Curriculum

Mathematics

Unit 2

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Unit 5

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Text lesson

Couples and the Resultant Couple

Mathematics

Couples and the Resultant Couple

Overview

This lesson covers lessons 3-1 and 3-2 of the statics unit "Couples". After it you should be able to:

define a couple and find its moment;

decide when two couples are equivalent or in equilibrium;
show that a system of forces is equivalent to a couple, and find the resultant of several couples.

🌳 Concept tree

Couples and the Resultant Couple

├── 3-1 (a) What is a couple?

│ ├── From the previous lesson

│ ├── Definition

│ └── Experiment

├── 3-1 (b) The moment of a couple

│ ├── Definition

│ ├── The moment is a constant vector

│ └── Observations

├── 3-1 (c) Equilibrium under coplanar couples

│ └── Condition

├── 3-1 (d) Equivalent couples

│ └── Definition

├── 3-2 (a) Systems equivalent to a couple

│ └── Two conditions

├── 3-2 (b) Forces represented by the sides of a triangle

│ ├── Rule

│ ├── Why

│ └── Generalisation to polygons

├── 3-2 (c) The three-point test

│ ├── Rule

│ └── Why

├── 3-2 (d) The resultant couple

│ └── Definition

└── 3-2 (e) Summary

├── Comparison

└── Key points

Couples and the Resultant Couple · 1 / 10

3-1 (a) What is a couple?

From the previous lesson

Two unlike parallel forces of unequal size have a resultant $F_{1} - F_{2}$ .
If the two forces become equal, the resultant force is zero, but the body does not stay still: it turns.
Such a pair cannot be replaced by a single force, nor by two non-parallel forces; it has its own name.

A couple is a system of two forces of equal magnitude, opposite directions and different lines of action.
Examples turning a tap, a screwdriver or the steering wheel with two hands, a key in a lock.
The perpendicular distance between the two lines of action is the arm of the couple.

Two forces of 15 N act in opposite directions at the ends of a graduated ruler. Take the sum of their moments about different points of the ruler.
You find the same number at every point the turning effect does not depend on the point chosen.

Couples and the Resultant Couple · 2 / 10

3-1 (b) The moment of a couple

Definition

The moment of a couple is the sum of the moments of its two forces about any point in space.
Its magnitude is the product of one force and the arm M=F×L where F is the magnitude of each force and L is the arm.
Its direction is perpendicular to the plane of the couple, given by the right-hand rule. In a plane we use a sign: anticlockwise positive, clockwise negative.
Units N·m, N·cm, kg·wt·cm, dyne·cm.

Couples and the Resultant Couple: a steering wheel pushed by two hands with equal forces of 10 N shows a couple and its moment: 4 N m.

The moment is a constant vector

Let the forces $\vec{F}$ and $- \vec{F}$ act at $A$ and $B$ , and let $O$ be any point.
$\vec{M} = \vec{O A} \times \vec{F} + \vec{O B} \times (- \vec{F}) = (\vec{O A} - \vec{O B}) \times \vec{F} = \vec{B A} \times \vec{F}$ .
The point O has vanished from the result the moment is the same about every point.

Observations

If the arm becomes larger, the same moment needs a smaller force.
Only the moment (size and sense) matters; the couple can be moved anywhere in its plane or into a parallel plane without changing its effect.
A couple can be rotated in its own plane, and its forces and arm can be changed, provided the moment stays the same.

