Is there any work done by static friction when accelerating a car?

[alkaspeltzar]

I'm asking for clarification, but it's my understanding, that of the thread below, and my college physics book Paul A tipler, that when walking or driving a car, the force of friction from the ground does no work. This makes sense in a car becuase the engine/fuel makes the power, driving the wheels and car forward given adequate friction. Essentially friction helps convert rotational energy to kinetic energy.

Can someone tell me if that is correct? I read a lot of other opinions and started to question it. Below seemed to makes sense. Thanks for help
https://www.physicsforums.com/threa...one-by-friction-when-a-man-is-walking.331641/

 

[Dale]

when walking or driving a car, the force of friction from the ground does no work.

That is correct under the usual “no slip” condition.

 

[alkaspeltzar]

That is correct under the usual “no slip” condition.

Thanks Dale! You've been great help. I was getting told on by others the ground did work. To do work something must have energy, which the ground doesn't. I've always understood that essentially the engine pushes the car thru the wheels due to friction. Simple as that. Thanks again!

 

[Dale]

Simple as that.

Yes, simple as that!

 

[Herman Trivilino]

I was getting told on by others the ground did work.

Different introductory physics textbooks define work differently. That is likely why you were told different things by different people.

 

[russ_watters]

I was getting told on by others the ground did work. To do work something must have energy, which the ground doesn't.

The key, to me, is that work is force times distance, and as far as the ground is concerned, the tire contact patch with the ground isn't moving horizontally, it is only moving up and down. It's easier to visualize when talking about a person walking. The ground doesn't know if you are walking forward or just leaning against a wall. It just feels a static force against a static contact patch.

 

[rcgldr]

The question is any work done by the ground in the case of an accelerating car. Define "contact patch" as the part of a tire in contact with the ground, and one that moves with the car (the tread of the tire and the ground "flow" through the contact patch due to rolling motion. The ground exerts a forwards force onto the contact patch of the tires, but since the frame of reference is the ground itself, the ground itself doesn't perform any work. However, the combined effect of the forward force from the ground and the contact patch that moves with the car provides the ability to perform work, increasing the kinetic energy of the car during acceleration. This is offset by the decrease in potential energy of the fuel consumed by the engine, so that the net work performed on the car is zero, since any gain of kinetic energy was at the expense of potential energy.

Assume a closed system of moon and car, there are no external forces. The momentum of the closed system is conserved, and the total kinetic energy of the system increases over time. Using a reference frame that is fixed with respect to the center of mass of the closed system, the car accelerates "forwards", while the moon mostly experiences a very small "backwards" angular acceleration due to friction force. The static friction force times the distance the car moves (with respect to the reference frame) corresponds to the increase in kinetic energy of the car, while the static friction force times the distance the moons surface moves (with respect to the reference frame) corresponds to the increase of kinetic energy of the moon (which will be very small).

I don't see how this differs much from a rocket in space. The expanding force of the spent fuel accelerates the fuel backwards and the rocket forwards. That expanding force is performing work on both the rocket and the fuel. To me, this is similar to the friction force between contact patch and ground performing work on both the car and the moon.

 

[alkaspeltzar]

See post #14 from Doc Al in the following.
https://www.physicsforums.com/threads/work-done-by-friction-by-wheels.841792/

The force of friction x the displacement of COM will help you calculate the Kinetic energy gained but it is not from the ground/friction. It is from the engine. Remember, it is only gaining that energy because the engine is powering the wheels, creating a frictional force. SO friction is accelerating the wheel and car, delivering the energy of the engine as kinetic energy. if the car where just sitting still, no engine, the ground has no energy to do any work.

Kinda like the engine is pushing the car and helping it convert the energy via the external friction force. Friction is doing the grunt work.

maybe others can explain more, but that is my understanding

 

[Dale]

I don't see how this differs much from a rocket in space. The expanding force of the spent fuel accelerates the fuel backwards and the rocket forwards. That expanding force is performing work on both the rocket and the fuel. To me, this is similar to the friction force between contact patch and ground performing work on both the car and the moon.

