A uniform beam’s weight acts through its centre of gravity. If the pivot is away from that point, the beam’s own weight produces a moment. Measure the perpendicular distance from the pivot to the line of action of the weight.
The 2027 SEC G3 Physics syllabus (K323) covers moments and centre of gravity. This example shows how to balance a horizontal beam when its weight and an added load act on the same side of the pivot.
- Mark the pivot and the point where the beam’s weight acts.
- Measure each distance from the pivot.
- Balance clockwise and anticlockwise moments.
Worked example | A 1.2 m beam balanced at 0.4 m
A uniform beam weighs 12 N. The pivot is at the 0.40 m mark, measured from the left end. A 6 N load hangs at the 1.00 m mark. Find the downward force to apply at the left end to balance the beam.
The beam’s weight acts at the 0.60 m mark. Its clockwise moment is 12 × (0.60 − 0.40) = 2.4 N m. The load is 0.60 m from the pivot and contributes 6 × 0.60 = 3.6 N m clockwise.
The total clockwise moment is 6.0 N m. A downward force at the left end has a 0.40 m moment arm and acts anticlockwise. It must be 6.0 ÷ 0.40 = 15 N. If you omit the beam’s weight, you would obtain 9 N and leave a clockwise moment unbalanced.
Check the support force separately
With only the three downward forces and the support force acting vertically, the support supplies 12 + 6 + 15 = 33 N upwards. It contributes zero moment about the pivot because its line of action passes through that point.
Zero moment about the pivot and zero total force answer different questions. For a beam in equilibrium, check both where the task requires it.
Move the pivot to the middle
Move the pivot to the 0.60 m mark while keeping the load at 1.00 m and the applied force at the left end. The beam’s weight now has zero moment. The load produces 6 × 0.40 = 2.4 N m, and the left force must be 2.4 ÷ 0.60 = 4.0 N.
For a non-uniform beam, use the centre of gravity supplied in the question. Its position may differ from the midpoint.
Frequently asked questions
Why is the beam’s moment zero at the midpoint in the second setup?
For this uniform beam, its weight acts through the pivot.
Does the reaction at the pivot have zero force?
It can be nonzero even though it has zero moment about that pivot.
Official information
SEAB’s 2027 SEC G3 Physics syllabus (K323), Turning Effects of Forces. Checked 9 November 2026.