Parallel Forces in Cartesian Coordinates

[ognik]

Not sure I understand this question, which says:

"Develop a condition when 2 forces are parallel, with & without using Cartesian co-ords."

I think there must be a common normal between the 2, so that $ F_1.\vec{n} = 0 = F_2. (- \vec{n}) $ or $ (F_1 + F_2).\vec{n} = 0 $ - is that correct, and is there any other 'condition' anyone can think of?

I wondered about torque about some point $\vec{r}$ , don't think there is enough info to go there ?
Similarly angular momentum about some point $\vec{r}$, Force is proportional to Torque so nothing new.

 

[Ackbach]

A common normal won't do, I'm afraid. What is more, your equation there assumes the forces have equal magnitude, which is by no means an assumption that's safe.

I would think about adding the two forces together. Think about that graphically. Can you think of a condition that would ensure the two forces are parallel?

Another possibility is to think about the possibility of doing some sort of operation on one of the force vectors to get the other. Can you think of one that would ensure they are parallel?

 

[ognik]

Firstly, what do you think they mean by 'a condition'? I took it to mean any eqtn describing a relationship between them ...

A common normal won't do, I'm afraid. What is more, your equation there assumes the forces have equal magnitude, which is by no means an assumption that's safe.

I didn't intend that they were same magnitude, they should be 2 vectors with unique magnitude and same direction - what makes them look equal in my eqtn.? Apart from that, was my eqtn at least correct?

I would think about adding the two forces together. Think about that graphically. Can you think of a condition that would ensure the two forces are parallel?

I can't see anything from adding them, however if parallel, their dot product should = 1, but that doesn't seem more interesting than the normal above? Is there more I'm not thinking of yet?

I could also find the eqtn for the plane they are on, but again can't see why that would be a good thing?

Another possibility is to think about the possibility of doing some sort of operation on one of the force vectors to get the other. Can you think of one that would ensure they are parallel?

Maybe using position vectors and the normal. If we allow $\vec{F1}$ to pass through the origin, then $\vec{F1}.\vec{n} = 0 $ and $\vec{F2}.\vec{n} = d $ (shortest distance between them) - so $\vec{F2}$ is displaced from $\vec{F1}$ by distance d - is that on the right track?

 

[Ackbach]

I didn't intend that they were same magnitude, they should be 2 vectors with unique magnitude and same direction - what makes them look equal in my eqtn.? Apart from that, was my eqtn at least correct?

If $(\mathbf{F}_1+\mathbf{F}_2)\cdot \mathbf{n}=0$, then it follows that $\mathbf{F}_1 \cdot \mathbf{n}=-\mathbf{F}_2\cdot \mathbf{n}$. Taking the magnitude of both sides yields $F_1 n \cos(\theta)=F_2 n \cos(\varphi)$. But now, if $\mathbf{F}_1$ and $\mathbf{F}_2$ really are parallel, then $\theta=\varphi$, and $F_1=F_2$. This is why your equation proves too much.

Firstly, what do you think they mean by 'a condition'? I took it to mean any eqtn describing a relationship between them ...

All equations are conditions imposed on a situation, aren't they? Equations are pieces of information, or clues.

I can't see anything from adding them, however if parallel, their dot product should = 1, but that doesn't seem more interesting than the normal above? Is there more I'm not thinking of yet?

Yes: think about the Triangle Inequality. It says that the length of one side of a triangle is less than the sum of the lengths of the other two sides. That is, $|a+b|\le |a|+|b|$. When do you get equality?

I could also find the eqtn for the plane they are on, but again can't see why that would be a good thing?

Maybe using position vectors and the normal. If we allow $\vec{F1}$ to pass through the origin, then $\vec{F1}\cdot\vec{n} = 0 $ and $\vec{F2}\cdot\vec{n} = d $ (shortest distance between them) - so $\vec{F2}$ is displaced from $\vec{F1}$ by distance d - is that on the right track?

As a $\LaTeX$ note, use

\cdot

to get the dot product, as in $\mathbf{F}_1 \cdot \mathbf{n}$.

I think that might be overthinking it.

If two vectors are parallel, would you have to rotate one of them to get the other? No? Well, what other operations could you perform on one vector to get a parallel vector? Could you then turn around and make that almost the definition of parallel vectors?

 

[ognik]

Thanks for those various points, making good sense, also the latex tips (now & previous). Sorry, I do sometimes overthink things - which I refer to as me 'missing the bleedin' obvious'

Yes: think about the Triangle Inequality. It says that the length of one side of a triangle is less than the sum of the lengths of the other two sides. That is, $|a+b|\le |a|+|b|$. When do you get equality?

When they are co-linear; but isn't that a special case of parallel?

If two vectors are parallel, would you have to rotate one of them to get the other? No? Well, what other operations could you perform on one vector to get a parallel vector?

Linear Translation? not sure how to say this correctly, but any $\vec{r}$ from $\vec{F_1}$ defines a point in that space, and a line through that point, in the same (or opposite) direction to $F_1$ would produce $F_2 $?

What about cross product $F_1 \times F_2 = 0$? I have heard that used as a definition of parallel vectors? (In 3-D only?)

 

[Ackbach]

Thanks for those various points, making good sense, also the latex tips (now & previous). Sorry, I do sometimes overthink things - which I refer to as me 'missing the bleedin' obvious'

Happens to us all!

