Step 1

We are given the force ${\bf{F}}$, the reaction at the bearing $C$ along $x$-axis is ${C_x} = 50\,{\rm{N}}$

We are asked to determine the magnitude of $F$.

 
Step 2

First draw the free body diagram for the given system.

Figure-(1)

We have given the distance between points $A$ and $B$ along $y$-axis as $a = 2\,{\rm{m}}$.

We have given the distance between points $A$ and $B$ along $x$-axis as $b = 1\,{\rm{m}}$.

We have given the distance between points $B$ and $C$ along $x$-axis as $c = 0.75\,{\rm{m}}$.

We have given the distance between points $B$ and $C$ along $z$-axis as $d = 2\,{\rm{m}}$.

 
Step 3

Find the component force of $F$ along $x$-axis using the following relation.

\[{F_x} = F\cos \theta \cos \phi \]

Substitute the known values in the above expression.

\[\begin{array}{c} {F_x} = F\cos 60^\circ \cos 30^\circ \\ = 0.4330F $ \end{array}\]
 
Step 4

Find the component force of $F$ along $y$-axis using the following relation.

\[{F_y} = F\cos \theta \sin \phi \]

Substitute the known values in the above expression.

\[\begin{array}{c} {F_y} = F\cos 60^\circ \sin 30^\circ \\ = 0.25F \end{array}\]
 
Step 5

Find the component force of $F$ along $z$-axis using the following relation.

\[{F_z} = F\sin \theta \]

Substitute the known values in the above expression.

\[\begin{array}{c} {F_z} = F\sin 60^\circ \\ = 0.8660F \end{array}\]
 
Step 6

Find the relation between ${C_y}$ and ${B_z}$ by taking moment along $x'$-axis under equilibrium condition using the following relation.

\[ - {C_y}\left( d \right) + {B_z}\left( a \right) - {F_z}\left( a \right) = 0\]

Substitute the known values in the above expression.

\[\begin{array}{c} - {C_y}\left( {2\,{\rm{m}}} \right) + {B_z}\left( {2\,{\rm{m}}} \right) - \left( {0.8660F} \right)\left( {2\,{\rm{m}}} \right) = 0\\ {C_y} = {B_z} - 0.8660F......\left( 1 \right) \end{array}\]
 
Step 7

Find the relation between ${C_x}$ and ${C_y}$ by taking moment along $z$-axis under equilibrium condition using the following relation.

\[ - {C_y}\left( {b + c} \right) - {C_x}\left( a \right) - {B_y}\left( b \right) - {F_x}\left( a \right) = 0\]

Substitute the known values in the above expression.

\[\begin{array}{c} - {C_y}\left( {1\,{\rm{m}} + 0.75\,m} \right) - \left( {50\,{\rm{N}}} \right)\left( {2\,{\rm{m}}} \right) - {B_y}\left( {1\,{\rm{m}}} \right) - \left( {0.4330F} \right)\left( {2\,{\rm{m}}} \right) = 0\\ - 1.75{C_y} - 100\;{\rm{N}} \cdot {\rm{m}} - {B_y} - 0.866F = 0\\ 1.75{C_y} + {B_y} + 0.866F = - 100\,{\rm{N}}\,......\left( 2 \right) \end{array}\]
 
Step 8

Find the relation between ${C_x}$ and ${B_z}$ by taking moment along $y$-axis under equilibrium condition using the following relation.

\[{B_z}\left( b \right) + {C_x}\left( d \right) = 0\]

Substitute the known values in the above expression.

\[\begin{array}{c} {B_Z}\left( {1\,{\rm{m}}} \right) + \left( {50\,{\rm{N}}} \right)\left( {2\,{\rm{m}}} \right) = 0\\ {B_z}\left( {1\,{\rm{m}}} \right) + 100\,{\rm{N}} \cdot {\rm{m}} = 0\,\\ {B_z} = - 100\;{\rm{N}} \end{array}\]
 
Step 9

Find the relation between ${B_y}$ and ${C_y}$ by adding all the forces along $y$-axis under equilibrium condition using the following relation.

\[{B_y} + {C_y} + {F_y} = 0\]

Substitute the known values in the above expression.

\[{B_y} + {C_y} + 0.25F = 0\]
 
Step 10

To find reaction ${C_y}$ use the following reaction.

\[{C_y} = {B_z} - 0.8660F\]

Substitute the known values in the above expression.

\[{C_y} = - 100\,{\rm{N}} - 0.8660F\]
 
Step 11

Find the ${B_y}$ relation using the following relation.

\[1.75{C_y} + {B_y} + 0.866F = - 100\,{\rm{N}}\]

Substitute the known values in the above expression.

\[\begin{array}{c} 1.75\left( { - 100\,{\rm{N}} - 0.866F} \right) + {B_y} + 0.866F = - 100\,{\rm{N}}\\ - {\rm{175}}\,{\rm{N}} - 1.5155F + {B_y} + 0.866F = - 100\,{\rm{N}}\\ {B_y} - 0.6495F = 75\,{\rm{N}}\\ {B_y} = 75\,{\rm{N}} + 0.6495F \end{array}\]
 
Step 12

Find the value of $F$ using the following relation.

\[{B_y} + {C_y} + 0.25F = 0\]

Substitute the known values in the above expression.

\[\begin{array}{c} 75\;{\rm{N}} + 0.6495F + \left( { - 100\,{\rm{N}} - 0.8660F} \right) + 0.25F = 0\\ - 25\,{\rm{N}} + 0.0335F = 0\\ F = \frac{{25\;{\rm{N}}}}{{0.0335}}\\ F = 746\,{\rm{N}} \end{array}\]
 
Step 13

To find ${C_y}$ use the following relation.

\[{C_y} = - 100\,{\rm{N}} - 0.8660F\]

Substitute the known values in the above expression.

\[\begin{array}{c} {C_y} = - 100\,{\rm{N}} - 0.8660\left( {746\,{\rm{N}}} \right)\\ = - 746.04\;{\rm{N}} \end{array}\]
 
Step 14

Now find ${B_y}$ using the following relation.

\[{B_y} = 75\,{\rm{N}} + 0.6495F\]

Substitute the known values in the above expression.

\[\begin{array}{c} {B_y} = 75\,{\rm{N}} + 0.6495\left( {746\,{\rm{N}}} \right)\\ = 559.5\;{\rm{N}} \end{array}\]