Two forces F and 2F are applied on a rod of length l and mass m, as shown. The angular acceleration of the rod is

Force equation $F-2F=ma\Rightarrow a=-\frac{F}{m}$

Hence, $\overrightarrow{a}=-\frac{F}{m}\widehat{i}$
Torque equation about center of mass

$\begin{array}{l}F\cdot \frac{l}{2}+\left(\frac{2}{3}l-\frac{l}{2}\right)=I_C\alpha \\ F\cdot \frac{l}{2}+F\frac{l}{6}=\frac{m{l}^2}{12}\cdot \alpha \\ \frac{2}{3}Fl=\frac{m{l}^2}{12}\cdot \alpha \Rightarrow \alpha =\frac{8F}{ml}\\ \text{ Hence, }\overrightarrow{\alpha }=-\frac{8F}{ml}\widehat{k}\end{array}$