Indeed, the two scenarios are related and can smoothly transform one into the other. As the mass of the fuel increases to infinity then the acceleration of the “exhaust” goes to zero in the center of momentum frame and the work done on/by the ground goes to zero. That is the limit under discussion previous to your post.

 

[rcgldr]

work done on/by the ground goes to zero.

I never mentioned work performed by the ground. I mentioned the work performed by the Newton third law pair of forces, the force exerted by the tires onto the ground, which coexists with the force exerted by the ground onto the tires. Those third law pair of forces are increasing the kinetic energy of the moon + car closed system in my example. The cars motor (since my example uses an electric motor), is what is supplying the energy to maintain the force as both moon and car are accelerated by the force. Being much more massive, the moon's acceleration is tiny compared to the car's acceleration, but it's non-zero.

 

[jbriggs444]

Years ago, someone (I think @Doc Al) explained the distinction between "real work" and "center of mass work". With that distinction in mind, much confusion evaporates.

For "real work", what one is considering is just the two interacting surfaces. One is not concerned with the entirety of the Earth or the entirety of the car. One is concerned only with the top surface of the pavement and its interaction with the contact patch on the bottom of the tire. The relevant motion is the motion of the two surfaces.

By contrast, for "center of mass work", one picks out objects of interest. For instance, the car and the Earth. One is concerned with the entire objects. Details of the interface or the internal motions of parts cease to be relevant. The relevant motion is the motion of the centers of masses of the objects.

For "real work" done on the tire by the ground, we focus on the motion of the contact patch. If we assume a frame of reference where the pavement and contact patch are at rest, the contact patch is at rest by definition and the work done is zero. [One could consider other reference frames. But let us not, at least for now]

For "center of mass work" done on the car by the ground, we focus on the motion of the center of mass of the car. Again, using a frame of reference where the pavement and contact patch are at rest, the center of mass of the car is moving and the work done is non-zero.

 

[rcgldr]

For "center of mass work" done on the car by the ground, we focus on the motion of the center of mass of the car. Again, using a frame of reference where the pavement and contact patch are at rest, the center of mass of the car is moving and the work done is non-zero.

"Contact patch" normally refers to the part of the tire in contact with the pavement, and moves with the same direction and velocity of the car. Both tire tread and pavement "flow" through the contact patch. This is different than "point of contact".

 

[Dale]

I never mentioned work performed by the ground. I mentioned the work performed by the Newton third law pair of forces

One of those forces in the third law pair gives the work done on the ground and the other gives the work done by the ground.

Being much more massive, the moon's acceleration is tiny compared to the car's acceleration, but it's non-zero.

The rate of momentum transfer does not go to zero, but the rate of energy transfer does. It always is zero when the velocity of the material at the contact patch is zero, and as the mass increases the contact patch remains at rest.

 

[alkaspeltzar]

Years ago, someone (I think @Doc Al) explained the distinction between "real work" and "center of mass work". With that distinction in mind, much confusion evaporates.

For "real work", what one is considering is just the two interacting surfaces. One is not concerned with the entirety of the Earth or the entirety of the car. One is concerned only with the top surface of the pavement and its interaction with the contact patch on the bottom of the tire. The relevant motion is the motion of the two surfaces.

By contrast, for "center of mass work", one picks out objects of interest. For instance, the car and the Earth. One is concerned with the entire objects. Details of the interface or the internal motions of parts cease to be relevant. The relevant motion is the motion of the centers of masses of the objects.

For "real work" done on the tire by the ground, we focus on the motion of the contact patch. If we assume a frame of reference where the pavement and contact patch are at rest, the contact patch is at rest by definition and the work done is zero. [One could consider other reference frames. But let us not, at least for now]

For "center of mass work" done on the car by the ground, we focus on the motion of the center of mass of the car. Again, using a frame of reference where the pavement and contact patch are at rest, the center of mass of the car is moving and the work done is non-zero.

