---
title: "Forces in cables AB and AC"
subject: "Applied Mechanics"
chapter: 2
key: "semester_1/applied_mechanics:cables-ab-ac-100n-load"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/cables-ab-ac-100n-load
site: BCT Engineering
---

# Forces in cables AB and AC

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

- Asked in: 2068 Shrawan, 5 marks
- Video: [Mechanical Engineering: Particle Equilibrium (7 of 19) Tension of Cables Attached to Hanging Object](https://www.youtube.com/watch?v=7ls-Dsu4ddY) — Michel van Biezen
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/cables-ab-ac-100n-load
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-2

## Question

Determine the forces in cables AB and AC as shown in figure below. [Figure: cables AB and BC fixed to the ceiling at A and C, meeting at B at angles 45° and 60° to the vertical, with a 100 N load hanging from B.]

## Short answer

Apply equilibrium at joint B. T_AB sin45° = T_AC sin60°, and vertical equilibrium gives T_AB = 89.66 N, T_AC = 73.21 N. Alternatively, using Lami’s theorem yields the same tensions.

## Answer

**Reading of the figure:** cable AB (from the support on the left) makes 45° with the vertical, the other cable (from the support on the right, joined at B) makes 60° with the vertical, and a load of 100 N hangs from B. The right-hand cable is the one called AC in the question (the supports are marked A and C), so I call its tension T_AC.

#### FBD of joint B
Forces: T_AB up-left (45° from vertical), T_AC up-right (60° from vertical), 100 N down.

ΣFx = 0: T_AC sin 60° − T_AB sin 45° = 0 …(1)
ΣFy = 0: T_AB cos 45° + T_AC cos 60° − 100 = 0 …(2)

From (1): T_AC = T_AB × (0.7071/0.8660) = 0.8165 T_AB

Substituting in (2): 0.7071 T_AB + 0.5 × 0.8165 T_AB = 100
1.1154 T_AB = 100
T_AB = 89.66 N
T_AC = 0.8165 × 89.66 = 73.21 N

(Check by Lami's theorem. The angle between T_AC and the load is 120°, between T_AB and the load is 135°, and between the two cables is 105°. So T_AB / sin 120° = T_AC / sin 135° = 100 / sin 105° = 103.53, which gives T_AB = 89.66 N and T_AC = 73.21 N.)

**Answer: T_AB = 89.66 N, T_AC (right cable) = 73.21 N.**

## Check yourself

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

1. The method used is:
   A. Taking the cables as rigid bars
   B. Moments about A
   C. Method of sections
   D. Equilibrium of the concurrent forces at joint B
   Answer: D. Three concurrent forces at B give SFx = 0 and SFy = 0. (Key point: Joint)

2. SFx = 0 gives:
   A. T_AC = T_AB
   B. T_AC cos 60° = T_AB cos 45°
   C. T_AB sin 45° = 100
   D. T_AC sin 60° = T_AB sin 45°
   Answer: D. Horizontal components balance. (Key point: Horizontal)

3. T_AB equals:
   A. 100 N
   B. 141.4 N
   C. 89.66 N
   D. 73.21 N
   Answer: C. 1.1154 T_AB = 100. (Key point: T_AB)

4. The cable at 45° from vertical carries more tension than the 60° one because:
   A. It is shorter
   B. It has less sag
   C. It is closer to vertical, so it supports more of the load
   D. It is on the left
   Answer: C. T_AB 89.66 N > T_AC 73.21 N. (Key point: Comparison)

## Related in Applied Mechanics

- [Truss indeterminacy; forces in BC, BG, HG, GD](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Atruss-indeterminacy-bc-bg-hg-gd) — Applied Mechanics, chapter 7
- [Truss member forces BC, BD, BE, DE, DG, EG](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Atruss-bc-bd-be-de-dg-eg-48kn) — Applied Mechanics, chapter 7
- [Use of trusses; cantilever truss member forces](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Acantilever-truss-bc-be-ef-ab-af-bf) — Applied Mechanics, chapter 7
- [FBD; tensions in strings AC, CD, BD and angle θ](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Afree-body-diagram-strings-ac-cd-bd-theta) — Applied Mechanics, chapter 2
- [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
- [Truss ideal assumptions and member forces](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Atruss-assumptions-member-forces-40kn-30kn) — Applied Mechanics, chapter 7, asked 2×

## 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
- [Head loss and velocity in loop pipes AB, BC, AC, CD](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Aloop-c73) — Water Supply Engineering, chapter 7
- [Design pipes AB, BC and AC, C = 110](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Apd-j66) — Water Supply Engineering, chapter 7
- [Design pipelines AB and BC, C = 100, 10 m head](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Apd-c70) — Water Supply Engineering, chapter 7