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

[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...

 

[Mister T]

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.

 

[jbriggs444]

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.

If we focus on you, we know by Newton's third law that the wall has been pushing on you with a force of 1N for those 60 seconds. We assume that you started the 60 seconds at rest. You've told us that you ended the 60 seconds at rest.

We can conclude that there were other forces acting on you whose integral over 60 seconds added up to an average of 1N in the opposite direction that the wall was pushing.

Possibly your feet were pushing the floor in a direction away from the wall and consequently, the floor was pushing you toward the wall.

 

[Dale]

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

Precisely. This is why the power at the contact patch is zero. No energy change means no power.

so the power output by the engine during the conversion of potential energy into mechanical energy is not zero

Obviously not, but that does not imply that any power is delivered to the car through the contact patch.

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

In this problem the work is given to be zero which, as you correctly point out here, implies that the distance is zero. In other words, the Earth is considered to be so massive that it does not move under the force.

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.

Yes, you got it!

 

[russ_watters]

If we focus on you, we know by Newton's third law that the wall has been pushing on you with a force of 1N for those 60 seconds. We assume that you started the 60 seconds at rest. You've told us that you ended the 60 seconds at rest.

We can conclude that there were other forces acting on you whose integral over 60 seconds added up to an average of 1N in the opposite direction that the wall was pushing.

Possibly your feet were pushing the floor in a direction away from the wall and consequently, the floor was pushing you toward the wall.

Yes, that's the scenario. My question is, do physicists really use momentum to describe such static forces?

 

[Dale]

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.

It is very useful. Since you had momentum flowing in from the wall and yet you didn’t accelerate it tells us that there must have been an opposite momentum flow somewhere else. This principle is used extensively in statics.

 

[russ_watters]

It is very useful. Since you had momentum flowing in from the wall and yet you didn’t accelerate it tells us that there must have been an opposite momentum flow somewhere else. This principle is used extensively in statics.

I honestly don't think I've ever heard the term "momentum flow" before. It never would have occurred to me to use the concept of momentum to describe a static force. I may need some time to digest...

 

[Dale]

I honestly don't think I've ever heard the term "momentum flow" before.

That may be my terminology. I think other sources say “transfer” or “rate of transfer”

 

[russ_watters]

That may be my terminology. I think other sources say “transfer” or “rate of transfer”

Well I googled it and it's definitely a thing. But regardless of the term used to label it, it's not something I would have expected to be useful.

 

[jbriggs444]

Well I googled it and it's definitely a thing. But regardless of the term used to label it, it's not something I would have expected to be useful.

I don't think it's anything more than a different mental bucket to use. A force called a "momentum flow" does the same thing as a force called a "force".

If one defines force using $F=\frac{dp}{dt}$ then the mental model of a flow fits well.

 

[A.T.]

Well I googled it and it's definitely a thing. But regardless of the term used to label it, it's not something I would have expected to be useful.

For example: When you think about forces as transfer (rate) of momentum, it becomes immediately clear how Newton's 3rd Law implies momentum conservation.

 

[rcgldr]

Change in momentum = net force · time. In the case of pushing against the wall, and assuming not slipping on the floor, then the sum of the forces exerted by the wall and floor onto the person is zero, so no net force.

power at the contact patch is zero.

Perhaps this is an issue with terminology. My interpretation is the same as the tire dynamics guys, the "contact patch" can be moving with respect to the ground (even though the tread isn't, since it's moves backwards through the contact patch), and power = force · speed. The contact patch static friction force from the ground times the speed of the contact patch equals the power that accelerates the car. The car's engine (or motor) is the source of that power and is responsible for the Newton 3rd law pair of static friction forces at the contact patch.

internal forces can't change energy

Internal forces can't change a systems linear velocity or energy, but can change angular velocity and energy (angular momentum is conserved).

 

[alkaspeltzar]

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?

i like this!

 

[jbriggs444]

power = force · speed

Right. But power times the speed of what, exactly? It it not speed of the contact patch. That speed is not relevant. It can be speed of the center of mass of the car. That speed is relevant. The figure you get for power using that speed will reflect the rate at which the bulk kinetic energy of the car increases due to the force.

The speed of the contact patch is only relevant to the extent that it matches the speed of the car.

 

[A.T.]

The contact patch static friction force from the ground times the speed of the contact patch equals the power that accelerates the car.

This only happens to be true in this special case, where:

speed of the contact patch = speed of the object (center of mass)

In general you cannot compute the power that accelerates the object based on the speed of the contact patch. Consider the Flintstones' car where the contact patch under the foot has a speed of zero, but the power that accelerates the car is not zero.

 

[A.T.]

The speed of the contact patch is only relevant to the extent that it matches the speed of the car.

