Egypt Secondary 3 Applied Mathematics, Week 13: The same idea for a closed polygon, using a kite as the model, and finding the forces from the couple’s moment

Systems equivalent to a couple

Week 13 · Resultant couple: a system equivalent to a couple, forces along the sides of a triangle, and of a closed polygon · 3 lessons

The same idea for a closed polygon, using a kite as the model, and finding the forces from the couple’s moment

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

GoalExtend the triangle rule to a closed polygon, using a rectangle and a kite as models, and find the forces from the moment of the couple.

New words

closed polygon · forces m × side along each side, in order round the figure: the resultant is zerokite · a quadrilateral with two pairs of equal adjacent sides, whose diagonals are perpendicular

Explanation

Take forces m × (side) along each side of a closed polygon, in order round it. Their sum is m × (the sum of the side vectors) = 0, so R = 0. For the moment, take a point O inside the polygon. The force on a side of length s has moment m × s × (distance from O to that side), and s × (distance) is twice the area of the triangle formed by O and that side. All these moments have the same sense, so they add: M = 2m × (sum of the areas of the triangles) = 2 × (area of the polygon) × m. The sign is anticlockwise when the vertices go round anticlockwise. The sides may differ in length but m is the same for all of them. A rectangle shows it: the two pairs of forces make two couples. A kite (two pairs of equal neighbouring sides, perpendicular diagonals) has area ½ × product of the diagonals, which makes the area quick to find.

Worked examples

A rectangle ABCD (anticlockwise) has AB = 10 cm and BC = 6 cm. Forces of 5 N, 3 N, 5 N, 3 N act along AB, BC, CD, DA in order. Show that the moment is 2 × area × m. Solution: m = 5 / 10 = 3 / 6 = 0.5 N per cm. Area = 60 cm², so 2 × 60 × 0.5 = 60 N·cm. Directly: the forces along AB and CD form a couple of 5 × 6 = 30 N·cm and those along BC and DA a couple of 3 × 10 = 30 N·cm, both anticlockwise: total +60 N·cm.

Kite ABCD has AB = AD = 10 cm and CB = CD = 17 cm. Its diagonals meet at right angles, AC = 21 cm and BD = 16 cm, and A, B, C, D go round anticlockwise. Forces proportional to the sides act along AB, BC, CD, DA in order and form a couple of 840 N·cm. Find the four forces. Solution: area = ½ × 21 × 16 = 168 cm². 2 × 168 × m = 840, so m = 2.5 N per cm. The forces are 2.5 × 10 = 25 N on AB, 2.5 × 17 = 42.5 N on BC, 42.5 N on CD and 25 N on DA.

Class activity

Cut a rectangle 10 cm by 6 cm from card. Mark on each side a small arrow in order round the figure, with lengths 5, 3, 5, 3 squares of a 1 cm grid (m = half a square per cm). Which pairs make couples? Add their moments (5 × 6 and 3 × 10) and compare with 2 × area × m.

Practice

A square ABCD of side 12 cm, anticlockwise, has forces of 6 N along each side in order. Find the moment of the couple.

A regular hexagon of side 10 cm (anticlockwise) has forces of 60 N along each side in order. Find the moment (exact and to 1 d.p.). Use area = (3√3 / 2) × side².

A rhombus ABCD (anticlockwise) has side 10 cm and angle ABC = 60°, with forces of 15 N along each side in order. Find the moment of the couple. Use area = side² × sin 60°.

A rectangle with sides 8 cm and 5 cm (anticlockwise) has forces proportional to the sides along them in order, forming a couple of 120 N·cm. Find the four forces.

A rectangle 10 cm by 6 cm has forces of 5 N along all four sides in order (anticlockwise). Is the system a couple? What is its moment? Does 2 × area × m apply?