Integrating Net Force: Limits on LHS?

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Discussion Overview

The discussion revolves around the integration of net force in a fluid dynamics context, specifically regarding the limits of integration applied to the left-hand side (LHS) of an equation. Participants explore the implications of not specifying limits on the LHS while integrating the right-hand side (RHS) in a scenario involving a container of water and the force exerted by water gushing out of a slit.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants question why the LHS of the integral is left without limits, suggesting that it should have limits corresponding to the initial and final forces.
  • Others propose that the limits on the LHS could be considered as 0 and the total force F, arguing that integrating 1 from 0 to F yields F.
  • A participant describes the physical setup involving a container of height h filled with water and a slit, detailing the force calculation based on the velocity of efflux and density of water.
  • Some participants express that the lack of limits on the LHS may stem from a lack of knowledge about the force value, implying that specifying limits may not provide additional useful information.
  • One participant draws a parallel to electromagnetism, discussing the total charge of a nonuniformly charged rod and how similar situations arise in different contexts.
  • There is a suggestion that while it may be technically incorrect to leave the LHS without limits, it does not significantly affect the overall solution to the problem.

Areas of Agreement / Disagreement

Participants express differing views on whether limits should be applied to the LHS of the integral. While some agree that limits could be defined, others suggest that it is unnecessary and does not impact the solution. The discussion remains unresolved regarding the necessity of limits on the LHS.

Contextual Notes

Participants acknowledge that the discussion involves assumptions about the values of forces and the context of integration, which may not be fully specified. There is also an indication that the integration process may vary depending on the specific physical scenario being analyzed.

andyrk
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We have: dF = (bdx)2gxρ. Now, x varies from h-l to h to calculate the entire force from x = h-l to x = h. So we put lower limit h-l and upper limit h while integrating RHS. But we don't put any limits on LHS and simply leave it as ∫dF. Shouldn't LHS also have some limits? If so, what would they be? And if not, then why?
 
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You can write the limits as 0 and the total force F, because that's exactly what you calculate (and integrating 1 from 0 to F will give F).
andyrk said:
(bdx)2gxρ
I have no idea what that means.
 
mfb said:
I have no idea what that means.
It means that there is a container of height h full of water up to the top. There is a slit of length l at the side wall of the container starting at the bottom of width b. The density of water can be taken as ρ. Then what would be the total force from the water gushing out of that slit? For this we used velocity of efflux as √(2gh) (v), which appears as 2gx (v2) in the expression I stated. Also, the force from water coming out of an area of cross section A would be Av2ρ. So, force for an element x length below the surface of water and of dx length would be : (bdx)*2gx*ρ = (bdx)2gxρ, which is what I wrote in the beginning.
 
Intuitively, I would think the limits would be the final force and the initial force. The total force would then just be that difference by the Fundamental Theorem of Calculus. The problem is if we knew those limits of integration, then there would be no need to calculate the other side of the integral. So, you're right in the sense that the indefinite integral on the left is not technically equal to the one on the right because the one on the right has limits of integration, but it doesn't really affect the solution to the problem. You bring up an interesting point though; I've seen this a few times before and asked myself the same question.
 
Apogee said:
Intuitively, I would think the limits would be the final force and the initial force. The total force would then just be that difference by the Fundamental Theorem of Calculus. The problem is if we knew those limits of integration, then there would be no need to calculate the other side of the integral. So, you're right in the sense that the indefinite integral on the left is not technically equal to the one on the right because the one on the right has limits of integration, but it doesn't really affect the solution to the problem. You bring up an interesting point though; I've seen this a few times before and asked myself the same question.
Yeah, this is interesting. So how would you explain why we didn't put any limits on the left hand side?
 
I would avoid the integral on the left side completely and directly write F = int ...
If you don't want to do this, see post #2.
 
andyrk said:
Yeah, this is interesting. So how would you explain why we didn't put any limits on the left hand side?

We didn't put any limits on the left hand side out of laziness to say the least hahaha. Because we do not know the value of the force in question, there is no point in putting limits on the left integral. If we really wanted to be precise, I guess you could do:

x?%5Cint_%7B0%7D%5E%7BF%7D%20dF%5E%7B%27%7D%20%3D%20%5Cint_%7Bh-l%7D%5E%7Bh%7D%202g%5Crho%20bxdx.gif


But we're splitting hairs at this point hahaha.
 
A good example of this situation comes up in electromagnetism. Suppose we want to find the total charge of a nonuniformly charged rod. Suppose the charge density of the rod is described by the equation:

gif.gif


If the rod is L meters long, then summing up all increments of the density function over length gives us the total charge of the rod:

gif.gif


In theory, it would have proper to say:

gif.gif


But this is entirely unnecessary. Sorry for my brief nerd-out and for the fact that I probably screwed up on my own problem somewhere along the way. I just wanted to give an example of when these situations occur. :smile:
 
Apogee said:
We didn't put any limits on the left hand side out of laziness to say the least hahaha. Because we do not know the value of the force in question, there is no point in putting limits on the left integral. If we really wanted to be precise, I guess you could do:

x?%5Cint_%7B0%7D%5E%7BF%7D%20dF%5E%7B%27%7D%20%3D%20%5Cint_%7Bh-l%7D%5E%7Bh%7D%202g%5Crho%20bxdx.gif


But we're splitting hairs at this point hahaha.
I would say the limits should be F1 and F2 rather than 0 and F. Or to be even more precise, they would be Fh-l and Fh.
 
  • #10
andyrk said:
I would say the limits should be F1 and F2 rather than 0 and F. Or to be even more precise, they would be Fh-l and Fh.

Hahaha, yeah but like I said: it isn't really necessary. Generally, if by writing something down, it doesn't tell us any more useful information about the system, then most of the time there isn't any need to write it down. I know someone is going to get really upset with me for saying this, but when you give it some thought, it makes sense intuitively.
 

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