Step 1

We are given that the weight of crate is $W = 150\;{\rm{lb}}$.

We are asked that the force developed in each member of truss.

 
Step 2

The free body diagram of the truss is shown as:

Here, ${F_{AB}}$ is the force in member AB, ${F_{AC}}$ is the force in member AC, ${F_{AD}}$ is the force in member AB, ${A_x}$ is the reaction force at point A, ${A_y}$ is the reaction force at point A, ${F_{BC}}$ is the force in member BC, ${F_{BD}}$ is the force in member BD, ${B_y}$ is the reaction force at point B and ${F_{DC}}$ is the force in member DC.

We have the distance between point D and C which is ${d_1} = 6\;{\rm{ft}}$.

We have the distance between point A and B which is ${d_2} = 6\;{\rm{ft}}$.

 
Step 3

To calculate the height of triangle ADB we use the formula:

\[h = \sqrt {d_1^2 - {{\left( {\frac{{{d_2}}}{2}} \right)}^2}} \]
 
Step 4

Substitute the known values in the formula:

\[\begin{array}{c} h = \sqrt {{{\left( {6\;{\rm{ft}}} \right)}^2} - {{\left( {\frac{{{\rm{6}}\;{\rm{ft}}}}{{\rm{2}}}} \right)}^2}} \\ = \sqrt {36\;{\rm{f}}{{\rm{t}}^2} - 9\;{\rm{f}}{{\rm{t}}^2}} \\ = \sqrt {27\;{\rm{f}}{{\rm{t}}^2}} \\ = 5.2\,{\rm{ft}} \end{array}\]
 
Step 5

The position vector of member AD is given as:

\[{\vec r_1} = - \left( {\frac{{{d_2}}}{2}} \right){\bf{\hat i}} + h{\bf{\hat j}}\]
 
Step 6

Substitute the known value in the equation:

\[\begin{array}{c} {{\vec r}_1} = - \left( {\frac{{6\;{\rm{ft}}}}{2}} \right){\bf{\hat i}} + \left( {5.2\;ft} \right){\bf{\hat j}}\\ = - \left( {3\,{\rm{ft}}} \right){\bf{\hat i}} + \left( {5.2\;{\rm{ft}}} \right){\bf{\hat j}} \end{array}\]
 
Step 7

The position vector of member BD is given as:

\[{\vec r_2} = \left( {\frac{{{d_2}}}{3}} \right){\bf{\hat i}} + h{\bf{\hat j}}\]
 
Step 8

Substitute the known value in the equation:

\[\begin{array}{c} {{\vec r}_2} = \left( {\frac{{6\;{\rm{ft}}}}{2}} \right){\bf{\hat i}} + \left( {5.2\;ft} \right){\bf{\hat j}}\\ = \left( {3\,{\rm{ft}}} \right){\bf{\hat i}} + \left( {5.2\;{\rm{ft}}} \right){\bf{\hat j}} \end{array}\]
 
Step 9

The position vector of member AC is given as:

\[{\vec r_3} = - \left( {\frac{{{d_2}}}{2}} \right){\bf{\hat i}} + {d_1}{\bf{\hat j}} + h{\bf{\hat k}}\]
 
Step 10

Substitute the known value in the equation:

\[\begin{array}{c} {{\vec r}_3} = - \left( {\frac{{6\;{\rm{ft}}}}{2}} \right){\bf{\hat i}} + \left( {6\;ft} \right){\bf{\hat j}} + \left( {5.2\;{\rm{ft}}} \right){\bf{\hat k}}\\ = - \left( {3\,{\rm{ft}}} \right){\bf{\hat i}} + \left( {6\;{\rm{ft}}} \right){\bf{\hat j}} + \left( {5.2\;{\rm{ft}}} \right){\bf{\hat k}} \end{array}\]
 
Step 11

The position vector of member BC is given as:

\[{\vec r_4} = \left( {\frac{{{d_2}}}{2}} \right){\bf{\hat i}} + {d_2}{\bf{\hat j}} + h{\bf{\hat k}}\]
 
