Modelling assumptions when friction is involved

[jbriggs444]

I meant to say that $f_k d$ is the total increase in mechanical energy by both frictional forces in the pair. The increase of thermal energy of the block is $f_k d_{eff}$ and the other $f_k (d - d_{eff})$.

When we speak of the interface, and power flows, etc. I believe with this framework everything works out if we use $f_k d$. However when considering the interacting bodies in isolation, we can no longer use $d$, as otherwise the FLT no longer works.

If you consider the bodies in isolation you have to worry about how to apportion the thermal energy to each body, yes. You can use a fudge factor to do this. But don't delude yourself by pretending that the fudge factor means anything.

 

[etotheipi]

I do not rea

If you consider the bodies in isolation you have to worry about how to apportion the thermal energy to each body, yes. You can use a fudge factor to do this. But don't delude yourself by pretending that the fudge factor means anything.

Yes, I think that's what it comes down to. I don't really buy the whole treatment with the "teeth" and so forth, and I certainly wouldn't ever make reference to $d_{eff}$ (I don't really even know what it's meaning is!).

To my mind, the distinction just clears a few points up that have been bugging me for days, since now the FLT, CM etc. are sort of consistent at least. Which one we use to solve a particular problem is another issue!

 

[Dale]

But isn't $P = \vec{f} \cdot \vec{v}$ referring to centre of mass power/work (I think you refer to it as net work)?

No, sorry I was not clear. $\vec v$ is the velocity of the material at the point of application of $\vec f$. That is only equal to the center of mass velocity for a point particle.

Sherwood's argument is that $-f_k d$ is not the work done by friction. The thermodynamic work done by friction is slightly less, and is dependent on stuff at the microscopic level.

The full amount $\vec f_k \cdot \vec d$ of mechanical work is done by friction. That generates heat at the interface. Sherwood's argument is simply dividing $Q$ by $f_k$ to get a distance that he calls the effective distance, but it is still mechanical power going out as normal and heat coming back in. Consider an ice cube sliding down a hot playground slide on a summer day. In that case $Q>f_k d$ so the "effective distance" would be greater than the actual distance. This shows that it is not a meaningful distance, it is just an idiosyncratic way of writing $Q$.

So, for the first law of thermo you have $Q+W=\Delta E$ or, written in terms of power $\dot Q + P = \dot E$. In your example, taking the block itself as the system then $P=\Sigma_i \vec f_i \cdot \vec v_i=\vec f_k \cdot \vec v + \vec f_{ext} \cdot \vec v$ and if the block is not accelerating then $\vec f_{ext}=-\vec f_k$ so $P=0$. Then $\dot E = \dot Q$ meaning that the amount of increase of internal energy from the block is equal to the heat received. That should be clear and feel quite natural given the situation. $Q$ must be determined thermodynamically, e.g. https://en.wikipedia.org/wiki/Heat_equation

 

[etotheipi]

So, for the first law of thermo you have $Q+W=\Delta E$ or, written in terms of power $\dot Q + P = \dot E$. In your example, taking the block itself as the system then $P=\Sigma_i \vec f_i \cdot \vec v_i=\vec f_k \cdot \vec v + \vec f_{ext} \cdot \vec v$ and if the block is not accelerating then $\vec f_{ext}=-\vec f_k$ so $P=0$. Then $\dot E = \dot Q$ meaning that the amount of increase of internal energy from the block is equal to the heat received. That should be clear and feel quite natural given the situation. $Q$ must be determined thermodynamically, e.g. https://en.wikipedia.org/wiki/Heat_equation

Thank you for this, it's really helpful. It's still going to take me a little while to put the pieces together but I could follow along with your reasoning. It appears my mistake was neglecting $Q$ (I always let $Q=0$), whilst it seems to be a fairly important part of the model. Your method is significantly more natural than that of Sherwood!

So if I understand correctly, the total power of friction by both frictional forces at an interface equals the mechanical power dissipated into thermal power at the contact patch. We don't yet necessarily know how the thermal power splits between the two bodies, though evidently if our system contains both bodies in question then we've in effect accounted for all of the possibilities

$-W_{f} = f_1 v_1 + f_2 v_2 = P_{th} = - P_{mech}$

It's more ambiguous if we take only one of the bodies as the system, since then $\dot{E} = P_{ext} + \dot{Q}$ and we generally don't know $\dot{Q}$. Though for instance, in the case of the block of ice, it is likely that due to the temperature gradient and hence $\dot{Q}$ for the ice block will be higher, i.e. greater rate of heat flow in this direction.