When they are co-linear; but isn't that a special case of parallel?

Yes, but with vectors that's the same thing, more-or-less. Don't forget that vectors are the same under translations.

Linear Translation?

As above, linear translation doesn't change the vector at all. So we can't start from two different vectors, translate one of them, and get the other.

not sure how to say this correctly, but any $\vec{r}$ from $\vec{F_1}$ defines a point in that space, and a line through that point, in the same (or opposite) direction to $F_1$ would produce $F_2 $?

Bit hard for me to follow, I'm afraid. Maybe a picture?

What about cross product $F_1 \times F_2 = 0$? I have heard that used as a definition of parallel vectors? (In 3-D only?)

Yes, if the cross product is zero, then the vectors are either parallel or anti-parallel. Problem: you're pretty much confined to 3D space. What if you're only working with vectors in 2D space? What if you're working in special or general relativity, and you need 4 dimensions? Or what if you're a string theorist working in 10 dimensions?

I think there are two conditions you could develop that are independent of the number of dimensions. One involves the Triangle Inequality, and I think we've more-or-less hit upon that one: when you get equality, the vectors are parallel or anti-parallel. You could supplement equality in the Triangle Equality with the additional constraint that the dot product be positive. That would rule out the anti-parallel case.

The other involves a transformation of one vector into the other. The list of all transformations you can perform on a vector are these: translation (which doesn't change the vector at all), rotation, stretching, and shrinking. Think about those last two; does anything suggest itself?

 

[ognik]

Fine with the triangle inequality, although this is not the first time I have constrained myself around vectors with the same direction. In many cases we can apparently talk about the direction of a vector and the vector itself, seemingly interchangeably - which something in my head doesn't like much. Is there some rule/principle which might help me with that?

The other involves a transformation of one vector into the other. The list of all transformations you can perform on a vector are these: translation (which doesn't change the vector at all), rotation, stretching, and shrinking. Think about those last two; does anything suggest itself?

But if we translated some copy of the vector, we would get a parallel vector?

There is also reflection about a parallel line?

Stretching & shrinking - I don't think linear combination applies - unless it's of the basis vectors - but that only produces a parallel vector in a special case ...
Not dissimilar to L/Combos, a scalar multiple of a vector would be parallel, but again - similar to my 1st point in this post - that is colinear to me, I keep thinking parallel must also include being able to be parallel and removed by some distance d?

 

[Ackbach]

Fine with the triangle inequality, although this is not the first time I have constrained myself around vectors with the same direction. In many cases we can apparently talk about the direction of a vector and the vector itself, seemingly interchangeably - which something in my head doesn't like much. Is there some rule/principle which might help me with that?

How about the unit vector? Let's suppose you have a vector $\mathbf{a}$, of length $a$. Then you can define the unit vector in the direction of $\mathbf{a}$ as $\hat{\mathbf{a}}=\mathbf{a}/a$, assuming $a\not=0$.

But if we translated some copy of the vector, we would get a parallel vector?

Technically, you'd get the same vector. Any vector is parallel to itself.

There is also reflection about a parallel line?

Sure, I suppose that would produce a parallel vector. There's probably a way to find out the matrix that does that, but I think that's overkill for your problem.

Stretching & shrinking - I don't think linear combination applies - unless it's of the basis vectors - but that only produces a parallel vector in a special case ...
Not dissimilar to L/Combos, a scalar multiple of a vector would be parallel, but again - similar to my 1st point in this post - that is colinear to me, I keep thinking parallel must also include being able to be parallel and removed by some distance d?

Well, in one sense, all parallel vectors and anti-parallel vectors are collinear: they can be translated to lie on the same line. And, since a translated vector is the same vector, I suppose the words "parallel" and "collinear", could easily be thought of as synonymous.

Now take the flip side. You wrote that you "keep thinking parallel must also include being able to be parallel and removed by some distance d". Can you see that the removal by a distance d is not necessary? It doesn't change the vector, does it?

In summary: we have two conditions that indicate whether two vectors are parallel:

1. Equality in the Triangle Inequality, and a positive dot product (if you need to rule out anti-parallel).
2. One vector is a scalar multiple of the other.

 

[ognik]

Thanks, I had pretty much gotten there, so I hope you didn't mind me exploring around a bit beyond that, I find this kind of interaction helps me quite a bit with those mental blocks one sometimes gets.

Now take the flip side. You wrote that you "keep thinking parallel must also include being able to be parallel and removed by some distance d". Can you see that the removal by a distance d is not necessary? It doesn't change the vector, does it?.

er...you are probably not going to believe this particular mental block, but this is the crux of it for me. 2 parallel lines some distance apart do strike me as being significantly different. Am I wrong to think a vector passing through some specific point(s) is part of it's 'vector-ness' - and the parallel vector doesn't pass through the same point(s)?

Yes, I have read that a vector only has magnitude and direction, but surely in practice, it matters where it is located as well? If the motion of a cars was described by a vector, I would prefer not be standing on a point in its way ...

 

[I like Serena]

Yes, I have read that a vector only has magnitude and direction, but surely in practice, it matters where it is located as well? If the motion of a cars was described by a vector, I would prefer not be standing on a point in its way ...

For a vector it doesn't matter where it is located.
That's why we need 2 vectors for the motion of a car: its position and its velocity. Together they form a line (that predicts where the car will be if it doesn't change speed or direction).