Step 1

We are given that the force applied to the pipe is $F = 80\;{\rm{N}}$.

We are asked to find the moment of force about point B.

 
Step 2

The co-ordinates of points are given as:

\[\begin{array}{l} B\left( {0,400,0} \right)\;{\rm{mm}}\\ C\left( {550,400, - 200} \right)\;{\rm{mm}} \end{array}\]
 
Step 3

The position vector of point B can be given as:

\[{\vec r_B} = \left( {0\hat i + 400\hat j + 0\hat k} \right)\;{\rm{mm}}\]

The position vector of point C can be given as:

\[{\vec r_C} = \left( {550\hat i + 400\hat j - 200\hat k} \right)\;{\rm{mm}}\]
 
Step 4

The position vector of point B relative to point C is calculated as:

\[{\vec r_{BC}} = {\vec r_C} - {\vec r_B}\]
 
Step 5

Substitute the known values in the equation:

\[\begin{array}{c} {{\vec r}_{BC}} = \left( {550\hat i + 400\hat j - 200\hat k} \right)\;{\rm{mm}} - \left( {0\hat i + 400\hat j + 0\hat k} \right)\;{\rm{mm}}\\ = \left( {\left( {550\hat i + 0\hat j - 200\hat k} \right)\;{\rm{mm}}} \right)\left( {\frac{{{{10}^{ - 3}}\;{\rm{m}}}}{{1\;{\rm{mm}}}}} \right)\\ = \left( {0.55\hat i - 0.20\hat k} \right)\;{\rm{m}} \end{array}\]
 
Step 6

The force vector applied at pipe is expressed as:

\[\vec F = F\left( {\left( {\cos 30^\circ \sin 40^\circ } \right)\hat i + \left( {\cos 30^\circ \cos 40^\circ } \right)\hat j + \left( { - \sin 30^\circ } \right)\hat k} \right)\]
 
Step 7

Substitute the known value in the equation:

\[\begin{array}{c} \vec F = \left( {80\;{\rm{N}}} \right)\left( {\left( {0.866} \right)\left( {0.643} \right)\hat i + \left( {0.866} \right)\left( {0.766} \right)\hat j - \left( {0.5} \right)\hat k} \right)\\ = \left( {80\;{\rm{N}}} \right)\left( {0.557\hat i + 0.663\hat j - 0.5\hat k} \right)\\ = \left( {44.56\hat i + 53.04\hat j - 40.0\hat k} \right)\;{\rm{N}} \end{array}\]
 
Step 8

To calculate the moment of force about point B we use the formula:

\[\vec M = {\vec r_{BC}} \times \vec F\]
 
Step 9

Substitute the known values in the formula:

\[\begin{array}{c} M = \left( {0.55\hat i - 0.2\hat k} \right)\;{\rm{m}} \times \left( {44.56\hat i + 53.04\hat j - 40.0\hat k} \right)\;{\rm{N}}\\ {\rm{ = }}\left( {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ {0.55}&0&{ - 0.2}\\ {44.56}&{53.04}&{ - 40.0} \end{array}} \right)\;{\rm{N}} \cdot {\rm{m}}\\ = \left\{ \begin{array}{c} \hat i\left( {\left( 0 \right)\left( { - 40.0} \right) - \left( { - 0.2} \right)\left( {53.04} \right)} \right) - \\ \hat j\left( {\left( {0.55} \right)\left( { - 40.0} \right) - \left( { - 0.2} \right)\left( {44.56} \right)} \right) + \\ \hat k\left( {\left( {0.55} \right)\left( {53.04} \right) - \left( 0 \right)\left( {44.56} \right)} \right) \end{array} \right\}\;{\rm{N}} \cdot {\rm{m}}\\ {\rm{ = }}\left( {10.6\hat i + 13.1\hat j + 29.2\hat k} \right)\;{\rm{N}} \cdot {\rm{m}} \end{array}\]