A moment describes the turning effect of a force about a chosen point. Its size is force multiplied by the perpendicular distance from that point to the force’s line of action. The key word is perpendicular: the length of a beam or handle is not automatically the correct distance.
These moments and equilibrium worked examples use constructed data and idealised stationary beams. They teach calculations rather than practical load testing. Unless a question supplies mass, loads are already stated as forces in newtons; do not multiply a force by gravitational field strength again.
For the basic moment rule and simple levers, see Physics Notes Form 2. This lesson develops complete beam equilibrium: include the beam’s weight, find support reactions, check force balance and confirm the answer about a second pivot.
Build a force map before writing an equation
Choose one body to analyse. Mark every external force on that body, including its own weight and the reactions at supports. A line of action is the straight line along which a force acts, extended if necessary. The perpendicular distance is the shortest distance from the chosen pivot to that line.
For a horizontal beam with vertical forces, the required distances are horizontal. For a tilted force, identify the right-angled distance or resolve the force before multiplying. State whether each force tends to turn the beam clockwise or anticlockwise about your chosen point.
- Moment magnitude = F × dperpendicular.
- Moment unit = N m, with force in N and distance in m.
- For rotational balance, total clockwise moment = total anticlockwise moment about the same point.
- For full static equilibrium, the resultant force must also be zero.
A balanced moment equation alone does not prove that an object can remain stationary. Equal forces in the same direction can have cancelling moments about a midpoint while still producing a nonzero resultant force.
Worked example 1: identify the actual lever arm
A force of 75 N acts perpendicular to a model lever at a point 28 cm from its pivot. Calculate the moment.
Convert 28 cm to 0.28 m. Then moment = 75 × 0.28 = 21 N m. The force is perpendicular to the lever, so the pivot-to-force distance is the perpendicular distance in this example.
Now a different force of 50 N acts at the end of a 0.40 m horizontal arm, at 30° above the arm. Its perpendicular component is 50 sin 30° = 25 N. The moment is 25 × 0.40 = 10 N m.
The same result follows from the perpendicular distance to the angled force’s line of action: d = 0.40 sin 30° = 0.20 m, giving 50 × 0.20 = 10 N m. The tempting calculation 50 × 0.40 = 20 N m wrongly assumes a 90° force.
With the arm extending rightwards from the pivot and the force angled upwards, this second moment is anticlockwise. The horizontal component acts along a line through the pivot and contributes no moment about it.
Worked example 2: balance first, then check support force
A uniform horizontal beam is 2.0 m long and weighs 10 N. It rests on a pivot at its centre. A downward 18 N load acts 0.40 m to the left of the pivot. A downward force F acts 0.60 m to the right. Find F and the upward pivot reaction R.
| Force | Position | Turning effect about pivot |
|---|---|---|
| 18 N downward | 0.40 m left | Anticlockwise |
| Beam weight 10 N downward | At pivot | Zero moment |
| R upward | At pivot | Zero moment |
| F downward | 0.60 m right | Clockwise |
Take moments about the centre. The clockwise and anticlockwise moments balance, so F × 0.60 = 18 × 0.40. Therefore F = 7.2/0.60 = 12 N.
Next balance vertical forces. R = 18 + 12 + 10 = 40 N upward. The beam’s weight disappears from this particular moment equation because it acts through the pivot, but it remains in the force equation.
Check both conditions independently: 18 × 0.40 = 12 × 0.60 = 7.2 N m, and total downward force = total upward force = 40 N. All horizontal forces are zero in the model.
The smaller load is farther from the pivot, which is physically sensible. If you had obtained 27 N on the right, you would have inverted the distance ratio: the right side would then produce too large a clockwise moment.
Worked example 3: a beam with two supports
A uniform 4.0 m beam weighs 60 N and is supported at its left and right ends. An additional downward 90 N load acts 3.0 m from the left end. Find the two upward support reactions, L and R.
The beam’s weight acts at its centre, 2.0 m from the left end. Choose the left support as the moment reference point. L then has zero moment about that point, leaving one unknown in the moment equation:
R × 4.0 = 60 × 2.0 + 90 × 3.0 = 120 + 270 = 390 N m.
Thus R = 97.5 N. Vertical force balance gives L + 97.5 = 60 + 90, so L = 52.5 N.
Check using moments about the right support instead. The left reaction produces 52.5 × 4.0 = 210 N m clockwise. The downward loads produce 60 × 2.0 + 90 × 1.0 = 210 N m anticlockwise. Both supports give the same physical result.
The larger reaction is at the right support, nearer the extra load. That is a useful sense check, although the equations are still needed to find its size. This model assumes a rigid beam, point supports and no additional applied turning couple.
Choose a sign convention and keep it
You can write equality of clockwise and anticlockwise totals, as above, or use signed moments. One common choice makes anticlockwise positive and clockwise negative. In example 2, that gives +18 × 0.40 − F × 0.60 = 0.
Either sign convention works if used consistently. Do not assign the sign by whether a force points up or down alone. A downward force to the left of the pivot turns anticlockwise; a downward force to the right turns clockwise. Position and direction together determine the sign.
Keep every distance measured from the same reference point within one equation. A distance measured from the left support cannot be combined with another measured from the centre unless each has first been converted to the chosen reference.
Independent practice with full solutions
Practice 1: convert centimetres
A perpendicular 120 N force acts 15 cm from a pivot. Calculate its moment and explain what happens if the force acts through the pivot instead.
Solution: 15 cm = 0.15 m, so moment = 120 × 0.15 = 18 N m. Through the pivot, the perpendicular distance becomes zero, so the moment becomes 0 N m, even though the force itself remains 120 N.
