Modelling assumptions when friction is involved

In summary, the conversation discusses the relationship between friction and dissipation of mechanical energy into thermal energy in mechanics problems. It is mentioned that in cases where friction exists, some of the thermal energy may be dissipated into other components, such as the wedge or surrounding air, but this is not relevant to the kinematic equations. The conversation also touches on the issue of determining which bodies gain thermal energy and how to calculate this energy, with the conclusion that it is only important to consider the kinetic energy of the block and not where the thermal energy goes.
  • #1
etotheipi
Whenever friction exists within a mechanics problem, there must be some dissipation of mechanical energy into thermal energy. However, I'm not sure how we determine which bodies can or cannot possesses thermal energy.

Suppose we consider the case of a block sliding down a rough wedge, which is fixed to the Earth. Treating the whole Earth-wedge-block system gives something along the lines of ##U_{1} = U_{2} + T_{block} + E_{th}##.

There are essentially two options, either the wedge gains all of this thermal energy, or it is split between the wedge and the block. If the block is specified to be rigid, the former applies and we might then say that ##E_{th} = f_{k}d## if ##f_{k}## is the frictional force applied by the block on the wedge. If not, I'm not entirely sure how it's possible to solve for the final velocity of the block.

My guess is that it is assumed in any mechanics problem that the quote-and-quote "main" object in the question is rigid (in this example, the block), and all of this thermal energy is dissipated in other components. I wondered whether anyone could think of any exceptions? On perhaps another level, does it matter/make any difference to the answer where this thermal energy is dissipated? I'm inclined to say yes, simply because I don't believe there is any way of computing ##E_{th}## unless it ends up in one body only.

Please do let me know if this is confused. Thanks!
 
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  • #2
etotheipi said:
On perhaps another level, does it matter/make any difference to the answer where this thermal energy is dissipated?

As far as the purely kinetic model is concerned it makes no difference how or where the energy is lost. It makes no difference whether it is thermal energy inside the bodies, thermal energy of the surrounding air or sound waves in the air - or even radiated away as EM radiation.

These dissipative processes themselves are outside the scope of the model.

As you improve your model, the form of the equations, e.g. involving coefficient of drag and dependence on the square of the velocity, may reflect a deeper knowledge of how the dissipative processes operate. But, the details of the processes themselves remain irrelevant to the kinematic equations.
 
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  • #3
PeroK said:
As far as the purely kinetic model is concerned it makes no difference how or where the energy is lost. It makes no difference whether it is thermal energy inside the bodies, thermal energy of the surrounding air or sound waves in the air - or even radiated away as EM radiation.

In the instance where only the block gains thermal energy, ##E_{th} = f_{k}d##. Since the mode of dissipation is then unimportant, is that to say that the total ##\Delta E_{th}## will always be ##f_{k}d## (even if, for instance, there are a pair frictional forces contributing to it)?

If we wanted to solve the system then the two routes would either be work-energy on the block ##W_{grav} + W_{fr} = \Delta T##, but this would only apply if the block were rigid.

Or if the block is no longer constrained to be rigid, we could try a more general energy conservation for the entire system, in which case we need to come up with a number for ##\Delta E_{th}## somehow. It's not immediately obvious how to do this, thought it seems sensible to also call it ##W_{fr}## so that the block ends up with the same final translational speed.
 
  • #4
etotheipi said:
Or if the block is no longer constrained to be rigid, we could try a more general energy conservation for the entire system, in which case we need to come up with a number for ##\Delta E_{th}## somehow. It's not immediately obvious how to do this, thought it seems sensible to also call it ##W_{fr}## so that the block ends up with the same final translational speed.
If what you want to know is "how far does the block slide before coming to rest", then, as @PeroK pointed out, the resulting thermal energy is irrelevant.

If what you want to know is "how hot does the block get" then you are out of luck. Conservation laws will not get you there.
 
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  • #5
jbriggs444 said:
If what you want to know is "how far does the block slide before coming to rest", then, as @PeroK pointed out, the resulting thermal energy is irrelevant.

I'm only really concerned about the kinetic energy of the block at the bottom of the ramp, and where everything fits into a conservation of energy expression.

The block is acted upon by weight and friction, the wedge by friction (and weight, but this is irrelevant). In the case of the block, the frictional force is responsible for reducing its kinetic energy and perhaps increasing its thermal energy. For the wedge, which I will assume is fixed, friction only increases its thermal energy.

