## Engineering Mechanics: Statics & Dynamics (14th Edition)

$a_B=7.59ft/s^2$
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 $a_B=7.59ft/s^2$