If the wedge is fixed, then $\Delta E_{th}$ for the entire system is $f_k d$. We might also write for each the block and the wedge

Block: $-f_k d + mgh + Q_{b} = \Delta E_{th, b} + \Delta T$
Wedge: $Q_{w} = \Delta E_{th, w}$

If we sum these equations,

$-f_k d + mgh + Q_{w} + Q_{b} = \Delta E_{th, b} + \Delta E_{th, w} + \Delta T$

However $\Delta E_{th, b} + \Delta E_{th, w}$ is nothing but $\Delta E_{th}$ for the system, which is then $f_k d$. So we can collect terms, and we're left with

$mgh + Q_{w} + Q_{b} = 2f_k d + \Delta T$

I have the feeling that $Q_{w} + Q_{b} = f_k d$, since then we get back to the good old equation $mgh - f_k d = \Delta T$, but don't know why this would be the case. I would have thought $Q_{w} + Q_{b} = 0$, since heat is transferred from one to the other?

I presume the answer is that heat is generated at the interface, so in fact both $\dot{Q}$'s will be positive and there is no constraint that $Q{w} = -Q_{b}$; in fact, they probably sum to the total change in thermal energy.

 

[Dale]

I have the feeling that $Q_{w} + Q_{b} = f_k d$,

Yes, that is correct.

I would have thought $Q_{w} + Q_{b} = 0$, since heat is transferred from one to the other?

I presume the answer is that heat is generated at the interface,

Yes, there may be some heat transferred from one to the other, e.g. in the case of the ice cube on the slide, but for objects that start at the same temperature then the important quantity is the amount of heat generated at the interface. This is no heat that is transferred from one to the other but heat that is actually produced (from the dissipated mechanical power) at the interface and then flows into both bodies per the normal heat equations.

We don't yet necessarily know how the thermal power splits between the two bodies

Correct. Nothing in the mechanical setup tells us this nor can be used to derive this. This question is a thermodynamic question and would require solving the heat equation with appropriate values for the materials.

 

[etotheipi]

Yes, that is correct.

Yes, there may be some heat transferred from one to the other, e.g. in the case of the ice cube on the slide, but for objects that start at the same temperature then the important quantity is the amount of heat generated at the interface. This is no heat that is transferred from one to the other but heat that is actually produced (from the dissipated mechanical power) at the interface and then flows into both bodies per the normal heat equations.

Correct. Nothing in the mechanical setup tells us this nor can be used to derive this. This question is a thermodynamic question and would require solving the heat equation with appropriate values for the materials.

Awesome, thank you for all your patience!

It's definitely easier to think in terms of power flows, though I hadn't encountered that approach before! So thank you!

Also, I think I understand this part now:

You might want to think about the region where two bodies are interacting as a kind of black box. You know how much force each object applies on its interface to the black box. You know the motion of each object at its interface with the black box. You can use this information to compute a quantity: the amount of mechanical energy input to the box. This number is the force at the interface (by Newton's third, the forces are equal) times the relative motion across the interface. This is the real work done on the black box. If energy is conserved and if the interaction region is vanishingly small, so that negligible energy can be stored there, then the real work input to the black box must be matched by some equal energy flow out of it.

Since we could give the vanishingly small interface an energy $E$, then

$f_k d + Q_{tot} = 0 \implies Q = -f_k d$

That is to say $f_k d$ of heat is transferred out of the interface into the two blocks.

 

[cianfa72]

Since we could give the vanishingly small interface an energy $E$, then

$f_k d + Q_{tot} = 0 \implies Q = -f_k d$

That is to say $f_k d$ of heat is transferred out of the interface into the two blocks.

I've been following this thread for some time. Just to be sure about my understanding:
As far as I can tell, formally we have to take in account both bodies each interacting with the interface (black box). Thus from the point of view of 'interface' as system the former equation actually should be

$2f_k d + Q_{tot} = 0 \implies Q = -2f_k d$

that from point of view of each body is really just $f_k d$

 

[etotheipi]

I've been following this thread for some time. Just to be sure about my understanding:
As far as I can tell, formally we have to take in account both bodies each interacting with the interface (black box). Thus from the point of view of 'interface' as system the former equation actually should be

$2f_k d + Q_{tot} = 0 \implies Q = -2f_k d$

that from point of view of each body is really just $f_k d$

I'm not sure that is correct, I think the issue is one concerning the definition of work done as displacement of the material at the point of application of the force. The interface I assume we take to be a system which contains a tiny surface layer of each of the interacting bodies (for pedagogical purposes, maybe the top layer of particles?).

