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Moments investigation

Key idea: Balance a loaded rule and compare clockwise and anticlockwise moments about its pivot. Measure each force and its distance from the pivot, including the rule's weight at its centre of gravity.

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  • SEC G3 Combined Science Physics component 2027
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A metre rule balanced on a knife edge, with loads hanging from it, is the simplest way to test the principle of moments. There is a catch: the rule has its own weight. Move the pivot away from the rule’s centre of gravity and that weight has a moment too, so the test works only if you include it. Practise the method itself in the laboratory, with a stable support and the loads secured.

Test the principle of moments

For a body in equilibrium, the total clockwise moment about any point equals the total anticlockwise moment about that point. A moment is the force multiplied by the perpendicular distance from the pivot to the force’s line of action.

A balanced metre rule with an off-centre pivotSide view of the third illustrative reading. The rule runs from 0 to 100 centimetres, with its pivot at the 40.0-centimetre mark. A 2.00-newton load hangs at 20.0 centimetres, the rule's 1.00-newton weight acts at its measured centre of gravity at 50.0 centimetres, and a 1.00-newton load hangs at 70.1 centimetres. Their perpendicular distances from the pivot are 0.200, 0.100 and 0.301 metres respectively. The upward support force acts through the pivot and has zero moment about it. The small difference between the two totals represents the illustrative reading discrepancy. Apparatus is schematic.Include the weight of the rule0102030405060708090100cmsupport forcethrough the pivotF₁ = 2.00 NW = 1.00 NF₂ = 1.00 Npivotcentre of gravity at 50.0 cmd₁ = 0.200 mdW = 0.100 md₂ = 0.301 m
A balanced metre rule with an off-centre pivotSide view of the third illustrative reading. The rule runs from 0 to 100 centimetres, with its pivot at the 40.0-centimetre mark. A 2.00-newton load hangs at 20.0 centimetres, the rule's 1.00-newton weight acts at its measured centre of gravity at 50.0 centimetres, and a 1.00-newton load hangs at 70.1 centimetres. Their perpendicular distances from the pivot are 0.200, 0.100 and 0.301 metres respectively. The upward support force acts through the pivot and has zero moment about it. The small difference between the two totals represents the illustrative reading discrepancy. Apparatus is schematic.Include the weight of the rule0102030405060708090100cmsupport forcethrough the pivotF₁ = 2.00 NW = 1.00 NF₂ = 1.00 Npivotcentre of gravity at 50.0 cmd₁ = 0.200 mdW = 0.100 md₂ = 0.301 m
Measure force distances from the pivot, not the rule’s zero. Include the rule’s weight: it gives a clockwise moment; the support gives zero moment.

A metre rule has weight too. That weight acts at the rule’s centre of gravity. If the pivot is not under that point, the rule’s weight has a moment and you must include it.

The positions, weights and readings below are an illustrative example, not measurements from a reported experiment. Find the balance point and weight of your own rule before using them in a calculation. Include the weight of each hanger in its load.

  1. Balance the rule alone on a narrow pivot (a knife edge). The balance point is its centre of gravity; in this example it is the 50.0 cm mark. Measure the rule’s weight W with a newton-meter: in this example, 1.00 N. Check the meter’s zero first.
  2. Move the pivot to the 40.0 cm mark. The rule’s weight now acts 0.100 m to the right of the pivot.
  3. Hang a load F₁ from a thin thread loop at the 20.0 cm mark, 0.200 m left of the pivot.
  4. Hang a load F₂ = 1.00 N on the right and slide it until the rule balances horizontally. Record its position and work out its distance d₂ from the pivot.
  5. Use a range of values of F₁. At each value, approach balance from both directions and repeat the position reading. Keep the pivot, left-hand load position and F₂ unchanged.

The pivot supplies an upward support force. It acts through the pivot, so it has zero moment about the pivot. The scale mark is a position; the moment uses the distance between that mark and the pivot.

F₁ / Nd₁ / mF₂ position / cmd₂ / manticlockwise moment F₁ d₁ / N m
1.000.20050.20.1020.200
1.500.20059.80.1980.300
2.000.20070.10.3010.400
2.500.20079.90.3990.500

Worked example 1

Does the principle of moments hold?

Problem

Use the third row (F₁ = 2.00 N) to test the principle of moments.

Show full solution

  1. Anticlockwise moment

    Method

    Only F₁ turns the rule anticlockwise.

    Reason

    It is the only downward force to the left of the pivot; the support force acts through the pivot.

    Working

    Answer

    2.00 × 0.200 = 0.400 N m

  2. Clockwise moments, including the rule

    Method

    Add the moment of F₂ and the moment of the rule’s weight.

    Reason

    Both forces act to the right of the pivot.

    Working

    Answer

    1.00 × 0.301 + 1.00 × 0.100 = 0.401 N m

  3. Compare

    Method

    0.400 N m and 0.401 N m agree to within 0.001 N m.

    Reason

    If repeated tipping tests bracket the balance position within ±0.2 cm, that position ambiguity alone corresponds to about ±0.002 N m for a 1.00 N load.

    Working

    Answer

    The small difference is consistent with the principle of moments. Check the force readings, pivot position and centre-of-gravity position too before claiming an overall uncertainty.

