Modelling assumptions when friction is involved

[jbriggs444]

I'm referring to translational plus rotational, though in this case just translational. I'm aware the decomposition into macroscopic (mechanical) and microscopic (thermal) KE is somewhat arbitrary, though it seems useful.

That still does not define what you mean by kinetic energy of a non-rigid block.

I think that what you are trying to get at is a sum of the kinetic energies of all of the component pieces of the non-rigid block down to a very tiny level, but stopping just short of where those kinetic energies become thermal energies. That would fit with the term "arbitrary".

 

[etotheipi]

That still does not define what you mean by kinetic energy of a non-rigid block.

In that case, the kinetic energy the block would possesses if we treat it as a point mass at the centre of mass.

 

[jbriggs444]

In that case, the kinetic energy the block would possesses if we treat it as a point mass at the centre of mass.

So in the case of an automobile with a flywheel, the rotational motion of the flywheel does not count.

It is easy to predict the kinetic energy of a sliding block in this case. Calculate using "center-of-mass" work, i.e. force times motion of the center of mass.

 

[etotheipi]

So in the case of an automobile with a flywheel, the rotational motion of the flywheel does not count.

I meant to say that it would count, though in this scenario it reduces to the centre of mass kinetic energy.

Centre-of-mass work would certainly work here.

$mgh - f_k d = \frac{1}{2}mv^{2}$

But now if I use the first law of thermodynamics,

$mgh - f_k d = \frac{1}{2}mv^{2} + \Delta E_{th}$

This is what's causing the problems, we're left with this bizarre $\Delta E_{th}$ left over if we consider thermodynamic work. This must turn out to be zero. Though the thermal energy of the wedge is also zero if we consider it as a system.

So something has to be wrong! I'm thinking that it has something to do with thermodynamic work vs centre of mass work. In the limiting case that the mass becomes a point particle, it has no thermal energy and both expressions are the same.

But if we scale it up to a fullsize block, either $f_k$ or $d$ has to decrease.

 

[jbriggs444]

I meant to say that it would count, though in this scenario it reduces to the centre of mass kinetic energy.

Centre-of-mass work would certainly work here.

$mgh - f_k d = \frac{1}{2}mv^{2}$

But now if I use the first law of thermodynamics,

$mgh - f_k d = \frac{1}{2}mv^{2} + \Delta E_{th}$

This is what's causing the problems, we're left with this bizarre $\Delta E_{th}$ left over if we consider thermodynamic work. This must turn out to be zero. Though the thermal energy of the wedge is also zero if we consider it as a system.

So something has to be wrong! I'm thinking that it has something to do with thermodynamic work vs centre of mass work. In the limiting case that the mass becomes a point particle, it has no thermal energy and both expressions are the same.

But if we scale it up to a fullsize block, either $f_k$ or $d$ has to decrease.

Center of mass work does not conserve energy. It's not about conservation of energy. It's about motion of the center of mass. The problem you wanted to solve wasn't about energy. You just wanted to know about the motion of the center of mass. And you ended up knowing the motion of the center of mass.

Why are you complaining that you didn't get total energy when you didn't ask for total energy?

 

[etotheipi]

Center of mass work does not conserve energy. It's not about conservation of energy. It's about motion of the center of mass. The problem you wanted to solve wasn't about energy. You just wanted to know about the motion of the center of mass. And you ended up knowing the motion of the center of mass.

Why are you complaining that you didn't get total energy when you didn't ask for total energy?

No complaints, I'm just trying to troubleshoot the other FLT method. The fact I can't get it to work (no pun intended) suggests a flaw in my understanding.

I'm having a read through Halliday & Resnick since I think they also cover energy quite well, but I can't find a comment on this precisely.

 

[jbriggs444]

It is hard to pick the right tool to solve a problem when you have not yet decided what problem you want to solve.

A problem well stated is half solved.

Successful problem solving is as much about knowing what can be safely ignored as about what needs careful attention.

 

[russ_watters]

I meant to say that it would count, though in this scenario it reduces to the centre of mass kinetic energy.

Centre-of-mass work would certainly work here.

