The resultant couple: the sum of several coplanar couples, with a cantilever, a lamina and a square as models
Applied Mathematics Sec 3 Term 1, Unit 3, Lesson 2, Part 2
GoalAdd several coplanar couples into one resultant couple, and use it with models such as a bracket, a lamina and a square.
New words
resultant couple · the single couple equivalent to several couples: its moment is M₁ + M₂ + …signed moments · anticlockwise couples count positive, clockwise couples negative
Explanation
Several couples in one plane act on a body. The moment of a couple is the same about every point and the forces of each couple add to zero, so the effect of all the couples together is a single couple whose moment is the sum of the moments: M = M₁ + M₂ + … + Mₙ, with signs. This is the resultant couple. It can be shown with any forces and arm giving that moment, for example with forces F and arm L = |M| / F, in the sense of the sign of M. If M = 0 the body is in equilibrium (last week). Method: (1) for each couple find F × L; (2) give it + or − from the way it turns; (3) add; (4) read the sense from the sign of the total.
Worked examples
Couple
Force
Arm
Sense
Moment
1
40 N
0.5 m
anticlockwise
+20 N·m
2
25 N
1.2 m
clockwise
−30 N·m
3
90 N
0.15 m
anticlockwise
+13.5 N·m
A bracket is acted on by three couples in its plane (anticlockwise positive): 40 N with arm 0.5 m anticlockwise, 25 N with arm 1.2 m clockwise, and 90 N with arm 0.15 m anticlockwise. Find the resultant couple and write it as a couple of 7 N forces. Solution: M = +40 × 0.5 − 25 × 1.2 + 90 × 0.15 = 20 − 30 + 13.5 = +3.5 N·m. The resultant couple is 3.5 N·m anticlockwise. With forces of 7 N the arm is 3.5 / 7 = 0.5 m.
Rectangle ABCD has AB = 50 cm and BC = 20 cm. Forces of 14 N act from A to B and from C to D, and forces of 6 N from C to B and from A to D. Find the resultant couple, and a couple of 5 N forces that would put the rectangle in equilibrium (anticlockwise positive). Solution: the 14 N forces push the bottom right and the top left (anticlockwise) with arm BC = 20 cm: +14 × 20 = +280 N·cm. The 6 N forces push the right side down and the left side up (clockwise) with arm AB = 50 cm: −6 × 50 = −300 N·cm. Resultant couple: 280 − 300 = −20 N·cm, that is 20 N·cm clockwise. To balance we need +20 N·cm: 5 × L = 20, so L = 4 cm, anticlockwise.
Class activity
Make three couples with your hands on a bottle cap: one with thumb and index finger turning it anticlockwise, one with the other hand’s thumb and index finger turning it clockwise, one anticlockwise using a larger or smaller arm. Which gives the biggest turning effect? What does the sum of the three tell you about the cap’s net turning?
Practice
1
Three couples act on a plate (anticlockwise positive): 12 N with arm 0.5 m anticlockwise, 8 N with arm 0.75 m clockwise, and 30 N with arm 0.1 m anticlockwise. Find the resultant couple.
Square ABCD has side 0.4 m. Forces of 30 N act from A to B and from C to D, and forces of 10 N from C to B and from A to D. Find the resultant couple (anticlockwise positive).
The 30 N pair is anticlockwise: +30 × 0.4 = +12. The 10 N pair is clockwise: −10 × 0.4 = −4. Total +8 N·m (anticlockwise).
3
A body has two couples, +40 N·cm and −75 N·cm. A third couple has forces F and arm 5 cm. Find F and its sense for equilibrium (anticlockwise positive).
40 − 75 = −35, so the third must be +35 N·cm: anticlockwise, 5F = 35, F = 7 N.
4
Two couples of moments +5 N·m and −5 N·m (anticlockwise positive) act on a rod. What is the resultant couple, and what does it mean for the rod?
5 − 5 = 0: no resultant couple, and since the forces also sum to zero, the rod is in equilibrium.
5
A resultant couple is +3.5 N·m (anticlockwise positive). Find the forces of an equivalent couple whose arm is 0.25 m.
F = 3.5 / 0.25 = 14 N, in the anticlockwise sense.
