---
title: "Tension in cable holding a smooth rod"
subject: "Applied Mechanics"
chapter: 2
key: "semester_1/applied_mechanics:ladder-cable-tension"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/ladder-cable-tension
site: BCT Engineering
---

# Tension in cable holding a smooth rod

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

- Asked in: 2074 Asoj, 6 marks
- Video: [Beam and cable problem:  tension in the cable using torque and hinge force at the end of the beam.](https://www.youtube.com/watch?v=NvERr6H77mM) — Zak's Lab
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/ladder-cable-tension
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-2

## Question

Determine the tension in the cable BC which holds a part AB of length 4 m from sliding. The part has a mass of 10 kg. Assume all the contact surfaces are smooth. [Figure: rod AB of length 4 m leaning with B on the floor and inclined at 70° to the horizontal, resting at D on a 3 m high vertical wall edge; cable BC horizontal along the ground from B to C at the foot of the wall.]

## Short answer

Take moments about B to eliminate T and R_B: N_D × (3/sin70°) = W × 2cos70°. With W=98.1 N, N_D=21.02 N. Horizontal equilibrium gives T = N_D sin70° = 19.75 N.

## Answer

**Given (read from the figure):** rod AB, L = 4 m, mass 10 kg, inclined at 70° to the horizontal. B is on the smooth floor, the rod rests on the smooth top edge D of a wall 3 m high, and the horizontal cable BC joins B to the foot C of the wall. Take g = 9.81 m/s².

#### Forces on the rod (FBD)
- Weight W = 10 × 9.81 = 98.1 N at the mid-point of AB (2 m from B).
- Reaction at B, R_B, vertical (floor smooth).
- Reaction at D, N_D, perpendicular to the rod (smooth contact at an edge).
- Tension T at B, horizontal along BC.

#### Geometry
BD = 3 / sin 70° = 3 / 0.9397 = 3.193 m

#### Moments about B (this removes T and R_B)
N_D × BD = W × 2 cos 70°
N_D = 98.1 × 2 × 0.3420 / 3.193 = 21.02 N

#### Horizontal equilibrium
N_D acts perpendicular to the rod, so its horizontal component is N_D sin 70° and it pushes the foot B away from the wall. The cable must balance it:
T = N_D sin 70° = 21.02 × 0.9397 = 19.75 N

(Check: vertically R_B = W − N_D cos 70° = 98.1 − 7.19 = 90.9 N, which is positive, so B stays on the floor.)

**Answer: Tension in cable BC ≈ 19.75 N (about 19.8 N).**

## Check yourself

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

1. Moments are taken about B because:
   A. B is the centre of gravity
   B. It removes the unknowns T and R_B from the equation
   C. The wall is at B
   D. It gives the largest moments
   Answer: B. Both T and R_B pass through B, so only N_D remains unknown. (Key point: Moment centre)

2. The reaction N_D at the smooth edge D acts:
   A. Perpendicular to the rod
   B. Horizontally
   C. Vertically
   D. Along the cable
   Answer: A. A smooth contact exerts a force normal to the surface of the rod. (Key point: Reaction direction)

3. The cable tension equals:
   A. N_D
   B. The weight of the rod
   C. N_D cos 70°
   D. The horizontal component of N_D, N_D sin 70°
   Answer: D. Horizontal equilibrium requires T = N_D sin 70° = 19.75 N. (Key point: Cable tension)

4. The length BD from B to the wall edge is found as:
   A. 3 x sin 70°
   B. 3 / sin 70° = 3.193 m
   C. 4 / 2
   D. 3 / cos 70°
   Answer: B. The wall is 3 m high and BD is inclined at 70°, so BD = 3/sin 70°. (Key point: Geometry)

## Related in Applied Mechanics

- [Dynamic equilibrium; moment of block B on rod](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Adynamic-equilibrium-block-b-rod-moment-about-o) — Applied Mechanics, chapter 9
- [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
- [Laws of dry friction; overturning and sliding](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Alaws-dry-friction-overturning-sliding) — Applied Mechanics, chapter 5
- [Force P to prevent block sliding down incline](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Ablock-50kg-force-p-prevent-sliding-down) — Applied Mechanics, chapter 5
- [Forces in cables AB and AC](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Acables-ab-ac-100n-load) — Applied Mechanics, chapter 2
- [Reactions on two cylinders in channel](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Atwo-cylinders-channel-four-contact-reactions) — Applied Mechanics, chapter 2

## Related in other subjects

- [Cable on piers: tension, length, pier thrust (180 m)](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Acable-180m-5m-above-60-degree-backstay) — Theory of Structures I, chapter 7, asked 2×
- [Cable on piers: tension, length, pier thrust (200 m)](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Acable-200m-5m-above-60-degree-backstay) — Theory of Structures I, chapter 7, asked 2×
- [Rod AB sliding over wheel C](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Arod-ab-over-wheel-c-velocity) — Applied Mechanics (Dynamics), chapter 4
- [Cable with UDL: length, H and max tension](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Acable-45m-udl-20knm-different-levels) — Theory of Structures I, chapter 7