Step 1

We are given the forces ${F_1} = 80\;{\rm{lb}}$ and ${F_2} = 50\;{\rm{lb}}$.

We are asked to express each force in the Cartesian vector form and also to determine the magnitude and coordinate direction angles of the resultant force.

 
Step 2

The coordinates of points of O, A, B and C can be written as:

\[\begin{array}{c} O\left( {0,\;0\;0} \right)\;{\rm{ft}}\\ A\left( {0,\;4,\;0} \right)\;{\rm{ft}}\\ B\left( {2,\;0,\; - 6} \right)\;{\rm{ft}}\\ C\left( { - 2.5,\;0,\;6} \right)\;{\rm{ft}} \end{array}\]

The position vector from origin to point A can be calculated as:

\[\begin{array}{c} {r_{OA}} = A - O\\ {r_{OA}} = \left[ {\left( {0\hat i + 4\hat j + 0\hat k} \right) - \left( {0\hat i + 0\hat j + 0\hat k} \right)} \right]\;{\rm{ft}}\\ {r_{OA}} = \left( {0\hat i + 4\hat j + 0\hat k} \right)\;{\rm{ft}} \end{array}\]

The position vector from origin to point B can be calculated as:

\[\begin{array}{c} {r_{OB}} = B - O\\ {r_{OB}} = \left[ {\left( {2\hat i + 0\hat j - 6\hat k} \right) - \left( {0\hat i + 0\hat j + 0\hat k} \right)} \right]\;{\rm{ft}}\\ {r_{OB}} = \left( {2\hat i + 0\hat j - 6\hat k} \right)\;{\rm{ft}} \end{array}\]

The position vector from origin to point C can be calculated as:

\[\begin{array}{c} {r_{OC}} = C - O\\ {r_{OC}} = \left[ {\left( { - 2.5\hat i + 0\hat j + 6\hat k} \right) - \left( {0\hat i + 0\hat j + 0\hat k} \right)} \right]\;{\rm{ft}}\\ {r_{OC}} = \left( { - 2.5\hat i + 0\hat j + 6\hat k} \right)\;{\rm{ft}} \end{array}\]
 
Step 3

In the Cartesian vector form, we can calculate the length of the rope AB as:

\[\begin{array}{c} {r_{AB}} = {r_{OB}} - {r_{OA}}\\ {r_{AB}} = \left[ {\left( {2\hat i + 0\hat j - 6\hat k} \right) - \left( {0\hat i + 4\hat j + 0\hat k} \right)} \right]\;{\rm{ft}}\\ {r_{AB}} = \left( {2\hat i - 4\hat j - 6\hat k} \right)\;{\rm{ft}} \end{array}\]

The magnitude of the length of the rope AB can be calculated as:

\[\begin{array}{c} \left| {{r_{AB}}} \right| = \sqrt {{{\left( {2\;{\rm{ft}}} \right)}^2} + {{\left( { - 4\;{\rm{ft}}} \right)}^2} + {{\left( { - 6\;{\rm{ft}}} \right)}^2}} \\ \left| {{r_{AB}}} \right| = 7.483\;{\rm{ft}} \end{array}\]

Similarly, in the Cartesian vector form, we can write the length of the rope AC as:

\[\begin{array}{c} {r_{AC}} = {r_{OC}} - {r_{OA}}\\ {r_{AC}} = \left[ {\left( { - 2.5\hat i + 0\hat j + 6\hat k} \right) - \left( {0\hat i + 4\hat j + 0\hat k} \right)} \right]\;{\rm{ft}}\\ {r_{AC}} = \left( { - 2.5\hat i - 4\hat j + 6\hat k} \right)\;{\rm{ft}} \end{array}\]

The magnitude of the length of the rope AC can be calculated as:

\[\begin{array}{c} \left| {{r_{AC}}} \right| = \sqrt {{{\left( { - 2.5\;{\rm{ft}}} \right)}^2} + {{\left( { - 4\;{\rm{ft}}} \right)}^2} + {{\left( {6\;{\rm{ft}}} \right)}^2}} \\ \left| {{r_{AC}}} \right| = 7.632\;{\rm{ft}} \end{array}\]
 
