# Homework Help: Friction and energy

1. May 25, 2014

### Karol

1. The problem statement, all variables and given/known data
There is an inclined path like in the upper drawing. a mass is released at height h (point A) and the mass slides until it stops on the horizontal surface at C. the energy required to drag it up again to A is 2mgh.
Now the path is curved like in the lower drawing but in this case we don't know what is the energy required to drag it up again to A. why?

2. Relevant equations
Friction force: $$F=mg\mu$$

3. The attempt at a solution
My explanation to the first situation with the straight inclined path is that there is a constant ratio between each meter of sliding along the path to the decrease in height, i mean every meter the mass advances consumes, lets say, 0.8 meter of vertical height. the first mgh is needed to overcome the friction and the second mgh in order to lift it.
But even if this ratio changes in case of the curved path, still the mass stops after it consumed energy oh height h, so why isn't the same answer 2mgh hold here?

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2. May 25, 2014

The total energy required to move the block up the plane = Mgh + Work done to overcome friction (Let's call it Wf)

Wf = ∫μmgx.dx. The displacement/distance covered in the first case is not equal to the one covered in the second one. More the displacement covered more will be the work done to overcome friction or less otherwise.

Hope this helps :)

3. May 26, 2014

### haruspex

A few things wrong there, I think.
Maybe you meant ∫μmg.dx, which is dimensionally correct, but fails to take into account that the normal force depends on the slope.

Karol, the work done to overcome friction depends on how the force is applied. It doesn't have to be parallel to the slope. If the angle of pull is steeper than the slope less work is needed. In the extreme, the pull is near vertical, and hardly any work is needed against friction.

So assume that the pull is parallel to the slope at all points in both cases. At a point where the slope is dy/dx = tan(θ), what is the normal force? What is the frictional force? In advancing dy vertically, how much work is done against friction and how much against gravity?

4. May 26, 2014

### Karol

I know that, the normal force at the slope of tag(θ) is mg$\cdot$cos($\theta$). and if i lift it vertically no force is done against friction.
My question is why in the second case where the path is curved we don't know the work that is needed to drag it upwards.

5. May 26, 2014

### haruspex

You're missing my point. You don't know the work required even on the straight slope unless you make an assumption about how the force is applied. You likewise need to make an assumption about that in the second case, but it's less obvious which assumption to make. I suggest assuming it is at all points parallel to the slope. This might let you determine the work done.... Or it might not.

6. May 26, 2014

### Karol

Yes, of course the angle at which the force is applied counts, no doubt. but the answer to the first case is 2mgh so it's always parallel.

7. May 26, 2014

### haruspex

You're still missing it. You asked why you don't know what the work required to overcome friction is in the second case. In the straight case you are just assuming it is parallel (you can't deduce that is parallel unless you are told the angle of the slope, the coefficient of friction and the extra work required). If you don't make an equivalent assumption in the curvy case you are working with less information.

Just had a thought - are you under the misapprehension that the total work is always 2mgh in the straight case? Where are you getting the 2mgh figure from?

8. May 26, 2014

### Nathanael

Isn't it false that you don't know the energy required? Wouldn't the energy be the same?
I wrote an equation for the work consumed by friction assuming the equation for the path (or at least it's derivative) is known, and it ended up "saying" that the path is irrelevant, and that as long as "B" and "h" are the same for both cases, the energy should be the same in both cases.
(This is assuming the path doesn't "back-track", or, in other words it passes the "vertical line test")

Does the frictional force actually depend on the path? (I've only been looking at friction because clearly the change in gravitational potential energy is the same.)

Edit:
I didn't notice this before, but you seem to be hinting at an intuitive understanding of why this is true. If you have more to say about it I'd love to hear it because I don't entirely understand why it's obvious that the path is irrelevant to the energy.

Last edited: May 26, 2014
9. May 27, 2014

### Karol

The question is from a text book but from an older release, 2008, maybe they changed the answer, i doubt.
The answer to the first case in the book is 2mgh and there is no word about direction of applying the dragging force.
To Nathanel: the points B needn't be in the same place! maybe i just drew them close, in the book it doesn't say they are in the same place.

10. May 27, 2014

### Nathanael

Oh, I just assumed that since they had the same letter representing them. If B is unknown, wouldn't that be why you don't know the required energy?

11. May 27, 2014

### haruspex

Ok, but the 2mgh is not generally true - they are telling you that it happens to be that in this case.
In posing the question, it's unclear to me whether the book expects you to assume the applied force is parallel to the slope; probably in the first case, maybe not in the second.
Anyway, suppose you are expected to assume that in both cases. The 2mgh allows you to write down an equation relating the angle of the slope to the coefficient of friction. Can you do that?
In the second case, can you answer the questions I posed in the last paragraph of my first post? That will give you a way to calculate the work done against friction. The answer is quite interesting.

