Integrating Net Force: Limits on LHS?

In summary, the conversation discussed the calculation of total force from a container of water with a slit at the side wall. The equation used was (bdx)2gxρ, and the limits for the integral on the right side were set as h-l and h. The question was raised about whether the left side of the equation should also have limits, and it was explained that while it would be more precise, it is not necessary as it does not provide any additional useful information. The conversation also touched on a similar situation in electromagnetism.
  • #1
andyrk
658
5
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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  • #2
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.
 
  • #3
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.
 
  • #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.
 
  • #5
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?
 
  • #6
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.
 
  • #7
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.
 
  • #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:
 
  • #9
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.
 

1. What is meant by "Integrating Net Force"?

Integrating net force refers to the process of determining the total force acting on an object by adding up all of the individual forces acting on that object.

2. What are the "Limits on LHS" in this context?

LHS stands for "left-hand side" and in this context, it refers to the left side of an equation. The limits on the LHS refer to any constraints or restrictions that must be taken into account when integrating the net force on an object.

3. Why is it important to consider limits on the LHS when integrating net force?

Limits on the LHS are important because they can affect the overall net force acting on an object, and therefore impact its motion. For example, constraints such as friction or a surface that is not perfectly smooth can change the net force and influence the object's movement.

4. How do you determine the limits on the LHS when integrating net force?

Limits on the LHS can be determined by considering all the factors that may affect the net force on an object. This can include things like friction, surface conditions, and any other constraints that may impact the forces acting on the object.

5. Can limits on the LHS change during the integration process?

Yes, limits on the LHS can change during the integration process if there are any changes to the conditions or constraints affecting the net force on an object. It is important to reassess these limits throughout the process to ensure an accurate calculation of the net force.

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