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

[A.T.]

...what force is responsible for the increase in mechanical kinetic energy ...

Depends on what "is responsible for" exactly means. If you want to gain clarity, you should try to avoid ambiguous terms.

Consider a simpler case:

A stick falls over from while the bottom end doesn't slide due to static friction.

- The gain in horizontal momentum comes from the impulse of the static friction.

- The gain in kinetic energy comes from energy the stick initially has, not from work by static friction.

There is no contradiction between these two statements, and they both also apply to the accelerating car. The changing contact location in the car case is irrelevant for the energy calculations.

 

[jbriggs444]

There is a sense in which one can attach useful meaning to the motion of the contact patch. If one views the contact patch as a "handle" which is rigidly attached to the car and through which the ground "grips" the car and if one ignores the motion of the tire surface entirely then the work done on this "handle" by the ground is identical to the bulk kinetic kinetic energy imparted to the [pretended-to-be] rigid car by the ground.

This includes the bulk rotational kinetic energy of the car as it rotates forward in the top-of-the-hill case or backward in a bottom of a valley case. It does not include any kinetic energy in the car-relative motion of the tires or drive train.

 

[Dale]

- The gain in horizontal momentum comes from the impulse of the static friction.

- The gain in kinetic energy comes from energy the stick initially has, not from work by static friction.

This is a very clear and simple example. Also, if you repeat the experiment but on a frictionless surface then you see that the KE is the same as with friction and only the momentum is changed.

 

[rcgldr]

There is no such thing as linear KE, and internal forces most certainly can increase KE.

Then restate this as energy related to change in linear movement. Internal forces can't change the center of mass, linear speed, or linear momentum of a closed system, while internal forces can can change angular velocity and angular energy, while angular momentum is conserved, of a closed system.

Although the accelerating car is converting potential energy into mechanical energy (so no "real" work is done), what I was getting at that for the static friction force and distance the force is applied, then

$$\int F \cdot ds = \Delta KE$$

 

[jbriggs444]

what I was getting at that for the static friction force and distance the force is applied, then

$$\int F \cdot ds = \Delta KE$$

Of course, that equation holds for point-like objects where there is little ambiguity about $ds$ or $\Delta \text{KE}$. Invoking it for non-rigid or rotating objects or where the point of application may otherwise not have a fixed position relative to the center of mass introduces ambiguity.

 

[jbriggs444]

I suggested using an inertial reference frame with the origin at the center of mass in my prior moon + car example. You could assume the moon to be of infinite mass and use it as the frame of reference. The distance traveled relative to this frame would not be ambiguous. I also made assumptions such as no losses, no drag (no atmosphere on this moon), and no rolling resistance.

It's not the lack of a frame of reference that makes $ds$ ambiguous. It is the lack of a clearly defined position for an extended object. Especially for one that is rotating or non-rigid.

 

[Doc Al]

It's not the lack of a frame of reference that makes $ds$ ambiguous. It is the lack of a clearly defined position for an extended object. Especially for one that is rotating or non-rigid.

One way to apply
$$\int F \cdot ds = \Delta KE$$
to a rotating or non-rigid object is to let $F$ be the net force and $ds$ be the displacement of the center of mass. In this case, $KE = 1/2mv^2$ where $m$ is the mass of the object and $v$ is the velocity of the center of mass.

Of course, this just underscores the fact that we are applying Newton's 2nd law, not conservation of energy.

 

[Dale]

Then restate this as energy related to change in linear movement.

OK, then internal forces can do that. Consider a spring that pushes two halves of an object apart. The KE has increased and it is related to a change in linear movement.

Internal forces can't change the center of mass, linear speed, or linear momentum of a closed system

I agree with center of mass and linear momentum, but not linear speed. That certainly can be changed, e.g. with the spring example.

Although the accelerating car is converting potential energy into mechanical energy (so no "real" work is done), what I was getting at that for the static friction force and distance the force is applied, then

$$\int F \cdot ds = \Delta KE$$

The $ds$ in this equation is defined by the center of mass, not the contact patch and $F$ is the net force, not an individual force. This is the usual "center of mass work", which has units of work and a similar formula, but does not represent the physical work done by any specific force, even when the net force is equal to that specific force. To calculate the physical work done by a given individual force the $ds$ is the displacement of the material at the point of application of that force and it does not have anything to do with $\Delta KE$

 

[rcgldr]

Consider a spring that pushes two halves of an object apart. The KE has increased and it is related to a change in linear movement. I agree with center of mass and linear momentum, but not linear speed. That certainly can be changed, e.g. with the spring example.

I agree with this, but a car is generally a rigid body with components that can rotate, but not translate very much (other than mostly vertical motion related to suspension).

The $ds$ in this equation is defined by the center of mass, not the contact patch and $F$ is the net force, not an individual force.

If F is force at the contact patch, then the ds would be the distance traveled by contact patch, or more generally it's the integral form of force at the point of application times the distanced moved of the point of application of force. Consider the force applied to the end of a long rod, much of the force goes into increasing the angular speed of the rod and only a fraction of the force displaces the center of mass of the rod to increase the linear speed of the rod. The rod's total KE can be separated into linear and angular components.

 

[Doc Al]

Consider the force applied to the end of a long rod, much of the force goes into increasing the angular speed of the rod and only a fraction of the force displaces the center of mass of the rod to increase the speed of the rod.

Seriously? Are you saying that the acceleration of an object's center of mass for a given force depends on where the force is applied?

