---
title: "Principle of transmissibility; resultant on plate ABCD"
subject: "Applied Mechanics"
chapter: 3
key: "semester_1/applied_mechanics:transmissibility-resultant-plate-abcd"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/transmissibility-resultant-plate-abcd
site: BCT Engineering
---

# Principle of transmissibility; resultant on plate ABCD

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

- Asked in: 2071 Chaitra, 3+8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/transmissibility-resultant-plate-abcd
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-3

## Question

Describe the principle of transmissibility. Determine the magnitude, direction and position with respect to center 'O' of the resultant of the forces acting on the rectangular plate ABCD as shown in figure below. [Figure: rectangular plate ABCD, DC = 4 m, BC = 3 m, centre O; 300 kN at C at 30°; 200 kN at D at 60°; 100 kN·m couple; 100 kN horizontal at A; 150 kN at B at 45°.]

## Short answer

Principle of transmissibility: a force's external effect on a rigid body does not change if moved along its line of action. Resultant: 98.13 kN at 34.84° below the horizontal, passing 0.87 m left of O and 0.61 m below O.

## Answer

#### Principle of transmissibility
The external effect of a force on a rigid body (its tendency to move or rotate the body) does not change if the force is moved to any other point on its line of action, with the same magnitude and direction. Only the line of action matters, not the point of application. For example, pushing a box at the back or pulling it with a rope at the front along the same line gives the same motion. It is valid only for rigid bodies and for external effects (not for internal stress or deformation).

#### Resultant of the forces on plate ABCD
**Reading of the figure:** A(−2, −1.5), B(2, −1.5), C(2, 1.5), D(−2, 1.5) with O at the centre (DC = 4 m, BC = 3 m). Forces:
- 300 kN at C, up and to the right at 30° to the horizontal.
- 200 kN at D, down and to the left, 60° from the vertical side DA (so 30° below the horizontal).
- 100 kN at A, horizontal, to the right.
- 150 kN at B, down and to the left at 45° to the vertical.
- Couple 100 kN·m, clockwise.

#### Components
| Force | Fx (kN) | Fy (kN) | Moment about O = x·Fy − y·Fx |
|---|---|---|---|
| 300 at C(2, 1.5) | 259.81 | 150.00 | 2(150) − 1.5(259.81) = −89.71 |
| 200 at D(−2, 1.5) | −173.21 | −100.00 | −2(−100) − 1.5(−173.21) = +459.81 |
| 100 at A(−2, −1.5) | 100.00 | 0 | 0 − (−1.5)(100) = +150.00 |
| 150 at B(2, −1.5) | −106.07 | −106.07 | 2(−106.07) − (−1.5)(−106.07) = −371.25 |
| Couple | | | −100.00 |
| **Sum** | **80.54** | **−56.07** | **+48.87** |

#### Resultant
R = √(80.54² + 56.07²) = **98.13 kN**
θ = tan⁻¹(56.07/80.54) = **34.84° below the horizontal** (towards the right).

**M_O = +48.87 kN·m (counter-clockwise).**

#### Position
Moment of R about O: x R_y − y R_x = M_O
- Cuts the horizontal through O (y = 0) at x = 48.87/(−56.07) = **−0.87 m** (0.87 m to the left of O).
- Cuts the vertical through O (x = 0) at y = −48.87/80.54 = **−0.61 m** (0.61 m below O).

**Answer: R = 98.13 kN at 34.8° below the horizontal, passing 0.87 m to the left of O on the horizontal through O (or 0.61 m below O on the vertical through O).**

## Check yourself

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

1. What does the principle of transmissibility state?
   A. A force can be moved anywhere along its line of action without changing its external effect on a rigid body
   B. A force can be moved to any parallel line with no change
   C. A force can be moved anywhere on the body
   D. Forces on a body are always transmitted equally to supports
   Answer: A. Only the line of action matters, not the point of application. (Key point: Principle statement)

2. When is the principle valid?
   A. For rigid bodies and external effects only
   B. For deformable bodies and internal stresses
   C. Only for couples
   D. Only for forces at the centre
   Answer: A. For deformable bodies or internal stresses, the point of application matters. (Key point: Validity)

3. How is the moment of the 300 kN force at C(2, 1.5) about O found?
   A. x·Fy - y·Fx = 2(150) - 1.5(259.81) = -89.71
   B. x·Fx + y·Fy
   C. F × 4 m
   D. 300 × 1.5
   Answer: A. Use the cross-product formula with components, taking counter-clockwise positive. (Key point: Moment about O)

4. How is the line of action of R located relative to O?
   A. x·R_y - y·R_x = M_O, giving x = -0.87 m (y = 0) or y = -0.61 m (x = 0)
   B. R acts through O, since all forces are on the plate
   C. At the corner nearest the largest force
   D. At a distance equal to R
   Answer: A. The resultant's moment about O must equal the +48.87 kN·m system moment. (Key point: Position of resultant)

## Related in Applied Mechanics

- [Resultant magnitude, direction and position of forces](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aresultant-force-position-rectangle-200-300-150) — Applied Mechanics, chapter 3
- [Resultant magnitude, direction, position about A](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aresultant-system-forces-about-a-beam) — Applied Mechanics, chapter 3
- [Resultant magnitude, direction and line of action](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aresultant-line-of-action-20n-100n-50n) — Applied Mechanics, chapter 3
- [Resultant of three forces on inclined block](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aresultant-three-forces-600n-400n-300n) — Applied Mechanics, chapter 3

## Related in other subjects

- [Truss ILDs for FG and BC; max forces from 100 kN](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Atruss-ild-fg-bc-100kn-roll) — Theory of Structures I, chapter 5