Step 12

Substitute the known value in the equation:

\[\begin{array}{c} {{\vec r}_4} = \left( {\frac{{6\;{\rm{ft}}}}{2}} \right){\bf{\hat i}} + \left( {6\;ft} \right){\bf{\hat j}} + \left( {5.2\;{\rm{ft}}} \right){\bf{\hat k}}\\ = \left( {3\,{\rm{ft}}} \right){\bf{\hat i}} + \left( {6\;{\rm{ft}}} \right){\bf{\hat j}} + \left( {5.2\;{\rm{ft}}} \right){\bf{\hat k}} \end{array}\]
 
Step 13

The position vector of member AB is given as:

\[{\vec r_5} = - {d_2}{\bf{\hat i}}\]
 
Step 14

Substitute the known value in the equation:

\[{\vec r_5} = - \left( {6\;{\rm{ft}}} \right){\bf{\hat i}}\]
 
Step 15

The position vector of member CD is given as:

\[{\vec r_6} = - {d_1}{\bf{\hat j}}\]
 
Step 16

Substitute the known value in the equation:

\[{\vec r_6} = - \left( {6\;{\rm{ft}}} \right){\bf{\hat j}}\]
 
Step 17

The vector for the weight of crate is given as:

\[\vec W = - \left( {150\;{\rm{lb}}} \right){\bf{\hat k}}\]
 
Step 18

The magnitude of position vector of AD is calculated as:

\[\begin{array}{c} {r_1} = \sqrt {{{\left( { - 3\;{\rm{ft}}} \right)}^2} + {{\left( {5.2\;{\rm{ft}}} \right)}^2}} \\ = \sqrt {9\;{\rm{f}}{{\rm{t}}^2} + 27.04\;{\rm{f}}{{\rm{t}}^2}} \\ = \sqrt {36.04\;{\rm{f}}{{\rm{t}}^2}} \\ = 6\;{\rm{ft}} \end{array}\]
 
Step 19

The magnitude of position vector of BD is calculated as:

\[\begin{array}{c} {r_3} = \sqrt {{{\left( {3\;{\rm{ft}}} \right)}^2} + {{\left( {5.2\;{\rm{ft}}} \right)}^2}} \\ = \sqrt {9\;{\rm{f}}{{\rm{t}}^2} + 27.04\;{\rm{f}}{{\rm{t}}^2}} \\ = \sqrt {36.04\;{\rm{f}}{{\rm{t}}^2}} \\ = 6\;{\rm{ft}} \end{array}\]
 
Step 20

The magnitude of position vector of AC is calculated as:

\[\begin{array}{c} {r_3} = \sqrt {{{\left( { - 3\;{\rm{ft}}} \right)}^2} + {{\left( {6\;{\rm{ft}}} \right)}^2} + {{\left( {5.2\;{\rm{ft}}} \right)}^2}} \\ = \sqrt {9\;{\rm{f}}{{\rm{t}}^2} + 36\;{\rm{f}}{{\rm{t}}^2} + 27.04\;{\rm{f}}{{\rm{t}}^2}} \\ = \sqrt {72.04\;{\rm{f}}{{\rm{t}}^2}} \\ = 8.49\;{\rm{ft}} \end{array}\]
 
Step 21

The magnitude of position vector of BC is calculated as:

\[\begin{array}{c} {r_4} = \sqrt {{{\left( {3\;{\rm{ft}}} \right)}^2} + {{\left( {6\;{\rm{ft}}} \right)}^2} + {{\left( {5.2\;{\rm{ft}}} \right)}^2}} \\ = \sqrt {9\;{\rm{f}}{{\rm{t}}^2} + 36\;{\rm{f}}{{\rm{t}}^2} + 27.04\;{\rm{f}}{{\rm{t}}^2}} \\ = \sqrt {72.04\;{\rm{f}}{{\rm{t}}^2}} \\ = 8.49\;{\rm{ft}} \end{array}\]
 
