Resultant magnitude, direction, position about A

Reading of the figure: a horizontal member from A with 3 m, 2 m and 3 m spans. Take A as the origin, x to the right, y up (counter-clockwise moments positive).

• 120 kN at A, acting down and to the right at 60° to the horizontal.
• 50 kN downward at x = 3 m.
• 100 kN·m couple at x = 5 m, counter-clockwise (as the arrow is drawn).
• 150 kN upward at x = 5 m.
• 80 kN horizontal (to the right) at the top of a vertical segment 2 m high at the right end, so at (8, 2).

Resultant force

ΣFx = 120 cos 60° + 80 = 60 + 80 = 140 kN ΣFy = −120 sin 60° − 50 + 150 = −103.92 − 50 + 150 = −3.92 kN

R = √(140² + 3.92²) = 140.05 kN Direction: θ = tan⁻¹(3.92/140) = 1.6° below the horizontal (pointing to the right).

Moment about A

120 kN  (passes through A)             =    0
50 kN   at x = 3:  −50 × 3             = −150
150 kN  at x = 5:  +150 × 5            = +750
Couple                                 = +100
80 kN   at height 2:  −80 × 2          = −160
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M_A                                    = +540 kN·m (counter-clockwise)

Position of the resultant

Moment of R about A must equal M_A: x · R_y − y · R_x = M_A.

• Where it cuts the line of the member (y = 0): x = M_A / R_y = 540/(−3.92) = −137.6 m, that is 137.6 m to the left of A.
• Where it cuts the vertical through A (x = 0): y = −M_A / R_x = −540/140 = −3.86 m, that is 3.86 m below A.

Answer: R = 140.05 kN at 1.6° below the horizontal, acting along a line 3.86 m below A (equivalently crossing the member axis 137.6 m to the left of A).