Step 1

We are given that force $F = 80\;{\rm{N}}$ is acting at the given coordinate system.

We are asked to replace these forces with an equivalent resultant force and a moment at point A.

 
Step 2

Draw a labeled diagram of the given force system.

Find the line vector $\overrightarrow {AC} $ that joins point A$\left( {0,\;0,\;0} \right)\;{\rm{m}}$ and point C$\left( {550,\;400,\; - 200} \right)\;{\rm{mm}}$:

\[\begin{array}{c} \overrightarrow {AC} = \left\{ {\left( {550 - 0} \right)\;\widehat i + \left( {400 - 0} \right)\;\widehat j + \left( { - 200 - 0} \right)\;\widehat k} \right\}\;{\rm{mm}}\\ \overrightarrow {AC} {\rm{ = }}\left\{ {550\;\widehat i + 400\;\widehat j - 200\;\widehat k} \right\}\;{\rm{mm}} \times \left( {\frac{{1\;{\rm{m}}}}{{1000\;{\rm{mm}}}}} \right)\\ \overrightarrow {AC} {\rm{ = }}\left\{ {0.55\;\widehat i + 0.4\;\widehat j - 0.2\;\widehat k} \right\}\;{\rm{m}} \end{array}\]

Calculate the equivalent resultant force:

\[{\overrightarrow F _R} = \left( {F \times \cos 30^\circ \times \sin 40^\circ } \right)\;\widehat i + \left( {F \times \cos 30^\circ \times \cos 40^\circ } \right)\;\widehat j - \left( {F \times \sin 30^\circ } \right)\;\widehat k\]

Substitute the value of $F$ in the above equation:

\begin{array}{c} {\overrightarrow F _R} = \left( {80\;{\rm{N}} \times \cos 30^\circ \times \sin 40^\circ } \right)\;\widehat i + \left( {80\;{\rm{N}} \times \cos 30^\circ \times \cos 40^\circ } \right)\;\widehat j - \left( {80\;{\rm{N}} \times \sin 30^\circ } \right)\;\widehat k\\ {\overrightarrow F _R} = \left\{ {44.53\;\widehat i + 53.07\;\widehat j - 40\;\widehat k} \right\}\;{\rm{N}} \end{array}
 
Step 3

Calculate the moment due to force $F$ about point A:

\[\begin{array}{c} {\overrightarrow M _A} = \overrightarrow {AC} \times {\overrightarrow F _R}\\ = \left| {\begin{array}{*{20}{c}} {\widehat i}&{\widehat j}&{\widehat k}\\ {0.55}&{0.4}&{ - 0.2}\\ {44.53}&{53.07}&{ - 40} \end{array}} \right|\;{\rm{N}} \cdot {\rm{m}}\\ = \left\{ \begin{array}{l} \left( {0.4 \times \left( { - 40} \right) - \left( { - 0.2} \right) \times 53.07} \right)\;\widehat i - \left( {0.55 \times \left( { - 40} \right) - \left( { - 0.2} \right) \times 44.53} \right)\;\widehat j\\ + \left( {0.55 \times 53.07 - 0.4 \times 44.53} \right)\;\widehat k \end{array} \right\}\;{\rm{N}} \cdot {\rm{m}}\\ = \left\{ {\left( { - 16 + 10.6} \right)\;\widehat i - \left( { - 22 + 8.9} \right)\;\widehat j + \left( {29.2 - 17.8} \right)\;\widehat k} \right\}\;{\rm{N}} \cdot {\rm{m}}\\ = \left\{ { - 5.4\;\widehat i + 13.1\;\widehat j + 11.4\;\widehat k} \right\}\;{\rm{N}} \cdot {\rm{m}} \end{array}\]