Step 1

We are given that the horizontal force applied to the scissors is $P = 600\;{\rm{N}}$, the stiffness constant of the spring is $k = 15\;{\rm{kN/m}}$, the angle made by links for equilibrium is $\theta = 60^\circ $ and the angle made by spring when it is unstretched is $\theta = 15^\circ $.

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

 
Step 2

The free body diagram of the system is shown as:

Here, ${F_S}$ is the spring force, ${E_x}$ is the reaction force at point E, ${E_y}$ is the reaction force at point E and ${D_x}$ is the reaction force at point D.

We have the length of links is $l = 200\;{\rm{mm}}$.

 
Step 3

From the given diagram,

\[\begin{array}{c} \sin \theta = \frac{{{x_A}}}{l}\\ {x_A} = l\sin \theta \end{array}\]
 
Step 4

Differentiate the above equation:

\[\begin{array}{c} \delta {x_A} = \delta \left( {l\sin \theta } \right)\\ = l\cos \theta \delta \theta \end{array}\]
 
Step 5

From the given diagram,

\[\begin{array}{c} \sin \theta = \frac{{{x_B}}}{{3l}}\\ {x_B} = 3l\sin \theta \end{array}\]
 
Step 6

Differentiate the above equation:

\[\begin{array}{c} \delta {x_B} = \delta \left( {3l\sin \theta } \right)\\ = 3l\cos \theta \delta \theta \end{array}\]
 
Step 7

From the given diagram,

\[\begin{array}{c} \sin \theta = \frac{{{x_C}}}{{8l}}\\ {x_C} = 8l\sin \theta \end{array}\]
 
Step 8

Differentiate the above equation:

\[\begin{array}{c} \delta {x_C} = \delta \left( {8l\sin \theta } \right)\\ = 8l\cos \theta \delta \theta \end{array}\]
 
Step 9

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

\[s = {\left( {2l\sin \theta } \right)_{\theta = \theta }} - {\left( {2l\sin \theta } \right)_{\theta = 15^\circ }}\]
 
Step 10

To calculate the force applied by spring we use the formula:

\[\begin{array}{c} {F_S} = ks\\ = k\left( {{{\left( {2l\sin \theta } \right)}_{\theta = \theta }} - {{\left( {2l\sin \theta } \right)}_{\theta = 15^\circ }}} \right) \end{array}\]
 
Step 11

According to principal of virtual work,

\[\begin{array}{c} \delta U = 0\\ {F_S}\delta {x_A} - {F_S}\delta {x_B} + P\delta {x_C} = 0 \end{array}\]
 
Step 12

Substitute the known values in the above equation:

\[\begin{array}{c} \left\{ \begin{array}{l} k\left( {{{\left( {2l\sin \theta } \right)}_{\theta = \theta }} - {{\left( {2l\sin \theta } \right)}_{\theta = 15^\circ }}} \right)\left( {l\cos \theta \delta \theta } \right) - \\ k\left( {{{\left( {2l\sin \theta } \right)}_{\theta = \theta }} - {{\left( {2l\sin \theta } \right)}_{\theta = 15^\circ }}} \right)\left( {3l\cos \theta \delta \theta } \right) + \\ P\left( {8l\cos \theta \delta \theta } \right) \end{array} \right\} = 0\\ k\left( {{{\left( {2l\sin \theta } \right)}_{\theta = \theta }} - {{\left( {2l\sin \theta } \right)}_{\theta = 15^\circ }}} \right)\left( { - 2l\cos \theta \delta \theta } \right) = - P8l\cos \theta \delta \theta \\ k = \frac{{4P}}{{2l\sin \theta - 2l\sin 15^\circ }} \end{array}\]
 
Step 13

Substitute the known value in the equation:

\[\begin{array}{c} k = \frac{{4\left( {600\;{\rm{N}}} \right)}}{{2\left( {200\;{\rm{mm}}} \right)\left( {\frac{{{{10}^{ - 3}}\;{\rm{m}}}}{{1\;{\rm{mm}}}}} \right)\sin 60^\circ - 2\left( {200\;{\rm{mm}}} \right)\left( {\frac{{{{10}^{ - 3}}\;{\rm{m}}}}{{1\;{\rm{mm}}}}} \right)\sin 15^\circ }}\\ = \frac{{2400\;{\rm{N}}}}{{\left( {0.4\;{\rm{m}}} \right)\left( {0.866} \right) - \left( {0.4\;{\rm{m}}} \right)\left( {0.259} \right)}}\\ = \frac{{2400\;{\rm{N}}}}{{0.2428\;{\rm{m}}}} \times \frac{{{\rm{1 kN}}}}{{1000{\rm{ N}}}}\\ = 9.885\;{\rm{kN/m}} \end{array}\]