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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!

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  • $\begingroup$ It seems that my Hello everybody at the beginning cannot be integrated to my message I don't know why ... Even after editing, it keeps deleting that first line so ... Hello everybody! $\endgroup$ – juh Feb 11 '17 at 11:27
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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 $$OA \times F = x b\,e_3 + y a\, e_3 = (xb + ya) \, e_3$$

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  • $\begingroup$ Thank you. Then, basically, the convention is defined by the right hand rule along the axis of ration. If positive moment is associated with clockwise rotation, then the right hand rule says that the rotational axis must point outward (away from me, along $-\mathbf{e_3}$ for the mathematical convention). Therefore, basis must be $Oe_2e_1e_3$ for the cross product of the first two basis vectors to point in the same direction as the moment axis. Is that it? $\endgroup$ – juh Feb 11 '17 at 12:56
  • $\begingroup$ @juh Yes, that's correct. $\endgroup$ – Futurologist Feb 12 '17 at 3:30
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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

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

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