Theorem of moments, forces along rectangle sides, and two-point conclusions
Week 3 · Moments: the theorem of moments for forces meeting at a point, forces along the sides of a rectangle or square, and equal or opposite moments about two points · 3 lessons
1
The theorem of moments: for forces acting at one point, the sum of their moments about any point equals the moment of their resultant, checked with coordinates
Applied Mathematics Sec 3 Term 1, Unit 1, Lesson 1, Part 2
GoalShow that for forces acting at one point the sum of their moments about any point equals the moment of their resultant, and use it with coordinates.
New words
theorem of moments · for forces acting at one point, the sum of their moments about O equals the moment of their resultant about Oresultant R · the single force equal to the vector sum of the forces: R = F₁ + F₂ + …line of action of R · the line through the point of action in the direction of R
Explanation
Suppose forces F₁, F₂, F₃ all act at the same point A(x, y), and R is their resultant, with components that are the sums of the components: Rx = ΣF_x and Ry = ΣF_y. The moment of each force about O is x Fy − y Fx, with the same x and y for every force. Add them: x(ΣF_y) − y(ΣF_x), and this is exactly x Ry − y Rx, the moment of R. So the moments add up just as the forces do: the sum of the moments of the forces about any point equals the moment of their resultant about that point. This is the theorem of moments for forces meeting at a point. It lets us replace many moments by one, or find the moment of a force we cannot measure directly. A special case is very useful: if the sum of the moments about O is zero, then the line of action of R passes through O (or R is the zero force). Sign convention: anticlockwise moments are positive and clockwise moments are negative.
Worked examples
Forces F₁ = (3, −1), F₂ = (−2, 4), F₃ = (1, 1) N act at A(2, −3). Find the moment of their resultant about B(−1, 1) (anticlockwise positive). The position of A from B is (3, −4). The moments are 3 × (−1) − (−4) × 3 = 9, 3 × 4 − (−4) × (−2) = 4 and 3 × 1 − (−4) × 1 = 7, total 20 N·m. Check with the resultant R = (2, 4): 3 × 4 − (−4) × 2 = 12 + 8 = 20 N·m. The two answers agree.
Two forces F₁ = (2, 3) N and F₂ = (−1, k) N act at A(1, 2). The moment of their resultant about O is +10 N·m. Find k. Moment of F₁: 1 × 3 − 2 × 2 = −1. Moment of F₂: 1 × k − 2 × (−1) = k + 2. Sum: k + 1 = 10, so k = 9. Check: R = (1, 12), and 1 × 12 − 2 × 1 = 10.
F₁ = (−4, 1) N and F₂ = (1, −3) N act at A(3, 2). About O their moments are 3 × 1 − 2 × (−4) = +11 and 3 × (−3) − 2 × 1 = −11, which add to zero. So the resultant R = (−3, −2) has zero moment about O: its line of action passes through O. Indeed R points from A straight towards O, since (−3, −2) is parallel to (3, 2).
Class activity
On squared paper mark a point A and draw two force arrows from it (1 cm for 1 N), then draw their resultant by completing the parallelogram. Choose any point O and measure the three perpendicular distances from O to the three lines. Check that F₁ × arm₁ ± F₂ × arm₂ matches R × arm, taking care with the senses of turning.
Practice
1
Two forces acting at one point have moments 9 N·m and −4 N·m about O (anticlockwise positive). What is the moment of their resultant about O?
9 + (−4) = 5 N·m, anticlockwise.
2
F₁ = (1, 2) N and F₂ = (3, −1) N act at A(2, 2). Find the moment of each about O and of their resultant (anticlockwise positive).
