---
title: "Rigid and deformable body; FBD principles"
subject: "Applied Mechanics"
chapter: 2
key: "semester_1/applied_mechanics:rigid-deformable-fbd-principles-static-equilibrium"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/rigid-deformable-fbd-principles-static-equilibrium
site: BCT Engineering
---

# Rigid and deformable body; FBD principles

Applied Mechanics · Chapter 2: Basic Concept in Statics and Static Equilibrium

- Asked in: 2067 Asar, 4 marks
- Video: [Equilibrium of Rigid Bodies (2D - Coplanar Forces) | Mechanics Statics | (Solved examples)](https://www.youtube.com/watch?v=jQDEOwrR4UU) — Question Solutions
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/rigid-deformable-fbd-principles-static-equilibrium
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-2

## Question

Define rigid and deformable body. Explain principles of free body diagram and static equilibrium while solving problems in statics. Support your answer with examples.

## Short answer

A rigid body does not deform under load; a deformable body changes shape. FBD principles: isolate the body, replace supports by reactions, show all external forces. Static equilibrium ΣF=0, ΣM=0 ensures rest, used to solve reactions and internal forces, e.g., ladder against smooth wall.

## Answer

#### Rigid body
A rigid body is an ideal body that does not deform under load: the distance between any two of its points is constant. Statics treats solid bodies as rigid.

#### Deformable body
A deformable body changes shape and size under load (stretching, bending, twisting). This is the real behaviour of bodies and is studied in strength of materials.

#### Principles of the free body diagram
1. Isolate the body or joint completely from its surroundings.
2. Show every external force on it: applied loads, weight (at the centre of gravity) and reactions.
3. Replace each support by its reaction: roller, one normal force; hinge, two components; fixed, two components and a moment; cable, a tension along the cable; smooth contact, a normal force.
4. Show only external forces. Internal forces cancel in pairs (Newton's third law).
5. Mark known values, label unknowns, and show dimensions and angles.

#### Principle of static equilibrium
A body at rest stays at rest when the resultant force and moment on it are both zero:
- Particle: ΣFx = 0, ΣFy = 0 (and ΣFz = 0 in 3D).
- Rigid body (2D): ΣFx = 0, ΣFy = 0, ΣM = 0.

#### Example
A ladder AB of length L and weight W rests on a smooth wall at A and on rough ground at B. It makes angle θ with the ground, and A is at height h above the ground. FBD: W at the middle, N_A horizontal at A, N_B vertical and friction F horizontal at B. Then ΣFx = 0 gives F = N_A; ΣFy = 0 gives N_B = W; and ΣM_B = 0 gives N_A × h = W × (L/2) cos θ. These give all the unknowns.

## Check yourself

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

1. A deformable body differs from a rigid body in that it:
   A. Obeys no force law
   B. Stays fixed
   C. Changes shape and size under load
   D. Has no mass
   Answer: C. It is studied in strength of materials. (Key point: Rigid vs deformable)

2. Internal forces are not shown on an FBD because they:
   A. Cancel in pairs by Newton's third law
   B. Do not exist
   C. Are too small
   D. Act at supports
   Answer: A. Only external forces matter. (Key point: External only)

3. For a rigid body in 2D static equilibrium:
   A. SFx = 0, SFy = 0, SM = 0
   B. SM = 0 only
   C. SFx = SFy = SFz = 0
   D. SFx = 0 only
   Answer: A. Three equations. (Key point: Equilibrium)

4. For the ladder, SFy gives:
   A. F = W
   B. F = N_A
   C. N_B = W
   D. N_A = W
   Answer: C. Vertical forces: N_B up, W down. (Key point: Ladder)

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