The principle of moments says: for an object in balance, the total clockwise moment equals the total anticlockwise moment about the same pivot. It lets you find an unknown force or distance on a beam, see-saw or lever.
This lesson builds directly on calculating a single moment and is part of moments and stability.
How do you decide which side is which?
Imagine the pivot as a fixed point and ask which way each force would try to spin the object. A weight on the right of the pivot, pulling down, turns it clockwise. A weight on the left, pulling down, turns it anticlockwise.
An upward force on the right would turn the object the other way. So the side a force is on is not enough. Think about the direction of the turn.
How do you set out a balance problem?
- Mark the pivot and draw every force acting on the object.
- Sort the forces into clockwise and anticlockwise about that pivot.
- Calculate each moment, using force × perpendicular distance from the pivot.
- Add each side to get total clockwise and total anticlockwise moments.
- Set them equal and solve for the unknown.
A force acting through the pivot has a distance of zero, so its moment is zero and it does not affect the sum.
Worked example
A light beam (its weight can be ignored) rests on a pivot.
On the left, a 20 N weight hangs 0.50 m from the pivot and a 10 N weight hangs 1.2 m from the pivot. On the right, a 55 N weight hangs a distance d from the pivot.
The beam balances. Find d.
Step 1, sort: the two left weights give anticlockwise moments. The right weight gives a clockwise moment.
Step 2, anticlockwise total: 20 × 0.50 = 10 N m, and 10 × 1.2 = 12 N m. The total is 10 + 12 = 22 N m.
Step 3, equate: clockwise moment = 55 × d = 22.
Step 4, solve: d = 22 ÷ 55 = 0.40 m.
Answer: 0.40 m. Check: 55 × 0.40 = 22 N m, which matches the anticlockwise total.
The force at the pivot is the upward support force. It balances the downward forces, so it equals 20 + 10 + 55 = 85 N. It has no moment about the pivot.
The mistake to watch for
A common slip is to balance the forces instead of the moments, or to forget one of the weights.
Mistaken working: 20 + 10 = 30 N on the left, so the right weight should be 30 N.
This ignores distance. The 30 N total says nothing about the turning effect.
The correction is to compute a moment for every force first. Then compare totals. A second habit worth building is to list each force on one line, with its distance and direction, so none is left out.
Check yourself
1. A child of weight 300 N sits 2.0 m from the pivot of a light see-saw. Another person balances it 1.5 m from the pivot on the other side. What is their weight?
Show answer
Moment of the child = 300 × 2.0 = 600 N m. Other person: W × 1.5 = 600, so W = 600 ÷ 1.5 = 400 N.
2. A light beam has 6.0 N hanging 0.20 m to the left of the pivot and 4.0 N hanging 0.35 m to the right. Does it balance? If not, which way does it turn?
Show answer
Anticlockwise: 6.0 × 0.20 = 1.2 N m. Clockwise: 4.0 × 0.35 = 1.4 N m. The moments are not equal, so it does not balance. It turns clockwise, because that moment is larger by 0.2 N m.
3. A uniform metre rule is pivoted at the 50 cm mark. A weight of 2.0 N hangs at the 20 cm mark. Where must a 1.5 N weight hang to balance the rule?
Show answer
Distance of the 2.0 N weight from pivot = 50 − 20 = 30 cm = 0.30 m. Moment = 2.0 × 0.30 = 0.60 N m. For 1.5 N: d = 0.60 ÷ 1.5 = 0.40 m = 40 cm, on the right, so at the 90 cm mark. The rule’s own weight acts at the pivot and has no moment.
Where this leads next
Next is locating a centre of mass conceptually, which explains why the weight of an object can be treated as a single force at one point. You can then apply everything in the moments and stability practice set.
If you can do a balance problem when the layout is familiar but hesitate on a new one, that is something a teacher in online one-to-one Physics tuition can build with you through fresh examples.