Putting it together
The whole chapter comes down to one decision: a single force about a pivot, or a couple?
| The situation | The moment | Distance measured |
|---|---|---|
| A single force on a pivoted body | force × perpendicular distance | Axis to the line of action |
| Two moments opposing each other | The difference between them | Each side separately |
| Two opposing moments that are equal | Zero — equilibrium | Each side separately |
| A couple | force × distance between the lines | Between the two lines of action |
Exercise 11.2 (ii)
The problem
Two forces act on a thin board pivoted at O. A force of 10 N acts downwards at one end and 10 N upwards at the other, one of them 0.4 m from O and the other 0.3 m from O. Find the moment of that couple.
Solution
The forces are equal (10 N), opposite, and on two different lines — so this is a couple.
perpendicular distance between the lines = 0.4 m + 0.3 m = 0.7 m
moment of a couple = force × distance between the lines
= 10 N × 0.7 m
= 7 N m
Do not add the two 10 N forces into 20 N. Only one force goes into the formula — it is the two distances that are added, not the two forces.
Mixed exercise (1)
The problem
The rod AB is 1.2 m long. It is suspended at its exact centre and held in equilibrium. A weight of 10 N is now hung at the end A. Find the force that must be applied 0.3 m from the centre of the rod to balance it.
Solution
The rod is 1.2 m long, so the distance from the centre to A is 0.6 m.
moment produced by the 10 N = 10 N × 0.6 m = 6 N m
Let the force needed be F, acting 0.3 m out.
F × 0.3 = 6
F = 6 ÷ 0.3 = 20 N
The distance is halved, so the force has to double — twice 10 N is 20 N.
Mixed exercise (2)
The problem
On a horizontal rod, a force of 10 N acts downwards 2 m from the end A, a force of 20 N acts upwards 4 m from A (at the point B), and a force of 10 N acts downwards 6 m from A. The point C lies 8 m from A. Find the resultant moment of these three forces about each of the points A, B and C.
Figure to be added
Solution — about A
Take clockwise as positive.
10 N down, 2 m: 10 × 2 = 20 N m clockwise
20 N up, 4 m: 20 × 4 = 80 N m anticlockwise
10 N down, 6 m: 10 × 6 = 60 N m clockwise
resultant = (20 + 60) − 80 = 0
Solution — about B
Distances are now measured from B.
10 N down, 2 m to the left of B: 20 N m anticlockwise
20 N up, at B itself: the distance is zero, so the moment is 0
10 N down, 2 m to the right of B: 20 N m clockwise
resultant = 20 − 20 = 0
Solution — about C
All three forces now lie to the left of C.
10 N down, 6 m from C: 60 N m anticlockwise
20 N up, 4 m from C: 80 N m clockwise
10 N down, 2 m from C: 20 N m anticlockwise
resultant = (60 + 20) − 80 = 0
All three answers are zero. If a body is in equilibrium, the resultant moment is zero about any point you care to choose.
Glossary
| Moment of force | බලයෙහි ඝූර්ණය |
| Turning effect of a force | බලයේ භ්රමණ ආචරණය |
| Couple of forces | බල යුග්මයක් |
| Axis of rotation | භ්රමණ අක්ෂය |
| Perpendicular distance | ලම්බක දුර |
| Equilibrium | සමතුලිතතාව |
Chapter summary
- The moment of a force is the tendency to turn that arises when a force is applied to a body.
- The moment of a force is calculated by multiplying the force applied by the perpendicular distance from a chosen axis to the line of action of that force. Its unit is N m.
- When opposing moments are equal the body does not rotate; it is then in equilibrium.
- A couple of forces is two forces, equal in magnitude and parallel, applied to a body in opposite directions in order to turn it.
- Applying a couple can rotate a body without any linear motion at all.