Answer
$d=1.56m$
Work Step by Step
We can determine the displacement of the cord as follows:
$\Sigma F_x=0$
$\implies -F+2F_{spring} cos\theta=0$..eq(1)
We know that
$F_{spring}=K\Delta x$
We plug in the known values to obtain:
$F_{spring}=500(\sqrt{(3)^2+d^2}-3)$
and given that $F=175N$
We plug in these values in eq(1) to obtain:
$-175+2(500(\sqrt{(3)^2+d^2}-3))=0$
This can be rearranged as:
$\theta=cos^{-1} (\frac{175}{1000\sqrt{(3)^2+d^2-3}})$
But from in the given scenario, $\theta=tan^{-1}\frac{3}{d}$
$\implies cos^{-1} (\frac{175}{1000\sqrt{(3)^2+d^2-3}})=tan^{-1}\frac{3}{d}$
This simplifies to:
$d=1.56m$
