# Homework Help: Pushing objects at constant velocity

1. Sep 10, 2011

### derivative_e

Say you're pushing a heavy desk (225kg) across a carpeted floor. You either push it at 3m/s (say you applied a 675N force for 1sec, so you apply the force for 1.5m), or at 6 m/s (you applied 1350N force for 1sec, so the force is applied over 3m). After that one second impulse, however, you continue pushing it at a constant velocity (a total distance of 6m).

After the initial second, are you just applying a force of FNuk to the object, or are you still pushing it at 675N or 1350N? I think that your initial 675N & 1350N forces are replaced by a new vector, equivalent but opposite, to the frictional force.

How much work do you do on the object? I believe you would

1) multiply 675N x 1.5 m & 1350N x 3m; then
2) add FNuk x [6m - (1.5m || 3m)] (the work done after the initial acceleration)

...to get the total work you impart on the object in your two separate trials.

However, due to friction, the force you apply is reduced, and the net K.E. for both objects is reduced by ukFN during part one. No work is done on the object during part two. This seems counter-intuitive, since you yourself are expending a lot of energy pushing an object a constant velocity, but no net work is being done on the object.

Could someone confirm this for me? Thanks.

2. Sep 10, 2011

### Filip Larsen

Welcome to PF!

Since the speed of the desk is constant it is not accelerating and, thus, the sum of all forces on the desk are zero and the net work on the desk is zero as you mention. The keyword here is "net". The work done by the desk on the carpet (which assumingly ends up as heat in the carpet) is exactly provided by your work on the desk. If you look at the two horizontal forces they will be opposite but of equal magnitude, hence there will be a transmission of work from you via the desk to the carpet.

3. Sep 10, 2011

What you think is obvious.There's a very fundamental mistake you have done here.

4. Sep 10, 2011

You have applied W=F*s.
But do you know that it is valid for point objects only??????

5. Sep 10, 2011

Can friction happen between point objects????

No.It can't because friction happens between rigid bodies having many particles.
It's generally very difficult to calculate work done by friction.One thing for sureIt is always LESS THAN mod(F*s).

6. Sep 10, 2011

If you have more doubts do see resnick halliday volume 1 where they have clearly explained this fundamental misunderstanding regarding frictional work.It's awesome..

7. Sep 10, 2011

### Filip Larsen

The usual simplifying assumption is to consider the object (the desk in this case) for a so-called rigid object and thereby ignoring the work going into various kinds of deformation of the object. In short, if the desk is rigid enough, the error in work will be small.

8. Sep 10, 2011

It is never small

9. Sep 10, 2011

If you have further doubts check this research paper.

www4.ncsu.edu/~basherwo/docs/Friction1984.pdf

10. Sep 10, 2011

### Filip Larsen

It is an interesting paper and it does illustrate that the term "work of frictional forces" can be model in more than one way if you dig into the particulars of friction. In short, while the work is the same, the force and distance may be modeled differently in the presence of deformation.

In this case however, it does not change anything for the original question. It is still valid to say that all the work W that a person apply to keep a desk sliding without accelerating is lost to friction.

I also think it is reasonable simplification in this case to "explain" the frictional force as the effective force Feff such that Feff*d = W instead of, as the paper suggest, using the "true" frictional force and introducing deff such that F*deff = W.

11. Sep 10, 2011

### Lsos

1994Bhaskar I'm not sure OP wants to go as deeply into this as you are leading him. I have a pretty good understanding of basic physics but even I'm confused as to what you're talking about...

12. Sep 10, 2011

I am talking about the very common misconception(it is given in many books too) that work done due to friction is W=F*s=µ*N*s .Where µ is coefficient of friction and s is displacement and N is normal force on body.
Consider this example.(Do see attachment)
A body is moving with constant velocity 'v' over a surface with coefficient of friction µ.
In the diagram we can clearly see N=mg.Pulled by force 'f' and a displacement of s.
If the body has to move with constant velocity then net acceleration must be zero.Thus net force must be zero.Thus f=-µmg .Do see the minus sign.
If we apply the work-energy theorem here then as per it:
Work done=Change in Kinetic energy + Change in internal Energy.
Here work done by f is W1=f*s. Work done by friction(here's the mistake)=W2=-µmg=-fs
Hence net work done=W1+W2=0.
Change in kinetic energy is zero as velocity is constant.
So we get change in internal energy as Zero.
This is completely wrong as it suggests that no heat is released which is not true practically. Heat is released in friction and is known to all.

