Step 1

We are given the force acts on the bracket is $F = 30\;{\rm{N}}$.

We are asked the moment of the force about the $a - a$ axis of the pipe, and coordinate directional angles of F to produce the maximum moment about the $a - a$ axis and the magnitude of maximum moment.

 
Step 2

The following is the diagram of the system:

We have the angle of force from x-axis is $\alpha = 60^\circ $.

We have the angle of force from y-axis is $\beta = 60^\circ $.

We have the angle of force from z-axis is $\gamma = 45^\circ $.

We have the x-component of position vector from $a - a$ axis to the line of action of force is ${r_1} = - 0.1\;{\rm{m}}$.

We have the y-component of position vector from $a - a$ axis to the line of action of force is ${r_2} = 0\;{\rm{m}}$.

We have the z-component of position vector from $a - a$ axis to the line of action of force is ${r_3} = 0.15\;{\rm{m}}$.

We have the unit vector of $a - a$ axis is ${U_a} = 0i + 1j + 0k$.

 
Step 3

The Cartesian form of the force acting on the bracket is,

\[F = F\left( {\cos \alpha i + \cos \beta j + \cos \gamma k} \right)\]
 
Step 4

Substitute the values in the above expression.

\[\begin{array}{l} F = \left( {30\;{\rm{N}}} \right)\left( {\cos 60^\circ i + \cos 60^\circ j + \cos 45^\circ k} \right)\\ F = \left( {15i + 15j + 21.21k} \right)\;{\rm{N}} \end{array}\]
 
Step 5

The position vector from line of action of force to $a - a$ axis is,

\[R = \left( {{r_1}i + {r_2}j + {r_3}k} \right)\]
 
Step 6

Substitute the values in the above expression.

\[R = \left( { - 0.1i + 0j + 0.15k} \right)\;{\rm{m}}\]
 
Step 7

The moment of force about $a - a$ axis is,

\[M = {U_a} \cdot \left( {F \times r} \right)\]
 
Step 8

Substitute the values in the above expression.

\[\begin{array}{l} M = \left( {0i + 1j + 0k} \right) \cdot \left[ {\left( {15i + 15j + 21.21k} \right)\;{\rm{N}} \times \left( { - 0.1i + 0j + 0.15k} \right)\;{\rm{m}}} \right]\\ M = \left| {\begin{array}{*{20}{c}} 0&1&0\\ {15}&{15}&{21.21}\\ { - 0.1}&0&{0.15} \end{array}} \right|\\ M = \left\{ \begin{array}{l} 0\left\{ {\left( {15 \times 0.15} \right) - \left( {21.21 \times 0} \right)} \right\} - 1\left[ {\left\{ {\left( {15 \times 0.15} \right)} \right\} - \left\{ {21.21 \times \left( { - 0.1} \right)} \right\}} \right]\\ + 0\left[ {\left\{ {15 \times 0} \right\} - \left\{ {15 \times \left( { - 0.1} \right)} \right\}} \right] \end{array} \right\}\\ M = - 4.37\;{\rm{N}} \cdot {\rm{m}} \end{array}\]
 
Step 9

The position vector of force F perpendicular to the position vector R is,

\[{U_F} = \frac{{\left( {0.15i - 0.1k} \right)\;{\rm{m}}}}{{\left| R \right|}}\]
 
Step 10

Substitute the values in the above expression.

\[\begin{array}{l} {U_F} = \frac{{\left( {0.15i - 0.1k} \right)\;{\rm{m}}}}{{\sqrt {{{\left( { - 0.1} \right)}^2} + {{\left( 0 \right)}^2} + {{\left( {0.15} \right)}^2}} \;{\rm{m}}}}\\ {U_F} = 0.832i + 0.555j \end{array}\]
 
Step 11

The coordinate directional angles of F to produce the maximum moment about the $a - a$ axis are,

From x-axis

\[\begin{array}{l} \alpha = {\cos ^{ - 1}}\left( {0.832} \right)\\ \alpha = 33.7^\circ \end{array}\]

From y-axis

\[\begin{array}{l} \beta = {\cos ^{ - 1}}\left( 0 \right)\\ \beta = 90^\circ \end{array}\]

From z-axis

\[\begin{array}{l} \gamma = {\cos ^{ - 1}}\left( {0.555} \right)\\ \gamma = 56.3^\circ \end{array}\]
 
Step 12

The magnitude of maximum moment is,

\[{M_{\max }} = F\left| R \right|\]
 
Step 13

Substitute the values in the above expression.

\[\begin{array}{l} {M_{\max }} = \left( {30\;{\rm{N}}} \right) \cdot \sqrt {{{\left( { - 0.1} \right)}^2} + {{\left( 0 \right)}^2} + {{\left( {0.15} \right)}^2}} \;{\rm{m}}\\ {M_{\max }} = 5.41\;{\rm{N}} \cdot {\rm{m}} \end{array}\]