2. As shown in Figure 1, in the Cartesian coordinate system $x O y$, $A(1,3), B\left(8 \frac{1}{3}, 1 \frac{2}{3}\right)$, $C\left(7 \frac{1}{3}, 4 \frac{2}{3}\right)$, $O A$ intersects the extension of $B C$ at point $D$, $M$ and $N$ are points on line segments $O D$ and $B D$ respectively, and $O M=M N=B N$. Then the length of line segment $M N$ is
2. $\frac{5 \sqrt{10}}{3}$.

As shown in Figure 4, the line segment $B D$ is translated downward by $1 \frac{2}{3}$ units to the line segment $B_{1} D_{1}$, where $B_{1}\left(8 \frac{1}{3}, 0\right)$. The point $O$ is translated to the right to point $P\left(1 \frac{2}{3}, 0\right)$, and the point $B_{1}$ is translated to the left by $1 \frac{2}{3}$ units to point $Q\left(6 \frac{2}{3}, 0\right)$. A perpendicular line to the $x$-axis is drawn from point $P$, intersecting $O D$ at point $M$, and a perpendicular line to the $x$-axis is drawn from point $Q$, intersecting $B D$ and $B_{1} D_{1}$ at points $N$ and $N_{1}$, respectively, where $M\left(1 \frac{2}{3}, 5\right)$.
It is easy to prove that quadrilateral $M P Q N_{1}$ is a square,
$\mathrm{Rt} \triangle O P M \cong \mathrm{Rt} \triangle N N_{1} M \cong \mathrm{Rt} \triangle B_{1} Q N_{1}$.
Thus, $O M=M N=B_{1} N_{1}=B N$.
Therefore, $M N=O M=\sqrt{\left(1 \frac{2}{3}\right)^{2}+5^{2}}=\frac{5 \sqrt{10}}{3}$.