This problem is related to Cross Product - Moments :: Dynamics Please look at that link for the background on the problem I am faced with right now, I have linked a pdf of the book that I am using and I have also supplied a screen shot of the page that I am on. I'll copy down the page just for clarity also. The diagram is at the bottom of the post.
Force $F$ is applied on $P$. The moment of this force with respect to point $O$ is null, because the support of this force with respect to point $Q$ is
$1.1$ $$ M^{F/Q} = P^{P/Q} \times F $$ $1.2$ $$= s(\cos\theta n_x + \sin\theta n_z) \times \frac{-F}{\sqrt{(s^2 + y_0^2)}}(s\cos\theta n_x + y_0n_y + s\sin\theta n_z)$$
$1.3$ $$= \frac{-Fsy_0}{\sqrt{(s^2 + y_0^2)}}(\sin\theta n_x - \cos\theta n_z)$$ Note that the moment of $F$ with respect to $Q$ can also be obtained (and more easily) by
$2.1$ $$M^{F/Q} = P^{O/Q} \times F$$ $2.2$ $$ = -y_0n_y \times \frac{-F}{\sqrt{(s^2 + y_0^2)}}(s\cos\theta n_x + y_0n_y + s\sin\theta n_z)$$
$2.3$ $$ = \frac{-Fsy_0}{\sqrt{(s^2 + y_0^2)}}(\sin\theta n_x - \cos\theta n_z)$$
The moment of vector $F$ with respect to the vertical axis $z$ passing through $Q$ is $3.1$ $$M^{F/z} = M^{F/Q} \cdot n_yn_y $$
$3.2$ $$= -\frac{Fsy_0}{\sqrt{(s^2 + y_0^2)}}(\cos\theta n_z),$$ and the moment of $F$ with respect to axis $Y$ is
$3.3$ $$M^{F/Y} = M^{F/Q} \cdot n_yn_y = 0$$
Questions : A.1] My first question comes from $1.2$ where it is to my understanding that $M^{F/Q}$ is $= P^{P/Q}$ because as the particle (assuming the scenario where the particle is in motion) travels further from point $Q$ the function of the particles distance [$s(\cos\theta n_x + \sin\theta n_z)$] is essentially point $Q$. However, where I seem to get lost is where does the $\sqrt{s^2 + y^2}$ come in, because it looks like in in $1.2$ he is turning $-F$ into it's unit vector by dividing the distance out of the vector thus leaving the unit vector, however, I'm having trouble understanding that if indeed that is true, how $\sqrt{s^2 + y^2}$ is the distance because obviously $a^2 + b^2 = c^2$ but $s^2$ is the hypotenuse if we were to draw a line normal from the particle to the axis parallel to the $X$ axis crossing through $Q$. Thus leading to my confusion because it seems to me $s^2 + y_0^2$ has no meaning. TL;DR What is the physical/Geometric meaning of $\frac{-F}{\sqrt{s^2 + y^2}}$, Also why do we choose to take whatever we're doing with $F$ in the $-F$.
A.2 (is just additional information regarding A.1)]Assuming that I am just retarded and we did create a unit vector in the direction of $-F$ it makes sense to me then that we multiply it by the position vector of particle $P$ because the force is bound to the particle, so obviously we combine both vectors to find the position of $F$ in the direction of $-F$ at the position of $P$
B.2] Next has to do with $1.2 \rightarrow 1.3$, How do we actually cross these vectors to obtain $1.3$ I'm literally looking for the computation from $1.2 \rightarrow 1.3$ because I can't figure out how to $\times$ those with eachother, I'm looking at it like $$ \begin{bmatrix} s\cos\theta n_x \\ 0 \\ s\sin\theta n_z \\ \end{bmatrix} $$ $$ \times \frac{-F}{\sqrt{(s^2 + y_0^2)}} \begin{bmatrix} s\cos\theta n_x \\ y_0n_y \\ s\sin\theta n_z \\ \end{bmatrix} $$
C.] Concluding remarks: So if someone could help give me some intuition on this, take the time out to thoroughly explain the processes here that Tenenbaum was going through it would help me greatly. I'm sure that once I get past this damn page I wont have a problem for a while. I'm trying to scrutinize every little thing here so I have a good foundation to continue learning on as these first couple of ideas are the most important in the whole book.
Here is a link to the PDF of the book Fundamentals of Applied Dynamics - Roberto Tenenbaum