In this scenario, work-energy doesn't really help us so our only option is conservation of energy, ##\Delta U + \Delta T + \Delta E_{th} = 0##. Then we've got to figure out what ##\Delta E_{th}## is.

I'm slightly at a loss on how to calculate that number, since there are two frictional forces contributing to it. My "instinct" is that it will still just be ##f_{k}d##, but I can't really rationalise this. If the block were rigid and all the thermal energy went to the wedge, then evidently the work done on the wedge would equal ##\Delta E_{th} = f_k d##. But now we've got two objects to worry about!

This is only a problem when the frictional forces are internal to the chosen system.
 
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  • #6
etotheipi said:
I'm only really concerned about the kinetic energy of the block at the bottom of the ramp, and where everything fits into a conservation of energy expression.
Then which object heats up is irrelevant. You can correctly calculate the KE of the block without concern about where the thermal energy goes.
 
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  • #7
Dale said:
Then which object heats up is irrelevant. You can correctly calculate the KE of the block without concern about where the thermal energy goes.

But if I take the whole configuration to be a system, then what do I take to be ##\Delta E_{th}##? In the case that both are capable of possessing thermal energy, I could take the block and wedge individually in turn as systems

Block: ##mgh - f_{k}d = \frac{1}{2}mv^{2} + \Delta E_{th, block}##.

Wedge: ##f_{k}d = \Delta E_{th, wedge}##

Adding these just causes the ##f_k d## terms to cancel,

##mgh = \frac{1}{2}mv^{2} + \Delta E_{th, tot}##
 
  • #8
etotheipi said:
But if I take the whole configuration to be a system, then what do I take to be ##\Delta E_{th}##? In the case that both are capable of possessing thermal energy, I could take the block and wedge individually in turn as systems

Block: ##mgh - f_{k}d = \frac{1}{2}mv^{2} + \Delta E_{th, block}##.

Wedge: ##f_{k}d = \Delta E_{th, wedge}##

Adding these just causes the ##f_k d## terms to cancel,

##mgh = \frac{1}{2}mv^{2} + \Delta E_{tot}##

I propose a different model, where energy is dissipated as EM radiation, rather than internal heat. Let's say that for every ##d## of relative motion between the surfaces, a total energy of ##E## is radiated as EM radiation. The ultimate source of this energy is the KE of the block (or at least the reduction in relative motion of the surfaces).

This results in an effective force on the block of ##f = E/d##. By experiment we find that, in fact, ##E/d = \mu N##, where ##\mu## is some constant determined by the surfaces and ##N## is the normal force between the surfaces.

Then you don't have to worry about thermal energy any more. Instead, if you want to check conservation of energy, you need to measure the energy of all the EM radiation. But that doesn't matter, because whether you measure it or not the energy is lost to the block. It doesn't really matter to the model whether that energy is still in our universe or not. Even if it is irretrievably gone, the simple kinetic model still works.
 
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  • #9
The friction energy goes to another dimension.
It is outside what you have chosen to call your system. Of course it is physically "inside" the materials in question, but it is outside the configuration space you have chosen to describe the salient features of your system
 
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  • #10
PeroK said:
This results in an effective force on the block of ##f = E/d##. By experiment we find that, in fact, ##E/d = \mu N##, where ##\mu## is some constant determined by the surfaces and ##N## is the normal force between the surfaces.

Ah okay, so first things first I should in theory replace ##E_{th}## with a more general term encapsulating any other medium that energy could be dissipated into.

But even if I limit that term to just thermal energy for simplicity, what I believe you are saying is that ##\Delta E_{th}## would equal the product of the frictional force and the distance moved.

That is indeed consistent with what you'd get if we take the model where only the wedge can gain thermal energy. It's a shame it's empirical but at least it makes sense.
 
  • #11
etotheipi said:
That is indeed consistent with what you'd get if we take the model where only the wedge can gain thermal energy.
Why only the wedge? The model says nothing about where the dissipated energy goes to.
 
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  • #12
A.T. said:
Why only the wedge? The model says nothing about where the dissipated energy goes to.

My impression was that if we let the block be rigid but allow only the wedge to possesses thermal energy, then we end up with the following system

Block: ##mgh - f_{k}d = \frac{1}{2}mv^{2}##.