Check your understanding 1 Compare two pivot positions

Show the steps Optional working
Unit: N cm
Unit: cm
Show the steps Optional working
Unit: m
Unit: N m
Unit: m
Unit: N
Show answer

For the centred uniform rule, 2.0 × 30.0 = 3.0 × d gives d = 20.0 cm to the right; the rule’s weight acts through the pivot. For the pivot at 40.0 cm, the weight acts 10.0 cm to its right and the load at 15.0 cm is 25.0 cm left: F = (1.00 × 0.100)/0.250 = 0.40 N. Scale marks are positions; subtract the pivot mark to obtain each perpendicular distance.

Guided practice 2

Predict the balance position

About 4 min

Problem

With the same set-up, F₁ is changed to 1.80 N. Predict where the 1.00 N load F₂ must hang for the rule to balance.

Try this before viewing the solution

Show the steps Optional working
Unit: N m
Unit: N m
Unit: m
Show the steps Optional working
Unit: m
Unit: cm
Unit: cm

Hints

Hint 1

set up the balance

Anticlockwise moment F₁ d₁ = clockwise moments F₂ d₂ + moment of the rule’s weight.

Hint 2

convert to a scale mark

d₂ is measured from the pivot at the 40.0 cm mark.

Show solution step by step

  1. Balance the moments

    Method

    Write clockwise = anticlockwise and solve for d₂.

    Reason

    The rule is in equilibrium.

    Working

    Answer

    1.80 × 0.200 = 1.00 × d₂ + 1.00 × 0.100 ⇒ d₂ = 0.260 m

  2. Find the scale mark

    Method

    Add d₂ to the pivot position.

    Reason

    F₂ hangs to the right of the pivot.

    Working

    Answer

    40.0 + 26.0 = 66.0 cm mark.

Worked example 3

Evaluating the moments experiment

Problem

In a repeat of the third-row set-up, the rule just tips in opposite directions when F₂ is at 69.9 cm and 70.3 cm. Explain this limitation and suggest a matching improvement.

Show full solution

  1. Source

    Method

    The rule stays balanced while F₂ is moved over a small range, so the exact balance position is hard to judge.

    Reason

    A pivot with some width and a little friction lets the rule rest over a range of positions.

    Working

    Answer

    The midpoint is 70.1 cm; half the 0.4 cm interval is 0.2 cm. This estimates the ambiguity of this balance setting, not the total experimental uncertainty.

  2. Effect

    Method

    d₂, and so the clockwise moment, is uncertain.

    Reason

    0.002 m × 1.00 N = 0.002 N m.

    Working

    Answer

    This contribution is small compared with the 0.400 N m moment in this row. Other force and distance errors still need checking.

  3. Improvement

    Method

    Use a narrow knife-edge pivot, find the positions where the rule just tips each way, and use the midpoint.

    Reason

    A narrow pivot can reduce the resting range. Approaching the balance from both directions reveals the range; its midpoint is a reasonable estimate.

    Working

    Answer

    Hang loads from thin thread loops, not thick string, so each distance is read at one clear point.

Finding the centre of gravity of a flat shape with a plumb line is covered in the centre-of-gravity investigation.

Check your understanding 2 Move the pivot past the rule’s centre of gravity

A rule weighs 1.00 N at its 50.0 cm mark and is pivoted at 60.0 cm. A 0.60 N load hangs at 35.0 cm. Find the mark for a balancing 1.00 N load on the right. Which way does the rule’s weight turn it?

Show the steps Optional working
Unit: N m
Unit: N m
Unit: m
Unit: cm
Moment of the rule’s weight
Show answer

The 0.60 N load is 0.250 m left of the pivot, giving 0.150 N m anticlockwise. The rule’s weight is now 0.100 m left and also gives 0.100 N m anticlockwise. Their moments add to 0.250 N m. The right-hand 1.00 N load needs d = 0.250/1.00 = 0.250 m, so its mark is 60.0 + 25.0 = 85.0 cm. Compare your sketch: three downward arrows at 35.0, 50.0 and 85.0 cm, upward support at 60.0 cm and distances from that pivot. The support balances the total 2.60 N weight but has zero moment about the pivot. The rule’s weight changes from clockwise in the earlier arrangement to anticlockwise here; its side and line of action decide the sign, not a memorised subtraction.

Check your understanding 3 Check the off-centre force sketch

Draw the rule from the previous question on paper, with the pivot at 60.0 cm and loads at 35.0 and 85.0 cm. Include all forces, their positions and the support-force magnitude. Explain which forces turn the rule anticlockwise.

Show answer

Draw downward forces of 0.60 N at 35.0 cm, 1.00 N at 50.0 cm and 1.00 N at 85.0 cm. Draw the upward 2.60 N support at 60.0 cm. The load at 35.0 cm and the rule’s weight at 50.0 cm both turn anticlockwise; the right-hand load turns clockwise. The support has zero moment because its line of action passes through the pivot. Compare the four arrows, their positions and directions, not just the balancing load mark.

Common mistakes

  • Using the load’s scale mark as its distance. Measure the distance from the pivot.
  • Using the length of a sloping thread instead of the perpendicular distance to the force’s line of action.
  • Leaving out the rule’s weight when its centre of gravity is not at the pivot.
  • Letting the rule rest on a wide or rough pivot, which hides a range of balance positions.

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