☰ Year plan · Applied Mathematics

The moment of a force: turning effect, coordinates and Varignon

Week 2 · Moments: the turning effect of a force about a point, moment arm and sign, the moment from coordinates, and Varignon’s principle · 3 lessons
3

Varignon’s principle: the moment of a force equals the sum of the moments of its two components, worked in two ways

Applied Mathematics Sec 3 Term 1, Unit 1, Lesson 1, Part 1
GoalUse Varignon’s principle: the moment of a force equals the sum of the moments of its components, and check it with the arm method.
New words
Varignon’s principle · the moment of a force about a point equals the sum of the moments of its components about that pointcomponent · one of two forces, usually along the axes, that together have the same effect as the force
Explanation

The formula M = x Fy − y Fx has two terms, and each one is the moment of one component: x Fy is the moment of Fy and −y Fx is the moment of Fx. So the moment of a force is the sum of the moments of its components. This is Varignon’s principle, and it gives two ways to find a moment. Way 1 (components): resolve F into Fx and Fy, find the moment of each about O with its own arm and sign, and add. Way 2 (arm): find the perpendicular arm L from O to the line of action, with trigonometry, and take F × L with the sign from the turning sense. The answers must agree, which is a good check. Choose the components when the point of action has easy coordinates, and the arm when a clear right-angled triangle is available. Be careful with signs: the two parts can turn the same way or opposite ways.

FFxFyLOC
Worked examples
FFxFyLOC
A rope pulls a shop-sign bracket at C with 80 N, 30° below the horizontal and pointing away from the wall. C is 1.5 m out and 1.0 m up from the foot O. Way 1: Fx = 80 cos 30° ≈ 69.28 N (rightward, above O: clockwise) and Fy = −80 sin 30° = −40 N (downward, to the right of O: clockwise). M = 1.5 × (−40) − 1.0 × 69.28 = −60 − 69.28 ≈ −129.28 N·m. Way 2: the arm is L = 1.5 sin 30° + 1.0 cos 30° ≈ 1.616 m, so |M| = 80 × 1.616 ≈ 129.28 N·m. Both say 129.28 N·m clockwise.
FFxFyLOC
A force of 60 N at 60° above the horizontal acts at C(1.0, 0.5) (metres from O). Fx = 30 N, Fy ≈ 51.96 N. M = 1.0 × 51.96 − 0.5 × 30 = 51.96 − 15 ≈ 36.96 N·m, anticlockwise. Here the two parts turn opposite ways, and Fy wins. The arm is L = 1.0 sin 60° − 0.5 cos 60° ≈ 0.616 m and 60 × 0.616 ≈ 36.96 N·m.
F4 m3 mOAB
A force of 50 N acts along the line from A(4, 0) to B(0, 3), pointing from A to B. Its components are (−40, 30), because the side lengths 4 and 3 give the ratio 4 : 3 : 5. Take the point of action at A(4, 0): y = 0, so only Fy matters: M = 4 × 30 − 0 × (−40) = 120 N·m, anticlockwise. The arm is 120 / 50 = 2.4 m, the height of the right-angled triangle OAB on its hypotenuse.
Class activity

Draw the bracket example on squared paper (1 square = 0.25 m): O, the point C, and the force arrow at 30° below the horizontal. Draw the perpendicular from O to the line of action and measure it. It should be close to 1.6 m. Then compute the moments of the two components and compare.

Practice
1
A force of 100 N at 30° above the horizontal acts at the point (2, 0) m. Find its moment about O (anticlockwise positive).
2
A force of 40 N at 45° below the horizontal, pointing right, acts at (1, 2) m. Find its moment about O (anticlockwise positive).
3
F = (5, 12) N acts at (2, 1) m. Find the moment about O by components (anticlockwise positive), then the arm.
4
A force of 20 N, at 60° above the horizontal and pointing right, acts at (0, 4) m. Find its moment about O (anticlockwise positive). Which component does not contribute?
5
The moment of a force about O is 30 N·m clockwise. Its horizontal component, acting at height 5 m above O, gives 20 N·m clockwise. What moment must the vertical component give?