Step 1

We are given the spring constant of springs BA and BC as $k = 500\;{\rm{N/m}}$, unstretched length of ${l_0} = 3\;{\rm{m}}$ and the force applied to the cord is $F = 175\;{\rm{N}}$.

We are asked to determine the displacement d of the cord from the wall.

 
Step 2

The free-body diagram of the cord at point B can be drawn as:

Here, ${F_{BA}}$ is the force in the spring BA, ${F_{BC}}$ is the force in the spring BC, $\theta$ is the angle made by the spring BA with the horizontal and $\alpha$ is the angle made by the spring BC with the horizontal.

On applying equilibrium condition along the vertical direction, we get:

\[\begin{array}{c} \sum {F_y} = 0\\ {F_{BA}}\sin \theta - {F_{BC}}\sin \alpha = 0\\ {F_{BA}}\left( {\frac{{3\;{\rm{m}}}}{{{L_{BA}}}}} \right) - {F_{BC}}\left( {\frac{{3\;{\rm{m}}}}{{{L_{BC}}}}} \right) = 0 \end{array}\]

Stretched lengths ${L_{BA}}$ and ${L_{BC}}$ are equal as both the springs have the same stiffness. So,

\[{F_{BA}} = {F_{BC}}\]

Similarly, on applying equilibrium condition along the horizontal direction, we get:

\[\begin{array}{c} \sum {F_x} = 0\\ {F_{BA}}\left( {\cos \theta } \right) + {F_{BC}}\cos \alpha = F\\ {F_{BA}}\left( {\frac{d}{{{L_{BA}}}}} \right) + {F_{BC}}\left( {\frac{d}{{{L_{BC}}}}} \right) - F = 0\\ F = 2 \times {F_{BC}}\left( {\frac{d}{{{L_{BC}}}}} \right) \end{array}\]

Substitute the values in the above expression, we get:

\[\begin{array}{c} \left( {175\;{\rm{N}}} \right) = 2{F_{BC}} \times \left( {\frac{d}{{{L_{BC}}}}} \right)\\ \left( {175\;{\rm{N}}} \right) = 2k{\delta _{BC}} \times \left( {\frac{d}{{{L_{BC}}}}} \right)\\ \left( {175\;{\rm{N}}} \right) = 2 \times \left( {500\;{\rm{N/m}}} \right) \times \left( {{L_{BC}} - 3\;{\rm{m}}} \right) \times \left( {\frac{d}{{{L_{BC}}}}} \right)\\ \left[ {d - \left( {3\;{\rm{m}}} \right)\left( {\frac{d}{{{L_{BC}}}}} \right)} \right] = \left( {0.175\;{\rm{N}}} \right) \end{array}\]
 
Step 3

On solving the above expression further, we get:

\[\begin{array}{c} \left[ {d - \left( {3\;{\rm{m}}} \right)\left( {\frac{d}{{\sqrt {{{\left( {3\;{\rm{m}}} \right)}^2} + {d^2}} }}} \right)} \right] = \left( {0.175\;{\rm{N}}} \right)\\ \left( {\frac{{d\sqrt {{{\left( {3\;{\rm{m}}} \right)}^2} + {d^2}} - \left( {3\;{\rm{m}}} \right) \times d}}{{\sqrt {{{\left( {3\;{\rm{m}}} \right)}^2} + {d^2}} }}} \right) = \left( {0.175\;{\rm{N}}} \right)\\ d\sqrt {{{\left( {3\;{\rm{m}}} \right)}^2} + {d^2}} - \left( {3\;{\rm{m}}} \right) \times d = \left( {0.175\;{\rm{N}}} \right) \times \sqrt {{{\left( {3\;{\rm{m}}} \right)}^2} + {d^2}} \\ 3d = \sqrt {9 + {d^2}} \left( {d - 0.175} \right) \end{array}\]

On squaring both sides, we get:

\[\begin{array}{c} 9{d^2} = \left( {9 + {d^2}} \right){\left( {d - 0.175} \right)^2}\\ 9{d^2} = \left( {9 + {d^2}} \right)\left[ {{d^2} + 0.030625 - 0.35d} \right]\\ 9{d^2} = \left[ \begin{array}{l} 9{d^2} + 0.2756 - 3.15d + {d^4}\\ + 0.030625{d^2} - 0.35{d^3} \end{array} \right]\\ \left[ \begin{array}{l} {d^4} - 0.35{d^3} + 0.030625{d^2}\\ - 3.15d + 0.2756 \end{array} \right] = 0\\ d = 1.56\;{\rm{m}} \end{array}\]