Practice 2: use both equilibrium conditions
A beam of negligible weight has a downward 30 N force 0.80 m left of a pivot. A downward force acts 0.50 m right of it. Find that force and the pivot reaction.
Solution: 30 × 0.80 = F × 0.50, giving F = 48 N. The upward reaction is 30 + 48 = 78 N. Both moments equal 24 N m. The larger force acts at the shorter distance, as expected.
Practice 3: check the result about a second point
A uniform 3.0 m beam weighs 40 N and is supported at both ends. A downward 60 N load acts 0.50 m from the left end. Find both reactions.
Solution: About the left support, R × 3.0 = 40 × 1.5 + 60 × 0.50 = 90, giving R = 30 N. Then L = 40 + 60 − 30 = 70 N. About the right support, 70 × 3.0 = 40 × 1.5 + 60 × 2.5 = 210 N m, confirming the answer.
Investigate centre of gravity and stability on paper
The retained beam examples test both force and moment balance. Stability asks a further question: after a small displacement, does the body return, move farther away or remain in its new position? In a nearly uniform gravitational field, the weight acts through the centre of gravity G, which coincides with the centre of mass.
Worked investigation 1: locate G from supplied suspension records
An original paper record describes a thin rigid lamina suspended freely from three different holes. Once at rest, its centre of gravity must lie vertically below the suspension point, so each supplied plumb-line trace passes through G. On a coordinate map in centimetres, the first trace is x = 4 and the second is y = 3. Their intersection is (4,3). A third independent trace, y = 7−x, also passes through (4,3), supporting the location. The traces represent different orientations transferred onto one lamina map; they are not three vertical lines in one fixed room view.
With real readings, traces may bound a small region rather than meet exactly. Repeating from another hole tests consistency. The location need not be a material point: a ring can have G in its empty centre. Uniform density is needed to infer G from symmetry alone, but not for the suspension principle. This is analysis of supplied records, not a hanging-load experiment.
Worked investigation 2: compare tipping thresholds
A rigid rectangular block rests on a tilting plane and does not slide. Its centre of gravity is midway across the base, at height h measured perpendicular to the base. Let b be half the base width along the slope. In the two-dimensional model, tipping begins when the vertical line through G reaches the downhill edge: tan θ = b/h. This comes from a right triangle; θ is the plane’s tilt from horizontal.
| Block | Full base width / cm | G height h / cm | b/h | Critical angle |
|---|---|---|---|---|
| A | 10 | 10 | 0.5 | 26.6° |
| B | 20 | 10 | 1 | 45.0° |
| C | 10 | 5 | 1 | 45.0° |
For A, tan θ = 5/10 = 0.5, giving θ ≈ 26.6°. At 20° the weight line remains inside the base; at 30° it is outside and gravity promotes tipping in this ideal model. B is wider at the same G height, while C has lower G at the same width. Both tolerate a larger tilt than A. The ratio has no length unit because cm/cm cancels. Do not use the full width in place of b.
The no-sliding assumption matters: a slippery block may slide before it reaches this angle. Adding mass does not by itself change the threshold if G and geometry remain unchanged. Moving that mass upward can raise G and reduce stability. This interpretation is not a safety limit for a vehicle, ladder or real structure.
Distinguish stable, unstable and neutral equilibrium
A slight tilt of a supported low block initially raises G and gravity tends to restore it: stable. Balancing a suitable body at a maximum of gravitational potential can give unstable equilibrium: a slight displacement lowers G and it moves farther away. A homogeneous sphere on an ideal level surface keeps G at the same height as it rolls: neutral equilibrium in that direction. Friction, constraints and geometry must be stated.
Independent practice and answers
- Task: Traces x=6 and y=8−x intersect where? Answer: (6,2) cm. A third trace should pass near that location within uncertainty.
- Task: A block has full width 12 cm and G height 8 cm. Find its ideal threshold. Answer: tan θ=6/8=0.75, so θ≈36.9°. Units cancel; sliding is excluded.
- Task: Why does a wider stance usually increase the tipping margin? Answer: The support boundary is farther from the weight line for the same G location; a larger displacement is needed to reach an edge.
- Task: Can zero total moment alone establish complete rest? Answer: No. Resultant force must also vanish, as the retained beam examples demonstrate.
Background: equilibrium and stability. Diagrams, data and computations are original.
Frequently asked questions
Is moment the same as force?
No. Force is measured in newtons; moment includes distance from a reference point and is measured in newton metres. The same force can produce different moments.
Why not call the moment unit a joule?
N m and J have the same base dimensions, but joule names energy or work. Write N m for a moment to preserve the physical meaning.
Where does a uniform beam’s weight act?
In a uniform gravitational field, its resultant weight acts through its centre. A nonuniform beam may have a different centre of gravity, which must be supplied or found.
Can I choose any point for moments?
For an equilibrium calculation, yes. Choosing a support often removes an unknown reaction from the equation. The final force and moment conditions must both hold.
Does a zero moment mean zero force?
No. A force whose line of action passes through the reference point has zero moment there, but can still contribute to the resultant force.
What if the problem gives a mass?
Convert mass to weight using W = mg and the stated gravitational field strength. Kilograms cannot be inserted as though they were newtons.
What does a negative support reaction mean?
It means the force direction is opposite to the assumed direction. If the support can only push upwards, the proposed equilibrium may be impossible because contact would be lost.
Reference and next revision step
See OpenStax’s explanation of moments and equilibrium for the standard torque relation. All numerical exercises here are original. The site’s existing Physics topical revision resource is login-gated; no particular member file is promised here.
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