Force-couple system about a point

Reading of the figure: a horizontal member with A at the left end (origin). B is 2 m from A, D (a support) is 6 m further, and E is 2 m beyond D, so B(2, 0), D(8, 0), E(10, 0). Loads: 60 kN downward at B, a 20 kN·m couple at B (counter-clockwise, as drawn), and a 50 kN force at E at 45° to the horizontal, acting towards E, that is down and to the left (arrowhead at E). Supports are not loads, so they are not included. Counter-clockwise moments are positive.

Resultant force (at A)

50 kN components:  Fx = −50 cos45° = −35.36 kN
                   Fy = −50 sin45° = −35.36 kN
ΣFx = −35.36 kN
ΣFy = −60 − 35.36 = −95.36 kN

R = √(35.36² + 95.36²) = 101.7 kN Direction: θ = tan⁻¹(95.36/35.36) = 69.6° below the negative x axis, i.e. down and to the left.

Moment about A

60 kN at x = 2 :  −60 × 2          = −120.00
Couple         :                   =  +20.00
50 kN at x = 10:  x·Fy = 10(−35.36)= −353.55   (Fx acts along the axis, no moment)
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M_A                                = −453.55 kN·m (clockwise)

Answer: the force-couple system at A is a force R = 101.7 kN (−35.36 i − 95.36 j kN) together with a clockwise couple M_A = 453.6 kN·m.

(If the 50 kN arrow is taken to act away from E, up and to the right, the same method gives R = 35.36 i − 24.64 j kN and M_A = +253.6 kN·m.)