The frictional force of the block on the wedge does no work on this interface since the particles in the top layer of the wedge are immobile. The frictional force of the wedge on the block does $f_k d$ of work. In another frame of reference these relative amounts of work might change, however the total work done on the interface is always $f_k d$ in any frame of reference.

So it is my understanding that $Q_{tot} = f_k d$ (or $-f_k d$, depending on whether you're measuring in/out heat flow).

 

[jbriggs444]

I've been following this thread for some time. Just to be sure about my understanding:
As far as I can tell, formally we have to take in account both bodies each interacting with the interface (black box). Thus from the point of view of 'interface' as system the former equation actually should be

$2f_k d + Q_{tot} = 0 \implies Q = -2f_k d$

that from point of view of each body is really just $f_k d$

No. That's double counting. The relative motion is what counts. Not the sum of the motion of the one surface in the rest frame of the other plus the motion of the other surface in the rest frame of the one.

Pick a frame and use it. You get nonsense results when you combine figures drawn from different reference frames willy nilly.

 

[Dale]

As far as I can tell, formally we have to take in account both bodies each interacting with the interface (black box). Thus from the point of view of 'interface' as system the former equation actually should be

$2f_k d + Q_{tot} = 0 \implies Q = -2f_k d$

that from point of view of each body is really just $f_k d$

Yes, you definitely do need to account for both interfaces, but your accounting is off here. In this case the $d$ for the other interface is zero. So we have $f_k d + f_k 0 + Q_{tot} = 0 \implies Q = -f_k d$

 

[etotheipi]

I'm aware we've smashed the 100 post mark, and don't want to prolong this thread much longer, though I just wondered if there was a concrete way of defining this interface we speak of? There must be a boundary of some sort, since we've applied the FLT to it without trouble.

A few posts back I suggested the top layer of particles on each interacting body, though I suspect it is actually a more abstract concept.

Since really, the wedge is doing work on the block and the block doing work on the wedge. We can "infer" the work done on this "interface" though I'm not entirely sure what constitutes the interface.

 

[jbriggs444]

A few posts back I suggested the top layer of particles on each interacting body, though I suspect it is actually a more abstract concept.

Abstract is how I like to think of it. We know what happens on the one side of the interface. Just a mechanical force. We know what happens on the other side of the interface. Just a mechanical force. We know there is a relative velocity. Conservation of energy tells us that if we have mechanical work going in, some other energy flow must come out. Experience tells us that there is thermal energy coming out [in principle, the outbound flow could be otherwise].

What more do we need to know? We can ignore the details and concentrate on the results.

 

[Dale]

I just wondered if there was a concrete way of defining this interface we speak of?

At the scale of interest it is simply the two dimensional surface where the two objects are touching. Note the phrase “at the scale of interest”. Obviously at small enough scales it will no longer be two dimensional, but at those scales you don’t have simple friction, you have surface irregularities, elastic and plastic deformation, contact welds, and so forth.

 

[A.T.]

I suspect it is actually a more abstract concept.

More abstract means more general, more widely applicable and thus more useful. The concept of using P = F * v at an interface doesn't just apply to friction:

Instead of surface irregularities that oppose relative motion, you could have many little ants or linear motors propelling the block, which you also don't want to model in detail. Here the sum of the work done by the force pair would be positive. This represents mechanical energy generated at the interface (converted from chemical or electrical).

The whole issue becomes even more fun if an object has two interfaces, with two other object that are in relative motion to each other.

Page 10-11:
https://www.aapt.org/physicsteam/2019/upload/USAPhO-2013-Solutions.pdf

Chapter 2.1:
https://backend.orbit.dtu.dk/ws/portalfiles/portal/3748519/2009_28.pdf

 

[cianfa72]

No. That's double counting. The relative motion is what counts. Not the sum of the motion of the one surface in the rest frame of the other plus the motion of the other surface in the rest frame of the one.

Pick a frame and use it. You get nonsense results when you combine figures drawn from different reference frames willy nilly.

I was wondering about the frame to use to do the calculation. I believe it has to be an inertial one thus avoiding to account for work done by fictitious (not real) forces existing there

In the problem of 'block + wedge' system assuming $M_w \gg M_b$ we've actually no doubt: we can definitely assume the frame at rest with wedge as inertial doing the work calculation there.

