---
title: "FBD; tensions in strings AC, CD, BD and angle θ"
subject: "Applied Mechanics"
chapter: 2
key: "semester_1/applied_mechanics:free-body-diagram-strings-ac-cd-bd-theta"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/free-body-diagram-strings-ac-cd-bd-theta
site: BCT Engineering
---

# FBD; tensions in strings AC, CD, BD and angle θ

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

- Asked in: 2069 Asar, 3+5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/free-body-diagram-strings-ac-cd-bd-theta
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-2

## Question

What is free body diagram? How is it drawn? Determine the tension on the strings AC, CD and BD and the angle θ in figure below. [Figure: string AC at 20° from point A to C; string CD between C and D at 85° at C; string DB from D to B; weight of 50 N hanging at C and 100 N hanging at D.]

## Short answer

At joint C, resolving forces horizontally and vertically gives T_AC = 54.96 N and T_CD = 18.87 N. At D, using T_CD and 100 N load yields T_BD = 100.14 N and the angle θ = 10.8° from the vertical.

## Answer

#### Free body diagram
A free body diagram is a sketch of a body (or joint) isolated from its surroundings, with all external forces acting on it shown.

**How it is drawn:**
1. Isolate the body or joint to be analysed.
2. Show the weight and all applied loads.
3. Replace each support or connected member by its force: a string or cable gives a tension pulling away from the body along the string.
4. Label known and unknown forces, with angles and axes.
5. Apply ΣFx = 0 and ΣFy = 0 (for a particle) or ΣM = 0 as well (for a rigid body).

#### Problem
**Reading of the figure:** string AC makes 20° with the vertical at C. String CD makes 85° with the vertical at C, so it is inclined 5° below the horizontal, going from C to D. Loads: 50 N hangs at C, 100 N hangs at D. String DB goes up to B at an angle θ to the vertical.

Let T₁ = T_AC, T₂ = T_CD, T₃ = T_BD.

#### Joint C (up and right positive)
T_AC pulls up-left, T_CD pulls right and slightly down, 50 N down.
```text
x:  −T1 sin20° + T2 sin85° = 0
y:   T1 cos20° − T2 cos85° − 50 = 0
```
From x: T1 = T2 (0.9962/0.3420) = 2.913 T2.
In y: 2.913 T2 × 0.9397 − 0.0872 T2 = 50 -> (2.737 − 0.0872) T2 = 50
**T₂ = T_CD = 18.87 N, T₁ = T_AC = 54.96 N**

#### Joint D
T_CD pulls D back toward C (left and slightly up), T_BD pulls up-right at θ to the vertical, 100 N down.
```text
x:  T3 sinθ = T2 sin85° = 18.87 × 0.9962 = 18.80 N
y:  T3 cosθ = 100 − T2 cos85° = 100 − 1.645 = 98.36 N
```
T₃ = √(18.80² + 98.36²) = 100.14 N
θ = tan⁻¹(18.80 / 98.36) = 10.82°

**Answer: T_AC = 54.96 N, T_CD = 18.87 N, T_BD = 100.14 N, θ = 10.8°.**

## Check yourself

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

1. A string acting on a joint gives:
   A. A reaction normal to the string
   B. A tension pulling away from the joint along the string
   C. A push along the string
   D. A moment
   Answer: B. A string can only pull. (Key point: String force)

2. String CD makes 85° with the vertical, so it is inclined to the horizontal at:
   A. 20°
   B. 15°
   C. 85°
   D. 5°
   Answer: D. 90° - 85° = 5°. (Key point: Angle)

3. Joint C is analysed first because:
   A. θ is known there
   B. It has the larger load
   C. It has only two unknowns, T_AC and T_CD
   D. It is at the ceiling
   Answer: C. Joint D has T_CD, T_BD and θ unknown. (Key point: Joint C)

4. At D, the horizontal equilibrium gives:
   A. T3 sin θ = 100
   B. T3 sin θ = T_CD sin 85°
   C. T3 cos θ = 100
   D. T3 = T_CD
   Answer: B. T_BD's horizontal component balances T_CD's, so θ ≈ 10.8°. (Key point: Joint D)

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