Answer
$B_y=69.7lb$
$B_x=73.9lb$
$N_A=120lb$
Work Step by Step
We can determine the required normal reactions and acceleration as follows:
$\Sigma M_B=\Sigma M_{kB}$
$\implies -N_A\times 13+W\times 10(\frac{12}{13})=ma\times (\frac{5}{13})$
$\implies -N_A\times 13+180\times 10(\frac{12}{13})=\frac{180}{32.2}\times 5\times 10(\frac{5}{13})$
This simplifies to:
$N_A=120lb$
We apply the equation of motion in x-direction
$\Sigma F_x=ma_x$
$\implies B_x-N_A(\frac{5}{13})=ma$
We plug in the known values to obtain:
$B_x-120\times (\frac{5}{13})=\frac{180}{32.2}\times 5$
This simplifies to:
$B_x=73.9lb$
Now, we apply the equation of motion in y-direction
$\Sigma F_y=ma_y$
$\implies B_y-W+N_A(\frac{12}{13})=0$
We plug in the known values to obtain:
$B_y-180+120(\frac{12}{13})=0$
This simplifies to:
$B_y=69.7lb$
