so actually some force is distributed to center of mass as linear force for whole body, how do i calculate that then? (i mean when i apply not-at-center force, ok i have a torque but what about linear force for the body (where i apply that to the center of mass of the body)

Fb += F

Tb += r x F

This is how we apply a force to a rigid body at non-center of mass, were Fb is force of body, and Tb is torque of body, F is force we are apply, and r as at this radius vector from center of mass. r = Point - COM (center of mass). A point applied to non-COM will generate more energy than one applied to COM due to the definition of work (force * distance). So, we just sum the linear forces, and sum r x F torques. Does this make sense?

More in-depth read: https://www.cs.cmu.edu/~baraff/sigcourse/notesd1.pdf

i am quite surprised that from what you have shown here whenever i 'kick' a penicl with a finger on a tip of it it will add same linear force as i would 'kick' it in the center , i though applying a force on one end of body will produce less linear force applied to its center, than applying the same force to its center

Moreover i just removed some great amount of comments from the physics function i wrote long time ago and i do exactly the same thing (add the same force) :X

i am quite surprised that from what you have shown here whenever i 'kick' a penicl with a finger on a tip of it it will add same linear force as i would 'kick' it in the center , i though applying a force on one end of body will produce less linear force applied to its center, than applying the same force to its center

I want to ask another thing when calculating drag of a body (rotational part) when i am counting through lets name that boxes, the actual velocity of the body is linear vel + rotational vel right?

and rotational vel since its in radians in my case i need to create a vector from actual box position = A to after returnNewPos(box_pos, rot_vel * deltaT) = B

the actual drag vector will be: Normalize( -(B-A) ); ?