Egypt Secondary 3 Applied Mathematics, Week 13: A system of coplanar forces equivalent to a couple: zero resultant but non-zero moment, and why zero resultant alone is not enough

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

A system of coplanar forces equivalent to a couple: zero resultant but non-zero moment, and why zero resultant alone is not enough

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

GoalRecognise when a system of coplanar forces is equivalent to a couple (zero resultant but non-zero moment) and find its moment.

New words

equivalent to a couple · the resultant is zero but the sum of moments is not zerosame about every point · when R = 0 the sum of the moments has the same value about every point

Explanation

For any system of coplanar forces we ask two questions. First, what is the resultant R = ΣF? Second, what is the sum of moments ΣM about a point? (1) R ≠ 0: the system is equivalent to a single force. (2) R = 0 and ΣM = 0: the body is in equilibrium. (3) R = 0 and ΣM ≠ 0: the system is equivalent to a couple of moment ΣM. In case (3) the body does not move as a whole but it turns, and the moment is the same about every point, so any point may be used to find it. Both conditions are needed: R = 0 alone is not enough (it still allows a couple), and ΣM = 0 about one point alone is not enough (a force through that point has no moment). Forces that all meet at one point and have R = 0 are always in equilibrium, because their moments about that point are all zero, so they can never form a couple. Sign convention: anticlockwise positive, and for forces along a rod take up as positive.

Worked examples

A light rod AB, 20 cm long, has 12 N upward at A, 5 N downward at C (AC = 6 cm) and 7 N downward at B. Show that the system is equivalent to a couple and find its moment (anticlockwise positive). Solution: R = 12 − 5 − 7 = 0 (up positive). Moments about A (distance to the right of A, upward force positive moment): 0 − 5 × 6 − 7 × 20 = −30 − 140 = −170 N·cm. Not zero, so it is a clockwise couple of 170 N·cm. Check about B: A is 20 cm to the left, C is 14 cm to the left: 12 × (−20) − 5 × (−14) = −240 + 70 = −170 N·cm, the same.

On a light rod AB of length 8 cm, 9 N acts upward at A, P newtons downward at C (AC = 3 cm) and Q newtons downward at B. The system is equivalent to a couple of 57 N·cm clockwise. Find P and Q. Solution: R = 0 gives P + Q = 9. Moments about A (anticlockwise positive): −3P − 8Q = −57. Substitute P = 9 − Q: −27 + 3Q − 8Q = −57, so −5Q = −30, Q = 6 and P = 3. Check: −3 × 3 − 8 × 6 = −9 − 48 = −57.

Class activity

Lay a ruler on the desk and press down with three fingertips: one at each end and one in the middle, so that the pushes balance in total (the two ends pushing down and the middle pushing up with twice as much). Does the ruler turn? Now move the middle fingertip a little towards one end and keep the total push balanced: does it turn? Say why in terms of R and ΣM.

Practice

A light rod AB, 24 cm long, has 15 N upward at A, 9 N downward at C (AC = 10 cm) and 6 N downward at B. Show that it is a couple and find the moment (anticlockwise positive).

The forces (2, 1) N at O(0, 0), (−1, 3) N at P(1, 0) and (−1, −4) N at Q(0, 2) act in a plane (m). Find R and the sum of moments about O (anticlockwise positive). What is the system equivalent to?

The forces (3, 1), (−1, 2) and (−2, −3) newtons all act at the point P(4, 5). Is the system a couple? What is it?

On a light rod AB, 15 cm long, 8 N acts upward at A, P newtons downward at C (AC = 5 cm) and Q newtons downward at B. The system is a clockwise couple of 90 N·cm. Find P and Q.

A system has R = 0 and its sum of moments about A is +40 N·cm (anticlockwise positive). What is its sum of moments about another point B? Why?