# Force distribution

1. Jul 22, 2015

### skrat

1. The problem statement, all variables and given/known data
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).

2. Relevant equations

3. 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.

Last edited: Jul 22, 2015
2. Jul 22, 2015

### Qwertywerty

What do you mean by normalisation ?

3. Jul 22, 2015

### skrat

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.

4. Jul 22, 2015

### 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 ?

5. Jul 22, 2015

### skrat

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

6. Jul 22, 2015

### Qwertywerty

7. Jul 23, 2015

### skrat

Ok. And did I?

If not, what is not clear?

8. Jul 23, 2015

### Qwertywerty

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

9. Jul 23, 2015

### 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 $$A=\sqrt {\frac L 2}.$$

10. Jul 23, 2015

### HallsofIvy

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

11. Jul 23, 2015

### Qwertywerty

Ok , my bad - this is beyond my scope .

I hope I haven't wasted your time .

12. Jul 23, 2015

### skrat

No problem. =)

13. Jul 23, 2015

### skrat

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 $L$.

14. Jul 23, 2015

### SammyS

Staff Emeritus
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\$

15. Jul 23, 2015

### 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.

16. Jul 23, 2015

### haruspex

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.

17. Jul 23, 2015

### haruspex

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.

18. Jul 24, 2015

### 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?

19. Jul 24, 2015

### 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 $x$.

20. Jul 24, 2015

### haruspex

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.