## Engineering Mechanics: Statics & Dynamics (14th Edition)

$d=1.56m$
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: $d=1.56m$