Exactly, but so does the speed of the rear view mirror. That doesn't make the rear view mirror key to power calculations.

 

[rcgldr]

The speed of the contact patch is only relevant to the extent that it matches the speed of the car.

This only happens to be true in this special case, where:
speed of the contact patch = speed of the object (center of mass)

Which by the usage of the term contact patch as I (and the tire dynamics guys) is always true (ignoring contact patch deformations due to load factors) (update and except as noted by A.T. if the road is not flat, but goes over a hill or through a dip).

In general you cannot compute the power that accelerates the object based on the speed of the contact patch. Consider the Flintstones' car where the contact patch under the foot has a speed of zero, but the power that accelerates the car is not zero.

In the case of the Flintstone's car, the average speed of the driver's feet is the same as the average speed of the car. In the prior thread, a wheel composed of feet for the contact patches was used as an example.

 

[A.T.]

Which by the usage of the term contact patch as I (and the tire dynamics guys) is always true (ignoring contact patch deformations due to load factors).

No it isn't always true. When you go over a hill, the contact patch moves slower than the center of mass.

In the case of the Flintstone's car, the average speed of the driver's feet is the same as the average speed of the car.

Power is an instantaneous quantity that can be computed at any instant, not just on average.

 

[rcgldr]

Which by the usage of the term contact patch as I (and the tire dynamics guys) is always true (ignoring contact patch deformations due to load factors).

No it isn't always true. When you go over a hill, the contact patch moves slower than the center of mass.

True, I was only considering a flat road. I assume that the power would sill be contact patch force times contact patch speed (which as you noted, would be slower than the cars's center of mass speed), since the contact patch is the point of application of the force that is generated by the engine.

Power is an instantaneous quantity that can be computed at any instant, not just on average.

True, but I was trying to average the overall effect of a "foot" powered car.

 

[jbriggs444]

Which by the usage of the term contact patch as I (and the tire dynamics guys) is always true (ignoring contact patch deformations due to load factors) (update and except as noted by A.T. if the road is not flat, but goes over a hill or through a dip).

In the case of the Flintstone's car, the average speed of the driver's feet is the same as the average speed of the car. In the prior thread, a wheel composed of feet for the contact patches was used as an example.

Again, irrelevant. Compute the integral of the work done by Fred's feet using the [erroneous] contact patch formulation...

Fred puts his right foot down and applies force F. The contact patch is stationary. The power delivered to the right foot is $0 \times F = 0$ while the right foot is on the ground.

Fred lifts his right foot. The foot and whatever contact patch you want to ascribe to it moves forward at 2v but the forward force on the contact patch by the road is zero. The power delivered to the right foot is $2v \times 0 = 0$ while the right foot is off the ground.

Similarly, the power delivered to the left foot is zero.

However, since the material at the contact patch in this case is moving at the same speed as the contact patch, this calculation of power does give a correct and relevant result -- the rate of energy transfer from the road to the soles of Fred's feet: Zero.

 

[rcgldr]

I see the points being made here. The static friction force at the contact patch is transmitted by the rolling tire, resulting in a forward force exerted by tire and wheel onto the axle of the tire, which does move at the same speed of the car (ignoring issued like hills and valleys), and the forward force on the axle times the speed of the axle (wrt grond) correlates to the power that accelerates the car and increases it's kinetic energy.

For the Fred Flinstone car, Fred transmits the forward force from the ground to the car, by moving his feet backwards relative to the car, and the forward force times the speed at where Fred exerts the forward force correlates to the power that accelerates the car and increases it's kinetic energy.

Going back to the rolling tire case, for a flat road, the speed of the contact patch matches the speed of the axle on a flat road. If going over a hill, the car rotates in the same direction as the tire, but at a much slower rate, but I suspect that this reduces the force, so that the forward force at the axle times the axle speed will equal the forward force at the contact patch times the contact patch speed.

Part of my issue with all of this, is I see the forward force as originating from the ground, and then transmitted to the car via the tire, wheel, and other components of the car. Maybe using power = contact patch force times contact patch speed is an over simplification, but my thinking is that it produces the same power result as power = forward force on the axle (or other component) times the speed of the axle. The only force external to the car is the forward force from the contact patch.

 

[A.T.]

True, I was only considering a flat road.

On a flat road you can also use the speed of the rear view mirror. The contact patch speed is no more special than that.

I assume that the power would sill be contact patch force times contact patch speed (which as you noted, would be slower than the cars's center of mass speed),

The acceleration power must match the increase in kinetic energy, for which the speed of the center of mass is relevant, not the speed of the contact patch.

Maybe using power = contact patch force times contact patch speed is an over simplification,

It's not a simplification, but an obfuscation.