Let the tension in the cord be $T$. Since the loads at the ends are identical ($mg$), the tension $T = mg$. Resolving forces vertically at the midpoint where force $F$ is applied: $$2T \cos \theta = F$$ $$F = 2mg \cos \theta$$ Let $d$ be the horizontal distance from the midpoint to the pulley (so total distance between pulleys is $2d$). From geometry: $\cos \theta = \frac{x}{\sqrt{x^2 + d^2}}$. $$F(x) = 2mg \frac{x}{\sqrt{x^2 + d^2}}$$

From the provided graph of $F$ vs $x$: 1. Asymptote: As $x \to \infty$, $\frac{x}{\sqrt{x^2+d^2}} \to 1$, so $F \to 2mg$. The graph levels off at $F = 10 \, \text{N}$. Therefore, $2mg = 10 \, \text{N}$.
2. Reference Point: Consider the point where $F$ is half its maximum value, $F = 5 \, \text{N}$. $$5 = 10 \cos \theta \implies \cos \theta = 0.5 \implies \theta = 60^\circ$$ At this angle, $\tan \theta = \frac{d}{x} = \tan 60^\circ = \sqrt{3}$. So, $d = x\sqrt{3}$.

We need to find the $x$ value from the graph where $F=5$. Observing the grid, the curve crosses $F=5$ at approximately $x=3 \, \text{m}$ (assuming standard scaling where the first major tick is 5).
Using $x = 3$: $$d = 3\sqrt{3} \approx 3 \times 1.732 = 5.196 \, \text{m}$$ The distance between the pulleys is $2d$: $$2d = 2 \times 5.196 \approx 10.4 \, \text{m}$$