✍️ Worked examples

Example 1:
Forces of 40 N form a couple with arm 4 cm in one case, and forces of 20 N form a couple with arm 8 cm in another.
Compare the moments.
Solution:
First couple 40×4=160 N·cm.
Second couple 20×8=160 N·cm.
The moments are equal in magnitude halving the force while doubling the arm leaves the turning effect unchanged.
Example 2:
Two forces of 20 N act in opposite directions at the ends of a rod 50 cm long, each making an angle of $3 0^{\circ}$ with the rod.
Find the moment of the couple they form.
Solution:
The arm is the perpendicular distance between the two lines of action: $L = 50 \sin 3 0^{\circ} = 25$ cm.
$M = 20 \times 25 = 500$ N·cm.
Example 3 The forces ${\vec{F}}_{1} = 2 \hat{i} + b \hat{j}$ at $A (- 1, 3)$ and ${\vec{F}}_{2} = a \hat{i} - 5 \hat{j}$ at $B (2, 2)$ form a couple. Find $a$ , $b$ and the moment.
Solution:
For a couple F⃗2=−F⃗1 a=−2 and b=5, so F⃗1=(2,5).
Moment about B BA→×F⃗1 with BA→=(−3,1).
(−3)(5)−(1)(2)=−17, so M⃗=−17k^ a clockwise couple of magnitude 17 units.
Example 4 Forces $(3 \hat{i} - 5 \hat{j})$ and $(- 3 \hat{i} + 5 \hat{j})$ N act at $A (6, 1)$ and $B (4, 1)$ m. Show that they form a couple and find its moment.
Solution:
The forces are equal and opposite, and the points differ so the lines of action are not the same: it is a couple.
$\vec{B A} = (2, 0)$ ; moment $= (2) (- 5) - (0) (3) = - 10$ .
$\vec{M} = - 10 \hat{k}$ N·m (clockwise, magnitude 10 N·m).

Couples and the Resultant Couple · 3 / 10

3-1 (c) Equilibrium under coplanar couples

Condition

A body acted on only by coplanar couples is in equilibrium if the sum of their moments is zero:
$M_{1} + M_{2} + \dots + M_{n} = 0$
The sum is algebraic anticlockwise couples positive, clockwise negative.
A couple can never be balanced by a single force; only by another couple.

✍️ Worked examples

Example 5:
A light rod is acted on by a pair of 10 N forces (arm 7 m), a pair of $40 \sqrt{3}$ N forces (making $6 0^{\circ}$ with a line of 3 m between them), and a pair of 250 N forces (arm 1 m).
Show that the rod is in equilibrium (given the senses the first two clockwise, the third anticlockwise).
Solution:
$M_{1} = - 10 \times 7 = - 70$ N·m.
$M_{2} = - 40 \sqrt{3} \times 3 \sin 6 0^{\circ} = - 40 \sqrt{3} \times 3 \times \frac{\sqrt{3}}{2} = - 180$ N·m.
$M_{3} = + 250 \times 1 = 250$ N·m.
Sum $= - 70 - 180 + 250 = 0$ , so the rod is in equilibrium.
Example 6:
A light rod $A B$ , 30 cm long, is pivoted at its midpoint.
Forces of 7.5 N act at its ends, one up and one down, and a string with tension 10 N pulls at $6 0^{\circ}$ to the rod at a point $C$ .
What force acting with these keeps the rod horizontal?
Solution:
The two 7.5 N forces form a couple of moment $- 7.5 \times 30 = - 225$ N·cm.
The additional force must form a couple with the tension. It equals the tension, $F = 10$ N, in the opposite direction, parallel, and the sum of moments must be $+ 225$ .
The arm is $C D \sin 6 0^{\circ}$ where $D$ is the point of application along the rod: $10 \times C D \sin 6 0^{\circ} = 225$ .
$C D = \frac{225}{10 \times 0.866} = 15 \sqrt{3} \approx 26$ cm, so $D$ is $15 \sqrt{3}$ cm from $C$ along the rod.
Example 7:
A thin square lamina $A B C D$ (side 60 cm, weight 200 gf at its centre) is hung from a pin at corner $A$ .
A couple of moment $3000 \sqrt{2}$ gf·cm acts in its plane.
Find the angle $A C$ makes with the vertical in equilibrium.
Solution:
The pin reaction and the weight form a couple (they are equal, opposite and parallel); it must balance the external couple.
Distance AM from the pin to the centre AM=302 cm.
If $θ$ is the angle of $A C$ with the vertical, the arm of the weight/reaction couple is $A M \sin θ$ .
$200 \times 30 \sqrt{2} \sin θ = 3000 \sqrt{2}$ , so $\sin θ = \frac{3000 \sqrt{2}}{6000 \sqrt{2}} = \frac{1}{2}$ .
Hence $θ = 3 0^{\circ}$ (or $15 0^{\circ}$ on the other side of the vertical).

