term. This is a squirm issue. The simplest model is a train

I agree. That's how I began with post #9. I'm merely trying to show that rolling resistance is NOT a friction force.

 

[hutchphd]

Not a crank. There is reason for confusion here. I have suggestions for how to think about this.

1) I am walking down the street. Is the road doing work on me? No. I am not a rigid body and it is the internal work of my muscles that does the work to deform me in such a way that the road friction pushes me forward
2) the automobil;e is similarly continuously deforming itself by forcing the rear wheels to rotate. It is not a rigid body and the road does no work. The work is done by the engine to deform the body (axles-wheels) to make friction propel me

I will not derive the equations because I am too damned lazy but the crux of the confusion is as stated. Maybe we should do a complete treatment

 

[erobz]

Not a crank. There is reason for confusion here. I have suggestions for how to think about this.

1) I am walking down the street. Is the road doing work on me? No. I am not a rigid body and it is the internal work of my muscles that does the work to deform me in such a way that the road friction pushes me forward
2) the automobil;e is similarly continuously deforming itself by forcing the rear wheels to rotate. It is not a rigid body and the road does no work. The work is done by the engine to deform the body (axles-wheels) to make friction propel me

I will not derive the equations because I am too damned lazy but the crux of the confusion is as stated. Maybe we should do a complete treatment

But I just got to the same result as #10 without considering friction at all via work/energy. That is usually worth something. That seems to be in alignment with what you are conveying here. However, considering the equations in #10 which appear to be equivalent, the net force on the car is the friction force. They seem to be different sides of the same coin.

 

[jack action]

But the engine is modeled as the applied torque $T$. Are the equations in #10 incorrect? #28? What are the proper EOM?

I asked you to do the FBD for the wheels and the body separately, which you haven't done. So here we go:

Rear [driven] wheels:

$$m_w a = F_{f_r} - F_{a_r}$$
$$W_r + m_w g = F_{N_r}$$
$$I_w \frac{a}{R} = T_w - F_{f_r} R - F_{N_r} d$$

Body:

$$m_b a = F_{a_r} - F_{a_f} - F_R$$
$$W = W_r + W_f$$
$$W L_r = W_f L$$

Front wheels:

$$m_w a = F_{f_f} - F_{a_f}$$
$$W_f + m_w g = F_{N_f}$$
$$I_w \frac{a}{R} = - F_{f_f} R - F_{N_f} d$$

So you have a system of 9 equations with 9 unknowns, namely:

$a$: body acceleration;
$F_{f_r}, F_{f_f}$: rear & front friction forces;
$F_{a_r}, F_{a_f}$: rear & front axle forces;
$W_r, W_f$: rear & front weights;
$F_{N_r}, F_{N_r}$: rear & front normal forces;

And the known values:

$m_w, m_b$: wheel & body mass;
$I_w$: wheel inertia;
$R$: wheel radius;
$T_w$: wheel input torque;
$d$: misalignment distance for rolling resistance (actually equals the rolling resistance coefficient times the wheel radius, see $F=\frac{Nb}{r}$ on Wikipedia);
$F_R$: Body resistance (drag, etc);
$W$: body weight;
$L$: wheelbase;
$L_r$: CG to rear axle distance.

There might be some mistakes (maybe some + or - signs) and it could be more detailed, but that should be what you are looking for.

 

[erobz]

I appreciate you elaborating on that model, but I was after the minimum physical model that captures the basic idea. A simple physics model. Not the model an engineer would use. I had already provided the FBD for that.

I think you misunderstood what I am concerned about. If you neglect all that “real” stuff, you get the most basic result that can be used to talk about the fundamental physics that drives the model. That is where this stuff is confusing for me, and probably where the OP is confused as well. Fundamentally, there appears to be no difference between what you did and what I did. Which is good news! The difference just lies in level of the accounting. That’s not what I’m worried about though.

 

[jack action]

If you neglect all that “real” stuff, you get the most basic result that can be used to talk about the fundamental physics that drives the model.

If you want to neglect the "real" stuff, then neglect the wheels as well. Then you have a simple block with the friction forces acting on it (including the one for rolling resistance). You can also increase the vehicle's mass to compensate for the rotating components' inertia when accelerating (equivalent mass concept).

 

[erobz]

If you want to neglect the "real" stuff, then neglect the wheels as well. Then you have a simple block with the friction forces acting on it (including the one for rolling resistance). You can also increase the vehicle's mass to compensate for the rotating components' inertia when accelerating (equivalent mass concept).

I feel like the wheels are a minimum model for which this concept of "work done by friction" can be looked at.

 

[jack action]

If you don't want to include the wheels, you do the FBD for the vehicle as a whole, with the friction forces where the wheels are supposed to be;

If you want the wheels included, you do a FBD for the body and one FBD for each axle, and analyze them together, as I did in post #28.

If you want the powertrain included as well, you do a FBD for the body, one FBD for each axle, and one for each component of the powertrain.

Your choice.

But using wheels that move relative to the body while still considering them as part of the body won't work.

 

[erobz]

So, what you are saying the result of #10 being equivalent to #24 is a happy accident? I must consider some unknown force as pushing on the axles to solve the problem (namely F_a in your analysis)?

 

[Lnewqban]

Picking up with Work/Energy replacing the LHS by the work done by a constant torque $2T$ ( $T$ is applied to each wheel).
...

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

This and the system of equations in #10 using Newtons Laws yield equivalent results.

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

 

[hutchphd]

However, considering the equations in #10 which appear to be equivalent, the net force on the car is the friction force. They seem to be different sides of the same coin.

They are in fact exactly that and they are consistent. There is a net external force on the car provided by the difference in the rear ground-tire friction and the front.

So, what you are saying the result of #10 being equivalent to #24 is a happy accident? I must consider some unknown force as pushing on the axles to solve the problem (namely F_a in your analysis)?

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 cantact 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. There is also some external "rotational friction" as expertly detailed by @jack action. But the motive energy does not come from an external force. It is internal energy from the long ago sun channeled through dead dinosaurs.
There is no unknown force required: this is not a rigid block.
This question has now been asked and answered. Finis.

 

[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.