#### Attached Files:

• ###### friction-1.GIF
File size:
2 KB
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13. Sep 10, 2011

I hope you understand the diagram.It's not that neat but that's best I could do.
By the way what's OP in your post???

14. Sep 10, 2011

### derivative_e

I'm the OP (Original Poster).

1994Bhaskar, I had no idea that the work due to friction couldn't be calculated by ukFN*d. Is there a more accurate way to calculate it?

It's great that 1994Bhaskar is introducing these more advanced concepts. My background is only in basic physics, so I'm not conversant with his arguments here, but I'm interested in learning more.

I wasn't aware that dW = f*dS was solely applicable to point objects. I'm assuming this has to do with some deformation that is wrought an on object as a force is being applied. How exactly would deformation of an object affect the acceleration imparted to it? A concrete example would be appreciated. Filip Larsen's comment on rigid bodies, however, seems to make sense to me. The structural integrity of the object should maintain the linear relationship between the applied force and object's acceleration. Naturally, the work done on an object should then be W=f*s, and, if an object is traveling at a constant velocity, the work done by friction (dissipated as heat) should be equivalent.

I'm not sure what 1994Bhaskar means by saying that no heat is released. My understanding is that heat IS released - and that that heat is the reason why the force imparted to an object moving at constant velocity doesn't accelerate it. The energy imparted to the object is dissipated as heat. Is this wrong?

Then again (and here, I'm trying to look at the problem from another angle), when you're moving the object at two different constant velocities (6m/s & 3 m/s), common sense says more heat should be dissipated by the object moving at the 6m/s, even though the force being imparted to it is the same (equal to ukFN). My understanding of heat dissipation, thus, might be erroneous.

Thanks for the support on my first post!

15. Sep 11, 2011

Yes it's true that heat is released and my aim is to show that using formula for work done by friction=F*s gives us practically wrong result that heat is not released as i showed earlier.And yes if you are talking Filip Larsen's comment that we should assume that objects are not deformable then it's wrong!!!
The basic reason for friction is that two surface of different objects have minor aberrations thus forming locks with each other.The force required to break these locks is the source of friction.If you assume that object is perfectly rigid(not deformable) then you can't break these locks.You can never move your object.And please don't tell me to assume that such locks don't exist.If you are being so much practical to assume the result of the locks that friction releases heat then why ignore it's source???
And talking about methods to calculate it correctly they are very complex and too much advanced.For simple and very symmetrical cases it's very easy.For one such example see the pdf file link which i had posted earlier for special cases where it's very easy to calculate.

16. Sep 11, 2011

### Filip Larsen

Different models have different purpose and validity. It is a art in itself to train up to construct models that are as simple as possible while still capturing the phenomenon that you are interested in and still being valid in the situations you are going to apply it. In that sense, rigid body models are a very useful abstractions that with care can be applied in many situations.

In the context of the OP's problem, you'd really have to know more in order to establish whether or not you need to make a detailed model of friction or if you can just concentrate on the mechanical interactions and gloss over micro-models of friction. As it is, I went for the simple description since that is what I expect such text book problems typically focus on.

To extrapolate a bit on your comments, I could equally well insist on your description of friction as being fundamentally wrong since the world really is quantum mechanical at the bottom and we really should model all mechanical interactions using quantum mechanics instead. However, such a model would hardly help anyone trying to grasp the basics of mechanical work.

17. Sep 11, 2011

### derivative_e

Thanks for the responses.

I'm looking to understand how forces work, so I'm going to postpone consideration of 1994Bhaskar's arguments right now.

I'm looking to make sure that my understanding of forces is correct. At the beginning, when you begin moving the objects, you're doing work on them, and you do more work on the faster moving object. However, once the objects begins moving at a constant velocity, the work done on the objects is zero, despite the fact you're straining harder to keep the faster moving object moving at a constant velocity than the slower moving object.

