Modelling assumptions when friction is involved

[A.T.]

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.

 

[etotheipi]

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!

 

[jbriggs444]

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]

 

[etotheipi]

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.

 

[jbriggs444]

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.

 

[etotheipi]

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.

That makes life a lot more complicated!

I assume by motion of the 'block material' or 'wedge material' you mean at a microscopic level.

How is it then possible to calculate anything that involves kinetic friction? Since in this case $W=f_{k} d$ no longer has any meaning.

 

[jbriggs444]

I assume by motion of the 'block material' or 'wedge material' you mean at a microscopic level.

No. I mean at an ordinary macroscopic level.

If you have a tire skidding down the pavement, the material of the tire and the material of the road are moving relative to one another. We do not need to delve into the microscopic details of that motion in order to calculate the energy that is dissipated. Instead, we do as @A.T. has suggested:

1. Pick a frame of reference.
2. Using this frame of reference, multiply the motion of the road material by the force of tire on road.
3. Using this same frame of reference, multiply the motion of tire material by force of road on tire.
4. Add the results together.

That's how much mechanical energy has been dissipated. No matter what frame of reference you choose, the result will be the same. The motion of the point of application relative to the frame of reference is irrelevant.

 

[A.T.]

That makes life a lot more complicated!

You seem to be making things more complicated than necessary.

 

[etotheipi]

No. I mean at an ordinary macroscopic level.

If you have a tire skidding down the pavement, the material of the tire and the material of the road are moving relative to one another. We do not need to delve into the microscopic details of that motion in order to calculate the energy that is dissipated. Instead, we do as @A.T. has suggested:

1. Pick a frame of reference.
2. Using this frame of reference, multiply the motion of the road material by the force of tire on road.
3. Using this same frame of reference, multiply the motion of tire material by force of road on tire.
4. Add the results together.

That's how much mechanical energy has been dissipated.

Okay, I finally see what you mean. My head sort of hurts, mainly because I've been living a lie.

Essentially we're saying that zero work is done on the wedge. That's... really odd.

Then everything you say makes sense, since the change in thermal energy would naturally be the total work done by friction which would indeed, in this case, be $f_k d$.

 

[jbriggs444]

Essentially we're saying that zero work is done on the wedge. That's... really odd.

If the wedge doesn't move then no work has been done on the wedge.

If you change to a different reference frame then friction may do work on the wedge. That work can be positive or negative depending on which way the wedge is moving in that reference frame.

Neither Work nor Energy are invariant quantities.

The energy dissipated by kinetic friction is an invariant quantity, however.

 

[etotheipi]

It does throw up a few new questions, the answers to which you might have already given earlier in the thread; I'd need to have a closer look at some of the posts above, since I presume the answer lies in my strange conviction that we must include a thermal energy term.

If the system is only the wedge, the first law of thermodynamics becomes $Q + W = \Delta E = 0$; that is to say the change in thermal energy of the wedge is zero.

My instinct is that the eventual change in thermal energy of the wedge must be due to a transfer of heat from the block to the wedge during its travel, and not due to the block working on the wedge.

If the wedge doesn't move then no work has been done on the wedge.

You and @A.T. have fully convinced me, it's just a little strange since I'd never thought of it like that before. My "mantra" was force multiplied by displacement of point of application. This, in hindsight, always works when the actual material at the point of contact is moving with the force (i.e. someone pushing a rigid body around an axis), but evidently doesn't work in scenarios like these.

Displacement of the material at the contact point. Got it.

Sorry about this, I am conscious of how annoying I'm being...

 

[jbriggs444]

My "mantra" was force multiplied by displacement of point of application. This, in hindsight, always works when the actual material at the point of contact is moving with the force.

Yes indeed. One place where this really becomes important is with rotation. Wheels on roads and such.

 

[Dale]

But if I take the whole configuration to be a system, then what do I take to be ΔEthΔEth\Delta E_{th}?

For the whole system that is just $f_k d$. That is all that you need to know to calculate the KE of the block using conservation of energy.

