246 - System with couples and forces on it

$M_A = 4Q_H - 2P_V - 3P_H - 80(2) + 100(\sqrt{2})$
 

 

$M_A = 4(100)(\frac{1}{\sqrt{5}}) - 2(200)(\frac{2}{\sqrt{13}}) - 3(200)(\frac{3}{\sqrt{13}}) - 160 + 100\sqrt{2}$

$M_A = -560.8 \, \text{ lb}\cdot\text{ft}$

$M_A = 560.8 \, \text{ lb}\cdot\text{ft clockwise}$           answer