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

[A.T.]

I don't see why you have an issue with the integral form of contact patch force times contact patch speed representing the power that is accelerating the vehicle

People come to PF to learn physics, not numerology. Physics is about finding a consistent generally applicable approach, not an approach which happens to give the same result in just one special case.

 

[rcgldr]

Change the situation to calculating power at a constant speed using some type of load so that there is no acceleration. Replace the drum of the dynamometer with a treadmill. Since the speed of the treadmill is constant, the inertial frame of reference can be changed from an Earth based frame of reference to a treadmill surface based frame of reference, which would result in the equivalent of the car on a road. In the Earth frame of reference, the treadmill is moving, but the contact patch isn't. In the treadmill surface frame of reference, the treadmill surface isn't moving, but the contact patch is. This is just a case of changing from one inertial frame of reference to another, and the calculated power will be the same for both frames of reference.

 

[A.T.]

... the calculated power will be the same for both frames of reference.

Which power? Power representing energy transfer is frame dependent, just like work.

 

[rcgldr]

Which power? Power representing energy transfer is frame dependent, just like work.

The energy transfer is only to the load, since neither the treadmill or the car are accelerating. Say the load is a generator, and power is calculated based on the output of the generator. In this case, the output of the generator will be the same regardless of which reference frame is used.

 

[A.T.]

The energy transfer is only to the load, since neither the treadmill or the car are accelerating. Say the load is a generator, and power is calculated based on the output of the generator. In this case, the output of the generator will be the same regardless of which reference frame is used.

Yes, there are different power quantities that one can compute. So why do you consistently fail to specify which one you mean in your posts.

 

[rcgldr]

Yes, there are different power quantities that one can compute. So why do you consistently fail to specify which one you mean in your posts.

Because I was trying to address the situation of an accelerating car, and realized that by eliminating the acceleration and replacing it with a load, then I can treat the two situations as a change in frame of reference. The contact patch force can be measured using the straps holding the car in place and is frame independent. For the Earth frame, the treadmill has a negative speed, for the treadmill frame, the contact patch has a positive speed, but after taking into account the sign differences, power = force x speed in either frame and = load generator power in an ideal zero loss situation. The key idea I've switched to is to attempt to make this a frame of reference issue.

I recall an old thread (may have been another forum) where dealing with a moving contact patch is just the consequence of rolling motion, despite the fact there is no movement between the surfaces. I haven't found it yet and I don't recall how the issue was addressed. (It produced the numerically same result, but is not the actual explanation).

 

[A.T.]

Because I was trying ...

Whatever you are trying to compute, you should define it precisely first.

 

[Dale]

In the Earth frame of reference, the treadmill is moving, but the contact patch isn't. In the treadmill surface frame of reference, the treadmill surface isn't moving, but the contact patch is.

So, if the contact patch velocity method can only be used in certain reference frames then it is not a valid law of physics. A valid law of physics works in all reference frames.

The dynamometer example shows that the contact patch method does not work in the reference frame of the dynamometer. Since it fails in that frame it does not work in all frames and therefore is not a law of physics.

This is just a case of changing from one inertial frame of reference to another, and the calculated power will be the same for both frames of reference.

It shouldn’t be. Mechanical power is a frame dependent quantity. Yet another strike against your method.

Your adherence to this method has far exceeded the bounds of rationality at this point.

 

[rcgldr]

For a inertial frame independent means of expressing this, using the treadmill + car with a load (generator) and straps on the car, and using the force measured on the straps as the static friction force, then power = force times ((speed of contact patch) - (speed of treadmill surface)). This will work for any inertial frame frame of reference, such as Earth (where the speed of contact patch is zero), treadmill, or some fraction of the speed of the treadmill.

 

[A.T.]

...then power =

Again you fail to define what that power is supposed to represent?

... force times ((speed of contact patch) - (speed of treadmill surface)). This will work for any inertial frame frame of reference ...

You are transforming into the rest frame of the treadmill surface with that subtraction, so it doesn't "work for any inertial frame". Who do you think you are fooling here?

 

[Dale]

the average speed of the entire tire tread surface is the same as the vehicle (on a flat road)

By the way, this is irrelevant. The power is still zero. At the contact patch the force is non-zero and the velocity is zero. Over the rest of the tread the force is zero and the velocity is non-zero. So overall the power is zero, as it should be by conservation of energy.

 

[rcgldr]

Again you fail to define what that power is supposed to represent?

The power is still zero.

To summarize the point I've been attempting to make:

I was stuck with the same apparent dilemma as Russ Waters, the only external force is the road, and internal forces can't accelerate a car. Russ's idea was that the tire and wheel (combined with the torque on the driven tire axles) act as a lever to transmit the external force from the road to the axle of the driven tires.