In the case that both are capable of possessing thermal energy, I could take the block and wedge individually in turn as systems

That is a different problem. If you are ONLY interested in the KE of the block, as you said earlier, then you wouldn’t do this approach. So this is the same scenario but a different problem. Now, you are interested in the division of thermal energy between the two objects. That is no longer just a mechanics problem, but a thermodynamics problem too.

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.)

That would be the thermodynamics part of the problem, but I highly doubt that it will be a simple split. I would assume that, for similar materials in both sides that the slower the block is moved the closer to 50/50 split you get, but the faster you go the larger fraction of the total energy will go into the ramp.

 

[Dale]

How is it then possible to calculate anything that involves kinetic friction? Since in this case W=fkdW=fkdW=f_{k} d no longer has any meaning.

It is actually easier to work with power: $P=f_k \cdot v$. Note that by definition the $v$ is different for the two sides during kinetic friction. This results in some amount of mechanical power that goes into the contact patch that does not go out. This is the mechanical power that is converted to thermal power.

 

[etotheipi]

That is a different problem. If you are ONLY interested in the KE of the block, as you said earlier, then you wouldn’t do this approach. So this is the same scenario but a different problem. Now, you are interested in the division of thermal energy between the two objects. That is no longer just a mechanics problem, but a thermodynamics problem too.

Right, sure. So I suppose the conclusion is that we can't find the final KE of a non-rigid block if we consider the block as a single system.

Whilst we can if we take the wedge-block system and note that the change in thermal energy is the negative of the total work done by internal kinetic frictional forces.

The algebra's a bit weird though. For the block-wedge system

$mgh = \Delta T + \Delta E_{th} = \Delta T + f_k d$

But for the block and wedge individually,

Block: $mgh - f_{k}d = \Delta T + \Delta E_{th, block}$
Wedge $0 = \Delta E_{th, wedge}$

Since the two models have to be consistent, we can only then conclude that $\Delta E_{th, block} = 0$. That can't be right, though, since we've somehow determined that the change in thermal energy of the block + the wedge is zero, whilst we know it to be $f_k d$.

 

[jbriggs444]

Right, sure. So I suppose the conclusion is that we can't find the final KE of a non-rigid block if we consider the block as a single system.

What do you mean by "KE of a non-rigid block"?
What are you using as inputs to an attempted calculation of "KE of a non-rigid block?"
What constraints are you imposing on yourself by "considering the block as a single system"?

I do not understand what you are going on about.

 

[Dale]

Right, sure. So I suppose the conclusion is that we can't find the final KE of a non-rigid block if we consider the block as a single system.

No. I am not sure where that conclusion came from.

 

[etotheipi]

What do you mean by "KE of a non-rigid block"?
What are you using as inputs to an attempted calculation of "KE of a non-rigid block?"
What constraints are you imposing on yourself by "considering the block as a single system"?

I do not understand what you are going on about.

I'm just trying to approach it via the first law of thermodynamics, such that any external work or heat will change the total energy content of a chosen system.

Granted 'non-rigid' is poor terminology, I mean to say that I am allowing the block to possesses thermal energy. As opposed to a rigid body which will only ever possesses mechanical energy.

So if I draw a system boundary around only the block, friction becomes an external force and the total energy of the block becomes its kinetic plus its thermal energy. Then, the work done by friction plus the work done by gravity should equal the total change in the block's energy.

It's fine if we take the block and the wedge as one system since then we have no external work and we can apply conservation of energy in one line.

 

[jbriggs444]

I'm just trying to approach it via the first law of thermodynamics, such that any external work or heat will change the total energy content of a chosen system.

Granted 'non-rigid' is poor terminology, I mean to say that I am allowing the block to possesses thermal energy. As opposed to a rigid body which will only ever possesses mechanical energy.

So if I draw a system boundary around only the block, friction becomes an external force and the total energy of the block becomes its kinetic plus its thermal energy. Then, the work done by friction plus the work done by gravity should equal the total change in the block's energy.

It's fine if we take the block and the wedge as one system since then we have no external work and we can apply conservation of energy in one line.

None of that defines what you mean by "kinetic energy of a non-rigid block".

 

[etotheipi]

None of that defines what you mean by "kinetic energy of a non-rigid block".

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.

 

[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?