That link i attached above is where doc Al breaks down. My college Tipler Physics book picks on this example, stating friction is what propels the vehicle forward, but does no work, as all the energy gained thru friction is really via the internal energy of the engine.

 

[Doc Al]

Years ago, someone (I think @Doc Al) explained the distinction between "real work" and "center of mass work". With that distinction in mind, much confusion evaporates.

Yep, I wrote quite a bit about that some years ago. I sometimes used the term "pseudowork" in contrast to "real work". I can dig up some of my rants on the topic, but you guys have nailed it.

 

[Herman Trivilino]

Yep, I wrote quite a bit about that some years ago. I sometimes used the term "pseudowork" in contrast to "real work". I can dig up some of my rants on the topic, but you guys have nailed it.

More information on the topic can be found in this AJP article written by Arnold Arons:

https://aapt.scitation.org/doi/abs/10.1119/1.19182?journalCode=ajp

 

[Dale]

it's acceleration is always non-zero, with respect to an inertial frame of reference.

Yes, but even when the acceleration is non-zero the power is zero when the velocity of the material at the contact patch is zero. Acceleration and power are different things.

 

[rcgldr]

Yes, but even when the acceleration is non-zero the power is zero when the velocity of the material at the contact patch is zero. Acceleration and power are different things.

Other than then initial start from rest, the "contact patch" velocity is non-zero. The "contact patch" moves at the same velocity as the car.

If the power output is constant, then the rate of increase of total kinetic energy of moon + car in my example is also constant. The initial state is an issue, as Δt and Δv approach 0 for time t = 0 to time = t + Δt, and velocity v = 0 to v + Δv, the associated power approaches 0. At some point in time, the acceleration is power limited instead of traction limited, and at that point, the situation can become a constant power case.

 

[Doc Al]

Other than then initial start from rest, the "contact patch" velocity is non-zero. The "contact patch" moves at the same velocity as the car. Over time, any point on the tire, including any point that is momentarily in contact with the pavement, has an average velocity the same as the car.

The "contact patch" is that part of the tire in momentary contact with the ground. As the car moves, a different piece of the tire becomes the next "contact patch". While the average velocity of any part of the tire must equal that of the car, the instantaneous velocity of the "contact patch" is always zero. That's what matters when considering the work done by the ground on the car.

 

[Dale]

Other than then initial start from rest, the "contact patch" velocity is non-zero. The "contact patch" moves at the same velocity as the car.

That is why I said “material at the contact patch” and not “contact patch”. When the material at the contact patch has zero velocity the power is zero regardless of the acceleration. This is always the case in the situation under consideration in the OP.

While I agree with @Doc Al’s use of the term I recognized from your previous comments that you understood it differently. So I used unambiguous terminology.

Over time, any point on the tire, including any point that is momentarily in contact with the pavement, has an average velocity the same as the car

Not relevant.

If the power output is constant, then the rate of increase of total kinetic energy of moon + car in my example is also constant.

Agreed, but the rate of KE increase of the system is not the same as the rate that work is done at the contact patch.

I am now getting quite lost in correcting the detailed mistakes of your posts. What is your point now?

 

[rcgldr]

What is your point now?

That if the tire is not slipping, then the ground is directly connected to the engine in the same manner as the tires, and neither ground nor tires are a source of energy, but instead transfer the force converted from the torque of the engine through the drivetrain, with that Newton 3rd law pair of forces I posted above (tires exert force onto ground, ground exerts force onto tires). There have been statements that the ground can't be the source of energy, but likewise, the tires can't be the source of energy either.

There was a prior thread about this that included an analysis of the car engine's power increasing the energy of both the planet (i used a moon in my example) and the vehicle.

 

[Dale]

There have been statements that the ground can't be the source of energy, but likewise, the tires can't be the source of energy either.