My prior point was that you get the same numerical result for ΔKE if you consider the road as exerting a force onto a contact patch that is moving at the same speed as the car (with respect to the earth) (on a flat road).

Treating the contact patch as moving at the same speed as the car is one way of noting that the point of application of force from the road is moving, even though there is no relative motion between the surfaces, due to the nature of rolling motion without slipping. In this view of the contact patch, the contact patch moves and the tire tread and road surfaces "flow" through the contact patch. The tire dynamic people that study contact patch issues also use a moving contact patch model, but for different reasons (deformation, heat dissipation, ... ).

 

[sophiecentaur]

To do work something must have energy, which the ground doesn't.

This correct, of course and that view makes sense to me. I always have a problem with the Work done by or Work done on question. PF constantly gets asked this question and that very fact means that trying to make a distinction between by and on is not a good idea. Answers can go either way. If anyone really wants to convince themselves and still use by or on then I suggest that Energy Flow direction would avoid the problem. Otherwise just say 'work is done'.

" something must have energy, which the ground doesn't" . To make life (needlessly?) more complicated, you could imagine driving a car on a road with a lossless rubber sheet on it. In that case, actual work could be said to be done on the sheet during acceleration and stored as it gets stretched at the leading edge of the tyre and done by the sheet on the trailing edge of the tyre as it slows down. Then there is a real Energy flow into and off the sheet (like an electrical Capacitor)
It's all relative but there will be no confusion if you stick to the calculations and follow the sign rules consistently,

 

[A.T.]

PF constantly gets asked this question and that very fact means that trying to make a distinction between by and on is not a good idea.

Quite the opposite. It's the failure to specify what work is meant that leads to confusion. Being explicitly precise is always a good idea.

 

[sophiecentaur]

Quite the opposite. It's the failure to specify what work is meant that leads to confusion. Being explicitly precise is always a good idea.

There is no problem for someone who understands what to do with the data and who can "specify" the situation. But that doesn't apply to people who pose the same question time and time again. Is there really any point in someone who 'understands' it all telling them that they just have to sort it out when the terms by and on just don't need to be used in the first place. Is there any way of reconciling the Energy Flow idea when there appear to be paradoxes of the sort that the OP has brought up?

We don't need a reiteration of the advice to be "explicitly precise" when the person receiving this advice is trying to make sense of something they are reading that isn't.
Work is done and the direction of the Energy flow (possibly shared in two directions) can be calculated without 'who is doing what' being involved.

 

[A.T.]

Work is done and the direction of the Energy flow (possibly shared in two directions) can be calculated without 'who is doing what' being involved.

If the work done by A on B comes out positive, then the direction of the energy transfer is from A to B. If it comes out negative then the direction of the energy transfer is from B to A.

But without stating the "by ... on ..." (as you suggest), it's not clear at all how the resulting sign corresponds to the direction of the energy transfer. I see no value in being deliberately ambiguous here. Especially beginners without a firm grasp should be very explicit about what they are calculating.

 

[sophiecentaur]

it's not clear at all how the resulting sign corresponds to the direction of the energy transfer.

Yes it's not easy but context usually tells you which end is the motor and which end is the load. Does the wheel do work on the road or does the road do work on the wheel? In that situation no energy can flow 'into' the road. Which description should the beginner use (they always feel obliged to involve by and on) ? The situation is totally "ambiguous" afaics. A wheel mounted on a massive support, driving a moving belt could have exactly the same forces and speeds involved as for the car. The Energy flows are different. Isn't that pretty confusing? On the basis of Force times displacement, what does work on what?
If we could rely on every description, explanation or problem that is available for every beginner could be stated in full and precisely then I reckon your view could be right. But experience tells us that life's not like that and that is why I don't think the on and by idea is helpful. I'm sure that you must have had similar problems at times in your life in Physics.

 

[Dale]

I was stuck with the same apparent dilemma as Russ Waters, the only external force is the road, and internal forces can't accelerate a car. Russ's idea was that the tire and wheel (combined with the torque on the driven tire axles) act as a lever to transmit the external force from the road to the axle of the driven tires.

This is completely correct. The external force does accelerate the car and internal forces cannot do that.

The dilemma lies in conflating the change in momentum with a change in energy. Momentum and energy are closely related, but they are not the same thing. You can have a force which transfers momentum and does not transfer energy, which is the case with the car.

 

[Dale]

The Energy flows are different. Isn't that pretty confusing? On the basis of Force times displacement, what does work on what?

$P=F\cdot v$ is the rate of mechanical work done on the system where $F$ is the force on the system and $v$ is the velocity of the system's material at the point of application of the force. Work done on a system increases its total energy.

It is a clear and unambiguous rule. If $P$ is negative then it represents negative mechanical work done on the system which is mechanical work being done by the system. I don't think that "on" or "by" is the culprit here. I think that the problem is ambiguity in what $F$ and $v$ are. In this thread $v$ has been particularly problematic. In other threads I have seen, the ambiguity is in defining what the system is.

