Answer
$a_B=7.59ft/s^2$
Work Step by Step
We can determine the required acceleration as follows:
We know that
$\Sigma F_x=m_A a_x$
$12-N_B sin 15=(\frac{8}{32.2})a_A~~~$eq(1)
$\Sigma F_y=m_Ba_y$
$\implies N_Bcos15-15=(\frac{15}{32.2})a_B~~~$eq(2)
As $S_B=S_Atan 15$
$\implies a_B=a_A tan15~~~$eq(3)
Solving eq(1), eq(2) and eq(3), we obtain:
$a_A=28.3ft/s^2$, $N_B=19.2lb$ and $a_B=7.59ft/s^2$
