2. In $\triangle A B C$, it is known that
$$
|\overrightarrow{A B}|=2,|\overrightarrow{A C}|=3,|\overrightarrow{B C}|=4 \text {. }
$$

Let $O$ be the incenter of $\triangle A B C$, and $\overrightarrow{A O}=\lambda \overrightarrow{A B}+\mu \overrightarrow{B C}$. Then $\lambda+\mu=$ $\qquad$
2. $\frac{7}{9}$.

Let $A O$ intersect $B C$ at point $D$.
By the Angle Bisector Theorem, we have $\frac{B D}{D C}=\frac{A B}{A C}=\frac{2}{3}$.
Thus, $\overrightarrow{A D}=\frac{3}{5} \overrightarrow{A B}+\frac{2}{5} \overrightarrow{A C}$.
Also, $\frac{A O}{O D}=\frac{A B}{B D}=\frac{A C}{C D}=\frac{A B+A C}{B D+C D}=\frac{5}{4}$, so
$$
\begin{aligned}
\overrightarrow{A O} & =\frac{5}{9} \overrightarrow{A D}=\frac{1}{3} \overrightarrow{A B}+\frac{2}{9} \overrightarrow{A C} \\
& =\frac{1}{3} \overrightarrow{A B}+\frac{2}{9}(\overrightarrow{A B}+\overrightarrow{B C}) \\
& =\frac{5}{9} \overrightarrow{A B}+\frac{2}{9} \overrightarrow{B C} .
\end{aligned}
$$

Therefore, $\lambda+\mu=\frac{7}{9}$.