Step 1

We are given the following data:

The magnitude of the first force is ${F_1} = 500\;{\rm{lb}}$.

The magnitude of the second force is ${F_2} = 500\;{\rm{lb}}$.

The horizontal distance $AF$ is $AF = 9\;{\rm{ft}}$.

We are asked to determine the force in each member of the truss, and state if the members are in tension or compression.

 
Step 2

We will draw the free-body diagram of the truss.

Here, ${A_y}$ and ${F_y}$ are the vertical component of supports at point $A$ and $F$ respectively.

From the above figure some useful values are:

The value of the distance ${x_1}$ is ${x_1} = 1.5\;{\rm{ft}}$.

The value of the distance ${x_2}$ is ${x_2} = 3\;{\rm{ft}}$.

The value of the distance ${x_3}$ is ${x_3} = 3\;{\rm{ft}}$.

The value of the distance ${x_4}$ is ${x_4} = 3\;{\rm{ft}}$.

The value of the distance ${x_5}$ is ${x_5} = 1.5\;{\rm{ft}}$.

The value of the distance ${y_1}$ is ${y_1} = 6\;{\rm{ft}}$.

The value of the distance ${y_2}$ is ${y_2} = 6\;{\rm{ft}}$.

 
Step 3

The given truss is symmetrically loaded and is symmetric about center. So, the forces in the corresponding members are same.

Force in member $DE$ and $BC$ are equal with each other.

Force in member $EF$ and $AB$ are equal with each other.

We will apply the equations of equilibrium along the $y$ axis.

\[\begin{array}{c} \sum {{F_y}} = 0\\ {A_y} + {F_y} = {F_1} + {F_2} \end{array}\]

Substitute all the known values in the above equation.

\[\begin{array}{c} {A_y} + {F_y} = \left( {500\;{\rm{lb}}} \right) + \left( {500\;{\rm{lb}}} \right)\\ {A_y} + {F_y} = 1000\;{\rm{lb}}\\ {A_y} = \left[ {\left( {1000\;{\rm{lb}}} \right) - {F_y}} \right] \end{array}\]
 
Step 4

We will take the moment about the joint $A$ to obtain ${F_y}$.

\[\begin{array}{c} \sum {{M_A}} = 0\\ \left[ \begin{array}{l} {F_1}\left( {{x_2}} \right) + {F_2}\left( {{x_2} + {x_3}} \right)\\ - {F_y}\left( {AF} \right) \end{array} \right] = 0 \end{array}\]

Substitute all the known values in the above equation.

\[\begin{array}{c} \left[ \begin{array}{l} \left( {500\;{\rm{lb}}} \right)\left( {3\;{\rm{ft}}} \right)\\ + \left( {500\;{\rm{lb}}} \right)\left( {3\;{\rm{ft}}\;{\rm{ + }}\;{\rm{3}}\;{\rm{ft}}} \right)\\ - {F_y}\left( {9\;{\rm{ft}}} \right) \end{array} \right] = 0\\ {F_y} = 500\;{\rm{lb}} \end{array}\]
 
Step 5

Substitute the value ${F_y} = 500\;{\rm{lb}}$ in the equation ${A_y} = \left[ {\left( {1000\;{\rm{lb}}} \right) - {F_y}} \right]$ to obtain ${F_y}$.

\[\begin{array}{c} {A_y} = \left[ {\left( {1000\;{\rm{lb}}} \right) - \left( {500\;{\rm{lb}}} \right)} \right]\\ {A_y} = 500\;{\rm{lb}} \end{array}\]
 
Step 6

The formula to calculate the angle ${\theta _1}$ is given by,

\[{\theta _1} = {\tan ^{ - 1}}\left( {\frac{{{y_2}}}{{{x_1}}}} \right)\]

Substitute all the known values in the above formula.

\[\begin{array}{c} {\theta _1} = {\tan ^{ - 1}}\left( {\frac{{6\;{\rm{ft}}}}{{1.5\;{\rm{ft}}}}} \right)\\ {\theta _1} = {\tan ^{ - 1}}\left( 4 \right)\\ {\theta _1} = 75.96^\circ \end{array}\]
 
Step 7

We will draw the free-body diagram of the joint $D$.

We will apply the equations of equilibrium along the $y$ axis.

\[\begin{array}{c} \sum {{F_y}} = 0\\ {F_{DE}}\sin {\theta _1} - {F_2} = 0 \end{array}\]

Substitute all the known values in the above equation.

\[\begin{array}{c} \left[ {\left( {{F_{DE}}\sin 75.96^\circ } \right) - \left( {500\,{\rm{lb}}} \right)} \right] = 0\\ {F_{DE}} = 515.39\;{\rm{lb}} \end{array}\]

Therefore, the force in the member $DE$ and $BC$ is ${F_{DE}} = {F_{BC}} = 515.39\;{\rm{lb}}$ and the nature of the force is compressive.

