Worked examples
The order of work in a moment problem is always the same: find force × distance on each side, then set the two equal.
Example 1
The problem
A uniform rod AB of length 1 m is suspended at its exact centre and balanced.
- A weight of 4 N is now hung at the end B. Find the initial (clockwise) moment it produces.
- With that 4 N still at B, what weight hung at a point C, 0.4 m from the centre of the rod, would balance the rod again?
Figure to be added
Solution — part one
The rod hangs from its exact centre, so the distance from the centre to B is 0.5 m.
clockwise moment = force × perpendicular distance
clockwise moment = 4 N × 0.5 m
clockwise moment = 2 N m
Solution — part two
Let the weight hung 0.4 m from the centre be x.
The moment it produces is anticlockwise, and to balance the rod it must equal the moment produced by the 4 N.
x × 0.4 = 4 × 0.5
x × 0.4 = 2
x = 2 ÷ 0.4 = 5 N
Check it: 5 N × 0.4 m = 2 N m, exactly the 2 N m found in part one.
Exercise 11.1 (1)
The problem
A rod AB is 0.8 m long. It is suspended at its exact centre and balanced, and a weight of 2 N is then hung at its end A. At what distance from the point of suspension must a weight of 4 N be hung on the other side to bring the rod back into equilibrium?
Figure to be added
Solution
The rod is 0.8 m long, so the distance from the centre to A is 0.4 m.
moment produced by the 2 N = 2 N × 0.4 m = 0.8 N m
Let the distance at which the 4 N must hang be x.
4 × x = 0.8
x = 0.8 ÷ 4 = 0.2 m
You could have expected the distance to halve, since the force doubled: half of 0.4 m is 0.2 m.
Where is the distance measured from?
Distances are measured from the point of suspension, never from the end of the rod. The 0.8 m is the whole length of the rod, not a distance that goes into a moment.