Couples and the Resultant Couple · 4 / 10

3-1 (d) Equivalent couples

Definition

Two couples in the same plane are equivalent if their moments are equal (in size and in sense).
Equivalent couples have the same effect on a rigid body even if their forces and arms are different.

✍️ Worked examples

Example 8:
$A B C D$ is a rectangle with $A B = 12$ cm and $B C = 16$ cm.
Equal forces of 96 N act along $A B$ and $C D$ (opposite directions).
Find the equal forces acting at $A$ and $C$ parallel to $B D$ whose couple is equivalent to this one.
Solution:
First couple arm =BC=AD=16 cm, moment 96×16=1536 N·cm (clockwise).
$B D = \sqrt{1 2^{2} + 1 6^{2}} = 20$ cm.
The distance of $A$ from $B D$ is $\frac{12 \times 16}{20} = 9.6$ cm, and the same for $C$ on the other side, so the arm of the second couple is $2 \times 9.6 = 19.2$ cm.
Equal moments F×19.2=1536, so F=80 N, acting clockwise.
Example 9:
A square $A B C D$ has side 10 cm.
Forces of 60 N act along $B A$ and $D C$ .
Find the equal forces at $A$ and $C$ parallel to $B D$ forming an equivalent couple.
Solution:
First couple arm =10, moment =600 N·cm.
Diagonal $B D = 10 \sqrt{2}$ ; the distance from $A$ to $B D$ is $5 \sqrt{2}$ , so the arm of the new couple is $10 \sqrt{2}$ .
$F \times 10 \sqrt{2} = 600$ , so $F = 30 \sqrt{2} \approx 42.4$ N.

Couples and the Resultant Couple · 5 / 10

3-2 (a) Systems equivalent to a couple

Two conditions

A system of coplanar forces is equivalent to a couple if both hold:
the resultant force is zero (the sum of components along any direction is zero);
the sum of the moments about a point is not zero.
If the first condition holds but the moments also sum to zero, the system is in equilibrium, not a couple.
If concurrent forces have a zero resultant, they are in equilibrium and cannot be a couple.

✍️ Worked examples

Example 10:
A light rod $A B$ carries a 15 N force at $A$ and opposite forces of 8 N at $C$ ( $A C = 5$ cm) and 7 N at $B$ ( $A B = 14$ cm).
Show that the system is equivalent to a couple and find its moment.
Solution:
Take the 15 N direction as positive R=15−8−7=0.
Moments about A −8×5−7×14=−40−98=−138 N·cm.
The resultant is zero but the moment is not, so it is equivalent to a clockwise couple of moment $- 138$ N·cm.
Check about B −15×14+8×9=−138, the same value, as it must be for a couple.
Example 11:
Points $A$ , $B$ , $C$ lie on a line in that order with $A B = 6$ cm and $B C = 4$ cm.
Parallel forces of 5 N up at $A$ , 9 N down at $B$ and 4 N up at $C$ act on a rod.
Show that the system is equivalent to a couple.
Solution:
Upward positive R=5−9+4=0.
Moments about $A$ (anticlockwise positive, force times distance to the right): $- 9 \times 6 + 4 \times 10 = - 14$ N·cm.
Moments about C 5×(−10)+(−9)(−4)=−50+36=−14 N·cm.
The resultant is zero and the moment is the non-zero constant $- 14$ N·cm: a clockwise couple of moment 14 N·cm.

Couples and the Resultant Couple · 6 / 10

3-2 (b) Forces represented by the sides of a triangle

Rule

If three coplanar forces acting on a rigid body are completely represented (in magnitude, direction and line of action) by the sides $\vec{A B}$ , $\vec{B C}$ , $\vec{C A}$ of a triangle taken in one cyclic order, the system is equivalent to a couple.
The magnitude of its moment is
$M = 2 \times (area of the triangle) \times m$
where $m$ is the force that represents one unit of length.

The vector sum $\vec{A B} + \vec{B C} + \vec{C A} = \vec{0}$ , so the resultant force is zero.
Replace ${\vec{F}}_{1}$ at $A B$ and ${\vec{F}}_{2}$ at $B C$ by their resultant $- {\vec{F}}_{3}$ at $B$ ; together with ${\vec{F}}_{3}$ at $C A$ they form a couple.
Its arm is the altitude $B D$ from $B$ to $A C$ , so $M = F_{3} \times B D = (m \cdot A C) B D = m \times 2 area$ .

