--- title: "Force-couple system about a point" subject: "Applied Mechanics" chapter: 3 key: "semester_1/applied_mechanics:force-couple-system-beam-60kn-20knm-50kn" source: https://www.bctengineer.com/q/semester_1/applied_mechanics/force-couple-system-beam-60kn-20knm-50kn site: BCT Engineering --- # Force-couple system about a point Applied Mechanics · Chapter 3: Forces acting on particle and rigid body - Asked in: 2068 Shrawan, 10 marks - Video: [Force-Couple Systems [Vector Statics #41]](https://www.youtube.com/watch?v=wtHwACL-Wik) — Simmy Sigma - 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-beam-60kn-20knm-50kn - Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-3 ## Question Determine force couple system about point 'A' for the given system of forces as shown in figure below. [Figure: horizontal member with A at left end; 60 kN downward load and 20 kN·m couple at B (2 m from A); support at D (6 m from B); 50 kN load at 45° at E, 2 m from D.] ## Short answer Resolve all forces into components, sum to get resultant force, and sum moments about A including the couple. Result: resultant force R = 101.7 kN (−35.36i −95.36j kN) and a clockwise couple M_A = 453.6 kN·m. ## Answer **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) ```text 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 ```text 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) ------------------------------------------------ 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.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With the arrowhead at E, which components does the 50 kN force have? A. Fx = -35.36 kN, Fy = -35.36 kN (down and left) B. Fx = +35.36 kN, Fy = -35.36 kN C. Fx = -50 kN, Fy = 0 D. Fx = +35.36, Fy = +35.36 Answer: A. The arrow points toward E, so the force acts down and to the left at 45°. (Key point: Direction of the 50 kN force) 2. Why is the support at D not included in the force sum? A. Supports provide reactions; they aren't applied loads B. Because D has no weight C. Because D is at the centre D. Because the force there cancels Answer: A. The force-couple system covers the applied loads only. (Key point: Supports) 3. How does the 50 kN force contribute to M_A? A. 10 × (-35.36) = -353.55; Fx acts along the axis, so no moment B. 50 × 10 C. -35.36 × 2 D. Both Fx and Fy contribute equally Answer: A. Fx has a line of action along the member so zero moment arm; Fy has arm 10 m. (Key point: Moment of 50 kN) 4. How is the 20 kN·m couple included? A. Added directly as +20 kN·m B. Multiplied by 2 m C. Ignored D. Added to the force Answer: A. A couple's moment is the same about any point. 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