Verification of the above interpretation would be appreciated.

18. Sep 11, 2011

### Filip Larsen

As discussed previously in this thread it really depends on how you define the system boundary of your object and what types of energy that passes that boundary.

In the classical pure Newtonian model usually employed in such text book problems, the system does not include internal energy and much of the work supplied to the system by the person is transferred out of the system again as heat and internal energy so that the net work done on the system is close to zero. Only during the acceleration phase does some of the applied work go into increasing the kinetic energy of the system.

In a more complete thermodynamical model, where the system also will include any internal energy in desk, only a fraction of the applied work is leaving the system during the constant speed slide with the remaining going into raising the internal energy of the system. Exactly how large a fraction that is will depend on the geometry and material properties of the desk and carpet; assuming (unrealistically, in this case) that the situation is symmetrical will give a fraction around a half purely from reasons of symmetry.

For the sake of argument, you could even define your system to include both mechanical and thermodynamical state of both the desk and the carpet (but, say, not including you), and in this situation all the work you apply goes into the system (ignoring any uneven heat transfer that may occur between the system and its surroundings). Such a model would of course not be useful if you want to understand what happens between the desk and the carpet, but it may be a valid model in other situations.

In practice you will want to employ a model (a specifically defined system boundary) that most simply will be able to provide answers to the questions you may have. As a rule of thumb you should make sure that the energy and momentum flows in and out of your system is correctly modeled and only after that focus on correct modeling of the (generalized) forces and displacement, if those are needed at all. So, in this case, if all you care about is work and mechanical energy, then the first model should be sufficient as long as you remember that it is a simplification that abstracts away certain physical aspects.

19. Sep 11, 2011

Did you even read my post??
If you assume rigid bodies then friction can never happen.The link which I gave you clearly states the microscopic reason of friction as breaking of minute surface interlocking.Just because a wrong point is given in majority of textbooks that does not mean that you ignore the divine truth of physics.

And if you have read the pdf link fully then in one example(similar as posted by OP) then work done by friction comes as to be ((F*s)/2)and not simply(F*s).
And talking about quantum mechanical world then can you explain that how friction is going to be related to quantum mechanical concepts like uncertainty, wave nature of mass etc.Even if it comes then we know that quantum mechanics comes in role for microscopic bodies and not for macroscopic bodies(as in here). And here It will be negligible--try any concept of quantum physics.

The concept which I have stated is completely practically true and is purely mechanical model and easily understandable-YET IGNORED.

For advanced level guys(like masters, PHD) friction can be related to second law of thermodynamics(concept of entropy comes here) but not quantum mechanics.

20. Sep 11, 2011

### Filip Larsen

Yes, I read your post - did you read mine? I think you are mixing the issue of making appropriate models with the issue of establishing ultimately physically accurate models. I still claim that rigid body models are very useful when applied to mechanical problems. You seem to imply that since rigid bodies doesn't seem to fit well with how a micro-model of friction "works", then rigid bodies are useless in all situations.

You don't necessarily need to understand the micro-model of friction in order to, say, model the energy flow going into friction. Would you get a more accurate model if you did? Probably, but it will most likely also be more complex. It seems you are in effect saying, that Newtonian mechanics, however basic, should not be taught without also teaching thermodynamics, or, that students of Newtonian mechanics should stay away from modeling any situation involving friction until they can utilize the laws of thermodynamics in their models.

And, by the way, "divine"? Really? I hope not.

Hardly surprising, given that they removed half the heat sink from the model. What the example points out is that you cannot necessarily make an arbitrary free-body diagram (i.e. arbitrary system boundary) and then expect that "force time displacement" on the boundary will sum to the total change in mechanical energy - especially if you end up with a result that gives friction without increase in internal energy of some part.

My point is that there are many areas in physics in general where quantum mechanical effects have a macroscopic effect or where the effect has no explanation in classical physics, but even so, you can still employ simple non-quantum mechanical models that may be sufficient and valid for other purposes.

I'm not sure how I can respond to that sentence without also pointing out the crackpot index score of it - so I wont. In fact, I've already repeated myself plenty here without bringing much new into the discussion so I'll refrain from any further posts in this thread that only would bring repeats.