14. In $\triangle A B C$, it is known that $G$ is the centroid, the opposite sides of the three interior angles $\angle A$, $\angle B$, and $\angle C$ are $a$, $b$, and $c$ respectively, and
56a $\overrightarrow{G A}+40 b \overrightarrow{G B}+35 c \overrightarrow{G C}=0$.
Then $\angle B=$ $\qquad$
14. $60^{\circ}$.

Since $\overrightarrow{G A}+\overrightarrow{G B}+\overrightarrow{G C}=0$, therefore, $40 b \overrightarrow{G A}+40 b \overrightarrow{G B}+40 b \overrightarrow{G C}=0$.

Thus, $(56 a-40 b) \overrightarrow{G A}+(35 c-40 b) \overrightarrow{G C}=0$. Since $\overrightarrow{G A}$ and $\overrightarrow{G C}$ are not collinear, we have $7 a-5 b=0, 7 c-8 b=0$.
Let $a=5 k$. Then $b=7 k, c=8 k$.
By the cosine rule,
$$
\cos B=\frac{25 k^{2}+64 k^{2}-49 k^{2}}{2 \times 5 k \times 8 k}=\frac{1}{2} \text {. }
$$

Therefore, $\angle B=60^{\circ}$.