Correct (I don't think anyone claimed otherwise here). The energy comes from the engine's fuel or battery. When the material at the contact patch is at rest then there is no work done either by the tire or the road and therefore no energy transferred from the car to the earth. All of the energy in the fuel goes into kinetic energy of the car and none to the earth.

There was a prior thread about this that included an analysis of the car engine's power increasing the energy of both the planet (i used a moon in my example) and the vehicle.

Yes, there is a range of velocities for the material at the contact patch such that this is true. It is not v=0, which is the case under consideration in this thread.

 

[rcgldr]

v=0, which is the case under consideration in this thread.

The title of this thread is "... accelerating a car", and the OP didn't mention if the initial state was v=0.

 

[Dale]

The title of this thread is "... accelerating a car", and the OP didn't mention if the initial state was v=0.

From the OP "my college physics book Paul A tipler, that when walking or driving a car, the force of friction from the ground does no work" indicates that the ground is at v=0 throughout the problem.

 

[rcgldr]

From the OP "my college physics book Paul A tipler, that when walking or driving a car, the force of friction from the ground does no work" indicates that the ground is at v=0 throughout the problem.

In the case of a person walking, there's no acceleration over the long term, but since momentum of the planet + person is conserved, and if in the initial state neither planet or person were moving (both at v=0), then once the person reached a steady average speed while walking then v ≠ 0 for either ground or person.

The confusion on my part is the title mentions acceleration, while I assume the referenced text refers to non-accelerating cases.

I found one of the prior threads. One of the posts makes a comparison using a spring between two blocks, and mentions the case where one of the blocks is much more massive than the other. Another post in that prior thread mentions contact point as a virtual point that moves along the ground.

https://www.physicsforums.com/threads/work-done-on-accelerating-car-is-zero.734203

I'm still looking for the analysis of energy output versus kinetic energy gained for a planet + vehicle closed system, that had to include the increased energy of the planet for the potential energy input to exactly equal the kinetic energy gained (it assumed no losses like drag or rolling resistance).

 

[Dale]

there's no acceleration over the long term

That is irrelevant. The specification that no work is done at the contact patch implies that v=0 for the material at the contact patch. If there is acceleration then that also implies that the problem is assuming that the mass of the person/car is negligible compared to the mass of the earth. Since the power is $P=F\cdot v$ then regardless of $F$ if $v\approx 0$ then $P\approx 0$. So the person can accelerate or decelerate arbitrarily provided that at all times $v\approx 0$ for the material at the contact patch, i.e. the ground.

 

[rcgldr]

From Russ Waters post in that prior thread:

"The contact points are virtual and change over time (move) and the wheel is always rotating about the contact point, even though the individual points are not translating. ... The effect of multiple contact points in different places is mathematically equal to one continuous force acting over a distance. "

https://www.physicsforums.com/threads/work-done-on-accelerating-car-is-zero.734203/post-4637801

The car's engine (or motor) is the source of the force from the ground (along with the coexistent force from the tires), and that force can perform work. This is different than claiming the ground is performing the work.

 

[jbriggs444]

From Russ Waters post in that prior thread:

I will not try to argue with what @russ_watters stated in a different post and a different context.

The tire material at the contact patch is not moving. No work is being done on it. This is a truth.

That work can be done on a moving car body by virtue of a mechanism that begins with an engine and ends with a rotating tire on an axle does not change that truth.

 

[alkaspeltzar]

In the end, I think this statement is correct. That it is really thru the engine, that creates the force of friction on the tires, which then pushes the car, so as the energy is transformed. In a general sense, work of the engine is begin done thru the help from friction. Also, i was never taught this fully either, so it was kinda a mind blow when i finally started to question it. I think there is a lot of confusion when it comes to work/energy, it is vaguely taught. I just know that the car engine produces the force in the end, and the power from the engine is what the car gains. When i step it down, you feel it! And in the end, I am okay with that understanding

"That work can be done on a moving car body by virtue of a mechanism that begins with an engine and ends with a rotating tire on an axle does not change that truth. "

 

[rcgldr]

What was stated by Russ is that the point of contact (through which the tire material flows through) is moving with respect to the ground.

