--- title: "Resultant magnitude, direction and line of action" subject: "Applied Mechanics" chapter: 3 key: "semester_1/applied_mechanics:resultant-line-of-action-20n-100n-50n" source: https://www.bctengineer.com/q/semester_1/applied_mechanics/resultant-line-of-action-20n-100n-50n site: BCT Engineering --- # Resultant magnitude, direction and line of action Applied Mechanics · Chapter 3: Forces acting on particle and rigid body - Asked in: 2070 Asar, 8 marks - Video: [How to Simplify a Force and Moment System and Find Line of Action](https://www.youtube.com/watch?v=mfQYzFpWusI) — Brian Bernard - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/resultant-line-of-action-20n-100n-50n - Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-3 ## Question 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.] ## 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. ## Answer **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 | Force | Fx (N) | Fy (N) | |---|---|---| | 20 N | 20 cos 60° = +10.00 | −20 sin 60° = −17.32 | | 100 N | −100 sin 70° = −93.97 | −100 cos 70° = −34.20 | | 50 N | +50.00 | 0 | | **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) ```text 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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The 100 N force makes 70° with the vertical BC and acts down-left. Its components are? A. Fx = -100 sin70° = -93.97 N, Fy = -100 cos70° = -34.20 N B. Fx = -100 cos70°, Fy = -100 sin70° C. Fx = +93.97 N, Fy = -34.20 N D. Fx = -100 N, Fy = 0 Answer: A. The angle is measured from the vertical, so the vertical component uses cos70° and horizontal uses sin70°. (Key point: Components of the 100 N force) 2. How is R found from the table? A. sqrt(33.97^2 + 51.52^2) = 61.71 N B. 33.97 + 51.52 = 85.49 N C. 51.52 N D. sqrt(33.97 × 51.52) Answer: A. R is the vector sum of the sums ΣFx and ΣFy. (Key point: Resultant magnitude) 3. Both ΣFx and ΣFy are negative. Which direction does R point? A. Down and to the left, 56.6° to the horizontal B. Up and to the right C. Down and to the right D. Up and to the left Answer: A. Signs of the components fix the quadrant: here the third quadrant (236.6° from +x). 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