 
Step 8

We will apply the equations of equilibrium along the $x$ axis.

\[\begin{array}{c} \sum {{F_x}} = 0\\ \left[ {{F_{CD}} - \left( {{F_{DE}}\cos {\theta _1}} \right)} \right] = 0 \end{array}\]

Substitute all the known values in the above formula.

\[\begin{array}{c} \left[ {{F_{CD}} - \left( {515.39\;{\rm{lb}}} \right)\left( {\cos 75.96^\circ } \right)} \right] = 0\\ {F_{CD}} = 125\;{\rm{lb}} \end{array}\]

The force in the member $CD$ is compressive in nature.

 
Step 9

The formula to calculate the angle ${\theta _3}$ is given by,

\[{\theta _3} = {\tan ^{ - 1}}\left( {\frac{{{y_2}}}{{{x_3} + {x_1}}}} \right)\]

Substitute all the known values in the above formula.

\[\begin{array}{c} {\theta _3} = {\tan ^{ - 1}}\left( {\frac{{6\;{\rm{ft}}}}{{3\;{\rm{ft}} + 1.5\;{\rm{ft}}}}} \right)\\ = {\tan ^{ - 1}}\left( {1.3333} \right)\\ = 53.13^\circ \end{array}\]
 
Step 10

We will draw the free-body diagram of the joint $C$.

We will apply the equations of the equilibrium along $x$ axis.

\[\begin{array}{c} \sum {{F_x}} = 0\\ \left[ {{F_{CD}} + \left( {{F_{CE}}\cos {\theta _3}} \right) - \left( {{F_{BC}}\cos {\theta _1}} \right)} \right] = 0 \end{array}\]

Substitute all the known values in the above equation.

\[\begin{array}{c} \left[ \begin{array}{l} \left( {125\,{\rm{lb}}} \right) + \left( {{F_{CE}}\cos 53.31^\circ } \right)\\ - \left( {515.39\;{\rm{lb}}} \right)\left( {\cos 75.96^\circ } \right) \end{array} \right] = 0\\ {F_{CE}} = 0\;{\rm{lb}} \end{array}\]
 
Step 11

We will draw the free-body diagram of the joint $A$.

The formula to calculate the angle ${\theta _4}$ is given by,

\[{\theta _4} = {\tan ^{ - 1}}\left( {\frac{{{x_5}}}{{{y_1}}}} \right)\]

Substitute all the known values in the above formula.

\[\begin{array}{c} {\theta _4} = {\tan ^{ - 1}}\left( {\frac{{1.5\;{\rm{ft}}}}{{6\;{\rm{ft}}}}} \right)\\ {\theta _4} = 14.036^\circ \end{array}\]
 
Step 12

We will apply the equations of the equilibrium along $y$ axis.

\[\begin{array}{c} \sum {{F_y}} = 0\\ \left[ {\left( {{F_{AB}}\cos {\theta _4}} \right) + {A_y}} \right] = 0 \end{array}\]

Substitute all the known values in the above equation.

\[\begin{array}{c} \left[ {\left\{ {{F_{AB}}\cos \left( {14.036} \right)^\circ } \right\} + \left( {500\;{\rm{lb}}} \right)} \right] = 0\\ {F_{AB}} = 515.38\;{\rm{lb}} \end{array}\]

Therefore, the magnitude of the force in the member $AB$ and $EF$ is ${F_{AB}} = {F_{EF}} = 515.38\;{\rm{lb}}$ and the nature of the force is compression.

 
Step 13

We will draw the free-body diagram of the joint $E$.

We will apply the equations of the equilibrium along $y'$ axis.

\[\begin{array}{c} \sum {{F_{y'}}} = 0\\ \left( {{F_{BE}}\cos \theta } \right) = 0\\ {F_{BE}} = 0\;{\rm{lb}} \end{array}\]
 
Step 14

The equation to calculate the angle ${\theta _5}$ is given by,

\[{\theta _5} = {\tan ^{ - 1}}\left( {\frac{{{y_1}}}{{{x_6}}}} \right)\]

Substitute all the known values in the above formula.

\[\begin{array}{c} {\theta _5} = {\tan ^{ - 1}}\left( {\frac{{6\;{\rm{ft}}}}{{7.5\;{\rm{ft}}}}} \right)\\ {\theta _5} = 38.66^\circ \end{array}\]
 
Step 15

We will draw the free-body diagram of the joint $F$.

We will apply the equations of the equilibrium along $y$ axis.

\[\begin{array}{c} \sum {{F_y}} = 0\\ \left[ {{F_y} + \left( {{F_{EF}}\cos {\theta _4}} \right) + \left( {{F_{BF}}\sin {\theta _5}} \right)} \right] = 0 \end{array}\]

Substitute all the known values in the above equation.

\[\begin{array}{c} \left[ \begin{array}{l} \left( {500\;{\rm{lb}}} \right) + \left( { - 515.38\;{\rm{lb}}} \right)\left( {\cos 14.036^\circ } \right)\\ + \left( {{F_{BF}}\sin 38.66^\circ } \right) \end{array} \right] = 0\\ \left( {{F_{BF}}\sin 38.66^\circ } \right) = 0\\ {F_{BF}} = 0\;{\rm{lb}} \end{array}\]