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
1
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.
m = 9 / 6 = 1.5 N per cm. M = 2 × 61.94 × 1.5 ≈ 185.8 N·cm.
2
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.
Half of AC is 12, so half of BD is √(13² − 12²) = 5 and BD = 10. Area = ½ × 24 × 10 = 120 cm². m = 26 / 13 = 2. M = 2 × 120 × 2 = +480 N·cm.
3
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).
R = 0 gives x = 12 and y = 7. M = 12 × 10 + 7 × 10 = +190 N·cm (anticlockwise).
4
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.
(a) both forces point the same way, R = 12 N: not a couple. (b) equal, opposite, on different lines: a couple. (c) equal and opposite on the same line: moment 0, equilibrium.
5
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.
Area = 36 cm². 2 × 36 × m = 144, so m = 2. Forces: 18 N on AB and CD, 8 N on BC and DA.
