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

[Dale]

So what about chassis dynamometers? The only force they use to calculate power is static friction force (times the speed of the tire/drum(s) surfaces at the contact patch). in this case, the static friction force is performing work on the drum(s), increasing their angular kinetic energy.

I already addressed this explicitly above in post 82. The “velocity of the material at the contact patch” correctly calculates the power for a dynamometer. The “contact patch velocity” does not.

 

[rcgldr]

The “velocity of the material at the contact patch” correctly calculates the power for a dynamometer. The “contact patch velocity” does not.

Just to be clear, static friction force can perform work on the drum(s) in a dynamometer, but not on car on a road?

 

[jbriggs444]

Just to be clear, static friction force can perform work on the drum(s) in a dynamometer, but not on car on a road?

Static friction can perform work on a box of car parts in the bed of your pickup truck when you accelerate after a stop.

Static friction from the highway can perform "center of mass" work on a car on the highway.

 

[Dale]

Just to be clear, static friction force can perform work on the drum(s) in a dynamometer, but not on car on a road?

Yes. On the drums the velocity of the material is non-zero so the work is non-zero. On the road the velocity of the material is zero so the work is zero.

 

[rcgldr]

Static friction from the highway can perform "center of mass" work on a car on the highway.

On the road the velocity of the material is zero so the work is zero.

This seems like a conflict, what am I missing here?

The only external force acting on a car on the road is the static friction force from the road, created as part of a Newton 3rd law pair of forces that originates with the car's engine or motor. If that is not the external force that performs work on an accelerating car, then what force is responsible for the increase in mechanical kinetic energy that was converted from potential energy of fuel or battery by the car's engine? Internal forces can't increase linear kinetic energy.

 

[jbriggs444]

This seems like a conflict, what am I missing here?

The need to be clear on what definition of "work" one is using. Which means the need to be clear about which velocity (or displacement) is being used.

 

[Dale]

This seems like a conflict, what am I missing here?

The "center of mass work" is not work since it can occur without any transfer of energy. It is a somewhat common misuse of language. I prefer $\Delta KE$.

 

[Doc Al]

This seems like a conflict, what am I missing here?

"Center of mass" work is just an application of Newton's 2nd law; the actual "real" work done (as in the first law of thermodynamics) is zero. As @jbriggs444 says, be careful what version of "work" you're using. (I would prefer that center of mass "work" not be called work at all.)

The only external force acting on a car on the road is the static friction force from the road, created as part of a Newton 3rd law pair of forces that originates with the car's engine or motor. If that is not the external force that performs work on an accelerating car, then what force is responsible for the increase in mechanical kinetic energy that was converted from potential energy of fuel or battery by the car's engine? Internal forces can't increase linear kinetic energy.

Sure, an external force is needed to convert internal energy into mechanical kinetic energy. Yet no work is done by that force.

 

[jbriggs444]

I agree with @Dale on this. Center of mass work is as much about momentum as about energy. You just apply the SUVAT equations and get something that looks like energy.

Note that when you apply it to the case of a car cresting a hill where the center of mass and the contact patch have significant differences in speed, you will find that the normal force on front and back tires have a significant discrepancy in magnitude and angle. That combines to produce a retarding horizontal force. Fail to account for that force and your measure of center-of-mass work will be off.

That is a case where using contact patch force times contact patch speed leads to a simple computation and a correct answer.

 

[Dale]

The only external force acting on a car on the road is the static friction force from the road, created as part of a Newton 3rd law pair of forces

Yes.

If that is not the external force that performs work on an accelerating car

There is no work on the car (under the usual assumptions). Note, I consistently use work meaning an actual transfer of energy, not fictitious “center of mass work”.

what force is responsible for the increase in mechanical kinetic energy

Forces are the rate of change of momentum, not energy. Power is the rate of change of energy. So this question is wrong. It incorrectly assumes that forces do something that power actually does.

The power that is responsible for the increase in KE is the engine/motor power.

Internal forces can't increase linear kinetic energy.

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

Force is the rate of change of momentum. The external force provides the momentum, not the kinetic energy. Both increase but they have different sources.

 

[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.