✍️ Worked examples

Example 12:
$A B C$ is isosceles with $A B = B C = 17$ cm and $A C = 16$ cm.
Forces of 340, 340 and 320 N act along $A B$ , $B C$ , $C A$ .
Show the system is a couple and find its moment.
Solution:
The ratios $\frac{340}{17} = \frac{340}{17} = \frac{320}{16} = 20$ are equal, so the forces are represented by the sides at $m = 20$ N per cm.
They act in one cyclic order, so the system is equivalent to a couple.
Altitude from B 172−82=15 cm; area =12×16×15=120 cm2.
$M = 2 \times 120 \times 20 = 4800$ N·cm.
Example 13 $A B C$ is right-angled at $B$ with $A B = 6$ cm and $B C = 8$ cm. Forces of 3, 4 and 5 N act along $\vec{A B}$ , $\vec{B C}$ , $\vec{C A}$ . Find the moment of the equivalent couple.
Solution:
$A C = 10$ , and $\frac{3}{6} = \frac{4}{8} = \frac{5}{1} 0 = 0.5$ N per cm, so $m = 0.5$ .
Area $= \frac{1}{2} \times 6 \times 8 = 24$ cm $^{2}$ .
$M = 2 \times 24 \times 0.5 = 24$ N·cm.

Generalisation to polygons

If a system of coplanar forces is completely represented by the sides of a closed polygon taken in one cyclic order, the system is equivalent to a couple of moment $2 \times (area of the polygon) \times m$ .

✍️ Worked examples

Example 14 A square $A B C D$ of side 10 cm has forces of 10 N along $A B$ , $B C$ , $C D$ and $D A$ (in that cyclic order). Find the moment.
Solution:
$m = \frac{10}{10} = 1$ N per cm and the area is $100$ cm $^{2}$ .
$M = 2 \times 100 \times 1 = 200$ N·cm.
Check by moments about A the forces along AB and DA pass through A (zero moment); the force along BC has arm 10 and the force along CD has arm 10, both in the same sense: 100+100=200.
Example 15:
$A B C D$ is a quadrilateral with $A B = A D = 20$ cm, $B C = C D = 10 \sqrt{7}$ cm and angle $A = 12 0^{\circ}$ .
Forces represented by $\vec{A B}$ , $\vec{B C}$ , $\vec{C D}$ , $\vec{D A}$ give a couple of moment $180 \sqrt{3}$ N·cm.
Find the forces.
Solution:
By symmetry $A C$ bisects the angle at $A$ , so ngle BAC = 60^°.
Cosine rule in triangle ABC (107)2=202+AC2−2(20)(AC)cos⁡60∘, so 700=400+AC2−20 AC.
$A C^{2} - 20 A C - 300 = 0$ gives $(A C - 30) (A C + 10) = 0$ , so $A C = 30$ cm.
Area of $A B C D = 2 \times \frac{1}{2} \times 20 \times 30 \times \sin 6 0^{\circ} = 300 \sqrt{3}$ cm $^{2}$ .
$2 \times 300 \sqrt{3} \times m = 180 \sqrt{3}$ , so $m = 0.3$ N per cm.
Forces AB→,DA→: 0.3×20=6 N; BC→,CD→: 0.3×107=37 N.

Couples and the Resultant Couple · 7 / 10

3-2 (c) The three-point test

Rule

If the sum of the moments of a system of coplanar forces about three non-collinear points in its plane has the same non-zero value $k$ , the system is equivalent to a couple of moment $k$ .

A coplanar system reduces to a single force, a single couple, or equilibrium.
It is not in equilibrium (the moments are non-zero).
If it reduced to a single force $F$ at distances $L_{1}, L_{2}, L_{3}$ from the three points, then $F L_{1} = F L_{2} = F L_{3}$ , so the three points would lie on a line parallel to the force; they do not. So it must be a couple.