In the general case assuming the 'system' considered as closed (no external work/heat) we can select the frame at rest with system CM doing the calculation there

In any case, as highlighted before in the thread, the displacement to account for in work calculation is actually the 'relative' displacement at the interface

 

[etotheipi]

More abstract means more general, more widely applicable and thus more useful. The concept of using P = F * v at an interface doesn't just apply to friction:

Instead of surface irregularities that oppose relative motion, you could have many little ants or linear motors propelling the block, which you also don't want to model in detail. Here the sum of the work done by the force pair would be positive. This represents mechanical energy generated at the interface (converted from chemical or electrical).

The whole issue becomes even more fun if an object has two interfaces, with two other object that are in relative motion to each other.

That Blackbird is a pretty crazy machine! The one other thing I got from it is that we can use either real work or centre-of-mass work at the interface, but to compute different quantities. If we choose centre-of-mass work,

$P_{CM, tot} = P_{CM, 12} + P_{CM, 21}$ and then also $P_{CM, tot} + \dot{E}_{int} = 0$ if the whole thing is isolated; the total centre-of-mass power equals the negative rate of change of internal energy (which in this case, means everything except translational kinetic) right off the bat.

If we choose to analyse real work, it's a little bit more convoluted.

$P_{RE, tot} = P_{RE, 12} + P_{RE, 21}$

on each body,

$P_{RE, 12} + \dot{Q}_{1} = \dot{E}_{1} = \dot{E}_{int, 1} + \dot{E}_{CM, 1}$
$P_{RE, 21} + \dot{Q}_{2} = \dot{E}_{2} = \dot{E}_{int, 2} + \dot{E}_{CM, 2}$

Of course since the system is isolated, $P_{RE, tot} = -\dot{Q}_{tot}$. I don't know how to derive it, but I assume it is the case (from the previous discussion) that the total real work done equals negative the change in thermal energy only, and not just internal as was the case when we considered COM work. In essence, the total real work done at the interface is the change in mechanical energy, whilst the total COM work done at the interface is the change in translational KE.

In the case of the block and wedge, the centre-of-mass work and real work turn out to be identical, and this is fine since there is only translational KE involved. For the Blackbird, there is rotational KE involved in the turbine, so we need to be a little more careful.

 

[jbriggs444]

In the problem of 'block + wedge' system assuming $M_w \gg M_b$ we've actually no doubt: we can definitely assume the frame at rest with wedge as inertial doing the work calculation there.

The relevant point is that every inertial frame gives the same result. We do not have to choose the right one. We just have to choose.

 

[cianfa72]

The relevant point is that every inertial frame gives the same result. We do not have to choose the right one. We just have to choose.

Sure, the fundamental point anyway is that the frame of reference chosen has to be inertial

 

[Dale]

I believe it has to be an inertial one thus avoiding to account for work done by fictitious (not real) forces existing there

I don’t think that is a requirement. Since we can do this analysis in the presence of gravity then by the equivalence principle we should be able to do it in a uniformly accelerating non inertial frame also.

 

[jbriggs444]

I don’t think that is a requirement. Since we can do this analysis in the presence of gravity then by the equivalence principle we should be able to do it in a uniformly accelerating non inertial frame also.

It would be nice for the non-inertial frame to at least have a definable potential. [Which a uniformly accelerating or rotating frame would have, of course].

 

[Dale]

It would be nice for the non-inertial frame to at least have a definable potential.

Yes, otherwise the conservation of energy wouldn’t work, but I don’t think that would change the friction part of the analysis.

 

[cianfa72]

Yes, otherwise the conservation of energy wouldn’t work, but I don’t think that would change the friction part of the analysis.

Here the system taken into account for the analysis is 'block+wedge' in the external gravity field. This way gravity, basically, is accounted as an external force doing external work on a system component (the block).

If we choose a non-inertial frame to do the analysis, as far as I can understand, we have to add the work done by inertial force as measured by the potential defined in that (non-inertial) frame

 

[jbriggs444]

Here the system taken into account for the analysis is 'block+wedge' in the external gravity field. This way gravity, basically, is accounted as an external force doing external work on a system component (the block).

If we choose a non-inertial frame to do the analysis, as far as I can understand, we have to add the work done by inertial force as measured by the potential defined in that (non-inertial) frame

Alternately, we might focus our attention on a region very close to the block+wedge interface so that the mass of the system under consideration is negligible and the choice of frame becomes irrelevant.