Wedge: ##f_{k}d = \Delta E_{th, wedge}##

And then if we take the system as both the block and the wedge, ##\Delta E_{th, tot} = \Delta E_{th, wedge} = f_k d##. So this setup let's us calculate the thermal energy explicitly by considering work done on the wedge, since that's the only term on the RHS.

If we give both of them thermal energies, then the total amount will be split between them. This was the part I was unsure about. Going with the idea of friction as a measure of energy dissipated per unit distance, it appears that this total thermal energy will be the same in all scenarios (e.g. 100%/0%, 50%/50% etc.)

It appears to be the case that ##\Delta E_{th, etc.}## dissipated due to the action of a pair of frictional forces will be ##f_{k}d##.
 
  • #13
etotheipi said:
It appears to be the case that ##\Delta E_{th, etc.}## dissipated due to the action of a pair of frictional forces will be ##f_{k}d##.
You can also sum the work done the two frictional forces (with the proper signs) to get the mechanical energy dissipated or generated at the interface.
 
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  • #14
A.T. said:
You can also sum the work done the two frictional forces (with the proper signs) to get the mechanical energy dissipated or generated at the interface.

Though wouldn't that approach just give ##f_{k}d - f_{k}d = 0##?
 
  • #15
PeroK said:
I propose a different model, where energy is dissipated as EM radiation, rather than internal heat. Let's say that for every dd of relative motion between the surfaces, a total energy of ##E## is radiated as EM radiation. The ultimate source of this energy is the KE of the block (or at least the reduction in relative motion of the surfaces).
Yes, a total energy of ##E## is radiated away as EM radiation but that happens eventually. I believe it is more appropriate to consider the details of power dissipation ##\dfrac{dK}{dt}##.
 
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  • #16
etotheipi said:
My impression was that if we let the block be rigid but allow only the wedge to possesses thermal energy, then we end up with the following system

Block: ##mgh - f_{k}d = \frac{1}{2}mv^{2}##.

Wedge: ##f_{k}d = \Delta E_{th, wedge}##

And then if we take the system as both the block and the wedge, ##\Delta E_{th, tot} = \Delta E_{th, wedge} = f_k d##. So this setup let's us calculate the thermal energy explicitly by considering work done on the wedge, since that's the only term on the RHS.

If we give both of them thermal energies, then the total amount will be split between them. This was the part I was unsure about. Going with the idea of friction as a measure of energy dissipated per unit distance, it appears that this total thermal energy will be the same in all scenarios (e.g. 100%/0%, 50%/50% etc.)

It appears to be the case that ##\Delta E_{th, etc.}## dissipated due to the action of a pair of frictional forces will be ##f_{k}d##.
A useful exercise would be to consider a block thrown onto the surface of a trolley that is free to roll without loss of energy.

Then compare this with a block thrown onto a table that is fixed to the Earth, effectively.

The other consideration you are missing is momentum. We do not have conservation of momentum in the case of friction on a fixed surface. Even if you deal with thermal energy to your satisfaction, you have conservation of momentum to worry about.
 
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  • #17
PeroK said:
A useful exercise would be to consider a block thrown onto the surface of a trolley that is free to roll without loss of energy.

Then compare this with a block thrown onto a table that is fixed to the Earth, effectively.

I suppose with rolling there is no motion of the contact point so the work done by friction is zero, and the trolley would just carry on forever (assuming no rolling resistance).

Whilst the block thrown onto the table would eventually come to rest wrt the surface of the Earth. If we make the approximation that the moment of inertia of the Earth is so large that we can assume the Earth is fixed, we might then just say the kinetic energy of the block is dissipated via some frictional mechanism until it comes to rest.

PeroK said:
The other consideration you are missing is momentum. We do not have conservation of momentum in the case of friction on a fixed surface. Even if you deal with thermal energy to your satisfaction, you have conservation of momentum to worry about.

If the system is the block then momentum isn't conserved since the external force is non-zero. In an inertial frame in space, momentum is definitely conserved. In the lab frame fixed wrt the table, momentum definitely isn't conserved, as you say, since the surface of the Earth is always at rest relative to us (this frame is non-inertial).
 
  • #18
etotheipi said:
I suppose with rolling there is no motion of the contact point so the work done by friction is zero,

That's not what would happen.
 
