Resolution of force vectors

Question

So, I have had this query for like the longest time, ever since I first studied this topic. So, take a force vector $F$ that is making an angle of say, 30 degrees with the horizontal axis. Now, I want to resolve the vector $F$ perpendicularly and resolve it in terms of $\sin$ and $\cos$ of the $F$ vector. The questions are:

1. What do we essentially mean by resolving them, I have solved a plethora of mechanics questions but still there is this slight discomfort when I want to explain it to a kid that is how can a Force vector that had no presence in the perpendicular directions, exert a force there? Is it a force field or what?
2. This second question is the one that has been plaguing me for a long time, now after resolving the force vector $F$ into perpendicular components, I plan to take one of those components say the horizontal component, and resolve it again into two perpendicular components, What would those two imply? Are those two secondary resolutions in any way related to the Force vector F, I mean not algebraically but intuitively.
3. If the above question makes any sense, then this is a follow-up to that same stuff. If I keep on taking perpendicular resolutions of the last horizontal resolution, I'll come to a point where one of my resulting resolutions would be such that it would be completely perpendicular to the starting force vector $F$. How am I supposed to reason this, since a force vector must not have any influence on the immediate perpendicular side?

All this is extremely confusing, so whoever answers it, a billion thanks in advance but even though I can solve mechanics problems without having to struggle with these existing blanks in my comprehension, I can't quite feel the essence of this topic. Kindly help.

Answer 1

"Resolving" a vector simply means expressing the vector as the sum of two or more other vectors. So if we have a vector $\vec A$ then we can resolve it by expressing it as

$\vec A = \vec B + \vec C$

and we call $\vec B$ and $\vec C$ the components of $\vec A$. Usually we assume that $\vec B$ and $\vec C$ are perpendicular to one another - so they might be, for example, horizontal and vertical components. In this case

$|\vec B| = |\vec A| \cos \theta$

where $\theta$ is the angle between $\vec A$ and $\vec B$, and also

$|\vec C| = |\vec A| \sin \theta$

In this case, if $\theta=0$ (so $\vec B$ is parallel to $\vec A$ and $\vec C$ is perpendicular to $\vec A$) then $\sin \theta=0$ and $|\vec C| = 0$, which is why we say informally that a vector has no component perpendicular to itself.

However, $\vec B$ and $\vec C$ do not have to be perpendicular to one another. If, for example, $\vec B$ is at an angle of $-45^o$ to $\vec A$ and $\vec C$ is at an angle of $90^o$ to $\vec A$ then

$|\vec B| = \sqrt 2 |\vec A|$

$|\vec C| = |\vec A|$

and now $|\vec C|$ is not zero, even though $\vec C$ is perpendicular to $\vec A$.

Also, we can resolve $\vec A$ into more than two components. For example, if we start with

$\vec A = \vec B + \vec C$

where $\vec B$ and $\vec C$ are perpendicular to one another, and then we resolve $\vec C$ into $\vec D$ and $\vec E$ where $\vec D$ is parallel to $\vec A$ and $\vec E$ is perpendicular to $\vec A$ then we have

$\vec A = \vec B + \vec C = \vec B + \vec D + \vec E$

Now $|\vec E| = |\vec C| \cos \theta =|\vec A| \sin \theta \cos \theta$ (you might like to work out the magnitude of $\vec D$ as well). If $\theta=45^o$ then $|\vec E| = \frac 1 2 |\vec A|$ and once again $\vec A$ has a non-zero component perpendicular to itself.

So, in short, the rule "a vector has no component perpendicular to itself" only applies if we resolve a vector into exactly two components which are perpendicular to one another.