Modelling assumptions when friction is involved

[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.