Lesson 2 of 4
Balancing moments
I can state the principle of moments for a body in equilibrium.Syllabus
Syllabus K323, 4(c). State the principle of moments for a body in equilibrium.
I can apply the principle of moments to new situations.Syllabus
Syllabus K323, 4(d). Apply the principle of moments to new situations.
Make a guess. You are not marked.
A 40 kg child sits 2.0 m from the pivot of a see-saw. A 20 kg child wants to balance it on the other side. Where should the lighter child sit?
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4.0 m from the pivot, twice as far out.
Balance needs equal moments about the pivot. Half the weight needs twice the distance: 200 N x 4.0 m matches 400 N x 2.0 m.
For a body in equilibrium, the total clockwise moment equals the total anticlockwise moment about the same point.
Use force x perpendicular distance for each moment. When a load is given as a mass, use W = mg first: a total load-and-hanger mass of 200 g is 0.200 kg and weighs 2.0 N where g = 10 N/kg.
For a stationary body, the forces must balance as well as the moments. Moment balance concerns turning; force balance concerns changes in velocity. Zero total moment alone does not establish zero resultant force.
Include the rule's own weight
A uniform horizontal metre rule weighs 1.2 N. Its weight acts at its midpoint, the 50.0 cm mark. This point is its centre of gravity. A pivot supports the rule at 40.0 cm.
A downward 2.0 N load acts at 10.0 cm, and an unknown downward load F acts at 70.0 cm. The load forces include their hangers.
Forces on a uniform, balanced metre rule
The loads include their hangers. The rule's own weight acts at its 50 cm mark.
The rule's weight has an arm of 50 - 40 = 10 cm = 0.10 m. The support passes through the pivot, so its moment about the pivot is zero.
2.0 x 0.30 = 1.2 x 0.10 + F x 0.30, so F = 1.6 N.
Worked example
Find the load that balances the rule
- Anticlockwise: the 2.0 N force is left of the pivot, so its downward pull gives 2.0 x 0.30 = 0.60 N m.
- Clockwise from the rule: its 1.2 N weight is 0.10 m right of the pivot, giving 1.2 x 0.10 = 0.12 N m.
- Clockwise from F: this force acts 0.30 m right of the pivot, giving F x 0.30.
- Equate the total moments: 0.60 = 0.12 + 0.30F.
- Solve: 0.48 = 0.30F, so F = 1.6 N downwards.
The upward force from the pivot has zero moment about the pivot because its line of action passes through that point. It still acts on the rule.
For vertical force balance, the support is 2.0 + 1.2 + 1.6 = 4.8 N upwards. This equals the total downward force.
Try it step by step
Work it out in steps: Anticlockwise moment in N m, then Clockwise moment needed for balance in N m, then Unknown force in N.
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Problem 1.
Anticlockwise moment in N m: Anticlockwise moment = 4.0 x 0.30 = 1.2 N m
Clockwise moment needed for balance in N m: For balance, clockwise moment = 1.2 N m
Unknown force in N: F x 0.20 = 1.2, so F = 1.2 / 0.20 = 6.0 N (2 sf)
Problem 2.
Anticlockwise moment in N m: Anticlockwise moment = 6.0 x 0.25 = 1.5 N m
Clockwise moment needed for balance in N m: Clockwise moment needed = 1.5 N m
Unknown force in N: F = 1.5 / 0.50 = 3.0 N (2 sf)
Problem 3.
Anticlockwise moment in N m: Anticlockwise moment = 9.0 x 0.40 = 3.6 N m
Clockwise moment needed for balance in N m: Clockwise moment needed = 3.6 N m
Unknown force in N: F = 3.6 / 0.60 = 6.0 N (2 sf)
Omitting the rule's own weight would give F = 2.0 N, which is too large. Its weight has a clockwise moment because the pivot is not directly below the midpoint.
What if the 70.0 cm load is reduced?
The clockwise moment becomes smaller while the anticlockwise moment stays at 0.60 N m. The rule initially turns anticlockwise, with its left end moving downwards. You can predict the direction without calculating a new balance position.
Investigate balancing forces and distances
First balance an unloaded rule to locate its own centre of gravity. Place the pivot there, so the rule's weight has zero moment about the pivot. Keep a 2.0 N load 0.20 m to one side: its moment is 0.40 N m.
| Opposing load / N | Distance from pivot / m |
|---|---|
| 1.0 | 0.40 |
| 2.0 | 0.20 |
| 2.5 | 0.16 |
The products are 1.0 x 0.40, 2.0 x 0.20 and 2.5 x 0.16. Each gives 0.40 N m. A larger balancing force needs a shorter lever arm when the required moment is fixed. These are idealised model values; real readings vary.
In a practical investigation, record the loads and positions with quantity/unit headings, then calculate the pivot distances and moments separately. Keep the fixed-side load and distance unchanged while changing the opposing load and finding its balancing position.
- Use a narrow pivot with low friction, and identify its actual position on the rule.
- Keep the rule horizontal so vertical loads have horizontal perpendicular lever arms. Read the force positions at eye level.
- Include hanger weights and allow loads to hang freely, clear of the bench.
- Secure the support and let oscillations settle before recording a balance position.
Pivot friction may let the rule appear balanced over a range of positions. Approach balance from both directions to judge that range and reduce the pivot friction where possible. Repeating a reading alone does not remove this source of bias.
Check your understanding
A uniform horizontal metre rule weighs 2.0 N and balances on a pivot at 40 cm. A downward 3.0 N load acts at 20 cm and a downward load F at 80 cm. What is F? All load forces include their hangers.
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1.0 N
The rule's weight acts at 50 cm. About the pivot: 3.0 x 0.20 = (2.0 x 0.10) + (F x 0.40). Thus 0.60 = 0.20 + 0.40F, giving F = 1.0 N.
Practise an exam question
Write your answer first, then tick the mark points it makes.
Exam-style question
A uniform plank of weight 200 N is balanced on a pivot at its centre. A child of weight 300 N sits 1.2 m to the left of the pivot. How far to the right of the pivot must an adult of weight 400 N sit so that the plank balances? [2]
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