$mgh - f_k d = \frac{1}{2}mv^{2}$

But now if I use the first law of thermodynamics,

$mgh - f_k d = \frac{1}{2}mv^{2} + \Delta E_{th}$

Your second equation is wrong. The friction and thermodynamic work terms are the same thing with different labels. It's double-counted.

It might help if you arrange the first equation as a before and after or input and output.

 

[etotheipi]

Your second equation is wrong. The friction and thermodynamic work terms are the same thing with different labels. It's double-counted.

It might help if you arrange the first equation as a before and after or input and output.

I think you're right, but I can't exactly determine why. The goal is to determine the final KE of the block using only a system which contains only the block.

I'll define the system to be the block. Initially, $E = T_1 + {E_{th, block}}_1 = 0 + 0 = 0$, since I'll say it starts at rest with no thermal energy.

The external thermodynamic work done by friction is $-f_k d$, and the external work by gravity $mgh$. There is no heat transfer.

The final energy is $E = {T_{2}} + {E_{th, block}}_2$

The first law of thermodynamics yields

$-f_k d + mgh = T_2 + {E_{th, block}}_2$

It definitely appear that we've double counted, but I can't see what is wrong with this line of reasoning. When the wedge and block are a single system, the frictional forces are internal and we need not include them on the LHS.

When solely the block is the system, friction now does external work and we need to include it.

 

[russ_watters]

I think you're right, but I can't exactly determine why. The goal is to determine the final KE of the block using only a system which contains only the block.

You keep saying that, but it isn't what you are doing, so my only conclusion can be that that isn't what you really want. Your real goal is to find the KE and the thermodynamic energy. If you really only cared about the KE you would be totally ignoring the thermodynamic energy, not throwing it into the equation where it doesn't belong.

The external thermodynamic work done by friction is $-f_k d$, and the external work by gravity $mgh$. There is no heat transfer.

Ding, ding, ding! Re-read the name you gave that first term. External thermodynamic work. Isn't it mechanical work? (hint: your equation either has too many terms or not enough)

When solely the block is the system, friction now does external work and we need to include it.

Mechanical work or thermodynamic work?

 

[etotheipi]

You keep saying that, but it isn't what you are doing, so my only conclusion can be that that isn't what you really want. Your real goal is to find the KE and the thermodynamic energy. If you really only cared about the KE you would be totally ignoring the thermodynamic energy, not throwing it into the equation where it doesn't belong.

Ding, ding, ding! Re-read the name you gave that first term. External thermodynamic work. Isn't it mechanical work? (hint: your equation either has too many terms or not enough)

Mechanical work or thermodynamic work?

Thanks for your reply, I think I've thoroughly confused myself so I think I'm going to come back to this in a little while just in case a fresh outlook might help

 

[russ_watters]

I probably shouldn't be just giving you the answer, but this has dragged on too long and you're not getting it.

This equation is correct for mechanical work, from an input and output standpoint:
$mgh - f_k d = \frac{1}{2}mv^{2}$

Period. Full Stop. Problem over.

But that's not what you really want. You want a description of the situation from a before and after thermodynamic energy standpoint. As I said before, your second equation was wrong, but for some reason you haven't tried to fix it, you just went off in another direction. Here's what you gave us:

$mgh - f_k d = \frac{1}{2}mv^{2} + \Delta E_{th}$

But as you pointed out in post 59, the starting thermodynamic energy is zero, and your equation doesn't show a zero in the first term. And you also included kinetic energy, which isn't thermodynamic energy. A correct equation for the mechanical to thermodynamic energy (work-energy balance) could be:

$0 + f_k d= E_{th2}$

It's zero to start, has an input of frictional heating energy and ends with the thermodynamic energy of the block.

But you want the mechanical energy and thermodynamic energy in the same equation. Let's make it this form:

mechanical1 + thermo1 = mechanical2 + thermo2

$mgh + 0 = \frac{1}{2}mv^{2} + E_{th2}$

There are several ways to describe this that work. The key is to pick one and stick to it! Decide what you want, and just do that. Don't mix-and-match.

 

[etotheipi]

There are several ways to describe this that work. The key is to pick one and stick to it! Decide what you want, and just do that. Don't mix-and-match.

Right, what you've said makes sense, though I was under the impression that a valid formulation was

$W_{ext} + Q = \Delta E = \Delta E_{th} + \Delta E_{mech}$

the interpretation being the total energy transferred across the system boundary equals the change in total energy (indiscriminate for now against mechanical/thermal).

