Moments: finding the weight of an unknown mass
Balance a metre ruler pivoted at its centre against an unmarked mass Q, using a 2.0 N load and then a 3.0 N load, and use the principle of moments to find Q's weight two independent ways.
WHAT YOU'LL LEARN
- ✓The principle of moments: force × distance from the pivot is equal on both sides at balance
- ✓Why a fixed geometric distance (y) doesn't need repeat measurement, but a balance point (x) does
- ✓Getting an independent second measurement — a different load — to check a result, not just repeating the same one
- ✓Narrowing down an exact balance point by overshooting in each direction, rather than hunting for it directly
ON YOUR BENCH
- □Metre ruler (mm scale)
- □Triangular pivot block
- □Object Q — an unmarked mass, weight not disclosed to the candidate
- □2.0 N load (a labelled 200 g mass)
- □3.0 N load (a labelled 300 g mass)
The protocol, step by step
The same guide is printed and waiting at your bench.
- 01
Set up the pivot and place Q
Balance the ruler across the triangular block so the block sits under the 50.0 cm mark. Place object Q on the ruler so its centre is directly over the 90.0 cm mark — check by eye that Q's edges sit equal distances either side of the mark. y, the distance from the pivot to Q, is now fixed at 40.0 cm by this set-up alone; you don't measure it again.
- 02
Balance against the 2.0 N load
Place the 2.0 N load on the other arm of the ruler and slide it until the ruler is as level as you can get it. Record x — the distance from the 50.0 cm mark to the centre of the load — to the nearest millimetre.
- 03
Calculate W from the first run
By the principle of moments, (weight of Q) × y = 2.0 N × x, so W = (x / y) × 2.0 N. This should come out close to 2.0 N, since a load with a similar weight to Q balances at a similar distance.
- 04
Repeat with the 3.0 N load
Remove the 2.0 N load without disturbing Q. Slide the 3.0 N load until the ruler balances again, and record its own x. Because it's heavier, this balance point sits closer to the pivot — expect somewhere around 26–27 cm rather than 40 cm.
- 05
Calculate the second W and compare
Calculate W = (x / y) × 3.0 N for this run. Compare it with your first value: if the two agree to within about 10%, you have good independent evidence for Q's real weight, obtained two different ways.
What you should see
- →With the 2.0 N load, the balance point lands close to x ≈ 40 cm — near-symmetric with Q's own 40 cm distance, since the two weights are similar.
- →With the heavier 3.0 N load, the same turning effect is reached at a shorter distance: x ≈ 26.7 cm, since 3.0 × 26.7 ≈ 2.0 × 40.
- →Both calculations should give W close to 2.0 N and agree with each other within about 10% — small disagreement is normal experimental scatter from how precisely you can judge 'level'.
- →x must always come out at or below 50 cm — the load has nowhere else to sit. A calculated x greater than 50 means something has gone wrong upstream, not that the ruler is longer than it is.
| load | y /cm | x /cm | W /N |
|---|---|---|---|
| 2.0 N | 40.0 | 39.6 | 1.98 |
| 3.0 N | 40.0 | 27.1 | 2.03 |
⚠ BEFORE YOU START
- •No hazards beyond normal care with a pivoted ruler and small masses — nothing hot, sharp or corrosive is involved.
- •Keep fingers clear of the pivot edge while sliding loads, and don't let a load slide off the end of the ruler onto the bench or a foot.
Try it, then run it for real.
Practise the whole thing on the virtual bench, then book real lab time and do it with your own hands.