Why is $F$ defined as $ma$? I am currently using classical mechanics as a tool to learn how to construct a theory as an end. Therefore, I have a few questions, and a few reasonings behind them (I only consider reality, not imagined world), and I would like your suggestions or corrections. 
To make things simple, I first consider when mass is a constant, so $F=\frac{\mathrm{d}p}{\mathrm{d}t}$ can be written as $F=ma$.
Why $F = m a$? Instead of, say, I define new constants or variables, and  claim:
$$
I~=~N \frac{\mathrm{d}^3x}{\mathrm{d}t^3}
\,.
\tag{1}
$$
Or using the same $F$ both literally and physically, and claim 
$$
F~=~N\left(x,\,\dot{x},\,\ddot{x},\,{\dots},\,x^{\left(n\right)};\,t\right) \frac{\mathrm{d}^3x}{\mathrm{d}t^3}
\,,
\tag{2}
$$or$$
F~=~Q \frac{\mathrm{d}x}{\mathrm{d}t}
\,.
\tag{3}
$$
I mean, force is not something that you can touch nor see, either is mass;
so, we pretty much define them via indirect observations, recording the data, and then comparing with other objects, in order to give operational definitions on mass and force (in case of "mass", we will have to define a unit mass first, and then use $F= ma$ as operational definition for mass).
Under this line of reasoning, why can't we work things in a simple way, e.g.$$
I~=~Q \dot{x}
\,?
$$
Questions:


*

*Assuming a reality in which a falling object $a=g$ [?]:$$
  \begin{align}
  I & ~=~ Q \frac{\mathrm{d}x}{\mathrm{d}t} \\
  \dot{x} & ~=~ gt \\
  I & ~=~t \\
  Q & ~=~ \frac{1}{g} \,,
  \end{align}
  $$
I define such way $I=Q \dot{x}$, it is just the law of falling.
I assume that, yes, we can define things such way, but it has nothing to do with the subject we are dealing right now - inertia. Lesson learned: Know what you want to deal with.

*Noting $\operatorname{Eq}{\left(2\right)}$, since I only consider our reality, that means $F=ma$ is truth.  So I will comparing $F=Na'$ with $F=ma$ with a few cases (constant $a$, and SHM [?]).
And it turns out that $F=Na'$ is far more complicated (often a variable) and $N$ is going to be ill defined given constant $a$.
Reality: A falling object $a=g$:
$$
  \begin{array}{ccccc}
  F & =    & N\dot{a}          & = & 0                  \\
    & \to & N\dot{a}           & = & mg                 \\
    & \to & N\phantom{\dot{a}} & = & \frac{mg}{\dot{a}} \\
  \end{array}
  $$
as $\dot{a}=0$, $N\left(x,t\right)$ is ill defined.
Reality: SHM:
$$m\ddot x=-Ax$$
$$\to \dddot x= -A\dot x/m$$
Given
$$Na'=ma$$
$$\to N= ma/a'=-m^2\ddot x/\dot x A=(-m^2/A)(d\dot x/dx)$$
Too complicated. (comparing with the fact that $m$ is just a constant.)
Lessons learned:


*

*Best work for most of the cases.

*When a constant will do, don't use a function.

*Theory is about productivity.


*$F=Q \frac{\mathrm{d}x}{\mathrm{d}t}$, or any other orders, actually, I am just asking: 
for $$F= \frac {d^nR(W,\dot W,...,W^{(n)},x,\dot x,...,x^{(n)};t)}{dt^n}= \frac {dP(m,\dot m,x,\dot x;t)}{dt}$$

*Is there exist a n, such that there exist a R makes things simpler than $F=\dot P$ ? Define $R(W,x;t)=W(x,\dot x,\ddot x,...x^{(n)};t) x$, ($R$, $W$ new variables, $x$ is position function of time)
if there doesn't exist such an $n$, then 

*Why $F= ma$ turns out to be the simplest way to work things out?
Please give me some criticisms to my reasonings. 
 A: 
I mean, force is not a observable, either is mass

If this were the case, then it would indeed be completely arbitrary how to write Newton's second law, and we wouldn't have such a law of physics at all.
Force and mass are both observable.
I can measure mass like this. Start with some arbitrary object to be my basic unit, and then making another object that balances against it on a double-pan balance. In this way I can make up as many of these unit masses as I need. Now if 7 of these balance against my dog, I know that my dog's mass is 7 units.
Force is something we measure directly using spring scales, bathroom scales, etc.
It is possible to consider F=ma as a definition of force, or a definition of mass, but you certainly can't consider it as defining both force and mass. Even if you do consider it to be the fundamental definition of force, you are constrained by the fact that force can be measured in other ways, such as by spring scales.
A: From a philosophical perspective, one may prefer the original $F=\dot{p}$ over your formulism simply by Occam's razor. $F=\dot{p}$ is sufficient to describe all classical dynamics, and by adding another derivative you have only added another layer of complexity. First, your equation needs to be integrated to recover Newton's Law, and in the process a constant of integration will appear which will need to be fine tuned in order to agree with Newton's law and therefore classical reality. From a Popperian perspective, Newton's theory would be prefered over yours because, while they both account for classical mechanics, yours adds another layer of complexity (for which you haven't provided a satisfactory motivation, and which solves no new problem).
From a mathematical perspective, if you're willing to begin classical mechanics beginning from a configuration space, then $F=\dot{p}$ arises naturally as a consequence of some differential geometric machinery on that space. But I don't see why this would make it preferred over your method except by the Popperian criteria I outlined above.
A: It is just convenient to define force from  perspective of modern physics,it is not necessary,but if we had excluded the force from our theory,we would get a little bit more complex theory,you may do it to get what happens,and since we are used to it ,it is not harmful to omit it from modern theory.
Force is not even like an idea in different but equivalent axiomizated  system,all idea is necessary or at least independent,force is not independent,and I don't think it is observable since what we observe are other quantity of object.
A: A theory is based on some primitive concepts you never define. What you define is the "equation of motion" describing the (algebraic) relations between all these primitive concepts. This is what a theory does, and it does not evaluate or explicit a form of the primitive concepts: this is what a model does.
Ex: In Newtonian mechanics, the primitive concepts are the ones of (absolute) space, time, inertial mass, and force. Now, Newton tells us that absolute change is equivalent to acceleration, and that acceleration is induced by an interaction process (ie, a force acting on some object and reacting back). The force has to respect some symmetries (basically, be invariant under Galilean transformation  and maybe more if you want). This tells us that $F = ma$.
Now, a model is either 
1) giving an explicit form of $a$ knowing $m$, and computing back the form of $F$ thanks to 2nd law,
2) giving an explicit form of $F$ (usually as a function of relative space-time position, and relative velocity of at least two bodies) knowing $m$, and computing back the acceleration of such a force on a given object. In particular, note that if you do things rigorously, 2nd law is a system of equations (each for each physical system you consider, because of 3rd law).
PS: you might find the first five part of that http://www.sure.zhln.eu/wp/?p=126 interesting. You'll notice that what you claim as being "too complicated" actually suffers from much more philosophical problems than just complication. It is yet only an unfinished draft, but you'll find all answers to what you ask there.
