---
title: "Two wrenches to single wrench; line of action"
subject: "Applied Mechanics"
chapter: 3
key: "semester_1/applied_mechanics:two-wrenches-single-wrench-line-of-action"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/two-wrenches-single-wrench-line-of-action
site: BCT Engineering
---

# Two wrenches to single wrench; line of action

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

- Asked in: 2069 Chaitra, 8 marks
- Video: [Problem related WRENCH | Applied Mechanics | Chapter 3 @EngineeringStudios1](https://www.youtube.com/watch?v=QQbo0wXUmpY) — Engineering Studios
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/two-wrenches-single-wrench-line-of-action
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-3

## Question

Replace the two wrenches as shown in figure by a single equivalent wrench and determine (a) the resultant force, (b) indicate its line of action. [Figure: axes x, y, z with origin O; a wrench along the y axis with 50 N·m couple and 30 N force; second wrench at point P, 8 m from O along z, with 40 N force and a 90 N·m couple.]

## Short answer

Resultant force: 50 N (40i+30j) at 36.9° to the x-axis. The single equivalent wrench has couple 234 N·m along the force; its axis passes through z = 3.24 m on the z-axis, parallel to R.

## Answer

**Reading of the figure:** wrench 1 at O along the y axis: force 30 N along +y with a 50 N·m couple along the y axis (drawn opposite to the force, so −y). Wrench 2 at P on the z axis, 8 m from O: force 40 N along +x and a 90 N·m couple along +x.

#### (a) Resultant force
R = 40 i + 30 j = (40, 30, 0) N
|R| = √(40² + 30²) = **50 N**, at tan⁻¹(30/40) = 36.9° to the x axis, parallel to the xy plane.

#### Resultant moment about O
```text
Couples: (0, −50, 0) + (90, 0, 0) = (90, −50, 0)
r × F of 40 N: (0,0,8) × (40,0,0) = (0, 320, 0)
M_O = (90, 270, 0) N·m
```
R·M_O = 40×90 + 30×270 = 11 700
M∥ = 11 700 / 50 = 234 N·m along R; pitch p = 11 700/2500 = 4.68 m.

#### (b) Line of action (axis of the wrench)
r₀ = (R × M_O)/R²
R × M_O = (30×0 − 0×270, 0×90 − 40×0, 40×270 − 30×90) = (0, 0, 8100)
r₀ = (0, 0, 8100/2500) = (0, 0, 3.24) m

The line of action passes through the point (0, 0, 3.24) m on the z axis (3.24 m from O) and is parallel to R = (40, 30, 0), so its equation is
(x, y, z) = (0, 0, 3.24) + t (40, 30, 0).

**Answer: the single equivalent wrench is a force of 50 N (40 i + 30 j) plus a couple of 234 N·m along the force; its axis passes through z = 3.24 m on the z axis, parallel to R.**

## Check yourself

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

1. What is the resultant of 40 N along x and 30 N along y?
   A. R = 50 N at 36.9° to the x axis
   B. R = 70 N
   C. R = 10 N
   D. R = 50 N at 53.1° to the x axis
   Answer: A. R = (40, 30, 0), |R| = sqrt(1600+900) = 50 N, tan(theta) = 30/40. (Key point: Resultant force)

2. Why is M_y = 270 N·m?
   A. -50 (couple opposite to the force) + 320 (40 N at z = 8 m) = 270
   B. 50 + 320 = 370
   C. 320 - 90 = 230
   D. -50 + 90 = 40
   Answer: A. (0,0,8) × (40,0,0) = (0,320,0); couples (90,-50,0) are added. (Key point: Moment about O)

3. How is the pitch found?
   A. p = R·M_O/R^2 = 11700/2500 = 4.68 m
   B. p = 50/234
   C. p = M_O/R
   D. p = 270/40
   Answer: A. Pitch is the ratio of the parallel couple M∥ (234) to R (50). (Key point: Pitch)

4. The axis passes through which point?
   A. (0, 0, 3.24) m, parallel to R
   B. (0, 0, 8) m
   C. (3.24, 0, 0) m
   D. The origin O
   Answer: A. r0 = R × M_O/R^2 = (0,0,8100/2500). (Key point: Axis of wrench)

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