Step 1

We are given the tensions at pulley $A$ as ${T_{A1}} = 65\;{\rm{N}}$, and ${T_{A2}} = 80\;{\rm{N}}$, and tension on the one side of pulley $B$ ${T_{B1}} = 50\;{\rm{N}}$, the angle $\theta = 45^\circ $.

We are required to determine the horizontal tension $T$, $x,y$ and $z$ components of reaction at the journal bearing $C$ and thrust bearing $D$.

 
Step 2

The free body diagram of the boom is given below.

 
Step 3

Take the moment about $x$ axis.

\[\left( \begin{array}{l} {T_{A1}} \times \left( {\left( {80\;{\rm{mm}}} \right) \times \left( {\frac{{1\;{\rm{m}}}}{{{{10}^3}\;{\rm{mm}}}}} \right)} \right) - {T_{A2}} \times \left( {\left( {80\;{\rm{mm}}} \right) \times \left( {\frac{{1\;{\rm{m}}}}{{{{10}^3}\;{\rm{mm}}}}} \right)} \right)\\ T \times \left( {\left( {150\;{\rm{mm}}} \right) \times \left( {\frac{{1\;{\rm{m}}}}{{{{10}^3}\;{\rm{mm}}}}} \right)} \right) - {T_{B1}} \times \left( {\left( {150\;{\rm{mm}}} \right) \times \left( {\frac{{1\;{\rm{m}}}}{{{{10}^3}\;{\rm{mm}}}}} \right)} \right) \end{array} \right) = 0\]
 
Step 4

Substitute all the values in the above equation.

\[\begin{array}{c} \left( \begin{array}{l} \left( {65\;{\rm{N}}} \right) \times \left( {\left( {80\;{\rm{mm}}} \right) \times \left( {\frac{{1\;{\rm{m}}}}{{{{10}^3}\;{\rm{mm}}}}} \right)} \right)\\ - \left( {80\;{\rm{N}}} \right) \times \left( {\left( {80\;{\rm{mm}}} \right) \times \left( {\frac{{1\;{\rm{m}}}}{{{{10}^3}\;{\rm{mm}}}}} \right)} \right)\\ T \times \left( {\left( {150\;{\rm{mm}}} \right) \times \left( {\frac{{1\;{\rm{m}}}}{{{{10}^3}\;{\rm{mm}}}}} \right)} \right)\\ - \left( {50\;{\rm{N}}} \right) \times \left( {\left( {150\;{\rm{mm}}} \right) \times \left( {\frac{{1\;{\rm{m}}}}{{{{10}^3}\;{\rm{mm}}}}} \right)} \right) \end{array} \right) = 0\\ T \times \left( {0.15\;{\rm{m}}} \right) = \left( \begin{array}{l} \left( {80\;{\rm{N}}} \right) \times \left( {0.08\;{\rm{m}}} \right)\\ + \left( {50\;{\rm{N}}} \right) \times \left( {0.15\;{\rm{m}}} \right)\\ - \left( {65\;{\rm{N}}} \right) \times \left( {0.08\;{\rm{m}}} \right) \end{array} \right)\\ T = \frac{{\left( {8.7\;{\rm{N}} \cdot {\rm{m}}} \right)}}{{\left( {0.15\;{\rm{m}}} \right)}}\\ = 58\;{\rm{N}} \end{array}\]
 
Step 5

Take the moment about $y$ axis.

\[\begin{array}{c} \left( \begin{array}{l} \left( {{T_{A1}} + {T_{A2}}} \right) \times \left( {\left( {450\;{\rm{mm}}} \right) \times \left( {\frac{{1\;{\rm{m}}}}{{{{10}^3}\;{\rm{mm}}}}} \right)} \right)\\ - {C_z} \times \left( {\left( {750\;{\rm{mm}}} \right) \times \left( {\frac{{1\;{\rm{m}}}}{{{{10}^3}\;{\rm{mm}}}}} \right)} \right)\\ - {T_{B1}}\sin \left( {45^\circ } \right) \times \left( {\left( {200\;{\rm{mm}}} \right) \times \left( {\frac{{1\;{\rm{m}}}}{{{{10}^3}\;{\rm{mm}}}}} \right)} \right) \end{array} \right) = 0\\ {C_z} \times \left( {0.75\;{\rm{m}}} \right) = \left( \begin{array}{c} \left( {{T_{A1}} + {T_{A2}}} \right) \times \left( {0.45\;{\rm{m}}} \right)\\ - {T_{B1}}\sin \left( {45^\circ } \right) \times \left( {0.2\;{\rm{m}}} \right) \end{array} \right)\\ {C_z} = \left( {\frac{\begin{array}{l} \left( {{T_{A1}} + {T_{A2}}} \right) \times \left( {0.45\;{\rm{m}}} \right)\\ - {T_{B1}}\sin \left( {45^\circ } \right) \times \left( {0.2\;{\rm{m}}} \right) \end{array}}{{\left( {0.75\;{\rm{m}}} \right)}}} \right) \end{array}\]
 
