---
title: "Resultant force and moment of 3D system at O"
subject: "Applied Mechanics"
chapter: 3
key: "semester_1/applied_mechanics:resultant-force-moment-box-system-origin"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/resultant-force-moment-box-system-origin
site: BCT Engineering
---

# Resultant force and moment of 3D system at O

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

- Asked in: 2073 Shrawan, 10 marks
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/resultant-force-moment-box-system-origin
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-3

## Question

Determine the resultant force and moment of the following system about the point 'O' as shown in the figure below. [Figure: box with origin O(0,0,0), x, y, z axes; dimensions 4 m, 5 m, 3 m, 3 m, 2 m, 2.5 m as printed; forces: 10 kN at B along Y, 70 kN, 200 kN, 100 kN, 80 kN, 10 kN, and another 200 kN with a 2.5 M couple as marked.]

## Short answer

Write each force as a vector, sum forces, and add r × F plus the 500 kN·m couple. Resultant R = (−70.71i + 66.57j − 162.43k) kN, magnitude 189.2 kN. Moment about O is M_O = (120.3i − 307.2k) kN·m, magnitude 329.9 kN·m.

## Answer

**Readings from the figure (assumed where unclear):** O is the origin, x = 5 m, y = 4 m, z = 3 m for the box. The forces are:
1. 10 kN along +y at B (0, 4, 0).
2. 70 kN along −y at (2, 4, 0) (2 m from B along the top edge).
3. 200 kN acting at (0, 0, 3) along the line towards B.
4. 100 kN acting at (5, 0, 3) along the line towards B.
5. 80 kN along −y at (5, 0, 3).
6. 10 kN along −y at (5, 0, 3) (the second 10 kN, read as vertical downward at the same corner).
7. A couple formed by two 200 kN forces 2.5 m apart on the right face (parallel to z), taken as a couple of 500 kN·m about +x.

#### Force components (kN) and moments about O (r × F)
| Force | Point | Components | Moment about O (kN·m) |
|---|---|---|---|
| 10 | (0,4,0) | (0, 10, 0) | (0, 0, 0) |
| 70 | (2,4,0) | (0, −70, 0) | (0, 0, −140) |
| 200 | (0,0,3) | 200(0,4,−3)/5 = (0, 160, −120) | (−480, 0, 0) |
| 100 | (5,0,3) | 100(−5,4,−3)/√50 = (−70.71, 56.57, −42.43) | (−169.71, 0, 282.84) |
| 80 | (5,0,3) | (0, −80, 0) | (240, 0, −400) |
| 10 | (5,0,3) | (0, −10, 0) | (30, 0, −50) |
| Couple | | | (500, 0, 0) |

#### Resultant force
ΣFx = −70.71, ΣFy = 10 − 70 + 160 + 56.57 − 80 − 10 = 66.57, ΣFz = −120 − 42.43 = −162.43
**R = (−70.71 i + 66.57 j − 162.43 k) kN**, |R| = √(70.71² + 66.57² + 162.43²) = 189.2 kN

#### Resultant moment about O
Mx = −480 − 169.71 + 240 + 30 + 500 = 120.29
My = 0
Mz = −140 + 282.84 − 400 − 50 = −307.16
**M_O = (120.3 i − 307.2 k) kN·m**, |M_O| = 329.9 kN·m

The method is the same if a reading of the figure differs: write each force as a vector, find r × F, and add.

## Check yourself

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

1. How is the 200 kN force along a line with components (0, 4, -3) resolved?
   A. 200 × (0, 4, -3)/5 = (0, 160, -120) kN
   B. 200 × (0, 4, -3) = (0, 800, -600) kN
   C. 200 × (0, 4, -3)/7 = (0, 114, -86) kN
   D. (0, 200, 0) kN
   Answer: A. The line has length sqrt(4^2 + 3^2) = 5, so the unit vector is (0, 0.8, -0.6). (Key point: Inclined force components)

2. How is the moment of each force about O found?
   A. r × F, with r from O to the point of application
   B. F × (distance along the box edge)
   C. The sum of the force components
   D. Force magnitude times the box volume
   Answer: A. M_O = Σ(r × F) for each force; the vector form handles 3D geometry. (Key point: Moment of each force)

3. How is the 500 kN·m couple included in M_O?
   A. It is added directly to the moment components, with no r × F
   B. It is multiplied by the distance from O
   C. It is ignored because it has no resultant force
   D. It is added to the resultant force
   Answer: A. A couple is a free vector; its moment is the same about every point, so it is simply added to M_O. (Key point: Couple in the sum)

4. What is the resultant moment about O?
   A. M_O = (120.3 i - 307.2 k) kN·m, with M_y = 0
   B. M_O = (-70.7 i + 66.6 j - 162.4 k) kN·m
   C. M_O = (500 i) kN·m only
   D. M_O = (120.3 i + 307.2 k) kN·m
   Answer: A. Summing the table gives Mx = 120.29, My = 0, Mz = -307.16; the other vector is the resultant force. (Key point: Resultant)

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