Egypt Secondary 3 Applied Mathematics, Week 12: Equivalent couples (equal moments, so different forces and arms can give the same turning effect) and exercises on couples from coordinates

Cancelling couples, hung bodies and equivalent couples

Week 12 · Couples: cancelling a couple, a hanging body with a couple acting on it, equivalent couples and the exercises of Lesson 3-1 · 3 lessons

Equivalent couples (equal moments, so different forces and arms can give the same turning effect) and exercises on couples from coordinates

Applied Mathematics Sec 3 Term 1, Unit 3, Lesson 1, Part 2

GoalReplace a couple by an equivalent one with a different force and arm, and find the moment of a couple from the coordinates of the points where its forces act.

New words

equivalent couples · two couples in the same plane with equal signed moments (same size and same sense)same effect · a body acted on by one couple behaves exactly as if it were acted on by an equivalent one

Explanation

The turning effect of a couple depends only on its moment M = F × L, with its sign. So any couple can be replaced by another in the same plane with the same signed moment: we may double the force and halve the arm, turn the couple through any angle in its plane, or move it anywhere in the plane (or to a parallel plane), and nothing changes for the body. This is useful when we want the forces along convenient lines or at convenient points. Note that equivalent couples are equal in moment, not equal in force. To compare two couples, calculate F × L for each with its sign. For forces given as (F_x, F_y) at points A and B, the moment is (x_A − x_B)F_y − (y_A − y_B)F_x, and the arm is |M| / |F|.

Worked examples

A couple has two forces of 24 N whose lines are 0.5 m apart, anticlockwise. (a) Find its moment. (b) Find the forces of an equivalent couple whose arm is 0.2 m, and of one whose arm is 0.3 m. Solution: (a) M = +24 × 0.5 = +12 N·m. (b) Equivalent couples have the same M and the same sense: F = 12 / 0.2 = 60 N for arm 0.2 m, and F = 12 / 0.3 = 40 N for arm 0.3 m. A shorter arm needs a bigger force.

A couple is F = (6, −8) N at A(5, 2) and −F at B(1, −1) (coordinates in m). Find its moment (anticlockwise positive) and its arm. Then find the forces of the equivalent couple (same sense) with arm 2 m. Solution: the vector from B to A is (4, 3), so M = 4 × (−8) − 3 × 6 = −32 − 18 = −50 N·m (clockwise). |F| = √(6² + 8²) = 10 N, so L = 50 / 10 = 5 m. With arm 2 m: F = 50 / 2 = 25 N, clockwise.

Class activity

On squared paper draw a couple of two forces of 3 squares with an arm of 4 squares, and below it its twin: forces of 6 squares with an arm of 2 squares. Check that both give the same moment (12 square-units). Which of the two would be easier to apply to a stiff bottle cap, and why?

Practice

A couple of 15 N forces with arm 0.2 m is anticlockwise. Is it equivalent to a couple of 5 N forces with arm 0.6 m, anticlockwise? Why?

Square ABCD has side 20 cm. Forces of 40 N act from A to B and from C to D (an anticlockwise couple). Replace it by an equivalent couple of two forces at A and C perpendicular to AC. Find the force (to 1 decimal place).

A couple is F = (0, 5) N at A(4, 0) and −F at B(1, 0) (m). Find its moment (anticlockwise positive).

A clockwise couple has moment −36 N·cm. Give two different equivalent couples (force and arm, clockwise).

Two couples in a plane have forces 8 N, arm 0.9 m clockwise and forces 12 N, arm 0.6 m anticlockwise. Are they equivalent? What do they give together (anticlockwise positive)?