Step 6

Substitute all the values in the above equation.

\[\begin{array}{c} {C_z} = \frac{{\left( \begin{array}{l} \left( {\left( {65\;{\rm{N}}} \right) + \left( {80\;{\rm{N}}} \right)} \right) \times \left( {0.45\;{\rm{m}}} \right)\\ - \left( {50\;{\rm{N}}} \right)\sin \left( {45^\circ } \right) \times \left( {0.2\;{\rm{m}}} \right) \end{array} \right)}}{{\left( {0.75\;{\rm{m}}} \right)}}\\ = \frac{{\left( {58.17\;{\rm{N}} \cdot {\rm{m}}} \right)}}{{\left( {0.75\;{\rm{m}}} \right)}}\\ = 77.57\;{\rm{N}} \end{array}\]
 
Step 7

Take the moment about $z$ axis.

\[\begin{array}{c} \left( \begin{array}{l} T \times \left( {\left( {200\;{\rm{mm}}} \right) \times \left( {\frac{{1\;{\rm{m}}}}{{{{10}^3}\;{\rm{mm}}}}} \right)} \right)\\ + {T_{B1}}\cos \left( {45^\circ } \right) \times \left( {\left( {200\;{\rm{mm}}} \right) \times \left( {\frac{{1\;{\rm{m}}}}{{{{10}^3}\;{\rm{mm}}}}} \right)} \right)\\ - {C_y} \times \left( {\left( {750\;{\rm{mm}}} \right) \times \left( {\frac{{1\;{\rm{m}}}}{{{{10}^3}\;{\rm{mm}}}}} \right)} \right) \end{array} \right) = 0\\ {C_y} \times \left( {0.75\;{\rm{m}}} \right) = \left( \begin{array}{c} T \times \left( {0.2\;{\rm{m}}} \right)\\ + {T_{B1}}\cos \left( {45^\circ } \right) \times \left( {0.2\;{\rm{m}}} \right) \end{array} \right)\\ {C_y} = \frac{{\left( \begin{array}{c} T \times \left( {0.2\;{\rm{m}}} \right)\\ + {T_{B1}}\cos \left( {45^\circ } \right) \times \left( {0.2\;{\rm{m}}} \right) \end{array} \right)}}{{\left( {0.75\;{\rm{m}}} \right)}} \end{array}\]
 
Step 8

Substitute all the values in the above equation.

\[\begin{array}{c} {C_y} = \frac{{\left( \begin{array}{l} \left( {58.0\;{\rm{N}}} \right) \times \left( {0.2\;{\rm{m}}} \right)\\ + \left( {35.35\;{\rm{N}}} \right) \times \left( {0.2\;{\rm{m}}} \right) \end{array} \right)}}{{\left( {0.75\;{\rm{m}}} \right)}}\\ = \frac{{\left( {18.67\;{\rm{N}} \cdot {\rm{m}}} \right)}}{{\left( {0.75\;{\rm{m}}} \right)}}\\ = 24.9\;{\rm{N}} \end{array}\]
 
Step 9

Since, no force is acting in $x$ direction, therefore the $x$ component of reaction at thrust bearing $D$ is zero.

\[{D_x} = 0\;{\rm{N}}\]
 
Step 10

Equate all the forces acting along $y$ axis.

\[\begin{array}{c} {D_y} + {C_y} - {T_{B1}}\cos \left( {45^\circ } \right) - T = 0\\ {D_y} = {T_{B1}}\cos \left( {45^\circ } \right) + T - {C_y} \end{array}\]
 
Step 11

Substitute all the values in the above equation.

\[\begin{array}{c} {D_y} = \left( {50\;{\rm{N}}} \right)\cos \left( {45^\circ } \right) + \left( {58.0\;{\rm{N}}} \right) - \left( {24.9\;{\rm{N}}} \right)\\ = 68.5\;{\rm{N}} \end{array}\]
 
Step 10

Equate all the forces acting along $z$ axis.

\[\begin{array}{c} {D_z} + {C_z} - {T_{A2}} - {T_{A1}} = 0\\ {D_z} = {T_{A1}} + {T_{A2}} - {C_z} \end{array}\]
 
Step 11

Substitute all the values in the above equation.

\[\begin{array}{c} {D_z} = \left( {65\;{\rm{N}}} \right) + \left( {80\;{\rm{N}}} \right) - \left( {77.57\;{\rm{N}}} \right)\\ = 32.1\;{\rm{N}} \end{array}\]