Egypt Secondary 3 Applied Mathematics, Week 14: Exercises of Lesson 3-2: which systems are couples, forces on rectangles, rhombuses and regular polygons, and the missing force

The three-point rule and resultant couples

Week 14 · Resultant couple: the three-point rule, the sum of several couples, and the exercises of Lesson 3-2 · 3 lessons

Exercises of Lesson 3-2: which systems are couples, forces on rectangles, rhombuses and regular polygons, and the missing force

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

GoalApply the couple rules to the exercises of Lesson 3-2: decide whether a system is a couple, find the moment for forces on rectangles, rhombuses and regular polygons, and find a missing force.

New words

checklist · (1) R = 0? (2) ΣM ≠ 0? If both yes, a couple of moment ΣMregular polygon · equal sides and equal angles; area = ½ × perimeter × apothem (the distance from the centre to a side)

Explanation

Checklist for a system of forces: (1) Is R = 0? Add the x-components and the y-components. (2) If R = 0, is the moment about one point not zero? If yes, it is a couple, and the moment is the same about every point. (3) If the forces lie along the sides of a closed figure, in order, and are proportional to the sides, use M = 2 × area × m, positive if the figure is traced anticlockwise. (4) For a missing force, use R = 0 for its components first and then the moment. Areas you will need: rectangle: length × width; rhombus: ½ × (product of the diagonals), or side² × sin of an angle; regular hexagon: (3√3/2) × side²; regular n-gon: ½ × perimeter × apothem, and the apothem is side / (2 tan(180°/n)).

Worked examples

Rhombus ABCD has diagonals AC = 16 cm and BD = 12 cm, and A, B, C, D go round anticlockwise. A force of 20 N acts along each side, in order AB, BC, CD, DA. Find the resultant couple. Solution: the diagonals bisect each other at right angles, so a side is √(8² + 6²) = 10 cm. m = 20 / 10 = 2 N per cm. Area = ½ × 16 × 12 = 96 cm². M = 2 × 96 × 2 = +384 N·cm (anticlockwise).

Rectangle ABCD has AB = 15 cm and BC = 8 cm. Forces of 6 N act from A to B and from C to D, a force of 4 N from B to C and a force of P newtons from D to A. The system is equivalent to a couple. Find P and the moment (anticlockwise positive). Solution: R = 0. The x-components: 6 − 6 = 0. The y-components: the 4 N force goes up (from B to C) and the P force goes down (from D to A): 4 − P = 0, so P = 4 N. The moment: the 6 N pair gives +6 × 8 = +48 N·cm (anticlockwise) and the 4 N pair gives +4 × 15 = +60 N·cm (the right side up, the left side down). Total +108 N·cm.

Class activity

Cut out a card rhombus with diagonals 8 cm and 6 cm. Measure its side (it should be 5 cm). Draw an arrow of 5 squares of a 1 cm grid along each side in order round it. Work out the couple 2 × area × m using area = ½ × 8 × 6 = 24 and m = 1.

Practice

A regular pentagon of side 6 cm (anticlockwise) has forces of 9 N along each side in order. Given that its area is about 61.94 cm², find the moment of the couple to 1 decimal place.

A rhombus ABCD (anticlockwise) has side 13 cm and the diagonal AC = 24 cm. Forces of 26 N act along each side in order. Find the moment of the couple.

Square ABCD has side 10 cm. Forces act along AB (A to B) of 12 N, along CD (C to D) of x N, along BC (B to C) of 7 N and along DA (D to A) of y N. The system is equivalent to a couple. Find x, y and the moment (anticlockwise positive).

On rectangle ABCD, which of these is a couple? (a) 6 N from A to B and 6 N from D to C; (b) 6 N from A to B and 6 N from C to D; (c) 6 N from A to B and 6 N from B to A.

Rectangle ABCD has AB = 9 cm and BC = 4 cm (anticlockwise). Forces proportional to the sides act along AB, BC, CD, DA in order and form a couple of 144 N·cm. Find the four forces.