---
title: "Tensions in tugboat ropes pulling vessel"
subject: "Applied Mechanics"
chapter: 3
key: "semester_1/applied_mechanics:vessel-tugboats-rope-tensions"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/vessel-tugboats-rope-tensions
site: BCT Engineering
---

# Tensions in tugboat ropes pulling vessel

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

- Asked in: 2070 Asar, 4 marks
- Video: [Two tugboats are towing a ship, Each exerts a force of 6000N and the angle b/w the two ropes is 60⁰](https://www.youtube.com/watch?v=Eft4vnlv9oQ) — Physics With Akhtiar Ali Bullo
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/vessel-tugboats-rope-tensions
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-3

## Question

A commercial vessel is being pulled into harbour for unloading by two tugboats as shown in figure knowing the vessel requires 150 kN along its axis to move it steadily. Compute the tensions in rope AB and BC when α = 40°. [Figure: vessel B pulled by ropes to tugboats A (above) and C (below); one rope at 30° to the vessel's axis, the other at α; resultant 150 kN along the axis.]

## Short answer

Using equilibrium, T_AB sin30° = T_BC sin40° and T_AB cos30° + T_BC cos40° = 150 kN. Solving gives T_AB = 102.6 kN and T_BC = 79.8 kN.

## Answer

**Reading of the figure:** the vessel B needs a pull of 150 kN along its axis. Rope BA (to tugboat A) makes 30° with the axis, and rope BC (to tugboat C) makes α = 40° with the axis, on the other side. The two rope tensions have resultant 150 kN along the axis.

Let T_AB and T_BC be the tensions, x along the axis.

ΣFy = 0 (perpendicular to the axis, the components must cancel):
T_AB sin 30° = T_BC sin 40°
T_BC = T_AB × (0.5/0.6428) = 0.7779 T_AB …(1)

ΣFx = 150 kN (along the axis):
T_AB cos 30° + T_BC cos 40° = 150 …(2)

Put (1) in (2): T_AB (0.8660 + 0.7779 × 0.7660) = 150
T_AB (0.8660 + 0.5959) = 150
T_AB = 150/1.4619 = 102.6 kN
T_BC = 0.7779 × 102.6 = 79.8 kN

(Check with the sine rule: 150/sin 110° = 159.6; T_AB = 159.6 × sin 40° = 102.6 kN; T_BC = 159.6 × sin 30° = 79.8 kN.)

**Answer: T_AB = 102.6 kN, T_BC = 79.8 kN.**

## Check yourself

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

1. What condition connects the two rope tensions perpendicular to the vessel's axis?
   A. T_AB sin30° = T_BC sin40° (components cancel)
   B. T_AB = T_BC
   C. T_AB cos30° = T_BC cos40°
   D. T_AB sin30° + T_BC sin40° = 150
   Answer: A. The resultant must lie along the axis, so sideways components cancel. (Key point: Sideways balance)

2. What equation uses the 150 kN requirement?
   A. T_AB cos30° + T_BC cos40° = 150
   B. T_AB sin30° + T_BC sin40° = 150
   C. T_AB + T_BC = 150
   D. T_AB cos30° - T_BC cos40° = 150
   Answer: A. Components along the axis add up to the required pull of 150 kN. (Key point: Axis balance)

3. Why is T_AB (102.6 kN) larger than T_BC (79.8 kN)?
   A. Its angle with the axis is smaller (sin30° < sin40°), so it needs more tension to cancel the sideways pull
   B. Because tugboat A is larger
   C. Because rope AB is longer
   D. Because 30° is greater than 40°
   Answer: A. From T_BC = 0.7779 T_AB, the tension at the smaller angle is greater. (Key point: Which tension is larger)

## Related in Applied Mechanics

- [Force P to move block A (blocks A and B)](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Atwo-blocks-force-p-move-block-a-30deg) — Applied Mechanics, chapter 5
- [Parallel axis theorem; MOI of trapezoid with cutout](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aparallel-axis-theorem-moi-trapezoid-semicircle) — Applied Mechanics, chapter 4
- [MOI of composite area about Y axis](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Amoi-composite-area-about-y-axis) — Applied Mechanics, chapter 4
- [Replace two wrenches by single wrench](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Atwo-wrenches-single-wrench) — Applied Mechanics, chapter 3
- [Force to force and couple; two wrenches resultant](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aforce-to-force-couple-two-pairs-wrench) — Applied Mechanics, chapter 3
- [Two wrenches to single wrench; line of action](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Atwo-wrenches-single-wrench-line-of-action) — Applied Mechanics, chapter 3

## Related in other subjects

- [Field of charged rod along its axis](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Acharged-rod-field-along-axis) — Engineering Physics, chapter 7
- [Two-compartment vessel gauges: reading of gauge C](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Atwo-compartment-gauges-a4bar-b2bar) — Fundamental of Thermodynamics & Heat Transfer, chapter 1
- [Rigid bar AB hinged at A: 100 kN](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Arigid-bar-ab-hinged-a-copper) — Strength of Materials, chapter 3
- [Moment distribution: frame, 150 kN, 60 kN/m](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Amoment-dist-frame-150kn-50kn-60knm) — Theory of Structures II, chapter 3