Answer
$21^{\circ}$
Work Step by Step
Please see the attached image first.
We'll get the minimum angle when static friction is greatest. At that time $F_{S}=\mu n_{1}$
Let's consider the static equilibrium of the board.
$\uparrow \sum \vec F=0$
$n_{1}-m_{1}g-m_{2}g=0$
$n_{1}=(m_{1}+m_{2})g-(1)$
$\rightarrow \sum\vec F=0$
$F_{S}-n_{1}=0$
$\mu n_{1}=n_{2}-(2)$
$(1)=\gt (2)$
$\mu (m_{1}+m_{2})g=n_{2}-(3)$
$\sum \tau_{A}\curvearrowright=0$
$m_{1}g\frac{L}{3}cos\theta - m_{2}\frac{L}{2}cos\theta\space -n_{2}Lsin\theta=0$
$\frac{m_{1}gLcos\theta}{3}+\frac{m_{2}gLcos\theta}{2}-\mu(m_{1}+m_{2})gLsin\theta=0$
$\frac{m_{1}}{3}+\frac{m_{2}}{2}=\mu (m_{1}+m_{2})tan\theta$
Let's plug known values into equation.
$\frac{224\space kg}{3}+\frac{77.3\space kg}{2}=0.982 (224+77.3)kg\space tan\theta$
$0.38=tan\theta=\gt \theta=21^{\circ}$