12. May 28, 2014

### Karol

I cannot do that since point B on the horizontal plane compensates for the energy remainder from the height h.
The normal force is $$mg\cdot cos \theta$$
The work done against friction when climbing dy is $$\frac{dy}{sin\theta}mg\mu \cos \theta$$
I don't see how it helpes

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13. May 28, 2014

### haruspex

Sorry, I don't understand what you mean.
But I've just realised something isn't clear in the OP. Is the 2mgh the energy required to drag it from B to A or from C to A? If it's C to A then where is C in the second picture? Is BC the same in each, or is the horizontal distance from A to C the same in each?
Anyway, if you solve the part below it can be applied to both scenarios.
Next, express tan theta in terms of dy and dx.

14. May 28, 2014

### Karol

What is an OP?
There is nothing said about points B and C, they don't have to be in the same place. The energy is required to drag from C to A. just now i saw i forgot to mark the point C on the second graph.
$\tan \theta =\frac{dy}{dx}$
So simple, what do you want by that? how does it advance us?

15. May 28, 2014

### Nathanael

You've basically done it. Just pretend you know θ and B, instead of θ and h.

16. May 28, 2014

### Nathanael

"OP" is "Original Post"

It advances you because your last formula could be re-written as dy*mgμ*cotangent(θ) which, with the equation in the quote, (1/(dy/dx)=cotangent(θ)) becomes dy*mgμ/(dy/dx) which can be simplified to get what (I presume) he was trying to show you.

However you could just pretend you know B (or dx) instead of h (or dy) and do what you did at first, that way you'll get straight to the answer. (Derivitave is unneeded)

17. May 28, 2014

### Karol

I don't understand you, Nathanael, at all, what do you mean to pretend? and if so, i'd better know θ and h

18. May 28, 2014

### Nathanael

By pretend, I just meant, write the formula in terms of B instead of h (or dx instead of dy, if you want to call it)

But h is quite unnecessary for the frictional work. If you wrote it in terms of B you find a relationship that simplifies it and shows you that there's only 1 important factor for the frictional work (other than mgμ)

19. May 28, 2014

### Nathanael

Look at the thumbnail you attatched in post #12, (specifically the top triangle,) you wrote x in terms of h (or "dy") but really h is not what matters, it is x that matters, so why is it any better to know it in terms of h than B?

20. May 28, 2014

### tms

Assuming that the question as presented is complete, I think it is just about the path-dependence of non-conservative forces, since no details at all are given about the exact shape of the curvy path.

21. May 29, 2014

### haruspex

Quite so. Karol, Nathanael has done most of the work for you. What does dy*mgμ/(dy/dx) simplify to?

22. May 29, 2014

### Karol

$$\frac{dy}{sin\theta}mg\mu \cos \theta=dy \cdot mg\mu \cdot \cot \theta=mg\mu\cdot dx$$
The friction depends only on the distance (and the normal force), but that's nothing new, it doesn't advance us in solving the question.
In the first case the path is straight. let's assume that in every projected distance to the y axis the mass consumes 0.8 of the potential energy for that height. so when the mass is at point B it consumed 0.8h energy and the rest 0.2h is consumed by friction till point C. but even in the curvy path where there is no fix relation still the remainder of the h energy which isn't consumed during the vertical descend is consumed by the horizontal portion so why are those cases different? why isn't also in the second case the energy 2mgh?

23. May 29, 2014

### tms

You said it yourself: the friction depends on the distance. What is the length of the curvy path? (I am assuming that the question as presented is complete.)

24. May 29, 2014

### Karol

But in the first case i also don't know the length yet i know the work. why is the second case different?

25. May 29, 2014

### haruspex

No, look at the equation again. X is the horizontal distance. And it applies to both cases, so if:
- we are right to assume that the applied force is parallel to the slope at all points, and
- the vertical displacements are the same for both cases, and
- the horizontal displacements are the same for both cases
then we could conclude that the work done is the same for both, namely, mg(Δy+μΔx).
If we don't know how the two horizontal displacements compare, then of course we don't know how the energies compare.

Edit: to be completely accurate, the differential equation is dE = mg(dy + μ|dx|). As long as dx does not change sign along the path, that integrates to ΔE = mg(Δy + μ|Δx|).
In summary, I feel it was a very interesting question, very poorly posed.

Last edited: May 30, 2014