Friction and Energy: Understanding Inclined Paths

[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:
\frac{1}{\sqrt{1+(dy/dx)^{2}}}

(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
dx\sqrt{1+(dy/dx)^{2}}

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

xmg\mu

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.

 

[ehild]

Nathanael is right. For any form of the slope, the work to overcome friction when moving slowly from B to A is W=\int_{x_B}^{x_A}{μmg|dx|} In case dx ≥0 during the whole path (not going backward) W=μmg(x_B-X_A).

ehild

 

[tms]

Sure, but does the constraint of not going "backwards" apply to the problem? The OP said the situation in the problem was "like" the one in the drawing.

 

[ehild]

It looks that the slope does not bend backwards.
ehild

 

[Nathanael]

(if you are always moving horizontally forward, which is the case in his picture)

I don't know why I was so persistent in the assumption that there was no back-tracking in the picture. There is that apparently vertical spot which is uncertain if it back tracks a bit or not. Tms is right, the special case (of no backwards movement) is not necessarily the case.

Since the amount of "back-tracking" is unknown, (even though it could be zero,) the work is unknown.

 

[tms]

It does, but I think the uploaded image is the OP's own drawing, not the original. Do we know that they are the same shape? Recall that the question in the problem says that the work on the curvy path is not known, which would not be the case under the no-backwards-motion constraint.

 

[Nathanael]

It does, but I think the uploaded image is the OP's own drawing, not the original. Do we know that they are the same shape? Recall that the question in the problem says that the work on the curvy path is not known, which would not be the case under the no-backwards-motion constraint.

It doesn't have to do with physics anymore, the original post was unclear in a few ways.

It's good that you brought up the point of not knowing if there was no backwards movement though, because it didn't even cross my mind that that assumption could be wrong.

 

[ehild]

The problem asks the work required to lift up the body back to A. I understand it as the minimum work necessary. But it does not say that the lifting force is tangential to the curve at every point. You can keep the body just a fraction of millimetre above the slope, and then there is no friction. Or you can carry it horizontally from B to just below A and then lift up. Then the work needed is mgh.

If the work of friction is asked while the body slights along the curved path, the force of friction is the normal component of mg multiplied by μ only in case when the speed of the body is practically zero. Otherwise the normal force along the curved path depends on the speed (because of the centripetal acceleration) .
I agree that the wording of the problem is quite confusing.

ehild

 

[haruspex]

Now i see in the question that the points B and C are in the same places.

Sorry, what do you mean they're in the same places? Do you mean that in each diagram there should be an A, a B and a C, in the same relative positions?
If so, under the assumptions outlined (applied force always parallel to surface, no horizontal backtracking) the work done will be the same for each, in contradiction of what you were asked to explain (right?).
If there is an overhang in the second picture, i.e. backtracking as it rises, then it is so subtle I doubt it was intended.
That leaves the question of whether we should assume the applied force is always parallel to the surface. If we can't then the whole question is silly. You would not be able to deduce the work done is the same even if hauled twice over the same track.

So, according to this, can i explain the 2mgh result, at least for the first case?

You can't explain '2mgh' specifically. You can explain that it should be more than mgh, but how much more depends on the distances and the coefficient of friction.

The work done against a conservative force does not depend on the path taken.

True, but there can still be a degree of path independence even with non-conservative forces, as illustrated here.
It's a shame that the wording of the question was so poor; it is really quite an interesting and (to me) surprising result.

 

[Nathanael]

The problem asks the work required to lift up the body back to A. I understand it as the minimum work necessary. But it does not say that the lifting force is tangential to the curve at every point. You can keep the body just a fraction of millimetre above the slope, and then there is no friction. Or you can carry it horizontally from B to just below A and then lift up. Then the work needed is mgh.

Technically it says the work required to "drag" it back to A, which to me implies going against friction

I agree that the wording of the problem is quite confusing.

Indeed

 

[Karol]

You can't explain '2mgh' specifically. You can explain that it should be more than mgh, but how much more depends on the distances and the coefficient of friction.

I think the 2mgh result is true for all shapes (i think the path doesn't go backwards).
Because the whole h energy is consumed by friction when sliding down, then when up again the same mgh.
And more, i think any shape of the path (not going backwards) and in any place the point B may be, the point C must be in the same place because the work done against friction depends on the horizontal distance, and the work will always be 2mgh because we consume mgh by going down

 

[haruspex]

I think the 2mgh result is true for all shapes (i think the path doesn't go backwards).
Because the whole h energy is consumed by friction when sliding down, then when up again the same mgh.
And more, i think any shape of the path (not going backwards) and in any place the point B may be, the point C must be in the same place because the work done against friction depends on the horizontal distance, and the work will always be 2mgh because we consume mgh by going down

Aargh! So sorry - just realized I haven't been reading the question aright. Please disregard all my earlier posts on this thread.

You are correct - the idea is that mgh of work was done against friction on the way down, so the same work is done against friction on the way back up. However, that argument is bogus. The abrupt transition from a slope to level will cost it all its vertical KE on the way down. If it is slowly dragged back up it will require less than 2mgh. (I think the difference is mgh(sin θ - μ cos θ).)

If the curvy path is smooth it is likely to be closer to 2mgh, but as ehild wrote, it still will not be 2mgh because of the centripetal contribution to the normal forces. (It's not immediately clear, but I suspect it will always be less than 2mgh.)

In each case, there is also, as ehild also wrote, the question of whether the dragging back is supposed to be gradual. It almost surely is, but in principle you could instead arrange that it is exactly like playing the film backwards. If so, the work done would be 2mgh for both (I think).

Conclusion: it's still a poorly specified question.