Step 22

Applying the equilibrium of force equation at point C:

\[{F_{CA}}\left( {\frac{{ - {{\vec r}_3}}}{{{r_3}}}} \right) + {F_{CB}}\left( {\frac{{ - {{\vec r}_4}}}{{{r_4}}}} \right) + {\vec F_{CD}}\left( {\frac{{{{\vec r}_6}}}{{{r_6}}}} \right) + W = 0\]
 
Step 23

Substitute the known values in the equation:

\[\begin{array}{c} \left\{ \begin{array}{l} {F_{CA}}\left( {\frac{{ - \left( { - 3{\bf{\hat i}} + 6{\bf{\hat j}} + 5.2{\bf{\hat k}}} \right)\;{\rm{ft}}}}{{8.49\;{\rm{ft}}}}} \right) + {F_{CB}}\left( {\frac{{ - \left( {3{\bf{\hat i}} + 6{\bf{\hat j}} + 5.2{\bf{\hat k}}} \right)\;{\rm{ft}}}}{{8.49\;{\rm{ft}}}}} \right)\\ + {F_{CD}}\left( {\frac{{ - \left( {6{\bf{\hat j}}} \right)\;{\rm{ft}}}}{{6\;{\rm{ft}}}}} \right) - \left( {150\;{\rm{lb}}} \right){\bf{\hat k}} \end{array} \right\} = 0\\ \left\{ \begin{array}{l} {F_{CA}}\left( {0.353{\bf{\hat i}} - 0.707{\bf{\hat j}} - 0.612{\bf{\hat k}}} \right) + \\ {F_{CB}}\left( { - 0.353{\bf{\hat i}} - 0.707{\bf{\hat j}} - 0.612{\bf{\hat k}}} \right) + \\ {F_{CD}}\left( { - {\bf{\hat j}}} \right) - \left( {150\;{\rm{lb}}} \right){\bf{\hat k}} \end{array} \right\} = 0\\ \left\{ \begin{array}{l} \left( {0.353{F_{CA}} - 0.353{F_{CB}}} \right){\bf{\hat i}} + \\ \left( { - 0.707{F_{CA}} - 0.707{F_{CB}} - {F_{CD}}} \right){\bf{\hat j}} + \\ \left( { - 0.612{F_{CA}} - 0.612{F_{CB}} - 150\;{\rm{lb}}} \right){\bf{\hat k}} \end{array} \right\} = 0 \end{array}\]…… (1)
 
Step 24

Comparing the coefficient of ${\bf{\hat i}}$ on both sides of the equation (1):

\[\begin{array}{c} 0.353{F_{CA}} - 0.353{F_{CB}} = 0\\ 0.353{F_{CA}} = 0.353{F_{CB}}\\ {F_{CA}} = {F_{CB}} \end{array}\]…… (2)
 
Step 25

Comparing the coefficient of ${\bf{\hat j}}$ on both sides of the equation (1):

\[\begin{array}{c} - 0.707{F_{CA}} - 0.707{F_{CB}} - {F_{CD}} = 0\\ {F_{CD}} = - 0.707{F_{CA}} - 0.707{F_{CB}} \end{array}\]
 
Step 26

Substitute the known value in above equation:

\[\begin{array}{c} {F_{CD}} = - 0.707{F_{CA}} - 0.707{F_{CA}}\\ = - 1.414{F_{CA}} \end{array}\]…… (3)
 
Step 27

Comparing the coefficient of ${\bf{\hat k}}$ on both sides of the equation (1):

\[\begin{array}{c} - 0.612{F_{CA}} - 0.612{F_{CB}} - 150\;{\rm{lb}} = 0\\ - 0.612{F_{CA}} - 0.612{F_{CB}} = 150\;{\rm{lb}} \end{array}\]
 
Step 28

Substitute the known value in the equation:

\[\begin{array}{c} - 0.612{F_{CA}} - 0.612{F_{CA}} = 150\;{\rm{lb}}\\ - 1.224{F_{CA}} = 150\;{\rm{lb}}\\ {F_{CA}} = - 122.55\;{\rm{lb}} \end{array}\]

Therefore, the compression force acts in member CA.

