Answer
$N_A=326.81N$
$N_B=1144.69N$
$a=4m/s^2$
Work Step by Step
We can determine the required normal force and the acceleration as follows:
We know that
$\Sigma F_x=ma_{Gx}$
$\implies P=ma$
This can be rearranged as:
$a=\frac{P}{m}$
We plug in the known values to obtain:
$a=\frac{600}{150}$
$\implies a=4m/s^2$
We also know that
$M_A=M_{kA}$
$\implies -1.25ma=1.25mg+0.5P-2N_B$
We plug in the known values to obtain:
$-1.25(150)(4)=1.25(150)(9.81)+0.5(600)-2N_B$
This simplifies to:
$N_B=1144.69N$
Now the equation of motion in y-direction is given as
$\Sigma F_y=0$
$\implies N_A+N_B-mg=0$
$\implies N_A=mg-N_B$
We plug in the known values to obtain:
$N_A=(150)(9.81)-1144.69$
This simplifies to:
$N_A=326.81N$