 

[russ_watters]

Right, what you've said makes sense, though I was under the impression that a valid formulation was

$W_{ext} + Q = \Delta E = \Delta E_{th} + \Delta E_{mech}$

the interpretation being the total energy transferred across the system boundary equals the change in total energy (indiscriminate for now against mechanical/thermal).

It is. But you're not being consistent in your system definition when doing it that way. You're counting the friction a heat transfer into the system, when it isn't (it's an internal conversion). Q = 0 in that equation.

 

[etotheipi]

It is. But you're not being consistent in your system definition. You're counting the friction a heat transfer into the system, when it isn't. Q = 0 in that equation.

I was taking friction to be part of $W_{ext}$, since I presumed according to how I defined my system (only the block) that the force of the wedge on the block was an external force.

Your approach gives the right answer, though you have treated the frictional force of the wedge on the block as internal. And since the work done by internal friction is the negative of the change of thermal energy of the system, you only end up with the one $f_k d$ term on the RHS which is what we're after.

I do sincerely apologise to everybody who are (rightfully!) fed up, but I really am trying to understand! Why doesn't friction count as an external force to the block system in this case?

 

[russ_watters]

I was taking friction to be part of $W_{ext}$, since I presumed according to how I defined my system (only the block) that the force of the wedge on the block was an external force.

That's wrong, for that formula. The friction is either an external mechanical work input(output) or an internal energy conversion. It can't be both at the same time. You have to pick one.

 

[russ_watters]

Pay attention to the sign on that friction, too. It's negative on the input side. That means it's removing energy from the system, not adding it. How can the block get hotter if friction is reducing the system's energy?

Or to say it another way: if it is a negative input, it is removing energy from the system. If it is a positive output, it can still be considered removing energy, or it is an in internal conversion if you want to keep it. You have to pick one side of the equation to put it on; you can't have both.

 

[etotheipi]

That's wrong, for that formula. The friction is either an external mechanical work input(output) or an internal energy conversion. It can't be both at the same time. You have to pick one.

I see!

I think that's clicked now.

It's almost exactly like not double-counting the work done by gravity and the gravitational potential energy in an energy balance.

If I understand correctly, the total (external + internal) work done by friction on or within a chosen system can be restated as the negative of the change of thermal energy of that system.

So a thermal energy term is really just introduced as a means of keeping track of the work done by friction in a system. In other words, if we include a thermal energy term then friction no longer counts as an external force since its contribution is already accounted for by the thermal energy term.

Thank you for your patience! This has got to take the cake for the longest thread I've been involved in...

 

[russ_watters]

So a thermal energy term is really just introduced as a means of keeping track of the work done by friction in a system. In other words, if we include a thermal energy term then friction no longer counts as an external force since its contribution is already accounted for by the thermal energy term.

Right. And as said, usually it's done the first way - ignoring the thermal energy - if what you really want is the final kinetic energy.

And actually, if what you are most interested in is the thermal energy, you don't need the potential or kinetic energy at all. That simple equation of friction energy equals heat is all you need. Essentially it is saying the box is stationary and a rough surface slides under it, transferring heat.

 

[etotheipi]

Awesome, I'm glad I finally got it.

There's nothing more soul-crushing than getting hopelessly stuck at a topic which one first encounters at maybe 12 years of age . Talk about getting knocked down a few pegs...

 

[Dale]

The fact I can't get it to work (no pun intended) suggests a flaw in my understanding.

I don’t know what is the issue here. If you want to solve for the mechanical quantities you choose a method that focuses on the mechanical quantities. If you want to solve for the thermodynamic quantities then you choose a method that focuses on the thermodynamic quantities. Why should you expect a method focusing on thermodynamic quantities be well suited for mechanical quantities or vice versa?

Choose what you want to know and then use the corresponding method.

I think you're right, but I can't exactly determine why. The goal is to determine the final KE of the block using only a system which contains only the block.

If that is the goal then the thermal energy is completely irrelevant.

