Answer
a) See image
b) $F_{TA}=49.5N$$\space\space\space\space\space F_{TB}=75.7N$
c) $\sum \tau=-3.93Nm$$\space\space\space I=0.589kgm^2$
Work Step by Step
a) See image
b) $\sum F_Ax=F_{TA}-F_{GA}\sin(32^o)=(8.0kg)(1.00\frac{m}{s^2})$
$F_{TA}=(8.0kg)(1.00\frac{m}{s^2})+(8.0kg)(9.8\frac{m}{s^2})\sin(32^o)=49.5N$
$\sum F_Bx=F_{TA}-F_{GB}\sin(61^o)=(10.0kg)(-1.00\frac{m}{s^2})$
$F_{TB}=(10.0kg)(-1.00\frac{m}{s^2})+(10.0kg)(9.8\frac{m}{s^2})\sin(61^o)=75.7N$
c) $\sum \tau=r(F_{TA}-F_{TB})=(0.15m)(49.5N-75.7N)=-3.93Nm$
$\alpha=\frac{a_{tan}}{r}=\frac{1.00\frac{m}{s^2}}{0.15m}=6.67\frac{rad}{s^2}$
$I=\frac{\tau}{\alpha}=\frac{-3.93Nm}{6.67\frac{rad}{s^2}}=0.589kgm^2$