--- title: "Resultant magnitude, direction, position about A" subject: "Applied Mechanics" chapter: 3 key: "semester_1/applied_mechanics:resultant-system-forces-about-a-beam" source: https://www.bctengineer.com/q/semester_1/applied_mechanics/resultant-system-forces-about-a-beam site: BCT Engineering --- # Resultant magnitude, direction, position about A Applied Mechanics · Chapter 3: Forces acting on particle and rigid body - Asked in: 2072 Kartik, 12 marks - Video: [Magnitude, Direction and Position of Resultant force system| civil engineering| Mechanics of civil](https://www.youtube.com/watch?v=JwNimY9NKZ4) — SimpleCivil withsmriti - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/resultant-system-forces-about-a-beam - Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-3 ## Question Determine the magnitude, direction and position of the resultant of the system of forces with respect to point A shown in figure below. [Figure: horizontal member from A, spans 3 m, 2 m, 3 m; 120 kN at 60° at A; 50 kN downward at 3 m; 100 kN·m couple; 150 kN upward at 5 m; 80 kN horizontal at the right end on a vertical segment of height 2 m.] ## Short answer Sum forces gives R = 140.05 kN at 1.6° below the horizontal. Moment about A is +540 kN·m. The resultant acts along a line 3.86 m below A, equivalently crossing the member axis 137.6 m to the left of A. ## Answer **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 ```text 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 ----------------------------------------------- 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).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is ΣFy = -3.92 kN? A. -120 sin60° - 50 + 150 = -103.92 - 50 + 150 B. 120 sin60° + 50 + 150 C. -120 cos60° - 50 + 150 D. -50 + 150 - 80 Answer: A. The 120 kN force acts down at 60°, so its vertical part is -103.92 kN; the 80 kN force is horizontal. (Key point: Vertical resultant) 2. Why does the 120 kN force contribute nothing to M_A? A. Its line of action passes through A, so its moment arm is zero B. It is a couple C. It is horizontal D. Its components cancel Answer: A. Moment = force × perpendicular distance, and the distance from A to a line through A is zero. (Key point: Moment of the 120 kN force) 3. In M_A = +540, how are the 100 kN·m couple and the 80 kN force at height 2 m treated? A. Couple +100 added directly; 80 kN gives -80 × 2 = -160 B. Couple multiplied by 5 m; 80 kN gives zero C. Couple ignored; 80 kN gives +160 D. Both given as +100 and +160 Answer: A. A couple has the same moment about any point; the 80 kN horizontal force at height 2 m turns clockwise about A. (Key point: Couple and horizontal force) 4. How is the position of the resultant found? A. Set x·R_y - y·R_x = M_A, then find x (at y = 0) or y (at x = 0) B. Divide R by M_A C. x = M_A × R D. Place the resultant at the centroid of the member Answer: A. The resultant must have the same moment about A as the system: x = 540/(-3.92) = -137.6 m, or y = -3.86 m. 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