 

[jbriggs444]

What is being stated by Russ, others, and I is that the point of contact (through which the tire material flows through) is moving with respect to the ground.

That point is utterly and completely irrelevant. Put granny in a rocking chair facing forward in an accelerating rail car. The fore and aft movement of the contact patch on the rockers under her chair is not relevant. At best, her center of mass is relevant.

 

[rcgldr]

Put granny in a rocking chair facing forward in an accelerating rail car. The fore and aft movement of the contact patch on the rockers under her chair is not relevant. At best, her center of mass is relevant.

You could put granny in a granny sized hamster ball, and she could cause the ball to accelerate from 0 to some small velocity and maintain that velocity. In that case, the contact patch is moving at the same velocity as the center of mass of granny and the granny sized hamster ball.

 

[jbriggs444]

You could put granny in a granny sized hamster ball, and she could cause the ball to accelerate from 0 to some small velocity and maintain that velocity. In that case the contact patch is moving at the same velocity as the center of mass of granny and the granny sized hamster ball.

You could. But I didn't. I put granny in a rocking chair. The center of mass of the rocking chair does not hover over the contact patch.

Try again.

The center of mass of the hamster plus ball also does not hover over the contact patch, by the way.

 

[rcgldr]

I put granny in a rocking chair.

and how does that relate to a discussion about the rolling movement of the tires on a car, and the forces between tires and ground that originate from the cars engine?

 

[jbriggs444]

and how does that relate to a discussion about the rolling movement of the tires on a car, and the forces between tires and ground that originate from the cars engine?

It demonstrates that focus on the motion of the contact patch is incorrect and improper.

[As opposed to a focus on the material at the contact patch -- that still works just fine]

 

[rcgldr]

It demonstrates that focus on the motion of the contact patch is incorrect and improper.

My focus was on the Newton 3rd law pair of static friction forces between ground and tires, the contact patch is where those forces are applied.

 

[jbriggs444]

My focus was on the Newton 3rd law pair of static friction forces between ground and tires, the contact patch is where those forces are applied.

And the motion of that patch has nothing to do with the work done by that force.

You have two meaningful choices for work done by a force:

1. Motion of the center of mass of the object to which the force is applied.
2. Motion of the material at the point where the force is applied.

The one, in conjunction with the work-energy theorem gets you the change in bulk kinetic energy of the object due to the applied force.

The other, in conjunction with the work-energy theorem gets you the change in energy of the object due to the applied force.

The motion of the contact patch, by contrast, in conjunction with the work-energy theorem tells you NOTHING!

 

[Dale]

the static friction force from the ground performs work on the car, while the static friction force from the tires performs work on the earth.

This is not true when the velocity of the material at the contact patch is 0 (as is the case considered here).

My focus was on the Newton 3rd law pair of static friction forces between ground and tires, the contact patch is where those forces are applied.

But the motion of the contact patch is completely irrelevant for the work being done by the 3rd law pair of forces. For the work being done by the 3rd law pair of friction forces between the ground and the tires the work done is entirely determined by the velocity of the material at the contact patch, not the change in location of the contact patch.

The contact patch is not an object. It is the designation of a region of material. The important thing is how that material moves, not how the designation changes.

 

[russ_watters]

Years ago, someone (I think @Doc Al) explained the distinction between "real work" and "center of mass work". With that distinction in mind, much confusion evaporates.

For "real work", what one is considering is just the two interacting surfaces. One is not concerned with the entirety of the Earth or the entirety of the car. One is concerned only with the top surface of the pavement and its interaction with the contact patch on the bottom of the tire. The relevant motion is the motion of the two surfaces.

