It can be much simpler and much less error prone (user errors) to transform forces and moments between two points by using a cross product instead of using transformation matrices like:

\begin{bmatrix}  1&0&0\\  0&\cos{\theta} & \sin{\theta}\\  0&-\sin{\theta}& \cos{\theta}  \end{bmatrix}

 The simple trick is to recognize that a cross product can be expressed as a matrix multiply as follows:

\begin{matrix}  a\times{b} & = & skew(a) \cdot{b}\\  \begin{bmatrix}  a_x &a_y &a_z  \end{bmatrix}\times{b}&=&\left[ \begin{matrix}0&-a_z&a_y\\a_z&0&-a_x\\-a_y&a_x&0 \end{matrix}\right]\cdot{b}  \end{matrix}


So to show how we would use the cross product to our advantage, see the the following example.  We want to transform the forces and moments applied to a rigid body at point A to point B:

\begin{matrix}  F_b&=&T_{Fa2b}\cdot{F_a} \\  \begin{bmatrix}  F_{bx}\\  F_{by}\\  F_{bz}\\  F_{brx}\\  F_{bry}\\  F_{brz}  \end{bmatrix} & = &  \begin{bmatrix}  I&0\\skew(ptb-pta)&I\end{bmatrix}\\  \end{matrix}\cdot{\begin{bmatrix}  F_{ax}\\  F_{ay}\\  F_{az}\\  F_{arx}\\  F_{ary}\\  F_{arz}  \end{bmatrix}}

It’s quite easy to see that the forces along the cartesian coordinates remain the same between the two points, but the moments change due to the moment arm (ptb-pta).

It is also useful that displacements can be easily transformed using the transpose of the transformation matrix above:

\begin{matrix}  X_b&=&T_{Fa2b}'\cdot{X_a} \\  \begin{bmatrix}  X_{bx}\\  X_{by}\\  X_{bz}\\  X_{brx}\\  X_{bry}\\  X_{brz}  \end{bmatrix} & = &  \begin{bmatrix}  I&skew(ptb-pta)\\0&I\end{bmatrix}\\  \end{matrix}\cdot{\begin{bmatrix}  X_{ax}\\  X_{ay}\\  X_{az}\\  X_{arx}\\  X_{ary}\\  X_{arz}  \end{bmatrix}}


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