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  • #19
Ps You can set up a problem for yourself: trollley of mass ##M##, block of mass ##m##, thrown onto trolley with initial velocity ##v## and coefficient of friction ##\mu##.

What is the final velocity of the truck?

Where have you seen this equation before? So, what type of event has taken place?
 
  • #20
PeroK said:
That's not what would happen.

Hmm. For the two to coalesce the initial collision would have to be inelastic, so right off the bat some energy has to be dissipated somewhere. I'm guessing thermal in the trolley/block.

Then we're left with an object rolling at some speed relative to the surface of the Earth. The rolling wheel encounters no friction on a flat surface. Air resistance against the trolley, maybe.
 
  • #21
PeroK said:
Ps You can set up a problem for yourself: trollley of mass ##M##, block of mass ##m##, thrown onto trolley with initial velocity ##v## and coefficient of friction ##\mu##.

What is the final velocity of the truck?

Where have you seen this equation before? So, what type of event has taken place?

Right I won't be a minute
 
  • #22
etotheipi said:
Hmm. For the two to coalesce the initial collision would have to be inelastic, so right off the bat some energy has to be dissipated somewhere. I'm guessing thermal in the trolley/block.

Then we're left with an object rolling at some speed relative to the surface of the Earth. The rolling wheel encounters no friction on a flat surface. Air resistance against the trolley, maybe.

This is the problem. We're not worried about air resistance or rolling resistance (or the mechanism of thermal energy dissipation). You are not seeing the wood for the trees.

Part of what you have said gives you a shortcut to the answer.
 
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  • #23
PeroK said:
This is the problem. We're not worried about air resistance or rolling resistance (or the mechanism of thermal energy dissipation). You are not seeing the wood for the trees.

Part of what you have said gives you a shortcut to the answer.

I lied, I might be a little more than a minute...

Initially went to momentum conservation, then realized it probably wasn't what you're looking for. Either way, this gives ##v = \frac{m}{M+m}u##.

I'm trying to think about how the coefficient fits into it. For the little mass

##-\mu N = m\frac{dv}{dt}##

whilst the relative velocity of the little mass wrt the trolley is ##v - v_{t}## where ##v_{t}## is also a function of time, such that

##\mu N = M\frac{dv_t}{dt}##

Then you've got something like

##\int_{v}^{v_{f}} m dv = \int_{0}^{v_{f}} -M dv_t \implies mv_{f} -mv = -Mv_{f} \implies v_{f} = \frac{m+M}{m}v##

Or

##v_{f} = \frac{m}{M+m}v##
 
  • #24
etotheipi said:
I lied, I might be a little more than a minute...

Initially went to momentum conservation, then realized it probably wasn't what you're looking for. Either way, this gives ##v = \frac{m}{M+m}u##.

I'm trying to think about how the coefficient fits into it. For the little mass

##-\mu N = m\frac{dv}{dt}##

whilst the relative velocity of the little mass wrt the trolley is ##v - v_{t}## where ##v_{t}## is also a function of time, such that

##\mu N = M\frac{dv_t}{dt}##

Then you've got something like

##\int_{v}^{v_{f}} m dv = \int_{0}^{v_{f}} -M dv_t \implies mv_{f} -mv = -Mv_{f} \implies v_{f} = \frac{m+M}{m}v##

I must have made a mistake somewhere...
Yes, obviously that term should be inverted. The point is that we have a totally inelastic collision. It doesn't matter if the collision is by friction, hitting part of the trolley or a combination of both. The final velocity is determined by conservation of momentum.
 
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  • #25
Cool, thanks for your help.

At the end of the day lots of this energy stuff can be figured out with common sense, but I'm trying to be intentionally pedantic just to try and formalise a lot of the stuff I've covered over the past few years.

I'm finding energy especially is one of those topics that's a constant cycle of learning something and then realising that interpretation is limited, then needing to sort of "re-learn" a new framework, and then that same thing happening again. I guess that applies to a lot of Physics though. "If it ain't broke then don't fix it" works for a little while and then I run into problems and need to spend a lot of time just relearning the topic from the ground up in a different way.
 
  • #26
etotheipi said:
Cool, thanks for your help.

At the end of the day lots of this energy stuff can be figured out with common sense, but I'm trying to be intentionally pedantic just to try and formalise a lot of the stuff I've covered over the past few years.

