The three-point rule and resultant couples
Answers are for parents and teachers.
1
The forces (3, 0) N at A(0, 0), (−3, 4) N at B(6, 0) and (0, −4) N at C(0, 8) act in a plane (m). Find the sum of moments about A, about B and about D(6, 8), and say what the system is (anticlockwise positive).
About A: 0 + (6 × 4 − 0) + 0 = 24. About B: 0 + 0 + [(−6)(−4) − 8 × 0] = 24. About D: [(−6)(0) − (−8)(3)] + [0 − (−8)(−3)] + [(−6)(−4) − 0] = 24 − 24 + 24 = 24. A, B, D are not on a line, so it is a couple of +24 N·m.
2
Three couples act on a plate (anticlockwise positive): 15 N with arm 0.2 m anticlockwise, 12 N with arm 0.5 m clockwise, and 8 N with arm 0.25 m anticlockwise. Find the resultant couple and its forces if the arm is 0.2 m.
M = 3 − 6 + 2 = −1 N·m, that is 1 N·m clockwise. With arm 0.2 m: F = 1 / 0.2 = 5 N, clockwise.
3
Rectangle ABCD has AB = 20 cm and BC = 10 cm. Forces of 5 N act from A to B and from C to D, and forces of Q newtons from C to B and from A to D. The resultant couple is 30 N·cm clockwise. Find Q.
+5 × 10 − 20Q = −30, so 20Q = 80 and Q = 4 N.
4
Rhombus ABCD (anticlockwise) has diagonals 14 cm and 48 cm and forces of 15 N along each side in order. Find the moment of the couple.
Side = √(7² + 24²) = 25 cm, m = 15 / 25 = 0.6 N per cm; area = ½ × 14 × 48 = 336 cm². M = 2 × 336 × 0.6 = +403.2 N·cm.
5
Rectangle ABCD has AB = 12 cm and BC = 5 cm. Forces: 9 N from A to B, x N from C to D, 4 N from B to C and y N from D to A. The system is equivalent to a couple. Find x, y and the moment (anticlockwise positive).
R = 0 gives x = 9 and y = 4. M = 9 × 5 + 4 × 12 = 45 + 48 = +93 N·cm.