 

[etotheipi]

I don’t know what is the issue here. If you want to solve for the mechanical quantities you choose a method that focuses on the mechanical quantities. If you want to solve for the thermodynamic quantities then you choose a method that focuses on the thermodynamic quantities. Why should you expect a method focusing on thermodynamic quantities be well suited for mechanical quantities or vice versa.

Yes, retrospectively a lot of what you were saying in earlier posts makes a lot more sense to me now.

I wasn't aware that work done by friction and thermal energy were identical/interchangeable terms.

 

[A.T.]

I wasn't aware that work done by friction and thermal energy were identical/interchangeable terms.

These aren't identical/interchangeable terms. Work done by friction represents a transfer of mechanical energy. The losses in this transfer represent dissipation of mechanical energy to thermal energy.

 

[russ_watters]

I wasn't aware that work done by friction and thermal energy were identical/interchangeable terms.

Equal does not mean identical.

 

[etotheipi]

Apologies, I was being sloppy. I intended to say that friction doesn't count as an external force in the energy balance/FLT even if it is an external force to the system.

That is to say, for any given system, $$\sum_i {\vec{f}_{k}}_{i} \cdot \vec{d} = - \Delta E_{th} = \Delta E_{mech}$$ where all of the $f_{k}$'s are individual frictional forces acting on that system. However, each of these could be either internal or external.

It also has the consequence that friction, even external friction, does not change the total energy of a given system. This is unlike other external non-conservative forces which, evidently, need to be in $W_{ext}$ in the FLT.

 

[Dale]

It also has the consequence that friction, even external friction, does not change the total energy of a given system.

This is not generally correct. It is only correct in the special case that 100% of the thermal energy produced at the interface goes back into the system.

In general, kinetic friction results in mechanical power being transformed into heat at the contact patch. If one surface has a higher temperature than the other then more of the heat will flow into the other surface. If one surface has a higher thermal conductivity than the other then more heat will flow into that surface. The question about where the heat goes is a thermodynamics question and cannot be determined based on the mechanics. All the mechanics tells us is how much mechanical power is dissipated at the contact patch, not where it goes after.

 

[etotheipi]

This is not generally correct. It is only correct in the special case that 100% of the thermal energy produced at the interface goes back into the system.

In general, kinetic friction results in mechanical power being transformed into heat at the contact patch. If one surface has a higher temperature than the other then more of the heat will flow into the other surface. If one surface has a higher thermal conductivity than the other then more heat will flow into that surface. The question about where the heat goes is a thermodynamics question and cannot be determined based on the mechanics. All the mechanics tells us is how much mechanical power is dissipated at the contact patch.

Sure, I'll have a closer look at this tomorrow and see if I can try and deal with scenarios where heat transfer is permitted between the different bodies.

Friction is just slightly weird. I wonder if we need to attach a note to the first law of thermodynamics, saying

$Q + W = \Delta E = \Delta E_{mech} + \Delta E_{th}$

where $W$ is the total external work crossing the system boundary excluding external work done by kinetic friction.

That's the only way I can reconcile everything in my mind.

 

[Dale]

I wonder if we need to attach a note to the first law of thermodynamics, saying

$Q + W = \Delta E = \Delta E_{mech} + \Delta E_{th}$

where $W$ is the total external work crossing the system boundary excluding external work done by kinetic friction.

Definitely not.

Friction is just slightly weird

I think that a large part of that feeling is due to the fact that people don’t generally learn how mechanical power flows. It is really quite straightforward if you focus on power which is $P=\vec f \cdot \vec v$. With that focus you can analyze something like a rope where mechanical power goes in one end and out the other, you can use the same analysis to understand when friction does positive work, and you can apply the same analysis to see that mechanical power goes into the contact patch and does not leave with dynamic friction. The mechanical power that disappears at the contact patch is the heat produced at the contact patch. It is all very straightforward if it is approached systematically with the same concept that works for all mechanical systems.

 

[etotheipi]

I found this which might be of interest:

https://brucesherwood.net/wp-content/uploads/2017/06/Friction1984.pdf

I haven't had time to fully read it through but it seems to suggest the effective displacement of the frictional force is lower. That allows us to use the FLT formulation sort of similar to the one in post #79.

