--- title: "Resultant magnitude, direction and position of forces" subject: "Applied Mechanics" chapter: 3 key: "semester_1/applied_mechanics:resultant-force-position-rectangle-200-300-150" source: https://www.bctengineer.com/q/semester_1/applied_mechanics/resultant-force-position-rectangle-200-300-150 site: BCT Engineering --- # Resultant magnitude, direction and position of forces Applied Mechanics · Chapter 3: Forces acting on particle and rigid body - Asked in: 2072 Chaitra, 10 marks - Video: [Magnitude Direction and position | Resultant of forces | Mechanics of civil | easy and simple](https://www.youtube.com/watch?v=dgqkCfcE8Jk) — 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-force-position-rectangle-200-300-150 - Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-3 ## Question Find the magnitude, direction and position of resultant force of the following system as shown in figure. [Figure: rectangle ABCD, AB = 5 m, BC = 3 m; 200 kN at D at 30° to the horizontal; 300 kN horizontal at C; 150 kN at B at 30° to the horizontal; 100 kN horizontal at A.] ## Short answer Resultant magnitude 234.9 kN, direction 48.16° above the horizontal. It practically passes through A, crossing AB 0.031 m left of A and AD 0.034 m above A. ## Answer **Reading of the figure:** rectangle ABCD with A(0,0), B(5,0), C(5,3), D(0,3) (m). Forces: - 200 kN at D, directed up and to the left at 30° to the horizontal. - 300 kN at C, horizontal, to the right. - 150 kN at B, directed up and to the right at 30° to the horizontal. - 100 kN at A, horizontal, to the left. #### Resultant force ```text ΣFx = −200 cos30° + 300 + 150 cos30° − 100 = −173.21 + 300 + 129.90 − 100 = +156.70 kN ΣFy = +200 sin30° + 150 sin30° = 100 + 75 = +175.00 kN ``` R = √(156.70² + 175² ) = **234.9 kN** θ = tan⁻¹(175/156.70) = **48.16°** above the horizontal (towards the right). #### Moment about A (counter-clockwise positive, M = x·Fy − y·Fx) ```text 200 kN at D(0,3): 0 − 3(−173.21) = +519.62 300 kN at C(5,3): 0 − 3(300) = −900.00 150 kN at B(5,0): 5(75) − 0 = +375.00 100 kN at A: = 0 M_A = −5.38 kN·m ``` #### Position x · R_y − y · R_x = M_A - On AB (y = 0): x = −5.38/175 = **−0.031 m**, that is 0.031 m to the left of A. - On AD (x = 0): y = −M_A/R_x = 5.38/156.70 = **0.034 m** above A. So the resultant of 234.9 kN at 48.2° practically passes through A (it cuts AB 0.031 m to the left of A, and cuts AD 0.034 m above A). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The 200 kN force at D acts up and to the left at 30°. Which are its components? A. Fx = -173.21 kN, Fy = +100 kN B. Fx = +173.21 kN, Fy = +100 kN C. Fx = -100 kN, Fy = +173.21 kN D. Fx = -173.21 kN, Fy = -100 kN Answer: A. Left is negative x, up is positive y, with cos30° on x and sin30° on y. (Key point: Sign of components) 2. Which gives the magnitude and direction of R? A. R = sqrt(156.70^2 + 175^2) = 234.9 kN at 48.16° above the horizontal B. R = 156.70 + 175 = 331.7 kN C. R = 234.9 kN at 41.84° above horizontal D. R = 175 kN horizontal Answer: A. tan(theta) = ΣFy/ΣFx = 175/156.70. (Key point: Resultant magnitude and direction) 3. Which formula is used for the moment of each force about A? A. M = x·Fy - y·Fx (counter-clockwise positive) B. M = x·Fx - y·Fy C. M = F × (AB + BC) D. M = Fx + Fy Answer: A. The 300 kN force at C(5,3) gives 0 - 3(300) = -900 kN·m, clockwise. (Key point: Moment formula) 4. What does M_A = -5.38 kN·m tell us? A. The resultant passes very close to A (about 0.03 m away) B. The resultant is zero C. The resultant is a pure couple D. The resultant passes through C Answer: A. The offset is M_A/R_y = 0.031 m, so the line almost goes through A. 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