Force Distribution Problem | Normalize to F0

[skrat]

Homework Statement

This is not really a homework problem, it's a problem that I am working on and am not able to come to an end of it.
The story is:
I have a certain amount of force, let's call it $F_0$, that I can distribute on a certain length $L$ just the way I want it to. Meaning that the $F(x)$ can describe any function you can imagine for $0\%<x<100\%$.
For example, we can set $F(x)$ to be a sum of two Heaviside functions $$F(x)=aH(x)+(b-a)H(x-50)$$ or we can say it is linear or we can say the distribution is Chi squared for $k=4$, see graphics on the link (https://upload.wikimedia.org/wikipe...i-square_pdf.svg/600px-Chi-square_pdf.svg.png).

Homework Equations

The Attempt at a Solution

My problem here is that I am missing a condition to satisfy. So I have $F_0$ available and any distribution I can think of.

But whichever distribution I choose, I have to normalize it to that $F_0$. And this is where I am weak. I don't know how to do it, I don't understand the physics behind it.

Because:
Let's assume it is a linear distribution for a moment, so $F(x)=kx$. But I have no idea how to normalize it to $F_0$. If I was to integrate it $$\int F(x)dx$$ than I am afraid that would be pressure and not force, but if I was to somehow sum the values $F(x)$ for discrete (with very small steps) $x$ than I am afraid this is wrong, because it massively depends on how small the steps are...

So. Yeah. I'm stuck here. I hope you understood my confused description of the problem, though. :)

ps: I am working in Mathematica, so if there is anything I can do there, just let me know and I can try.

 

[Qwertywerty]

What do you mean by normalisation ?

 

[skrat]

What do you mean by normalisation ?

Yes, a good question.

I am trying to say that I have only $F_0$ of force available for the distance $L$. And no matter what force distribution I use, I can not exceed or be short of that $F_0$. Anything else is not important.

This doesn't really answer your question. Because if I knew exactly how to normalise it, I probably wouldn't open this thread.

 

[Qwertywerty]

Would finding a point on the rod where effect of a single force would be the same as that of the distributed F₀ answer your question ?

 

[skrat]

I don't see how could this help me. Can you try to explain?

 

[Qwertywerty]

This doesn't really answer your question. Because if I knew exactly how to normalise it, I probably wouldn't open this thread.

I didn't ask you how to normalize - I asked you to explain your question .

 

[skrat]

Ok. And did I?

If not, what is not clear?

 

[Qwertywerty]

What is normalization ? Is it a mathematical term ? If so , could you explain it to me ? Have you made it up ?

 

[skrat]

I don't get it... but ok.

For an electron in the infinite potential well the wave function is $$\psi (x)= A\sin(kx)$$ and normalization yields $$<\psi|\psi>=|A|^2\int _0 ^L \sin^2(kx)=1.$$Therefore the normalization constant is $$A=\sqrt {\frac L 2}.$$

 

[HallsofIvy]

It sounds to me that you are simply saying that you want the total force to be $F_0$: $\int F(x)dx= F_0$.

 

[Qwertywerty]

Ok , my bad - this is beyond my scope .

I hope I haven't wasted your time .

 

[skrat]

No problem. =)

 

[skrat]

It sounds to me that you are simply saying that you want the total force to be $F_0$: $\int F(x)dx= F_0$.

Exactly, BUT I believe that the approach with integral is wrong. Why? Two reasons for that:
1. Even the units don't match. $\int _0 ^L Fdx$ is (in one dimension) by definition work done by force $F$ on distance $L$ which are Joules, and not Newtons.
2. Integration gives me surface under the curve. Which is not what I am interested in. I only have to make sure that I distribute that $F_0$ on the rod (or whatever you wish) with length $L$.

 

[SammyS]

Exactly, BUT I believe that the approach with integral is wrong. Why? Two reasons for that:
1. Even the units don't match. $\int _0 ^L Fdx$ is (in one dimension) by definition work done by force $F$ on distance $L$ which are Joules, and not Newtons.
2. Integration gives me surface under the curve. Which is not what I am interested in. I only have to make sure that I distribute that $F_0$ on the rod (or whatever you wish) with length $L$.

The overall question seems somewhat vague.

If $\ F(x)\$ is a force distribution function, then it should have units such as force per unit length or force per unit area. In this case the integral $\displaystyle \ \int F(x)dx \$ should indeed give $\ F_0\ .$

On the other hand, if $\ F(x)\$ is in units of force, then you need to divide the result by an integrated value of the variable you are using for the force integral.

