Answer
$F_{R}$ = 1.80KN ↓
($M_{R}$)$_{A}$ =4.20KN.m (clockwise)
Work Step by Step
Summing the force along the y axis to find equivalent Resultant Force and couple Moment at Point A.
↑+($F_{R}$)$_{y}$ = Σ$F_{y}$ :
$F_{R}$ = -400(3) -$\frac{1}{2}$(400)(3)
$F_{R}$ = -1800N
$F_{R}$ = 1.80KN ↓
Now Summing the moment about point A.
↺+($M_{R}$)$_{A}$ = Σ$M_{A}$ :
($M_{R}$)$_{A}$ = -400(3)(1.5) - $\frac{1}{2}$(400)(3)(4)
($M_{R}$)$_{A}$ = -4200N.m
($M_{R}$)$_{A}$ =4.20KN.m (clockwise) Ans
