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Resultant magnitude, direction and line of action

Short answer

Magnitude 61.7 N, direction 56.6° below the negative x-axis (down-left). Its line of action intersects member AB at 3.98 m from A.

Determine magnitude, direction and line of action of the resultant of forces acting in the system shown in figure below. [Figure: points A, B, C; A to B horizontal with 3 m + 2 m; 20 N at 60° at the point 3 m from A; B to C vertical 3 m + 3 m; 100 N at C at 70°; 50 N horizontal at C.]

Reading of the figure (A at the origin, x right, y up): member A–B horizontal, 3 m + 2 m; B–C vertical downward, 3 m + 3 m. So B(5, 0), C(5, −6).

  • 20 N at (3, 0), acting down and to the right at 60° to the horizontal.
  • 100 N at the mid-point of BC, (5, −3), acting down and to the left, 70° from the vertical (BC).
  • 50 N at C, horizontal, to the right.

Components

ForceFx (N)Fy (N)
20 N20 cos 60° = +10.00−20 sin 60° = −17.32
100 N−100 sin 70° = −93.97−100 cos 70° = −34.20
50 N+50.000
Sum−33.97−51.52

Magnitude and direction

R = √(33.97² + 51.52²) = 61.71 N α = tan⁻¹(51.52/33.97) = 56.6° with the horizontal. Both components are negative, so R points down and to the left (236.6° from +x).

Moment about A (M = x·Fy − y·Fx)

20 N  at (3, 0):     3(−17.32) − 0               = −51.96
100 N at (5, −3):    5(−34.20) − (−3)(−93.97)    = −171.0 − 281.9 = −452.9
50 N  at (5, −6):    5(0) − (−6)(50)             = +300.00
M_A                                              = −204.88 N·m (clockwise)

Line of action

x R_y − y R_x = M_A

  • On the line AB (y = 0): x = M_A / R_y = −204.88/−51.52 = 3.98 m from A (to the right).
  • On the line BC (x = 5): y = (5 R_y − M_A)/R_x = (−257.6 + 204.88)/(−33.97) = +1.55 m, i.e. 1.55 m above B.

Answer: R = 61.7 N, acting down-left at 56.6° to the horizontal, along a line crossing AB at 3.98 m from A.

✓ Answer checked by Claude Opus 5.5 (high effort)

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