Step 1

We are given the force on member AC is $F = 150\;{\rm{lb/ft}}$.

We are asked the resultant forces at pins B and C on member ABC of the four-member frame.

 
Step 2

The free body diagram of the member ABC:

We have the length of member AB is $AB = 5\;{\rm{ft}}$.

We have the length of member BC is $BC = 2\;{\rm{ft}}$.

We have the length of member CD is $CD = 4\;{\rm{ft}}$.

 
Step 3

The magnitude of load acting on the member ABC is,

\[W = F\left( {AB + BC} \right)\]
 
Step 4

Substitute the values in the above expression.

\[\begin{array}{l} W = \left( {150\;{\rm{lb/ft}}} \right)\left( {5\;{\rm{ft}} + 2\;{\rm{ft}}} \right)\\ W = 1050\;{\rm{lb}} \end{array}\]
 
Step 5

The magnitude of angle $\theta$ is,

\[\tan \theta = \frac{{CD}}{{AB - BC}}\]
 
Step 6

Substitute the values in the above expression.

\[\begin{array}{c} \tan \theta = \frac{{4\;{\rm{ft}}}}{{5\;{\rm{ft}} - 2\;{\rm{ft}}}}\\ \theta = 53.13^\circ \end{array}\]
 
Step 7

The moment of all the forces about point A is,

\[\begin{array}{l} \Sigma {M_A} = 0\\ \left[ \begin{array}{l} - \left( W \right)\left( {\frac{{AB + BC}}{2}} \right) - {F_{CD}}\left( {AB + BC} \right)\\ + {F_{BE}}\sin \theta \left( {AB} \right) \end{array} \right] = 0 \end{array}\]
 
Step 8

Substitute the values in the above expression.

\[\begin{array}{c} \left[ \begin{array}{l} - \left( {1050\;{\rm{lb}}} \right)\left( {\frac{{5\;{\rm{ft}} + 2\;{\rm{ft}}}}{2}} \right) - {F_{CD}}\left( {5\;{\rm{ft}} + 2\;{\rm{ft}}} \right)\\ + {F_{BE}}\sin 53.13^\circ \left( {5\;{\rm{ft}}} \right) \end{array} \right] = 0\\ - \left( {7{F_{CD}}} \right)\;{\rm{ft}} + \left( {4{F_{BE}}} \right)\;{\rm{ft}} = 3675\;{\rm{lb}} \cdot {\rm{ft}} \end{array}\] \[ - 7{F_{CD}} + 4{F_{BE}} = 3675\;{\rm{lb}}\]   ... (1)
 
Step 9

The free body diagram of the member FED:

We have the length of member FE is $FE = 2\;{\rm{ft}}$.

We have the length of member ED is $ED = 5\;{\rm{ft}}$.

 
Step 10

The moment of all the forces about point F is,

\[\begin{array}{c} \Sigma {M_F} = 0\\ - {F_{BE}}\sin \theta \left( {FE} \right) + {F_{CD}}\left( {FE + ED} \right) = 0 \end{array}\]
 
Step 11

Substitute the values in the above expression.

\[\begin{array}{c} - {F_{BE}}\sin 53.13^\circ \left( {2\;{\rm{ft}}} \right) + {F_{CD}}\left( {2\;{\rm{ft}} + 5\;{\rm{ft}}} \right) = 0\\ - \left( {1.6{F_{BE}}} \right)\;{\rm{ft}} + \left( {7{F_{CD}}} \right)\;{\rm{ft}} = 0 \end{array}\] \[ - 1.6{F_{BE}} + 7{F_{CD}} = 0\]   ... (2)
 
Step 12

On adding equations (1) and (2):

\[\begin{array}{c} - 7{F_{CD}} + 4{F_{BE}} - 1.6{F_{BE}} + 7{F_{CD}} = 3675\;{\rm{lb}}\\ {\rm{2}}{\rm{.4}}{F_{BE}} = 3675\;{\rm{lb}}\\ {F_{BE}} = 1531.25\;{\rm{lb}} \end{array}\]
 
Step 13

Substitute the value of ${F_{BE}}$ in the equation (1).

\[\begin{array}{c} - 7{F_{CD}} + 4\left( {1531.25\;{\rm{lb}}} \right) = 3675\;{\rm{lb}}\\ {F_{CD}} = 350\;{\rm{lb}} \end{array}\]