By contrast, for "center of mass work", one picks out objects of interest. For instance, the car and the Earth. One is concerned with the entire objects. Details of the interface or the internal motions of parts cease to be relevant. The relevant motion is the motion of the centers of masses of the objects.

From Russ Waters post in that prior thread:

"The contact points are virtual and change over time (move) and the wheel is always rotating about the contact point, even though the individual points are not translating. ... The effect of multiple contact points in different places is mathematically equal to one continuous force acting over a distance. "

https://www.physicsforums.com/threads/work-done-on-accelerating-car-is-zero.734203/post-4637801

The car's engine (or motor) is the source of the force from the ground (along with the coexistent force from the tires), and that force can perform work. This is different than claiming the ground is performing the work.

I will not try to argue with what @russ_watters stated in a different post and a different context.

The tire material at the contact patch is not moving. No work is being done on it. This is a truth.

That work can be done on a moving car body by virtue of a mechanism that begins with an engine and ends with a rotating tire on an axle does not change that truth.

I'm comfortable saying I remain somewhat uncomfortable with this, and while the statement I made yesterday was a self-contained telling of the physics view, the previous one reflects that broader discomfort. We have two statements:

1. An element of tire rubber in the contact patch doesn't move with respect to the road, therefore there is no work done on it by the road (or the road by it).
2. The car accelerates, therefore there must have been an external force applied to accelerate it.

These two statements would appear to me contradict each other, and considering the point doing some sort of "virtual" movement 6 years ago was my way around it, as an engineer, looking for a simple way to visualize a calculation that actually works. It's almost certainly not the "physics way" (as can be seen, the physicists disagreed with me). The only way I can think of around the contradiction from a physics standpoint would be to assume the tire/wheel is a massless collection of levers that are external to the car and ground, and it is these levers that push the car forward, not the ground. In other words: the tire isn't part of the car, so the car is being acted on by an outside force pushing on it and the point of application of the force is moving with the car.

As an engineer, I tend to gloss over such details where they aren't essential to solving a problem (in my defense, for practical purposes, it's nonsense: of course the tire is part of the car) depending on the needs of the problem or level of depth. And I think my actual understanding has changed over the years of speaking with physicists more about such details and conventions, rather than the potentially sloppy, but simple and useful: ground applies a forward force, car moves forward, work is done on the car.

I kind of like this idea of "real work" vs "center of mass work" (or I might say "virtual work"), even if it feels like a loophole...

 

[Herman Trivilino]

I kind of like this idea of "real work" vs "center of mass work" (or I might say "virtual work"), even if it feels like a loophole...

I urge you to read the article I referenced in Post #16. There is nothing wrong with saying that the road exerts a force on the car, that the car moves a certain distance, and that the product of the two (ignoring other forces exerted on the car) equals the change in the car's kinetic energy. The problem arises when we make certain assertions about the associated transfers of energy involved in the process.

Some college-level introductory textbooks will assert that that product of force and distance is work, and constitutes an example of the work-energy theorem. That is a perfectly valid dynamical assertion. But it is not a valid thermodynamical assertion.

 

[Dale]

1. An element of tire rubber in the contact patch doesn't move with respect to the road, therefore there is no work done on it by the road (or the road by it).
2. The car accelerates, therefore there must have been an external force applied to accelerate it.

These two statements would appear to me contradict each other,

They don’t really contradict each other.

A force is a rate of transfer of momentum. So there is no doubt that the force of the road accelerated the car and leads to its increase in momentum.

But power is a rate of transfer of energy, and the car’s energy does not change, so the power is zero. A force has to transfer momentum, but it doesn’t have to transfer energy.

 

[hutchphd]

In my automobile, all work is being done by the four gerbils who live inside my engine block and who push the pistons through a distance when stimulated by the distributor. Those gerbils (and the similar stopping gerbils in the wheels) do all the "work", positive and negative. The rest of the machine simply redirects the force vector...just like on a bicycle.
And force from the road causes acceleration...why is this confusing?