✍️ Worked examples

Comparison aspect	Main idea	Keywords
Example 16	A square $A B C D$ of side 10 cm is acted on by the forces in Example 14. Use the test to confirm that it is a couple.	squareABCDsideacted
•	Solution	Solution
Moments about A	+200 (computed above)	computedabove
Moments about B	the forces along AB and BC pass through B CD has arm 10 and DA has arm 10 both in the same sense: +200	forcesalongpassthrough
Moments about C	similarly +200	similarly
•	Equal non-zero values at three non-collinear points, so the system is a couple of moment $200$ N·cm.	Equalzerovaluesthree

Couples and the Resultant Couple · 8 / 10

3-2 (d) The resultant couple

Definition

The sum of two or more coplanar couples is a single couple whose moment is the algebraic sum of the moments:
$M = M_{1} + M_{2} + \dots + M_{n}$
It is called the resultant couple. If $M = 0$ the couples are in equilibrium.

✍️ Worked examples

Example 17:
Three couples act on a plate 200 N forces with arm 0.8 m (anticlockwise), 500 N forces with arm 2.3 m (clockwise), and 100 N forces with arm 4 cm (clockwise).
Find the resultant couple.
Solution:
$M_{1} = + 200 \times 0.8 = 160$ N·m.
$M_{2} = - 500 \times 2.3 = - 1150$ N·m.
$M_{3} = - 100 \times 0.04 = - 4$ N·m.
M=160−1150−4=−994 N·m a clockwise couple of magnitude 994 N·m.
Example 18:
A square $A B C D$ of side 10 cm is acted on by 40 kg·wt forces along $A D$ and $C B$ (a couple) and 70 kg·wt forces along $A B$ and $C D$ (a second couple, opposite in sense).
Find the resultant couple.
Solution:
$M_{1} = + 40 \times 10 = 400$ kg·wt·cm.
$M_{2} = - 70 \times 10 = - 700$ kg·wt·cm.
$M = - 300$ kg·wt·cm (clockwise, magnitude 300).
Example 19 What force applied in the opposite sense with arm 12 cm balances a resultant couple of 300 kg·wt·cm?
Solution:
The balancing couple must have moment $+ 300$ , so $F \times 12 = 300$ and $F = 25$ kg·wt for each force.

Couples and the Resultant Couple · 9 / 10

3-2 (e) Summary

Point of comparison	Single force	Couple
Resultant force	the force itself	zero
Effect	translates (and may rotate)	pure rotation
Moment	depends on the point chosen	the same about every point
Can it be balanced?	by an equal opposite force on the same line	only by another couple of opposite moment

Key points

A couple is two equal, opposite, non-collinear parallel forces; its moment is $F \times L$ .
A couple’s moment is a constant vector, so its position in the plane does not matter.
Couples are equivalent when their moments are equal, in equilibrium when their moments sum to zero, and combine by adding moments.
A system is equivalent to a couple if its resultant force is zero while its moment is not zero.

Couples and the Resultant Couple · 10 / 10

Curriculum

Mathematics

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Unit 3

Unit 4

Unit 5

Unit 6

Unit 7

Couples and the Resultant Couple

define a couple and find its moment;

🌳 Concept tree

Couples and the Resultant Couple

│ ├── From the previous lesson

│ ├── Definition

│ └── Experiment

├── 3-1 (b) The moment of a couple

│ ├── Definition

│ ├── The moment is a constant vector

│ └── Observations

├── 3-1 (c) Equilibrium under coplanar couples

│ └── Condition

├── 3-1 (d) Equivalent couples

│ └── Definition

├── 3-2 (a) Systems equivalent to a couple

│ └── Two conditions

│ ├── Rule

│ ├── Why

│ └── Generalisation to polygons

├── 3-2 (c) The three-point test

│ ├── Rule

│ └── Why

├── 3-2 (d) The resultant couple

│ └── Definition

└── 3-2 (e) Summary

├── Comparison

└── Key points

From the previous lesson

Definition

The moment is a constant vector

Observations

✍️ Worked examples

Condition

✍️ Worked examples

Definition

✍️ Worked examples

Two conditions

✍️ Worked examples

Rule

✍️ Worked examples

Generalisation to polygons

✍️ Worked examples

Rule

✍️ Worked examples

Definition

✍️ Worked examples

Key points

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