Short question about moment force

Question

I have a question about moment Forces. Let $\mathbf{e_1}$, $\mathbf{e_2}$ be the unit vectors defining a Cartesion coordinate system $Oxy$. Let $\mathbf{F}$ be the force applied at point $A$.We have: $$\mathbf{F} = F_x \ \mathbf{e_1} + F_y \ \mathbf{e_2}$$ where $$\begin{cases} F_x &= a\\F_y &=-b \end{cases}$$

The moment of force $\mathbf{F}$ about point $O$ at point $A$ is, by definition: $$\mathbf{\mathcal{M}_{/O}}\left(\mathbf{F}(A)\right) = \mathbf{OA} \times \mathbf{F}$$

Therefore the magnitude $M$ of this moment force is $$M = \left(\begin{array}{c} x \\y\end{array}\right) \times \left(\begin{array}{c} F_x \\F_y\end{array}\right)= \left(\begin{array}{c} x \\y\end{array}\right) \times \left(\begin{array}{c} a \\-b\end{array}\right) = -(xb+ya)$$

However, we can also have $$M = \left(\begin{array}{c} y \\x\end{array}\right) \times \left(\begin{array}{c} F_y \\F_x\end{array}\right)= \left(\begin{array}{c} y \\x\end{array}\right) \times \left(\begin{array}{c} -b \\a\end{array}\right) = xb+ya$$

How to define which one to use? What is the convention used for the 1st and 2nd equations? It should depend on the convention used for a positive moment but I can't figure out how it's done.

Edit: Added my intuitive answer

I'll post my intuition just below but ... this is not really solid as it is only intuitive. I'd still like a solid proof.

1. Convention used: Moments are positive when rotation is clockwise (opposite of the geometrical convention)
2. For a positive moment, as rotation is clockwise, The vector along the rotation axis must be pointing outward (away) (defined by $\mathbf{e_3}$)
3. Therefore, $\mathbf{u_r} = \mathbf{e_2} \times \mathbf{e_1} = - (\mathbf{e_1} \times \mathbf{e_2})$ . Then, coordinates of $\mathbf{F}$ and $\mathbf{OA}$ are defined by $(\mathbf{e_2},\mathbf{e_1})$ and not $(\mathbf{e_1},\mathbf{e_2})$
4. Equation 2 for $M=xb+ya$ is correct for this convention (opposite to the geometrical/mathematical one)

Is this correct? How to demonstrate it? Thanks!

Answer 1

There is a strict convention for cross product in three space. Your plane $Oe_1e_2$ is viewed as sitting inside the three space $Oe_1e_2e_3$ with orthonormal basis vectors $e_1, e_2, e_3$ and you have the cross product between two vectors $OA = x \, e_1 + y \, e_2$ and $F= a \, e_1 - b \, e_2$. Then the cross product is linear $$OA \times F = ( x \, e_1 + y \, e_2) \times (a \, e_1 - b \, e_2) = $$ $$= (x\,e_1) \times (a \, e_1) - (x\,e_1) \times (b \, e_2) + (y\,e_2) \times (a \, e_1) - (y\,e_2) \times (b \, e_2)= $$ $$= x a \, (e_1 \times e_1) - x b\,(e_1 \times e_2) + y a\,(e_2\times e_1) - y b\,(e_2 \times e_2)$$ However $e_1 \times e_1 = e_2 \times e_2 = 0$ and $e_1 \times e_2 = e_3$ while $e_2 \times e_1 = - e_3$ so finally $$OA \times F = - x b\,e_3 - y a\, e_3 = -(xb + ya) \, e_3$$ This is the mathematical convention. In your convention, you have ordered your basis differently: $O e_2 e_1 e_3$ and you get $$OA \times F = x b\,e_3 + y a\, e_3 = (xb + ya) \, e_3$$

Answer 2

For 2D problems with cross products always make it a 3D problem with the z-coordinate 0. The cross product is uniquely defined for 3D problems.

This results in the following 2D cross products

$$ \begin{align} \omega \times (x,y) = \begin{pmatrix} 0\\0\\ \omega \end{pmatrix} \times \begin{pmatrix} x \\ y \\ 0 \end{pmatrix} & = (-\omega y, \omega x) \\ (v_x,v_y) \times z = \begin{pmatrix} v_x \\ v_y \\ 0 \end{pmatrix} \times \begin{pmatrix} 0 \\ 0 \\ z \end{pmatrix} & = (v_y z, -v_x z) \\ (x,y) \times (F_x,F_y) = \begin{pmatrix} x\\ y \\ 0 \end{pmatrix} \times \begin{pmatrix} F_x \\ F_y \\ 0 \end{pmatrix} &= F_y x - F_x y \end{align} $$