Skip to main content

Force-couple system at A; couple is free vector

Short answer

The force-couple system at A is R = 280 kN (components -141.4i+226.3j-84.9k kN) with M_A = 918 kN·m (-320.3i+424.3j+748.5k kN·m). A couple is two equal opposite parallel forces producing pure rotation; its moment is a free vector because its value is independent of the reference point.

Find the resultant of force couple system at point A as shown in figure below. Take F1 = 100 kN, F2 = 300 kN. Define a couple and show that couple is a free vector. [Figure: box with origin O, axes x, y, z, dimensions 4 m, 5 m, 3 m, 2 m, 1 m, 1.5 m; forces 50 kN, 50 kN, 200 kN, 200 kN, 300 kN, 300 kN, 20 kN (two), F1 and F2 as shown, A at the lower corner on the z axis.]

Reading of the figure: the box is x = 5 m, y = 4 m, z = 3 m with O at the back corner and A = (0, 0, 3). The loads are:

  • F₁ = 100 kN along the space diagonal from O to (5, 4, 3), and F₂ = 300 kN along the diagonal from (5, 0, 3) to (0, 4, 0).
  • 20 kN up at (5, 4, 0) and 20 kN down at A (a couple).
  • A pair of 50 kN vertical forces 2 m apart on the left top edge: couple with moment 100 kN·m, −x sense.
  • A pair of 200 kN horizontal forces (along x) 1 m apart: couple 200 kN·m, −z sense.
  • A pair of 300 kN forces (along z) 1.5 m apart on the right face: couple 450 kN·m, −x sense.

Forces

F₁ = 100 (5, 4, 3)/√50 = (70.71, 56.57, 42.43) kN F₂ = 300 (−5, 4, −3)/√50 = (−212.13, 169.71, −127.28) kN The two 20 kN forces cancel (zero resultant) but form a couple.

Resultant force at A: R = F₁ + F₂ = (−141.42, 226.27, −84.85) kN, |R| = 280 kN

Moment about A (r × F)

F1 acts at O:       r = (0,0,−3)   -> ( 169.71, −212.13,   0.00)
F2 acts at (5,0,3): r = (5,0,0)    -> (   0.00,  636.40, 848.53)
20 kN pair:         r = (5,4,−3)   -> (  60.00,    0.00, 100.00)
50 kN pair                           -> (−100.00,   0.00,   0.00)
200 kN pair                          -> (   0.00,   0.00, −200.00)
300 kN pair                          -> (−450.00,   0.00,   0.00)
------------------------------------------------------------------
M_A                                  = (−320.3,  424.3,  748.5) kN·m

|M_A| = 918 kN·m

Answer: the force-couple system at A is R = (−141.4 i + 226.3 j − 84.9 k) kN together with M_A = (−320.3 i + 424.3 j + 748.5 k) kN·m.

Couple

A couple is a pair of equal, opposite and parallel forces whose lines of action do not coincide. It produces pure rotation, no translation, because the resultant force is zero. Its moment is M = F × d, where d is the perpendicular distance between the forces (the arm of the couple). It is a vector perpendicular to the plane of the couple.

A couple is a free vector

Let forces F at A and −F at B form a couple. Take any point O, with position vectors r_A and r_B.

Moment about O = r_A × F + r_B × (−F)
               = (r_A − r_B) × F
               = r_AB × F

The result contains only the vector r_AB between the two points of application. It does not contain the position of O. So the moment of a couple is the same about every point. The couple can be moved anywhere in its plane, to a parallel plane, or turned in its plane, without a change in its effect on a rigid body, provided magnitude and direction of the moment vector remain the same. Hence its moment is a free vector.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Applied Mechanics chapter 3 questions →