 
Step 29

Substitute the known value in equation (2):

\[\begin{array}{c} - 122.55\;{\rm{lb}} = {F_{CB}}\\ {F_{CB}} = - 122.55\;{\rm{lb}} \end{array}\]

Therefore, the compression force acts in member CB.

 
Step 30

Substitute the known value in equation (3):

\[\begin{array}{c} {F_{CD}} = - \left( {1.414} \right)\left( { - 122.55\;{\rm{lb}}} \right)\\ = 173.28\;{\rm{lb}} \end{array}\]

Therefore, the tension force acts in member CD.

 
Step 31

Applying the equilibrium of force equation at point B:

\[{F_{BA}}\left( {\frac{{ - {{\vec r}_5}}}{{{r_5}}}} \right) + {F_{BC}}\left( {\frac{{ - {{\vec r}_4}}}{{{r_4}}}} \right) + {\vec F_{BD}}\left( {\frac{{{{\vec r}_2}}}{{{r_2}}}} \right) + {B_y} = 0\]
 
Step 32

Substitute the known values in the equation:

\[\begin{array}{c} \left\{ \begin{array}{l} {F_{BA}}\left( {\frac{{ - \left( { - 6{\bf{\hat i}}} \right)\;{\rm{ft}}}}{{6\;{\rm{ft}}}}} \right) + {F_{BC}}\left( {\frac{{\left( {3{\bf{\hat i}} + 6{\bf{\hat j}} + 5.2{\bf{\hat k}}} \right)\;{\rm{ft}}}}{{8.49\;{\rm{ft}}}}} \right)\\ + {F_{BD}}\left( {\frac{{\left( {3{\bf{\hat i}} + 5.2{\bf{\hat k}}} \right)\;{\rm{ft}}}}{{6\;{\rm{ft}}}}} \right) + {B_y}{\bf{\hat j}} \end{array} \right\} = 0\\ \left\{ \begin{array}{l} {F_{BA}}\left( {{\bf{\hat i}}} \right) + \\ {F_{BC}}\left( {0.353{\bf{\hat i}} + 0.707{\bf{\hat j}} + 0.612{\bf{\hat k}}} \right) + \\ {F_{BD}}\left( {0.5{\bf{\hat i}} + 0.867{\bf{\hat k}}} \right) - {B_y}{\bf{\hat j}} \end{array} \right\} = 0\\ \left\{ \begin{array}{l} \left( {{F_{BA}} + 0.353{F_{BC}} + 0.5{F_{BD}}} \right){\bf{\hat i}} + \\ \left( {0.707{F_{BC}} + {B_y}} \right){\bf{\hat j}} + \\ \left( {0.612{F_{BC}} + 0.867{F_{BD}}} \right){\bf{\hat k}} \end{array} \right\} = 0 \end{array}\]…… (4)
 
Step 33

Comparing the coefficient of ${\bf{\hat i}}$ on both sides of the equation (4):

\[{F_{BA}} + 0.353{F_{BC}} + 0.5{F_{BD}} = 0\]
 
Step 34

According to the free body diagram, ${F_{BC}} = - {F_{CB}}$. Substitute the known value in above equation:

\[\begin{array}{c} {F_{BA}} + 0.353\left( {{\rm{ - 122}}{\rm{.55}}\;{\rm{lb}}} \right) + 0.5{F_{BD}} = 0\\ {F_{BA}} = 43.26\;{\rm{lb}} - 0.5\;{F_{BD}} \end{array}\]…… (5)
 
Step 35

Comparing the coefficient of ${\bf{\hat j}}$ on both sides of the equation (4):

\[\begin{array}{c} 0.707{F_{BC}} + {B_y} = 0\\ {B_y} = - 0.707{F_{BC}} \end{array}\]
 
Step 36

Substitute the known value in above equation:

\[\begin{array}{c} {B_y} = - \left( {0.707} \right)\left( { - 122.55\;{\rm{lb}}} \right)\\ = 86.64\;{\rm{lb}} \end{array}\]
 
