Resultant of vertical forces

$R = \Sigma F_V$

$R = 60 + 20 - 30$

$R = 50 \, \text{ kN downward}$

Moment about point A

$M_A = 60(4) - 20(1)$

$M_A = 220 \, \text {kN}\cdot\text{m clockwise}$

Location of R as measured from point A

$Rd = M_A$

$50d = 220$

$d = 4.4 \, \text{ m}$

Magnitude of horizontal couple at B and C

$2F = M_A$

$2F = 220$

$F = 110 \, \text{ kN}$

Thus the given system is equivalent to downward force of 50 kN at point A and clockwise couple of 220 kN·m. The couple is represented by 110 kN horizontal forces at B and C. The force at B is to the right and the force at C is to the left, producing the clockwise couple. *answer*