---
title: "Force to force and couple; two wrenches resultant"
subject: "Applied Mechanics"
chapter: 3
key: "semester_1/applied_mechanics:force-to-force-couple-two-pairs-wrench"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/force-to-force-couple-two-pairs-wrench
site: BCT Engineering
---

# Force to force and couple; two wrenches resultant

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

- Asked in: 2070 Chaitra, 3+8 marks
- Video: [Engineering Mechanics: Statics Theory | Force Reduction (Wrench)](https://www.youtube.com/watch?v=SrQeoXIbV8s) — Dr. Clayton Pettit
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/force-to-force-couple-two-pairs-wrench
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-3

## Question

How can you reduce a force into a force and couple? Obtain the resultant of the two pairs of wrench shown in the figure. Indicate its line of action. [Figure: first wrench along the y axis with 30 N·m couple and 40 N force; second wrench along the z side with 60 N force and 40 N·m couple; distance 4 m from O along the axes.]

## Short answer

To reduce a force at A to O, introduce equal opposite forces at O to form a couple; equivalent system is F at O plus M_O = r×F. The two wrenches reduce to a single wrench: force 72.1 N (60i+40j N), couple 149.8 N·m along the force, axis through (0,0,2.115) m.

## Answer

#### Reducing a force into a force and a couple
A force F acting at point A can be replaced by an equal force F acting at another point O together with a couple.
1. At O, add two equal and opposite forces F and −F, both parallel to the original force. This does not change the effect.
2. The original F at A and the −F at O form a couple of moment M_O = r × F, where r is the vector from O to A.
3. What remains is the force F at O and the couple M_O.

So F at A is equivalent to F at O plus the couple M_O = r × F (magnitude F × d, where d is the perpendicular distance from O to the line of action).

#### Resultant of the two wrenches
**Reading of the figure:**
- Wrench 1 at O along the y axis: force 40 N along +y and couple 30 N·m (vector along −y, drawn opposite to the force).
- Wrench 2 at A on the z axis, 4 m from O: force 60 N along +x and couple 40 N·m along +x.

R = (60, 40, 0) N, |R| = √(60² + 40²) = 72.11 N
Direction: tan⁻¹(40/60) = 33.7° from the x axis towards y.

M_O = couples + r × F = (40, −30, 0) + (0, 0, 4) × (60, 0, 0) = (40, −30, 0) + (0, 240, 0) = (40, 210, 0) N·m

R·M_O = 60×40 + 40×210 = 10 800
Parallel part M∥ = 10 800/72.11 = 149.8 N·m, pitch p = 10 800/5200 = 2.08 m

Line of action: r₀ = (R × M_O)/R²
R × M_O = (0, 0, 60×210 − 40×40) = (0, 0, 11 000)
r₀ = (0, 0, 11 000/5200) = (0, 0, 2.115) m

**Answer:** the two wrenches reduce to one wrench: force R = 72.1 N (60 i + 40 j) with a couple of 149.8 N·m along R. The line of action of R passes through (0, 0, 2.115 m), on the z axis 2.115 m from O, and is parallel to R, lying in the horizontal plane z = 2.115 m.

## Check yourself

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

1. How is a force F at A replaced by a force at O?
   A. Add F and -F at O; the original F at A and -F at O form a couple M_O = r × F
   B. Move F to O with no change
   C. Add F at O only
   D. Double the force at O
   Answer: A. What remains is F at O plus the couple r × F. (Key point: Force to force and couple)

2. What is the resultant force of the two wrenches?
   A. R = (60, 40, 0) N, |R| = 72.11 N
   B. R = (100, 0, 0) N
   C. R = (40, 60, 0) N
   D. R = 60 + 40 = 100 N along y
   Answer: A. Sum 60 N along +x and 40 N along +y. (Key point: Resultant force)

3. How does M_O = (40, 210, 0) arise?
   A. Couples (40, -30, 0) plus r × F = (0, 240, 0) from the 60 N force
   B. The couples added with no r × F term
   C. 30 + 40 + 240 along y
   D. (0,0,4) × (40,0,0)
   Answer: A. The y component is -30 (couple opposite to force) + 240 = 210. (Key point: Moment about O)

4. Which expression locates the line of action?
   A. r0 = (R × M_O)/R^2 = (0, 0, 11000/5200) = (0, 0, 2.115) m
   B. r0 = (R · M_O)/R^2
   C. r0 = M_O/R
   D. r0 = (0, 0, 4) m
   Answer: A. This gives the point on the axis nearest to O, parallel to R. (Key point: Axis position)

## Related in Applied Mechanics

- [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
- [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
- [Couple, wrench, moment of P about points and lines](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Acouple-free-vector-wrench-cube-force-p) — Applied Mechanics, chapter 3
- [Resultant magnitude, direction and line of action](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aresultant-line-of-action-20n-100n-50n) — Applied Mechanics, chapter 3