Step 1

We are given the force $F = \left[ {50i - 40j + 20k} \right]\;{\rm{lb}}$.

We are asked to determine the moment of force about the AB axis and represent the result in Cartesian form.

 
Step 2

The coordinates of points are as follows:

\[\begin{array}{l} A\left( {4, - 0,0} \right)\;{\rm{ft}}\\ B\left( {0,3,0} \right)\;{\rm{ft}}\\ C\left( {4,0,2} \right)\;{\rm{ft}} \end{array}\]

To find the position vector of AB we will use the relation.

\[\begin{array}{l} AB = \vec B - \vec A\\ AB = \left[ {\left( {0i + 3j + 0k} \right) - \left( {4i - 0j + 0k} \right)} \right]\;{\rm{ft}}\\ AB = \left( { - 4i + 3j + 0k} \right)\;{\rm{ft}} \end{array}\]

The unit vector of AB is calculated as:

\[\begin{array}{l} {u_{AB}} = \frac{{AB}}{{\left| {AB} \right|}}\\ {u_{AB}} = \frac{{\left( { - 4i + 3j + 0k} \right)\;{\rm{ft}}}}{{\sqrt {{{\left( { - 4\;{\rm{ft}}} \right)}^2} + {{\left( {3\;{\rm{ft}}} \right)}^2} + {{\left( {0\;{\rm{ft}}} \right)}^2}} }}\\ {u_{AB}} = - 0.8i + 0.6j + 0k \end{array}\]

To find the position vector of AC we will use the relation.

\[\begin{array}{l} AC = \vec C - \vec A\\ AC = \left[ {\left( {4i + 0j + 2k} \right) - \left( {4i - 0j + 0k} \right)} \right]\;{\rm{ft}}\\ AC = \left( {0i + 0j + 2k} \right)\;{\rm{ft}} \end{array}\]
 
Step 3

To find the moment about AB axis we will use the relation,

\[\begin{array}{l} {M_{AB}} = {u_{AB}}\left( {F \times AC} \right)\\ {M_{AB}} = \left( { - 0.8i + 0.6j + 0k} \right)\left[ {\left( {50i - 40j + 20k} \right)\;{\rm{lb}} \times \left( {0i + 0j + 2k} \right)\;{\rm{ft}}} \right]\\ {M_{AB}} = \left| {\begin{array}{*{20}{c}} { - 0.8}&{0.6}&0\\ {0\;{\rm{m}}}&{0\;{\rm{m}}}&{2\;{\rm{m}}}\\ {50\;{\rm{lb}}}&{ - 40\;{\rm{lb}}}&{20\;{\rm{lb}}} \end{array}} \right| \end{array}\]

On further solving the above relation we get,

\[\begin{array}{l} {M_{AB}} = \left( { - 2\;{\rm{ft}}} \right)\left[ {\left( { - 40\;{\rm{N}} \times - {\rm{0}}{\rm{.8}}} \right) - \left( {50\;{\rm{N}} \times {\rm{0}}{\rm{.6}}} \right)} \right]\\ {M_{AB}} = - 4\;{\rm{lb}} \cdot {\rm{ft}} \end{array}\]

To determine the Cartesian form of the result we will use the relation.

\[\begin{array}{l} {\left( {{M_{AB}}} \right)_C} = {u_{AB}} \times {M_{AB}}\\ {\left( {{M_{AB}}} \right)_C} = \left( { - 0.8i + 0.6j + 0k} \right) \times \left( { - 4\;{\rm{lb}} \cdot {\rm{ft}}} \right)\\ {\left( {{M_{AB}}} \right)_C} = \left( {3.2i - 2.4j} \right)\;{\rm{lb}} \cdot {\rm{ft}} \end{array}\]