Friction and Energy: Understanding Inclined Paths

[Karol]

Homework Statement

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?

Homework Equations

Friction force: $$F=mg\mu$$

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, let's 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?

 

[Adithyan]

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, let's 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?

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

W_f = ∫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 :)
-Adithyan

 

[haruspex]

W_f = ∫μmgx.dx.

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?

 

[Karol]

Answer

I know that, the normal force at the slope of tag(θ) is mg[itex]\cdot[/itex]cos([itex]\theta[/itex]). 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.

 

[haruspex]

I know that, the normal force at the slope of tag(θ) is mg[itex]\cdot[/itex]cos([itex]\theta[/itex]). 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.

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.

 

[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.

 

[haruspex]

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.

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?

 

[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:

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?

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.

 

[Karol]

The question is from a textbook 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.

 

[Nathanael]

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.

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?

 

[haruspex]

The question is from a textbook 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.

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.

 

[Karol]

The 2mgh allows you to write down an equation relating the angle of the slope to the coefficient of friction. Can you do that?

I cannot do that since point B on the horizontal plane compensates for the energy remainder from the height h.

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.

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

 

[haruspex]

I cannot do that since point B on the horizontal plane compensates for the energy remainder from the height h.

Sorry, I don't understand what you mean.
But I've just realized 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.

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

Next, express tan theta in terms of dy and dx.

 

[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?

 

[Nathanael]

The work done against friction when climbing dy is $$\frac{dy}{sin\theta}mg\mu \cos \theta$$
I don't see how it helpes

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

 

[Nathanael]

What is an OP?

"OP" is "Original Post"

##\tan \theta =\frac{dy}{dx}##
So simple, what do you want by that? how does it advance us?

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)

 

[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

 

[Nathanael]

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

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)

and if so, i'd better know θ and h

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)

 

[Nathanael]

and if so, i'd better know θ and h

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?

 

[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.

 

[haruspex]

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.

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

 

[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?

 

[tms]

The friction depends only on the distance (and the normal force), ##\ldots## .

##\ldots## so why are those cases different? why isn't also in the second case the energy 2mgh?

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.)

 

[Karol]

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

 

[haruspex]

$$\frac{dy}{sin\theta}mg\mu \cos \theta=dy \cdot mg\mu \cdot \cot \theta=mg\mu\cdot dx$$
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.

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.

 

[Karol]

Now i see in the question that the points B and C are in the same places.
So, according to this, can i explain the 2mgh result, at least for the first case? i explained it by the first mgh is consumed by friction and going up again we add mgh

 

[tms]

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.

You seem to be saying that the work is the same no matter what path is taken, that it is path-independent. But friction is not a conservative force. Unless I am misunderstanding you.

 

[Nathanael]

You seem to be saying that the work is the same no matter what path is taken, that it is path-independent. But friction is not a conservative force. Unless I am misunderstanding you.

I don't know what "conservative force(s)" are, but I came to the same conclusion.

(frictional work depends only on dx, gravitational work depends only on dy)

 

[tms]

I don't know what "conservative force(s)" are, but I came to the same conclusion.

(frictional work depends only on dx, gravitational work depends only on dy)

The work done against a conservative force does not depend on the path taken. It is path-independent. That is, if you move something from A to B against a conservative force, the work is the same whether you go in a straight line from A to B, or whether you take a detour through C, D, E, and F. Another way to say it is that the work only depends on the positions of A and B, not on any point passed through on the way. An example is gravity.

A non-conservative force, such as friction, is different in that the work done does depend on the path taken. For friction, assuming uniformity, the longer the path, the more the work.

For an example, consider points A, B, and C on a line, in that order. For friction, there is more work if you go from A to C and then back to B than there is if you go from A to B and stop, although the ending positions are the same, For gravity (assuming it acts along the line), the work would be the same for both paths.

 

[Karol]

If the friction work depends only on the horizontal distance covered, and the work is 2mgh, then any shape the path will take then point C will remain in the same place, no?

 

[tms]

The work done against friction does not depend on the horizontal distance, it depends on the total distance traveled over which friction is acting. If one path is straight from point (0, 0) to (1, 0), and another is from (0, 0) to (0, 1) to (1, 1) to (1, 0), and friction acts everywhere, the work done on those two paths is different, even though they start and end at the same places.

 

[Nathanael]

The work done against friction does not depend on the horizontal distance, it depends on the total distance traveled over which friction is acting. If one path is straight from point (0, 0) to (1, 0), and another is from (0, 0) to (0, 1) to (1, 1) to (1, 0), and friction acts everywhere, the work done on those two paths is different, even though they start and end at the same places.

For an example, consider points A, B, and C on a line, in that order. For friction, there is more work if you go from A to C and then back to B than there is if you go from A to B and stop, although the ending positions are the same

Well we both said there is one constraint: You don't move backwards. In your example you would move backwards. (I said it in an earlier post, I forgot to mention it in the post you replied to.)

This is the math that led me to think this. Suppose you know the function of the path y(x) (the height at a given distance x)

dy/dx would be the instantaneous slope of the path, or the tangent of the instantaneous angle the path makes. So the instantaneous force of friction would be mgμ*cos(θ) where tan(θ) = dy/dx
If tan(θ)=dy/dx then you can rewrite cos(θ) as:
[itex]\frac{1}{\sqrt{1+(dy/dx)^{2}}}[/itex]

(mgμ is constant so you factor it out of the integral) And so you would need to integrate that (from the beggening of the path to the end, we'll say it's a horizontal distance x) against the instantaneous distance at each x value which from pythagoreans theorem is
[itex]dx\sqrt{1+(dy/dx)^{2}}[/itex]

and so, regardless of the path, (IF you DONT "back-track") the work done from friction simplifies to:

[itex]xmg\mu[/itex]

where x is horizontal distance. Thus, it only depends on horizontal distance (if you are always moving horizontally forward, which is the case in his picture).


P.S:
I know I'm awful at explaining the math but hopefully it was understood

 

[tms]

Well we both said there is one constraint: You don't move backwards. In your example you would move backwards. (I said it in an earlier post, I forgot to mention it in the post you replied to.)

I never said anything about any constraints. I have been talking about the general case.

 

[Nathanael]

I never said anything about any constraints. I have been talking about the general case.

The special case applies to the OP

 

[tms]

The special case applies to the OP

Does it? Is the drawing the one in the actual problem, or something similar to it drawn by the OP? Remember that the original question was: "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?", which contradicts your special case.