If we could rely on every description, explanation or problem that is available for every beginner could be stated in full and precisely then I reckon your view could be right. But experience tells us that life's not like that and that is why I don't think the on and by idea is helpful.

I don't think that eliminating "on" and "by" will fix this. I think that the more useful approach is to simply consistently apply the rule above and to teach students how to apply the rule.

 

[rcgldr]

$P=F\cdot v$ is the rate of mechanical work done on the system where $F$ is the force on the system and $v$ is the velocity of the system's material at the point of application of the force. Work done on a system increases its total energy.

Why can't $v$ be the velocity of the point of application of the force? A common definition for power is $P=F\cdot v$, where $F$ is the force on an object, and $v$ is the velocity of that object. Why can't the object be somewhat "abstract", such as the "contact patches" of the driven tires?

 

[A.T.]

But without stating the "by ... on ..." (as you suggest), it's not clear at all how the resulting sign corresponds to the direction of the energy transfer.

Yes it's not easy but context usually tells you ...

If it's "not easy", requires guessing based on intuition ("context tells you") and only works "usually ", then it's definitely not the right approach for beginners.

 

[A.T.]

Why can't $v$ be the velocity of the point of application of the force? A common definition for power is $P=F\cdot v$, where $F$ is the force on an object, and $v$ is the velocity of that object. Why can't the object be somewhat "abstract", such as the "contact patches" of the driven tires?

See post #74.

 

[jbriggs444]

Why can't vvv be the velocity of the point of application of the force?

Because, moving the point of application independent of the motion of the body does not transfer energy.

Nor, without a corresponding movement of the body, does it, along with velocity of the body, establish a time frame for the transfer of momentum.

Both versions of the work-energy theorem fail.

As I have pointed out more than once, you can model the dynamics of a car in terms of an opaque, rigid black box with a particular mass, center of mass and moment of inertia. If you apply an external force at such and such a position relative to that black box, the internal mechanical details of the car become irrelevant to the resulting position, velocity, momentum and angular momentum of the box.

In particular, it will turn out not to matter whether the car is accelerated by a wheel embedded in the road that pushes the car forward or a wheel embedded in the car that pushes the ground backward.

It will also turn out not to matter where the actual contact point is! Any point along the same line of action will do just as well. [To be perfectly clear, when I say it does not matter, I am speaking of not mattering for the position, velocity, momentum or angular momentum of the opaque blob that we call the car]

 

[rcgldr]

See post #74.

I see your point. In post #74, the object is the board, so power = force exerted on board · velocity of board. The vehicle is a real object though, so power = force exerted on vehicle · velocity of vehicle. It doesn't matter that the road is not moving, since this doesn't affect the ability for the road to exert a force on a moving vehicle, due to the rolling motion of the driven tires.

 

[A.T.]

so isn't power = ...

Again you fail to define what that power is supposed to represent.

 

[jbriggs444]

The vehicle is a real object though

The distinction between a "real object" such a car and a non-real object such as a board is not clear. How can I look at a board and tell that it is not real. It certainly feels real if it hits me in the shins as I am pushing a car out of snow bank.

 

[rcgldr]

Again you fail to define what that power is supposed to represent.

That $\int_{t0}^{t1} P(t) dt = \Delta KE$

The distinction between a "real object" such a car and a non-real object such as a board is not clear.

I should have clarified this in my prior post. The board is a real object, the spinning wheel accelerating the board is a real object, the vehicle is a real object, the road is a real object, the contact patch is an abstract object, that creates an issue because it's an interface between two real objects.

 

[sophiecentaur]

If it's "not easy", requires guessing based on intuition ("context tells you") and only works "usually ", then it's definitely not the right approach for beginners.

I wonder how many beginners you have dealt with / tried to educate. Many eventually successful Scientists are actually very Right Brained individuals.

 

[Dale]

Why can't vvv be the velocity of the point of application of the force?

Because that doesn’t give the rate of mechanical energy transferred to the system by the force, as we have shown again and again and again and again

A common definition for power is P=F⋅vP=F⋅vP=F\cdot v, where FFF is the force on an object, and vvv is the velocity of that object.

Note that this works when it is equivalent to the above definition, I.e. when the velocity of the object is the same as the the velocity of the material at the point of application of the force.

Why can't the object be somewhat "abstract", such as the "contact patches" of the driven tires?

Why should it be? You want to know the energy transferred to a material object, so why would you think you could use a non-material velocity? The assumption that you should be able to do that seems baseless, and it demonstrably leads to incorrect values.

 

[Dale]

To be perfectly clear, when I say it does not matter, I am speaking of not mattering for the position, velocity, momentum or angular momentum of the opaque blob that we call the car

But without that extra information you cannot determine the change in the energy of the system.