Step 37

Comparing the coefficient of ${\bf{\hat k}}$ on both sides of the equation (4):

\[\begin{array}{c} 0.612{F_{BC}} + 0.867{F_{BD}} = 0\\ 0.867{F_{BD}} = - 0.612{F_{BC}}\\ {F_{BD}} = - 0.706{F_{BC}} \end{array}\]
 
Step 38

Substitute the known value in the equation:

\[\begin{array}{c} {F_{BD}} = - \left( {0.706} \right)\left( { - 122.55\;{\rm{lb}}} \right)\\ = 86.6\;{\rm{lb}} \end{array}\]

Therefore, the tension force acts in member BD.

 
Step 39

Substitute the known value in equation (5):

\[\begin{array}{c} {F_{BA}} = 43.26\;{\rm{lb}} - \left( {0.5} \right)\left( {86.6\;{\rm{lb}}} \right)\\ = 43.26\;{\rm{lb}} - 43.26\;{\rm{lb}}\\ = 0\;{\rm{lb}} \end{array}\]
 
Step 40

Applying the equilibrium of force equation at point A:

\[{F_{AB}}\left( {\frac{{ - {{\vec r}_5}}}{{{r_5}}}} \right) + {F_{AC}}\left( {\frac{{ - {{\vec r}_3}}}{{{r_3}}}} \right) + {\vec F_{AD}}\left( {\frac{{{{\vec r}_1}}}{{{r_1}}}} \right) + \vec A = 0\]
 
Step 41

Substitute the known values in the equation:

\[\begin{array}{c} \left\{ \begin{array}{l} \left( 0 \right)\left( {\frac{{ - \left( {6{\bf{\hat j}}} \right)\;{\rm{ft}}}}{{6\;{\rm{ft}}}}} \right) + {F_{AC}}\left( {\frac{{\left( { - 3{\bf{\hat i}} + 6{\bf{\hat j}} + 5.2{\bf{\hat k}}} \right)\;{\rm{ft}}}}{{8.49\;{\rm{ft}}}}} \right)\\ + {F_{AD}}\left( {\frac{{\left( { - 3{\bf{\hat i}} + 5.2{\bf{\hat k}}} \right)\;{\rm{ft}}}}{{6\;{\rm{ft}}}}} \right) + \left( {{A_x}{\bf{\hat i}} + {A_y}{\bf{\hat j}}} \right) \end{array} \right\} = 0\\ \left\{ \begin{array}{l} {F_{AC}}\left( { - 0.353{\bf{\hat i}} + 0.707{\bf{\hat j}} + 0.612{\bf{\hat k}}} \right) + \\ {F_{AD}}\left( { - 0.5{\bf{\hat j}} + 0.867{\bf{\hat k}}} \right) + \left( {{A_x}{\bf{\hat i}} + {A_y}{\bf{\hat j}}} \right) \end{array} \right\} = 0\\ \left\{ \begin{array}{l} \left( { - 0.353{F_{AC}} + {A_x}} \right){\bf{\hat i}} + \\ \left( {0.707{F_{AC}} - 0.5{F_{AD}} + {A_y}} \right){\bf{\hat j}} + \\ \left( {0.612{F_{AC}} + 0.867{F_{AD}}} \right){\bf{\hat k}} \end{array} \right\} = 0 \end{array}\]…… (6)
 
Step 42

Comparing the coefficient of ${\bf{\hat k}}$ on both sides of the equation (6):

\[\begin{array}{c} 0.612{F_{AC}} + 0.867{F_{AD}} = 0\\ 0.867{F_{AD}} = - 0.612{F_{AC}}\\ {F_{AD}} = - 0.706{F_{AC}} \end{array}\]
 
Step 43

According to the free body diagram, the force in member AC is ${F_{AC}} = - {F_{CA}}$.

Substitute the known value in the equation:

\[\begin{array}{c} {F_{AD}} = - \left( {0.706} \right)\left( { - 122.55\;{\rm{lb}}} \right)\\ = 86.52\;{\rm{lb}} \end{array}\]

Therefore, the tension force acts in member AD.