Step 1

We are given with the unstretched length of spring which is $DB = 2\;{\rm{m}}$ and the mass of crate is $m = 40\;{\rm{kg}}$.

We are asked to determine the stiffness constant of the spring.

 
Step 2

The free body diagram of forces at point D is shown as:

 
Step 3

The unstretched length of the spring can be calculated as:

\[\begin{array}{c} DB' = \sqrt {{{\left( {3\;{\rm{m}}} \right)}^2} + {{\left( {2\;{\mathop{\rm m}\nolimits} } \right)}^2}} \\ = \sqrt {9\,{{\rm{m}}^2} + 4\;{{\rm{m}}^2}} \\ = \sqrt {13} \;{\rm{m}} \end{array}\]
 
Step 4

According to the given diagram, the angle made by force DB with the horizontal is calculated as:

\[\begin{array}{c} \tan \alpha = \frac{{2\;{\rm{m}}}}{{3\;{\rm{m}}}}\\ \alpha = {\tan ^{ - 1}}\left( {0.67} \right)\\ = 33.7^\circ \end{array}\]
 
Step 5

According to the given diagram, the angle made by force CD with the horizontal is calculated as:

\[\begin{array}{c} \tan \beta = \frac{{2\;{\rm{m}}}}{{2\;{\rm{m}}}}\\ \beta = {\tan ^{ - 1}}\left( 1 \right)\\ = 45^\circ \end{array}\]
 
Step 6

The net force acting on point D along x-axis is zero. Therefore, the equation of equilibrium of forces is given as:

\[\begin{array}{c} \sum {F_x} = 0\\ {F_{CD}}\cos \beta - {F_{DB}}\cos \alpha = 0\\ {F_{CD}}\cos \beta = {F_{DB}}\cos \alpha \\ {F_{CD}} = \frac{{\cos \alpha }}{{\cos \beta }}{F_{DB}} \end{array}\]
 
Step 7

Substitute the known values in the equation:

\[\begin{array}{c} {F_{CD}} = \left( {\frac{{\cos 33.6^\circ }}{{\cos 45^\circ }}} \right) \times {F_{DB}}\\ = \left( {\frac{{0.833}}{{0.707}}} \right) \times {F_{DB}}\\ = 1.18{F_{DB}} \end{array}\]
 
Step 8

The net force acting at point D along y-axis is also zero. Therefore, the equation of equilibrium along y-axis can be written as:

\[\begin{array}{c} \sum {F_y} = 0\\ {F_{CD}}\sin \beta + {F_{DB}}\sin \alpha - mg = 0\\ {F_{CD}}\sin \beta + {F_{DB}}\sin \alpha = mg \end{array}\]
 
Step 9

The equation can be further solved as: \[\begin{array}{c} \left( {1.18} \right){F_{BD}}\sin 45^\circ + {F_{DB}}\sin 33.7^\circ = \left( {40\;{\rm{kg}}} \right) \times \left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right) \times \left( {\frac{{1\;{\rm{N}}}}{{1\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}} \right)\\ \left( {1.18} \right){F_{DB}}\left( {0.707} \right) + {F_{DB}}\left( {0.555} \right) = 392\;{\rm{N}}\\ {\rm{1}}{\rm{.389}}{F_{DB}} = 392\;{\rm{N}}\\ {F_{DB}} = 282.2\;{\rm{N}} \end{array}\]
 
Step 10

To calculate the distance of spring stretched we can use the formula:

\[x = DB' - DB\]
 
Step 11

Substitute the known values in the formula:

\[\begin{array}{c} x = \sqrt {13} \;{\rm{m}} - 2\;{\rm{m}}\\ {\rm{ = 3}}{\rm{.61}}\;{\rm{m}} - 2\;{\rm{m}}\\ = 1.61\;{\rm{m}} \end{array}\]
 
Step 12

To calculate the stiffness constant of the spring we can use the formula:

\[k = \frac{{{F_{DB}}}}{x}\]
 
Step 13

Substitute the known values in the formula:

\[\begin{array}{c} k = \frac{{\left( {282.2\;{\rm{N}}} \right)}}{{\left( {1.61\;{\rm{m}}} \right)}}\\ = 175.3\;{\rm{N/m}} \end{array}\]