 

[rcgldr]

But power is a rate of transfer of energy, and the car’s energy does not change, so the power is zero.

Ignoring losses, the total energy of the car remains constant, but the potential energy of the car's fuel or batteries is being converted into mechanical kinetic energy, and the car's mechanical kinetic energy is increasing during acceleration, so the power output by the engine during the conversion of potential energy into mechanical energy is not zero.

As for "contact patch", the way I recall this explained in tire dynamics articles, is that it's a dynamic situation, the tread flows into the contact patch at the "front" of the contact patch, and away from the contact patch at the back of the contact patch. The center of area of the contact patch advances forwards as the tire moves forward, so although the tread is not moving with respect to the ground, the contact patch is moving with respect to the ground.

Again assuming no losses, then the gain in kinetic energy of the car equals the force exerted by the ground times the distance the car / contact patch travel, with respect to an inertial frame of reference. From that same inertial frame of reference, the Earth also gains a tiny amount of kinetic energy from the force exerted by the tires time the distance that the ground moves with respect to that same inertial frame of reference. Ignoring the very tiny amount of acceleration of the earth, the Earth could be use as an approximately inertial frame of reference, in which case all of the energy converted by the cars engine or motor goes into increasing the kinetic energy of the car.

In the case of an accelerating car, the torque exerted by the engine slightly exceeds the opposing torque related to the force from the ground times the effective radius of the rolling tires. The excess torque from the engine is being used up by the angular acceleration of the drive train and tires, so that the net torque on the tire corresponds to the angular acceleration of the tire (due to acceleration of the car) divided by the angular inertia of the tire (and connected components).

 

[jbriggs444]

distance the car / contact patch travel

If you view the location of the contact patch as a proxy for the location of the car's center of mass, that's fine. Technically, the motions of the two are distinct. Practically, their velocities are identical.

 

[russ_watters]

They don’t really contradict each other.

A force is a rate of transfer of momentum. So there is no doubt that the force of the road accelerated the car and leads to its increase in momentum.

But power is a rate of transfer of energy, and the car’s energy does not change, so the power is zero. A force has to transfer momentum, but it doesn’t have to transfer energy.

I thought the whole point of this discussion was that the force at the contact patch is a static force, transferring nothing?

Rate of change of momentum isn't applicable to static forces, is it?

I thought the reason the statements don't contradict is that they are describing two separate models that just can't be mixed?

 

[jbriggs444]

Rate of change of momentum isn't applicable to static forces, is it?

Sure it is. $\Delta p = F \Delta t$. Nothing in there requires that the surface on which the force is applied needs to be moving.

Though if it continues at rest, one can conclude that some other forces are acting to make it so.

 

[Dale]

I thought the whole point of this discussion was that the force at the contact patch is a static force, transferring nothing?

Rate of change of momentum isn't applicable to static forces, is it?

All forces must transfer momentum, including static forces. If not how would the car accelerate, particularly an electric car with no exhaust? The battery doesn’t have momentum, so that has to come from outside.

 

[russ_watters]

Sure it is. $\Delta p = F \Delta t$. Nothing in there requires that the surface on which the force is applied needs to be moving.

How do you apply it in such a case? Just to be clear; I said moving, but I also mean accelerating.

 

[jbriggs444]

How do you apply it in such a case? Just to be clear; I said moving, but I also mean accelerating.

How long has the force been acting? How much force was there? Multiply the two together. That's the momentum change from this force. Repeat for all the other forces.

Just because there is a momentum change for a system as a whole does not mean that the surface where the external force is applied has to move. I can walk just fine, thank you.

 

[russ_watters]

How long has the force been acting? How much force was there? Multiply the two together. That's the momentum change from this force. Repeat for all the other forces.

I'm leaning against a wall, pushing with a force of 1N, for 60 seconds, so that's 60 kg-m/s. How does this tell us anything useful? I'm not moving and didn't specify my mass.