---
title: "Resultant force system at A; equivalent at origin"
subject: "Applied Mechanics"
chapter: 3
key: "semester_1/applied_mechanics:resultant-force-system-at-a-equivalent-at-origin"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/resultant-force-system-at-a-equivalent-at-origin
site: BCT Engineering
---

# Resultant force system at A; equivalent at origin

Applied Mechanics · Chapter 3: Forces acting on particle and rigid body

- Asked in: 2069 Asar, 6+4 marks
- Video: [Engineering Mechanics: Statics Theory | Force Reduction (Force-Couple System)](https://www.youtube.com/watch?v=NMCwwj8WWLQ) — Dr. Clayton Pettit
- 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-system-at-a-equivalent-at-origin
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-3

## Question

Compute the resultant force system of the applied loads at position A. What is the equivalent force system at the origin for the force F = −50i − 70j + 55k kN going through position P whose position vector is 10i − 8j + 8k m. [Figure: 3D L-shaped bar assembly with A at one end, loads 5 kN, 4 kN, 1 kN, 1 kN and dimensions 2 m, 2 m, 4 m, 6 m, 2 m.]

## Short answer

Resultant at A: force 4.97 kN (2.04i+1.96j-4.08k kN), moment 27.5 kN·m (-24.5i-2.0j-12.2k kN·m). Equivalent at origin for F = -50i-70j+55k kN through P(10,-8,8): same force and couple M_O = 120i-950j-1100k kN·m.

## Answer

**Reading of the figure:** the member AB is 6 m long along y from the wall at A, with a 2 m × 2 m cross-section. At the free end B: 4 kN along +y; 5 kN along the diagonal of a 2 m × 2 m × 4 m block directed towards B (down, from the top far corner); and a vertical pair of 1 kN forces 2 m apart along x, giving a couple. Axes: x towards the viewer, y to the right, z up. Take A as the origin, B = (0, 6, 0).

#### Force vectors
5 kN along the diagonal (−2 in x, +2 in y, +4 in z, taken towards B, so reversed): direction (2, −2, −4)/√24
F₅ = 5(2, −2, −4)/4.899 = (2.04, −2.04, −4.08) kN
F₄ = (0, 4, 0) kN
The two 1 kN forces (up and down) have zero sum.

**Resultant force:** R = (2.04, 1.96, −4.08) kN, |R| = **4.97 kN**

#### Moment about A
```text
M from 4 kN and 5 kN at B:  r × F = (0,6,0) × (2.04, 1.96, −4.08)
                           = (6(−4.08) − 0, 0 − 0, 0 − 6(2.04))
                           = (−24.49, 0, −12.25) kN·m
Couple of 1 kN × 2 m about y: (0, −2, 0) kN·m (up-force on the near side)
-----------------------------------------------------------------
M_A = (−24.49, −2.00, −12.25) kN·m,  |M_A| = 27.5 kN·m
```
So at A: R = (2.04 i + 1.96 j − 4.08 k) kN and M_A = (−24.5 i − 2.0 j − 12.2 k) kN·m.

#### Equivalent system at the origin for F = −50i − 70j + 55k kN through P(10, −8, 8) m
Move F to O: the force stays the same and a couple M_O = r × F is added.
```text
M_O = (10, −8, 8) × (−50, −70, 55)
    = ( (−8)(55) − (8)(−70),  (8)(−50) − (10)(55),  (10)(−70) − (−8)(−50) )
    = ( −440 + 560,  −400 − 550,  −700 − 400 )
    = ( 120, −950, −1100 ) kN·m
```
**Answer: at the origin, F = (−50 i − 70 j + 55 k) kN (|F| = 102.1 kN) with a couple M_O = (120 i − 950 j − 1100 k) kN·m.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why is the 5 kN force written as 5(2, -2, -4)/4.899?
   A. Dividing by the diagonal length sqrt(24) converts the direction vector to a unit vector
   B. 4.899 is the force magnitude
   C. To convert N to kN
   D. The 5 kN is shared between three axes equally
   Answer: A. Force vector = magnitude × (direction vector)/(its length). (Key point: Unit vector)

2. What do the two 1 kN vertical forces contribute?
   A. Zero resultant force but a couple
   B. A 2 kN resultant
   C. Nothing at all
   D. A moment equal to 1 kN·m only about z
   Answer: A. Equal and opposite forces cancel in the force sum but give a couple about y. (Key point: Pair of 1 kN forces)

3. How do we replace F through P at the origin?
   A. Keep F and add the couple M_O = r × F
   B. Keep F only
   C. Move F with no couple
   D. Replace by r × F only
   Answer: A. Shifting a force off its line produces a moment r × F. (Key point: Equivalent at origin)

4. What is the x component of (10, -8, 8) × (-50, -70, 55)?
   A. (-8)(55) - (8)(-70) = 120 kN·m
   B. (-8)(55) + (8)(-70) = -1000
   C. (8)(-50) - (10)(55) = -950
   D. (10)(-70) - (-8)(-50) = -1100
   Answer: A. x component = r_y F_z - r_z F_y. Others are the y and z components. (Key point: Cross product)

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