Review: couples, equivalent couples, a system equivalent to a couple and the resultant couple
Applied Mathematics Sec 3 Term 1, Units 1–3, Review
GoalReview couples: the moment of a couple, equivalent couples, a system equivalent to a couple and the resultant couple, with new problems.
New words
couple · two equal, opposite forces on different lines of action; moment M = F × L, the same about every pointresultant couple · the single couple equivalent to several couples: the sum of their signed moments
Explanation
Weeks 11 to 14 in one page. A couple is two equal and opposite forces on different lines of action; its moment is M = F × L with L the arm, positive if anticlockwise, and it is the same about every point. Couples with equal signed moments are equivalent. A body under several couples is in equilibrium when the signed moments add to zero, and the couples together are equivalent to one resultant couple whose moment is their sum. A hung body with a couple: the weight and the pin reaction form a couple W × AG × sin θ. A system of coplanar forces is equivalent to a couple when R = 0 and ΣM ≠ 0, and then the moment is the same about every point; equal non-zero moments about three points not on a line also mean a couple. Forces along the sides of a triangle or closed polygon in order, proportional to the sides (m newtons per unit length), form a couple of moment 2 × area × m.
Idea
Condition
Result
Couple
equal, opposite, different lines
M = F × L
Equilibrium of couples
ΣM = 0
no turning
System equals a couple
R = 0 and ΣM ≠ 0
moment ΣM, same about every point
Triangle (polygon) rule
forces ∝ sides, in order
M = 2 × area × m
Worked examples
Triangle ABC has AB = 9 cm, BC = 12 cm, CA = 15 cm (right angle at B), anticlockwise. Forces of 6 N, 8 N and 10 N act along AB, BC, CA in order. Find the moment of the couple, and check with the moment of the 8 N force about A. Solution: m = 6 / 9 = 8 / 12 = 10 / 15 = 2/3 N per cm. Area = ½ × 9 × 12 = 54 cm². M = 2 × 54 × 2/3 = +72 N·cm. Check: the 6 N and 10 N forces pass through A; the 8 N force at B is perpendicular to AB at 9 cm from A: +8 × 9 = +72 N·cm.
A square plate ABCD of side 0.3 m has these couples (anticlockwise positive): forces of 20 N from A to B and from C to D; forces of 12 N from C to B and from A to D; and a third couple of 16 N forces with arm 0.15 m, anticlockwise. Find the resultant couple, and an equivalent couple of 24 N forces. Solution: first pair: anticlockwise, +20 × 0.3 = +6 N·m. Second pair: clockwise, −12 × 0.3 = −3.6 N·m. Third: +16 × 0.15 = +2.4 N·m. Total +4.8 N·m, anticlockwise. With forces of 24 N the arm is 4.8 / 24 = 0.2 m.
Class activity
Take a cardboard triangle with sides 3, 4 and 5 cm. Draw an arrow along each side, 3, 4 and 5 squares long, going round the triangle. Add the moments by counting squares and check that the total is 2 × area × m (m = 1 square per cm, area = 6 squares).
Practice
1
A light rod AB, 20 cm long, has 16 N upward at A, 6 N downward at C (AC = 5 cm) and 10 N downward at B. Show that it is a couple and find its moment (anticlockwise positive).
A uniform rod of length 80 cm and weight 10 N is hinged at one end and hangs in a vertical plane under an anticlockwise couple of 160 N·cm. Find the angle with the downward vertical (nearest 0.1°).
AG = 40 cm: 10 × 40 × sin θ = 160, so sin θ = 0.4 and θ ≈ 23.6°.
3
Triangle ABC has sides AB = 5 cm, BC = 5 cm, CA = 6 cm (anticlockwise), with forces 10 N, 10 N and 12 N along them in order. Find the moment of the couple.
m = 2 N per cm. Height on the base 6 cm is √(25 − 9) = 4 cm, so the area is ½ × 6 × 4 = 12 cm². M = 2 × 12 × 2 = +48 N·cm.
4
A couple of two 18 N forces has arm 0.3 m and is anticlockwise. Find the arm of the equivalent couple with forces of 27 N.
M = 18 × 0.3 = 5.4 N·m, so L = 5.4 / 27 = 0.2 m.
5
In two sentences, say why a couple cannot be balanced by a single force, and what balances it.
A couple has resultant zero, so one extra force would leave a resultant not equal to zero. Only another couple of equal and opposite moment balances it.