 

[rcgldr]

Are you saying that the acceleration of an object's center of mass for a given force depends on where the force is applied?

No. Using the rod as an example again, the center of mass of the rod's acceleration = force / (rod's mass). However, for the same impulse (force x time), if the point of application is at the end of the rod, then the distance traveled at the point of application will be greater than if the force was applied at the center of mass of the rod, and the resulting increase in KE will be greater, due to the rod experiencing both linear and angular acceleration.

This is what I was getting at with my small moon + car on stilts example, where the car's center of mass is 25% further away from the center of the moon than the contact patch. The increase in kinetic energy will be equal to the static friction force time the distance traveled at the point of application, not at the center of mass of the car.

 

[Dale]

If F is force at the contact patch,

In that equation F is not the force at the contact patch, it is the net force.

Consider the force applied to the end of a long rod, much of the force goes into increasing the angular speed of the rod and only a fraction of the force displaces the center of mass of the rod to increase the linear speed of the rod.

This is completely false.

The increase in kinetic energy will be equal to the static friction force time the distance traveled at the point of application, not at the center of mass of the car.

Are you sure? I haven’t worked it out yet, but I think not.

 

[rcgldr]

Consider the force applied to the end of a long rod, much of the force goes into increasing the angular speed of the rod and only a fraction of the force displaces the center of mass of the rod to increase the linear speed of the rod. The rod's total KE can be separated into linear and angular components.

This is completely false.

I worded that badly. The force is the same, but assume the force and the impulse for both cases are the same, and that the impulse takes place over some finite period of time. During the impulse, the distance or average speed of the application of force if applied to the center of the rod will be less than the distance or average speed of the application of force if applied to the end of the rod, corresponding to the second case involving more work, resulting in an increase in both linear and angular kinetic energy.

 

[Dale]

I worded that badly. The force is the same, but assume the force and the impulse for both cases are the same, and that the impulse takes place over some finite period of time. During the impulse, the distance or average speed of the application of force if applied to the center of the rod will be less than the distance or average speed of the application of force if applied to the end of the rod, corresponding to the second case involving an increase in both linear and angular kinetic energy.

Yes, that is not in dispute because in that case the contact patch has the same velocity as the material at the contact patch.

Do you not see the obvious fact that the dynamometer example completely refutes your approach. In the dynamometer the contact patch is stationary and yet work (transfer of energy at the patch) is actually done. The total energy of the car decreases and the total energy of the dynamometer increases. It is as clear as day. The only way to explain that transfer of energy is through using the velocity of the material at the contact patch, not the velocity of the contact patch.

 

[rcgldr]

Do you not see the obvious fact that the dynamometer example completely refutes your approach. In the dynamometer the contact patch is stationary and yet work (transfer of energy at the patch) is actually done. The total energy of the car decreases and the total energy of the dynamometer increases.

If the car and the dynamometer are considered as components of a closed system, then you're back to the case where energy is converted and transferred within a closed system, which would be similar to my example of a small moon and car (another closed system).

 

[Dale]

If the car and the dynamometer are considered as components of a closed system

And if they are considered separate systems with the static friction as an external force between the systems? Do you not see that your idea to use the velocity of the contact patch instead of the velocity of the material at the contact patch is completely disproved?

If your contact patch velocity idea were right then it should be able to handle the dynamometer case with the contact patch as an interface between systems. But it fails.

 

[rcgldr]

It's not clear to me if OP is asking if the static friction force can perform "real" work on the car or if static friction force is responsible for the increase in KE of the car.

If your contact patch velocity idea were right then it should be able to handle the dynamometer case with the contact patch as an interface between systems. But it fails.

Change the car scenario to a electrical track powered vehicle. The electrical power source is external to the vehicle, and it is picked up from the track through the wheels. The vehicle converts the power and generates a Newton third law pair of forces: the vehicle wheels exert a "backwards" force onto the track, and the track exerts a "forwards" force onto the vehicle. In this case, both the source of power and the force from the track that accelerates the vehicle are external to the vehicle. In this case, is "real" work being performed on the vehicle?

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, or with integral from of contact patch force times contact patch distance equating to the increase of KE of the vehicle (assuming no losses). I consider this to be a consequence of rolling motion of wheels.

 

[Dale]

Change the car scenario

Sure. We can change the car scenario as you like, but please first address the dynamometer scenario. It is clear, simple, and completely invalidates the contact patch velocity approach. Do you agree? I get the sense from your avoidance in the last few posts that you do recognize it.

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

Because the same procedure fails in so many other cases.

 

[rcgldr]

We can change the car scenario as you like, but please first address the dynamometer scenario. It is clear, simple, and completely invalidates the contact patch velocity approach.

In the dynamometer case, the surfaces are moving with respect to an inertial reference frame, in the car on a road case, the contact patch is moving with a respect to an inertial reference frame. When accelerating, both cases result in an increase of kinetic energy. Using imperial units to calculate horsepower for both cases, power (horsepower) = force (lbs) · speed (miles / hour) / 375.

In the second case, if the focus is the tire tread surface, the average speed of the entire tire tread surface is the same as the vehicle (on a flat road). It's 0 at the middle of the contact patch, and twice the speed of the vehicle at the top of the contact patch.

 

[Dale]

Using imperial units to calculate horsepower for both cases, power (horsepower) = force (lbs) · speed (miles / hour) / 375.

And for the dynamometer the speed of the contact patch is zero.

 

[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 defintion 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 defintion 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 defintion 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.