I'm finding energy especially is one of those topics that's a constant cycle of learning something and then realising that interpretation is limited, then needing to sort of "re-learn" a new framework, and then that same thing happening again. I guess that applies to a lot of Physics though. "If it ain't broke then don't fix it" works for a little while and then I run into problems and need to spend a lot of time just relearning the topic from the ground up in a different way.

Possibly one interesting detail that as come out of this is: If we let the second object move, then the frictional force accelerates it (no loss of energy directly). But, the force decelerates the incoming object (which is where the energy loss comes from).

If you imagine the friction with the wedge accelerating the wedge-Earth a tiny amount, then that interaction incurs no energy loss. That's why you don't double count the friction force and get twice the energy loss.
 
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  • #27
etotheipi said:
Though wouldn't that approach just give ##f_{k}d - f_{k}d = 0##?
No, because the displacement is different for the two contact patches, if they are in relative motion (sliding).
 
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  • #28
A.T. said:
No, because the displacement is different for the two contact patches, if they are in relative motion (sliding).

I'm not sure I follow. We might consider making the block arbitrarily small, the point of application of both frictional forces (block on wedge and wedge on block) would be the same point. Then the displacements of both forces should be equal.
 
  • #29
etotheipi said:
I'm not sure I follow. We might consider making the block arbitrarily small, the point of application of both frictional forces (block on wedge and wedge on block) would be the same point. Then the displacements of both forces should be equal.
No, the displacements are different, because the physical points of force application (the material parts of each body were the force is applied) are moving relative to each other.
 
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  • #30
A.T. said:
No, the displacements are different, because the physical points of force application (the material parts of each body were the force is applied) are moving relative to each other.

I've realized I've drawn cylinders instead of a block (please ignore this part, I didn't intend for any rolling, just sliding!). Though I can't see why the two displacements would be any different.

IMG_4841.JPG
 
  • #31
etotheipi said:
Though I can't see why the two displacements would be any different.
Displacements are integrals of velocity, but you can also consider the velocities and thus the instantaneous power, instead of work. Since the velocities of the contact patches are different, the equal but opposite kinetic friction forces are not doing equal but opposite work/time (power). The discrepancy represents the dissipated energy.
 
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  • #32
A.T. said:
Since the velocities of the contact patches are different

I might need to come back to this in the morning in the case that I'm missing something obvious since it seems I'm stuck in a bit of a pantomime-style "Oh yes it is!", "Oh no it isn't!" situation, but I really struggling to figure this out.

To me it seems there is only one common point of contact between our arbitrarily small block and the wedge, and the velocity of the contact point is obviously equal to the velocity of the block. At all points in the motion, the two opposite frictional forces act at the same point (the contact point). Both undergo the same displacement by a vector ##\vec{d}## as measured in the Earth frame in which the wedge is fixed.

I do apologise if I'm missing the point!
 
  • #33
etotheipi said:
I do apologise if I'm missing the point!
The work done by wedge on block is determined by the force of wedge on block multiplied by the motion of the block material at the contact point.

The work done by block on wedge is determined by the force of block on wedge multiplied by the motion of the wedge material at the contact point.

The forces are equal and opposite, of course. That's Newton's third law. But no matter what frame of reference we choose, the motion of the block material and the motion of the wedge material at the contact point will differ. Otherwise we are not dealing with kinetic friction.

[One should draw sliding objects as rectangles rather than as circles to avoid the implication that rolling and static friction are involved]
 
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  • #34
jbriggs444 said:
The forces are equal and opposite, of course. That's Newton's third law. But no matter what frame of reference we choose, the motion of the block material and the motion of the wedge material at the contact point will differ. Otherwise we are not dealing with kinetic friction.

Ah, okay. I was working under the assumption that work done by a force in a certain frame of reference was the dot product of the force and the displacement of the point of application of the force.

The two bodies are certainly in relative motion. Though the work done as calculated by the definition I gave as opposed to the ##P = \vec{F} \cdot \vec{v}## calculation are then apparently different. I don't understand how this can be the case.
 
  • #35
etotheipi said:
I was working under the assumption that work done by a force in a certain frame of reference was the dot product of the force and the displacement of the point of application of the force.
Indeed, that is a common misunderstanding. It is the displacement of the target material at the point of application that matters.

The point of application cannot be in a fixed position relative to both of two slipping surfaces.
 
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