newtonian mechanics - 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:

1. 
   

   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.

   
   



2. 
   

   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:

   
   
   
   
   1. 
      

      Best work for most of the cases.

      
      
   
   
   2. 
      

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

      
      
   
   
   3. 
      

      Theory is about productivity.

      
      
      
   
   
   
   


3. 
   

   $F=Q \frac{\mathrm{d}x}{\mathrm{d}t}$, or any other orders, actually, I am just asking:

   
   

   for $$F= \frac {d^nR(W,\dot W,...,W^{(n)},x,\dot x,...,x^{(n)};t)}{dt^n}= \frac {dP(m,\dot m,x,\dot x;t)}{dt}$$

   
   


4. 
   

   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

   
   
   


5. 
   

   Why $F= ma$ turns out to be the simplest way to work things out?

   
   

   Please give me some criticisms to my reasonings.