☰ Year plan · Applied Mathematics

Systems equivalent to a couple

Week 13 · Resultant couple: a system equivalent to a couple, forces along the sides of a triangle, and of a closed polygon · 3 lessons
2

Three forces represented by the sides of a triangle taken in order: a couple of moment 2 × area × force per unit length

Applied Mathematics Sec 3 Term 1, Unit 3, Lesson 2, Part 1
GoalFind the couple formed by three forces represented by the sides of a triangle taken in order, and use M = 2 × area × m.
New words
force per unit length m · forces proportional to the sides: the force along a side of length s is m × staken in order · along AB from A to B, along BC from B to C, along CA from C to A
Explanation

Take a triangle ABC and let three forces act along its sides: m × AB along AB (from A to B), m × BC along BC and m × CA along CA, where m is a fixed number of newtons per unit length. The first two forces meet at B. Their resultant is m(AB + BC) = m × AC as a vector, that is m × AC in the direction A to C, acting at B. The third force is m × CA, in the direction C to A, along the line CA. So the system is equal and opposite forces of size m × AC on parallel lines whose distance apart is the height BD of the triangle from B to AC. That is a couple, and its moment is m × AC × BD. But AC × BD = 2 × area of triangle, so M = 2 × (area) × m. The sign is anticlockwise (+) when A, B, C go round anticlockwise, and clockwise (−) when they go round clockwise. The rule needs forces proportional to the sides.

F₁F₂F₃ABC
Worked examples
5 cm12 cm13 cm10 N24 N26 NABC
In triangle ABC, AB = 5 cm, BC = 12 cm, CA = 13 cm (right angle at B) and A, B, C go round anticlockwise. Forces of 10 N, 24 N and 26 N act along AB, BC and CA in order. Find the resultant couple. Solution: m = 10 / 5 = 24 / 12 = 26 / 13 = 2 N per cm. Area = ½ × 5 × 12 = 30 cm². M = 2 × 30 × 2 = +120 N·cm (anticlockwise). Check about A: the 10 N and 26 N forces pass through A (no moment); the 24 N force at B, 5 cm from A, perpendicular to AB: 5 × 24 = 120.
13 cm14 cm15 cmF₁F₂F₃ABC
Triangle ABC has AB = 13 cm, BC = 14 cm, CA = 15 cm and A, B, C go round anticlockwise. Forces along AB, BC, CA in order, proportional to the sides, form a couple of 1680 N·cm. Find the three forces. Solution: area (Heron, s = 21) = √(21 × 8 × 7 × 6) = 84 cm². 2 × 84 × m = 1680, so m = 10 N per cm. The forces are 10 × 13 = 130 N, 10 × 14 = 140 N and 10 × 15 = 150 N.
Class activity

On squared paper draw a triangle with sides 3, 4 and 5 squares (right angle at one corner). Draw along each side an arrow whose length in squares equals the side (so m = 1 square of arrow per square of side). Check that the three arrows, put head to tail, make a closed triangle (zero resultant). Then count the area of the triangle in squares and compute 2 × area × 1.

Practice
1
An equilateral triangle ABC of side 8 cm (anticlockwise order) has forces of 12 N along each side in order. Find the moment of the resultant couple (give the exact value and a decimal to 1 d.p.).
2
Triangle ABC has sides 6 cm, 8 cm, 10 cm (AB, BC, CA), anticlockwise, with forces 3 N, 4 N and 5 N along them in order. Find the moment of the couple.
3
An isosceles triangle has sides 10 cm, 10 cm and 12 cm (anticlockwise order, base 12 cm) with forces 5 N, 5 N and 6 N along them in order. Find the moment of the couple.
4
The triangle of Example 1 (5 cm, 12 cm, 13 cm, with 10 N, 24 N, 26 N) has its forces taken in the order A→C, C→B, B→A, so they go round clockwise. What is the moment (anticlockwise positive)?
5
In that triangle the forces along AB, BC, CA are 10 N, 24 N and 20 N. Is the system a couple? Why?