Week 11 · Couples: two equal opposite forces, the moment of a couple, its independence of the point, and equilibrium under several couples · 3 lessons
2
The moment of a couple is the same about every point, shown with vectors, and finding the constants of two forces that form a couple and its moment from coordinates
Applied Mathematics Sec 3 Term 1, Unit 3, Lesson 1, Part 1
GoalShow that the moment of a couple is the same about every point, and find the constants of two forces that form a couple and the moment of the couple from coordinates.
New words
moment about O · for a force (Fx, Fy) acting at (x, y), M = x·Fy − y·Fx about the origin (anticlockwise positive)forces that form a couple · F₂ = −F₁, acting at two different points not on one line of action
Explanation
A force (Fx, Fy) at the point (x, y) has the moment x·Fy − y·Fx about the origin (Week 2). Let a couple be F at A(x₁, y₁) and −F at B(x₂, y₂). About the origin the two moments are x₁Fy − y₁Fx and −(x₂Fy − y₂Fx). Adding them gives M = (x₁ − x₂)Fy − (y₁ − y₂)Fx. The position of the origin has disappeared: only the vector from B to A is left. So a couple has the same moment about every point, and we can get it quickly as the moment of one force about the point where the other force acts. Two forces F₁ and F₂ form a couple when F₂ = −F₁ (equal and opposite components) and they act at different points. Comparing components gives unknown constants. The arm is L = |M| / |F|. Sign convention: anticlockwise positive.
Worked examples
The forces F₁ = (3, b) N at A(2, −1) and F₂ = (a, −4) N at B(−1, 3) (coordinates in m) form a couple. Find a and b, the moment of the couple (anticlockwise positive) and its arm. Solution: F₂ = −F₁, so a = −3 and b = 4. The moment of F₁ about B: the vector from B to A is (2 − (−1), −1 − 3) = (3, −4), so M = 3 × 4 − (−4) × 3 = 12 + 12 = 24 N·m, anticlockwise. Since |F| = √(3² + 4²) = 5 N, the arm is L = 24 / 5 = 4.8 m. Check about the origin: F₁ gives 2 × 4 − (−1) × 3 = 11 and F₂ = (−3, −4) gives (−1)(−4) − 3 × (−3) = 13, and 11 + 13 = 24.
F₁ = (p, 2) N at A(4, 1) and F₂ = (−1, q) N at B(0, 5) form a couple (coordinates in m). Find p, q and the moment (anticlockwise positive), then check it with moments about C(3, −2). Solution: F₂ = −F₁ gives p = 1 and q = −2, so F₁ = (1, 2). The vector from B to A is (4, −4), so M = 4 × 2 − (−4) × 1 = 8 + 4 = 12 N·m. Check about C: F₁ at A, vector from C (1, 3): 1 × 2 − 3 × 1 = −1. F₂ = (−1, −2) at B, vector from C (−3, 7): (−3)(−2) − 7 × (−1) = 13. Sum: −1 + 13 = 12 N·m, the same.
Class activity
On squared paper (1 square = 1 unit) mark A(2, 0) and B(0, 3). Draw a force of 2 units upward at A and a force of 2 units downward at B. For each of the points O(0, 0), P(5, 1) and Q(−2, −2), add the moments of the two forces using x·Fy − y·Fx with x and y measured from that point. You should get the same total (4) every time.
Practice
1
A couple is F = (0, 6) N at A(3, 0) and −F at B(−1, 0) (m). Find its moment (anticlockwise positive) and its arm.
M = (3 − (−1)) × 6 − 0 = 24 N·m, anticlockwise. Arm L = 24 / 6 = 4 m.
2
F₁ = (a, 4) N at A(5, 2) and F₂ = (−6, b) N at B(1, −1) form a couple (m). Find a, b and the moment (anticlockwise positive).
F₂ = −F₁ gives a = 6 and b = −4. The vector from B to A is (4, 3), so M = 4 × 4 − 3 × 6 = −2 N·m (clockwise).
3
For the couple of question 1, find the sum of the moments of the two forces about the point P(1, 5), anticlockwise positive. What do you notice?
F at A: vector from P (2, −5), 2 × 6 − (−5) × 0 = 12. −F = (0, −6) at B: vector from P (−2, −5), (−2)(−6) − (−5) × 0 = 12. Sum 24 N·m, the same as before: the moment of a couple does not depend on the point.
4
Two equal and opposite forces act along the same line of action. Do they form a couple? Explain.
No. The arm is 0, so M = F × 0 = 0: they have no turning effect and just cancel (equilibrium).
5
A couple has moment +18 N·m and its arm is 0.45 m. Find the size of each force.
