Week 14 · Resultant couple: the three-point rule, the sum of several couples, and the exercises of Lesson 3-2 · 3 lessons
1
The rule of three points: equal non-zero sums of moments about three points not on a line mean the system is a couple of that moment
Applied Mathematics Sec 3 Term 1, Unit 3, Lesson 2, Part 2
GoalUse the rule of three points: equal non-zero sums of moments about three points not on a line mean that the system is a couple of that moment.
New words
three-point rule · MA = MB = MC = M ≠ 0 for three points not on a line means a couple of moment Mnot on a line · the points A, B, C form a triangle (not collinear)
Explanation
Sometimes it is hard to see whether the resultant of a set of forces is zero. A test that avoids finding R: take the sum of moments about three points A, B, C that are not on one line. If all three sums are equal to the same non-zero number M, the system is a couple of moment M. Why? If R were not zero, the system would be one single force. The moment of a single force about a point is its size times the distance from the point to its line of action, and it is positive or negative according to the side of the line. Three equal moments would mean three points at the same distance, on the same side of that line, so they would lie on a line parallel to the force. But A, B, C are not on a line, so R must be zero, and a system with R = 0 and moment M ≠ 0 is a couple of moment M. Notice that two equal moments are not enough: they only say that R is parallel to AB or that R = 0. Sign convention: anticlockwise positive.
Worked examples
ABCD is a rectangle with AB = 8 cm and BC = 6 cm. Forces of 10 N along AC (from A to C), 8 N along CD (from C to D) and 6 N along DA (from D to A) act. Use the moments about A, B and D to show the system is a couple (anticlockwise positive). Solution: take A at the origin, B(8, 0), C(8, 6), D(0, 6), and write the forces as (8, 6) at A, (−8, 0) at C and (0, −6) at D. About A: 0 + [0 × 0 − 6 × (−8)] + 0 = 48. About B: the first force gives (−8)(6) − 0 = −48, the second gives 0 − 6 × (−8) = +48 and the third gives (−8)(−6) − 0 = 48: total 48. About D: first force 0 − (−6)(8) = 48, second 0, third 0: total 48. The sums about A, B, D (not on a line) are all +48 N·cm: a couple of +48 N·cm (anticlockwise).
System
MA
MB
MC
1
30
30
30
2
20
20
8
3
0
0
0
Three systems of coplanar forces have the sums of moments about three points A, B, C (not on a line) shown in the table, in N·cm. What can you say about each? Solution: system 1: 30, 30, 30: equal and not zero, so a couple of +30 N·cm. System 2: 20, 20, 8: A and B give equal moments but C does not, so it is not a couple; if R is not zero, it is a single force parallel to AB. System 3: 0, 0, 0: R must be zero (a force cannot pass through three points not on a line), so the system is in equilibrium.
Class activity
Draw a triangle on squared paper and mark its corners A, B, C. Draw a force of 3 squares along each side in turn (all three going round the triangle the same way: the sides are 3, 4 and 5 squares, so they will not close - this is a case with R not zero). Compute the sum of moments about A, B and C by counting squares: they are not all equal. Then change the forces to 3, 4, 5 squares along the sides and check that all three sums become equal.
Practice
1
A system has MA = 18, MB = 18 and MC = 18 N·m about three points not on a line. What is the system? What if MC = 6 instead?
All equal and non-zero: a couple of 18 N·m (anticlockwise). With MC = 6 it is not a couple: the equal moments about A and B mean a single force parallel to AB.
2
The forces (8, 0) N at A(0, 0), (−8, 6) N at B(4, 0) and (0, −6) N at C(0, 3) act in a plane (m). Find the sums of moments about A, about B and about D(4, 3), and decide what the system is (anticlockwise positive).
About A: 0 + [4 × 6 − 0] + 0 = 24. About B: 0 + 0 + [(−4)(−6) − 3 × 0] = 24. About D: [(−4)(0) − (−3)(8)] + [0 × 6 − (−3)(−8)] + [(−4)(−6) − 0] = 24 − 24 + 24 = 24. A, B, D are not on a line, so the system is a couple of +24 N·m.
3
The sums of moments (anticlockwise positive) about three points not on a line are 15, 15 and 9 N·cm. Can the system be a couple? What can it be?
It cannot be a couple (the three are not equal). The equal moments about two of the points mean a single force parallel to the line joining them.
4
Why must the three points not be on a line? Give an example of a single force with equal moments about three points on a line.
A force parallel to the line is at the same distance from every point on the line, so it has equal moments about all of them, yet it is a single force, not a couple. For example 5 N in the +x direction along the line y = 2 has moment −10 about every point of the x-axis (anticlockwise positive).
5
In rectangle ABCD, AB = 8 cm and BC = 6 cm, forces of 6 N act from A to B and from C to D, and forces of 4 N from B to C and from D to A. Find the sums of moments about A and about C (anticlockwise positive). What is the system?
About A: 4 × 8 (force at B) + 6 × 6 (force at C) = 32 + 36 = 68. About C: 6 × 6 (force at A) + 4 × 8 (force at D) = 36 + 32 = 68. Both are +68 N·cm, and the resultant is zero: a couple of 68 N·cm anticlockwise.