$\displaystyle \ \frac{\displaystyle\int_a^b F(x)dx}{\displaystyle\int_a^b dx}=F_0\$

 

[skrat]

Is it just me or does this $$\displaystyle \ \frac{\displaystyle\int_a^b F(x)dx}{\displaystyle\int_a^b dx}=F_0\$$ look very similar to weighted mean value from statistics?

Either way it is the second case we are talking about, meaning $F(x)$ has units of force.

 

[haruspex]

What do you mean by normalisation ?

Normalizing generally means multiplying by a constant in order to meet some constraint. In this case, the constraint is the value F₀. In probability, if you know the relative probabilities of all the outcomes then multiplying by the right constant will give a total probability of 1, as required.

 

[haruspex]

Is it just me or does this $$\frac{\int_a^b F(x)dx}{\int_a^b dx}=F_0$$ look very similar to weighted mean value from statistics?

Either way it is the second case we are talking about, meaning $F(x)$ has units of force.

Your original question admits two main interpretations:
1. F(x) is the force experienced at point x. Integrating over the length and dividing by the length yields the average force.
2. F(x).dx is the force experienced over element dx. This makes F a force per unit length, and integrating yields the total force.

 

[skrat]

To make things clear, what I need it $F(x)$ being the force at a point x.

So, let's assume that my $F(x)=ax^2$, than $$a\frac{\int _0^Lx^2dx}{L}=F_0$$ therefore $$aL^2=3F_0.$$ From here I can extract the normalization constant $a$, and this should be it. Right?

 

[skrat]

I attached a picture, just to make things a bit more clear.

At the picture I have drawn discrete forces $F_i$ and the condition says that $$\sum_{i=1}^NF_i=F_0.$$ In other words, what I did here in discrete case is that I have split the available force $F_0$ into $N$ forces that sum up to back to $F_0$. But I don't want to work with discrete values, because it is unrealistic, I would like to find a continuous function that meets the same condition ( $F_0$).

If it is possible. Anyway I am completely open to any other suggestions or any other ideas on how to solve this problem. It doesn't matter how I get it, in the end all I need is $F(x)$, where $F(x)$ describes the amount of force (in Newtons) exerted on a rod at length $x$.

 

[haruspex]

I attached a picture, just to make things a bit more clear.
View attachment 86337
At the picture I have drawn discrete forces $F_i$ and the condition says that $$\sum_{i=1}^NF_i=F_0.$$ In other words, what I did here in discrete case is that I have split the available force $F_0$ into $N$ forces that sum up to back to $F_0$. But I don't want to work with discrete values, because it is unrealistic, I would like to find a continuous function that meets the same condition ( $F_0$).

If it is possible. Anyway I am completely open to any other suggestions or any other ideas on how to solve this problem. It doesn't matter how I get it, in the end all I need is $F(x)$, where $F(x)$ describes the amount of force (in Newtons) exerted on a rod at length $x$.

I consider your scenario as really being a force-per-unit-length. In the discrete form you get discrete bits of force each discrete bit of length, but in the limit it becomes a continuous function.

 

[skrat]

I consider your scenario as really being a force-per-unit-length. In the discrete form you get discrete bits of force each discrete bit of length, but in the limit it becomes a continuous function.

Ok than, so following SammyS my condition than says $$\int _0 ^L f(x)dx=F_0$$ where $f(x)=a\frac{F(x)}{L}$ for $L$ being the total length of my rod, $a$ normalization constant and $F(x)$ actual force.
Right?

 

[lightgrav]

yes, so this new f = F/L ... just like you said back in #14

 

[haruspex]

Ok than, so following SammyS my condition than says $$\int _0 ^L f(x)dx=F_0$$ where $f(x)=a\frac{F(x)}{L}$ for $L$ being the total length of my rod, $a$ normalization constant and $F(x)$ actual force.
Right?

That's not how I'd put it.
In the discrete view, you have forces F_i, i=0...n, at regular intervals S, nS=L.
Suppose we now double n and halve S. We have to fill in extra forces. To get the same total, we must roughly halve each existing force and fill in new forces halfway between them by, say, taking the average of two neighbours. This keeps the total about constant, and also the local force per unit length: half the force at half the distance.
Repeating this process, in the limit you have a smooth curve f(x). At an original point i, x=iS, f(x) ~ F_i/S.

 

[skrat]

Repeating this process, in the limit you have a smooth curve f(x). At an original point i, x=iS, f(x) ~ Fi/S.

Also repeating this process $F_i\rightarrow 0$.