Need clearing up on vector decomposition in motion physics I've been pursuing physics on my own, and I need something cleared up. Say I have two arbitrary objects, I have their velocities, I know when the collide, I have their normal vectors, etc. I know where a force is applied, how much, which direction(lever arm, etc). I THINK that what I'm trying to do is vector decomposition. I know that torque is the force applied perpendicular to the normal(tangent?), and that linear force is the force applied directly through the center of mass. I believe I intuitively understand these interactions, but I'm having an issue... what is the mass with the splitting of the forces? How 'much' goes through the center of mass, and how much is perpendicular, or torque? I feel like trigonometry may be a simple answer, but I'm unsure of the implementation. My variables I have are the force vector(x, y coordinates scaled to magnitude), and vector(distance vector?) which represents it's offset position from the center of mass. I'm not asking how do I calculate the torque on the CM, but how do I simply determine the force that is perpendicular, not yet crossed by the distance vector(?), and how much of the force goes through the CM, and converts into linear velocity? P.S. This is in a 2D coordinate plane, in a computer simulation I'm making. Labeled as homework as I want to intuitively understand, not actually assigned by a school.
 A: Here's a picture of a rigid object. The center of mass (CM) is labeled. The arrow represents a force being exerted on the surface of the object.

To figure out the component of the force parallel and perpendicular to the surface, first draw the vector from the center of mass to the point of application of the force.

Now draw the two vectors with their tails next to each other. Label the angle between them $\theta$.

At this point it's trigonometry. Use $\cos\theta$ and $\sin\theta$ to extract the components of $\vec{F}$ that are parallel or perpendicular to $\vec{r}$. Be careful about signs. I would just take the absolute value and add the sign in manually if you care about it.
As a side note, I want to stress that I'm not aware of any time in physics that the component of a force that's parallel to $\vec{r}$ is useful. The perpendicular component can certainly be handy when dealing with torque, but I'm not sure what the parallel component is used for. In particular, it is not what you need to calculate the acceleration of the center of mass of the object. For that, you need the entire force $\vec{F}$. See my other answer for more info.
A: There are a lot of ideas in your question above. But I think I see what your major concern is. Here is my rephrasing of what I believe you are asking:

When a rigid object (like a bar or disk, not a point-particle) is subjected to a force $\vec{F}$ how much of this force goes into the linear acceleration of the center of mass, and how much goes into torque (causing angular acceleration)?

The short answer is that a force $\vec{F}$ can fully affect both the center-of-mass acceleration and the angular acceleration of an object without diminishing its effect on either.
The longer answer is that Newton's second law (N2L for short),
$$\vec{F}_\text{net}=m\vec{a},\tag{common form of N2L}$$
is more amazing than it seems. It doesn't matter where or how a given force or forces are exerted on the object; you sum them up, and you can calculate the center-of-mass (CM) acceleration of the object. I personally like the more explicit form
$$\vec{F}_\text{net}=m\vec{a}_\text{CM},\tag{more explicit form of N2L}$$
which reminds me which acceleration N2L refers to. There are indeed consequences of this that may seem counterintuitive. For example, applying a force to the edge of a rigid object will affect the CM acceleration just as much as if the force were applied directly inline with the CM.
Any force $\vec{F}$ on a rigid object will also cause a torque $\vec{\tau}=\vec{r}\times\vec{F}.$ (The cross product in this equation has, built into it, your idea of how far off the CM the torque is applied.) The usual view is that torques cause angular acceleration (or put in a more precise term, it changes the angular momentum $\vec{L}$:
$$\vec{\tau}_\text{net}=\vec{r}_1\times\vec{F}_1+\vec{r}_2\times\vec{F}_2+\cdots=I\vec{\alpha}.\tag{for rigid objects}.$$
(There are more precise forms of the above equation, but this will suffice.) To properly use this rotational analog of Newton's second law, you have to use the full force(s) $\vec{F}$.
The big idea here is that that the full force $\vec{F}$ goes into calculating both the linear acceleration $\vec{a}_\text{CM}$ and the angular acceleration $\vec{\alpha}$.
I do want to comment that one can often interpret $\left|\vec{r}\times\vec{F}\right|=rF\sin\theta=rF_\perp$ as involving only the component of the force that is perpendicular ($\perp$ means perpendicular) to the "lever arm", but this is only an interpretation. (One could also interpret it as $r_\perp F$.) This does not diminish the force's affect on the CM acceleration.
