# 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.

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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.

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Wait, hold on a second, are you saying that linear force is the same as force applied? If so, that kind of makes sense of what your last answer said, but doesn't that violate some laws of conservation of momentum and such? – user1413725 Jan 1 '14 at 23:05
I've not heard the term "linear force" before. But the force applied $\vec{F}$ is what you need to figure out the acceleration of the center of mass of an object. You need tho whole thing, not any particular component of the force. The perpendicular component of the force, though, is useful for angular acceleration. – BMS Jan 2 '14 at 1:26
.. no, the above wouldn't violate any conservation laws. This could be another question if you phrase it well. – BMS Jan 2 '14 at 1:47
(hopefully)last question, so I have the angle of the two vectors, how do you exactly... multiply the force, or a vector in general, by an angle? I tried googling, didn't find anything but dot products, which doesn't seem to fit the bill of cross products, unless you do the cross first and...? Can you explain the process of the vector math(again, more calculus, sorry), behind this? – user1413725 Jan 2 '14 at 16:16
You're trying to find the component of a vector. Look up "component decomposition of vectors" or something similar. – BMS Jan 2 '14 at 16:43

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.

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I'm having trouble following you, although what I think your trying to say is that the force in the torque and angular acceleration is.. the same variable? I don't see how that works with the law of conservation of momentum. What is the upside-down T figure mean? What is 'N2L'? My question is how do you get the component of the force that is perpendicular to the lever arm, and the component that passes through the center of mass? – user1413725 Jan 1 '14 at 7:05
I edited to explain N2L and $\perp$. As for calculating the component of the force perpendicular to lever arm, it involves trigonometry. For the component "through" the CM, I would ask why, since this isn't needed for the linear CM acceleration (as your question seems to imply). Is my boxed "question" at the top not related at all to what you want? Does this link help? – BMS Jan 1 '14 at 7:13
Well, it kind of is, but not entirely. I figured that it was some kind of trig, and I found at some discrete place on google that it was a sin of of an angle and a cosine of an angle, which kind of doesn't tell me much. Trig of what? Magnitudes? How do you get direction from that? I see in your answer that you show the equation, where rF sin 0, but that doesn't make any sense. I'm not exactly a mathematician, so I'm guessing that it means lever cross force ... ? I know trig, not really with vectors and trig. And, would the CM be F cos 0? My issue is I'm having a hard time visualizing it. – user1413725 Jan 1 '14 at 7:19
That link(which as I tend to do, overlooked), seems to help. Not entirely sure if I understand it correctly. The order of operations here seem flexed. Where do cross-products fit into this? I imagine that it is r x (F sin 0), but that is multiplying by a scalar which you can't cross, I typed out another 300 or so characters of how I think it could go, and it's more of a mathematical question, but it doesn't really make any sense. – user1413725 Jan 1 '14 at 7:26