--- title: "Force-couple system at A; couple is free vector" subject: "Applied Mechanics" chapter: 3 key: "semester_1/applied_mechanics:force-couple-system-point-a-box-couple-free-vector" source: https://www.bctengineer.com/q/semester_1/applied_mechanics/force-couple-system-point-a-box-couple-free-vector site: BCT Engineering --- # Force-couple system at A; couple is free vector Applied Mechanics · Chapter 3: Forces acting on particle and rigid body - Asked in: 2071 Shrawan, 8+3 marks - Video: [7.1 Replacing a Force with an equivalent Force-Couple system](https://www.youtube.com/watch?v=CFVwNRgCy3Q) — Vector In Motion - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/force-couple-system-point-a-box-couple-free-vector - Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-3 ## Question 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.] ## 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. ## Answer **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) ```text 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. ```text 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**. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is F1 = 100 kN along the diagonal from O to (5, 4, 3) written as a vector? A. 100 × (5, 4, 3)/sqrt(50) B. 100 × (5, 4, 3) C. 100 × (5, 4, 3)/12 D. 100 × (1, 1, 1) Answer: A. Force vector = magnitude × direction vector divided by its length (sqrt(25+16+9)). (Key point: Unit vector for F1) 2. What is the effect of the two 20 kN forces (one up, one down)? A. They add nothing to R but form a couple B. They double R C. They cancel completely with no effect D. They add 40 kN to R Answer: A. Equal and opposite forces sum to zero but, being on different lines, give a couple. (Key point: Pair of 20 kN forces) 3. How are the couple pairs added to M_A? A. Their moment vectors are added directly, since a couple's moment is the same about any point B. Each is multiplied by its distance from A C. Only the largest couple is included D. They are added to R Answer: A. A couple is a free vector, so its moment doesn't depend on A. (Key point: Adding couples) 4. What step shows a couple's moment doesn't depend on the reference point O? A. r_A × F + r_B × (-F) = (r_A - r_B) × F, which has no O B. F is the same for both forces C. r_A and r_B are both zero D. The sum of forces is nonzero Answer: A. The result r_AB × F contains only the vector between the two points of application. 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