Step 4

To calculate the force vector ${F'_1}$ along the length of a rope AC in the Cartesian vector form, we have:

\[{F'_1} = {F_1}\frac{{{r_{AC}}}}{{\left| {{r_{AC}}} \right|}}\]

Substitute the values in the above expression, we get:

\[\begin{array}{c} {{F'}_1} = \left( {80\;{\rm{lb}}} \right) \times \frac{{\left( { - 2.5\hat i - 4\hat j + 6\hat k} \right)\;{\rm{ft}}}}{{7.632\;{\rm{ft}}}}\\ {{F'}_1} = \left( { - 26.2\hat i - 41.93\hat j + 62.89\hat k} \right)\;{\rm{lb}} \end{array}\]

Similarly, to calculate the force vector ${F'_2}$ along the length of a rope AB in the Cartesian vector form, we have:

\[{F'_2} = {F_2}\frac{{{r_{AB}}}}{{\left| {{r_{AB}}} \right|}}\]

Substitute the values in the above expression, we get:

\[\begin{array}{c} {{F'}_2} = \left( {50\;{\rm{lb}}} \right) \times \frac{{\left( {2\hat i - 4\hat j - 6\hat k} \right)\;{\rm{ft}}}}{{7.483\;{\rm{ft}}}}\\ {{F'}_2} = \left( {13.36\hat i - 26.73\hat j - 40.09\hat k} \right)\;{\rm{lb}} \end{array}\]
 
Step 5

In the cartesian vector form, we can calculate the resultant force as:

\[\begin{array}{c} {F_R} = {{F'}_1} + {{F'}_2}\\ {F_R} = \left[ {\left( { - 26.2\hat i - 41.93\hat j + 62.89\hat k} \right) + \left( {13.36\hat i - 26.73\hat j - 40.09\hat k} \right)} \right]\;{\rm{lb}}\\ {F_R} = \left( { - 12.84\hat i - 68.66\hat j + 22.8\hat k} \right)\;{\rm{lb}} \end{array}\]

The magnitude of the resultant force can be calculated as:

\[\begin{array}{c} {F_R} = \sqrt {{{\left( { - 12.84\;{\rm{lb}}} \right)}^2} + {{\left( {68.66\;{\rm{lb}}} \right)}^2} + {{\left( {22.8\;{\rm{lb}}} \right)}^2}} \\ {F_R} = 73.47\;{\rm{lb}} \end{array}\]

For the coordinate direction angles of the resultant force along the x-axis, we have:

\[\cos \alpha = \frac{{{F_x}}}{{{F_R}}}\] … (1)

Here, ${F_x}$ is the magnitude of the resultant force along x direction.

For the coordinate direction angles of the resultant force along the y-axis, we have:

\[\cos \beta = \frac{{{F_y}}}{{{F_R}}}\] … (2)

Here, ${F_y}$ is the magnitude of the resultant force along y direction.

For the coordinate direction angles of the resultant force along the z-axis, we have:

\[\cos \gamma = \frac{{{F_z}}}{{{F_R}}}\] … (3)

Here, ${F_z}$ is the magnitude of the resultant force along z direction.

Substitute the values in equation (1), we get:

\[\begin{array}{c} \cos \alpha = \frac{{ - 12.84\;{\rm{lb}}}}{{73.47\;{\rm{lb}}}}\\ \alpha = {\cos ^{ - 1}}\left( {0.175} \right)\\ \alpha = 100.07^\circ \end{array}\]

Substitute the values in equation (2), we get:

\[\begin{array}{c} \cos \beta = \frac{{ - 68.66\;{\rm{lb}}}}{{73.47\;{\rm{lb}}}}\\ \beta = {\cos ^{ - 1}}\left( { - 0.935} \right)\\ \beta = 159.23^\circ \end{array}\]

Substitute the values in equation (3), we get:

\[\begin{array}{c} \cos \gamma = \frac{{22.8\;{\rm{lb}}}}{{73.47\;{\rm{lb}}}}\\ \gamma = {\cos ^{ - 1}}\left( {0.310} \right)\\ \gamma = 71.94^\circ \end{array}\]