It would really become $$mgh - f_k d_{eff}= \Delta E = \Delta E_{mech} + \Delta E_{th}$$ where $d_{eff} < d$

Here is his treatment of this exact problem:

So I believe we can always analyse mechanical power flows using the centre of mass displacement; that is, the total evolved thermal energy at any interface will be the product of the frictional force and relative displacement (i.e. total centre of mass work at that interface). But this won't equal the true work done by friction, on either of the bodies.

It seems slightly overkill, but it makes sense to me.

 

[A.T.]

I found this which might be of interest:

https://brucesherwood.net/wp-content/uploads/2017/06/Friction1984.pdf

I haven't had time to fully read it through but it seems to suggest the effective displacement of the frictional force is lower.

How is that effective displacement defined?

 

[etotheipi]

How is that effective displacement defined?

I can't see how it's defined at all. For their first example of pulling a block across another block, it's half the actual displacement. But I don't think that generalises at all, they in fact worked it out by using the CM equation first.

I think the point is that we can't actually determine the work done by friction. So we can instead use the CM equation to find velocity, or treat the system as a whole and note that the mechanical energy dissipated at the interface equals the total CM work done by friction at the interface. But neither of these is actually giving us the actual/thermodynamic work done by friction.

That's what I've concluded from reading it through.

 

[jbriggs444]

I can't see how it's defined at all

Then you cannot use it.

 

[Dale]

I think the point is that we can't actually determine the work done by friction.

That is wrong. We can determine the work done by friction exactly like we determine the work done by any force, as I described above.

 

[etotheipi]

That is wrong. We can determine the work done by friction exactly like we determine the work done by any force, as I described above.

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)?

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.

 

[jbriggs444]

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

Until you define exactly what $\vec{v}$ denotes, it could be either real work or center of mass work. The formula for both is identical.

For center of mass work, the $\vec{v}$ refers to the velocity of the center of mass. For real work, the $\vec{v}$ refers to the velocity of the target material at the point where the force is applied.

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.

Without reading the argument in detail, I hesitate to say what I think about such a claim. It is somewhat plausible that two contacting surfaces would lose only part of the input mechanical energy as irrecoverable thermal output and leave some fraction as available energy (ordered vibrations, thermal gradients and such) and that this is what Sherwood is trying to get at.

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.

 

[etotheipi]

There is another example he gives which provides something else to chew on. We consider a block being pulled across a rough surface, at a constant velocity, by an external force $P$. We take the system to be just the block.

If the force of friction is $f_k$, the constant velocity condition gives $f_k = P$. The centre of mass equation gives

$Pd - f_k d = 0 = \frac{1}{2} m \Delta v^{2}$, as expected

However, we might then wonder about the heat dissipated at the interface, since the block will heat up! If we apply the FLT to only the block, then we're left wondering about where this missing term is. Instead, the actual work done by friction must be less:

$Pd - f_k d_{eff} = \frac{1}{2} m \Delta v^{2} + \Delta E_{th}$

 

[etotheipi]

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.

I think this is indeed the increase in thermal energy, but not the real work done. I might well be wrong!

 

[jbriggs444]

There is another example he gives which provides something else to chew on. We consider a block being pulled across a rough surface, at a constant velocity, by an external force $P$. We take the system to be just the block.

If the force of friction is $f_k$, the constant velocity condition gives $f_k = P$. The centre of mass equation gives

$Pd - f_k d = 0 = \frac{1}{2} m \Delta v^{2}$, as expected

However, we might then wonder about the heat dissipated at the interface, since the block will heat up! If we apply the FLT to only the block, then we're left wondering about where this missing term is. Instead, the actual work done by friction must be less:

$Pd - f_k d_{eff} = \frac{1}{2} m \Delta v^{2} + \Delta E_{th}$

This amounts to a redefinition of "the work done by friction".

Instead of "the mechanical work being drained into thermal energy at the interface", it now means "the energy removed from the block at the interface". [And instead of being an invariant, it is now a frame-relative quantity]

This clarifies the meaning of "effective displacement". It is a fudge factor used to make the energy balance work out -- devoid of physical meaning.

 

[jbriggs444]

I think this is indeed the increase in thermal energy, but not the real work done. I might well be wrong!

Conservation of energy. The two are equal.

 

[etotheipi]

Conservation of energy. The two are equal.

I meant to say that $f_k d$ is the total decrease 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.

 

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