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Integrating Net Force

  1. Mar 28, 2015 #1
    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?
     
  2. jcsd
  3. Mar 28, 2015 #2

    mfb

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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).
    I have no idea what that means.
     
  4. Mar 28, 2015 #3
    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.
     
  5. Mar 29, 2015 #4
    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.
     
  6. Mar 30, 2015 #5
    Yeah, this is interesting. So how would you explain why we didn't put any limits on the left hand side?
     
  7. Mar 30, 2015 #6

    mfb

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    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.
     
  8. Mar 30, 2015 #7
    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.
     
  9. Mar 30, 2015 #8
    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:
     
  10. Mar 30, 2015 #9
    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.
     
  11. Mar 31, 2015 #10
    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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