Answer
$N_A=113.42N$
$N_B=325.18N$
Work Step by Step
We can determine the required normal reactions as follows:
According to the equation of motion in x-direction
$\Sigma F_x=ma_x$
$\implies -300 cos 30=60a$
This can be rearranged as:
$a=\frac{-259.807}{60}=-4.33m/s^2$ The negative sign shows that it is directed leftward.
Now the sum of moment about $A$ is given as
$\Sigma M_A=ma(0.3)$
$\implies -W(0.3)+N_B(0.5)+Pcos 30(0.4)-Psin 30(0.08)=60\times 4.33(0.3)$
$\implies -60(9.81)(0.3)+0.5N_B+300cos 30(0.4)-300sin 30(0.08)=77.94$
This simplifies to:
$N_B=325.18N$
We apply the equation of motion in y-direction
$\Sigma F_y=ma_y$
$\implies N_A+N_B-W+Psin30=ma_y$
We plug in the known values to obtain:
$N_A+325.18-(60\times 9.81)+300sin30=600(0)$
This simplifies to:
$N_A=113.42N$