Turning effect of forces

CORE VS EXTENDED GUIDE

• Core students study only the unlabelled sections.
• Extended students must study everything, including Extended Extended points.
• Extended = Core + Supplement.

The Moment of a Force

The moment of a force is a measure of its turning effect about a pivot [a fixed point around which rotation occurs].

Every time a force causes or tends to cause rotation, it produces a moment. Opening a door, pushing down on a spanner to tighten a bolt, and pressing a lever on a pair of scissors are all everyday examples of turning effects. In each case, a force acts at some distance from a pivot, and the object rotates.

Two factors determine the size of a moment: the magnitude of the force and the perpendicular distance from the force's line of action to the pivot. A larger force produces a larger moment. A greater perpendicular distance from the pivot also produces a larger moment.

This is why pushing a door near the handle (far from the hinge) requires less force than pushing near the hinge — the perpendicular distance is greater, so the same moment results from a smaller force.

Moments can act in two senses of rotation about a pivot: clockwise or anticlockwise. The direction depends on which way the force would cause the object to turn.

Calculating the Moment of a Force

Key Equations

Moment of a force:

$$M=F\times d$$

Variables:

• $M$ = moment, in newton-metres (N m)
• $F$ = force, in newtons (N)
• $d$ = perpendicular distance from the pivot to the line of action of the force, in metres (m)

Rearrangements:

$$F=\frac{M}{d}$$

$$d=\frac{M}{F}$$

ProportionalityMoment is directly proportional to force when perpendicular distance is constant. Moment is directly proportional to perpendicular distance when force is constant. Doubling either the force or the distance doubles the moment.

Moment: the product of a force and the perpendicular distance from the pivot. 📌

MisconceptionStudents often write "distance" instead of "perpendicular distance from the pivot." The distance must be measured at right angles from the pivot to the line of action of the force. If the force acts at an angle, the perpendicular distance is shorter than the distance along the beam. Exam cue: always write "perpendicular distance from the pivot" in definitions and calculations.

Representation note — the equation: In $M=F\times d$, the symbol $\times $ means multiplication. Each letter represents a physical quantity. Substituting measured values with correct units into the equation gives the numerical answer.

Worked Example

A spanner is used to turn a nut. A force of 25 N is applied at a perpendicular distance of 30 cm from the centre of the nut. Calculate the moment of the force.

Finding the moment

Equation used

$$M=F\times d$$

Given

$$F=25\text{ N}$$

$$d=30\text{ cm}$$

Working

Converting cm to m:

$$d=\frac{30}{100}$$

$$d=0.30\text{ m}$$

Substituting:

$$M=25\times 0.30$$

$$M=7.5\text{ N m}$$

Answer

$$M=7.5\text{ N m}$$

The Principle of Moments

The principle of moments states that for an object in equilibrium, the sum of the clockwise moments about any pivot equals the sum of the anticlockwise moments about that pivot.

Consider a beam balanced on a pivot. A force on the left side produces an anticlockwise moment, and a force on the right side produces a clockwise moment. The beam balances — remains in equilibrium — when these two moments are equal. In equation form for one force on each side:

$${F}_{1}\times {d}_{1}={F}_{2}\times {d}_{2}$$

where ${F}_{1}$ and ${d}_{1}$ are the force and perpendicular distance on one side, and ${F}_{2}$ and ${d}_{2}$ are the force and perpendicular distance on the other side.

This principle applies to seesaws, balance scales, and any beam loaded on both sides of a pivot. If one person on a seesaw sits closer to the pivot, a greater force (heavier person) is needed on that side to balance.

Examiner InsightCIE frequently presents a diagram of a balanced beam with three known values and one unknown. Set up the equation so that the sum of clockwise moments equals the sum of anticlockwise moments, substitute, and solve for the unknown. Exam cue: always identify which moments are clockwise and which are anticlockwise before writing the equation.

Worked Example

A uniform beam is balanced on a pivot at its centre. A 6.0 N weight hangs 0.25 m to the left of the pivot. Calculate the distance at which a 10 N weight must hang to the right of the pivot to balance the beam.

Finding the distance

Applying the principle of moments:

$$\text{anticlockwise moment}=\text{clockwise moment}$$

Equation used

$${F}_{1}\times {d}_{1}={F}_{2}\times {d}_{2}$$

Rearranging for ${d}_{2}$:

$${d}_{2}=\frac{{F}_{1}\times {d}_{1}}{{F}_{2}}$$

Given

$${F}_{1}=6.0\text{ N}$$

$${d}_{1}=0.25\text{ m}$$

$${F}_{2}=10\text{ N}$$

Substituting:

$${d}_{2}=\frac{6.0\times 0.25}{10}$$

$${d}_{2}=\frac{1.5}{10}$$

$${d}_{2}=0.15\text{ m}$$

Answer

$${d}_{2}=0.15\text{ m}$$

Equilibrium

An object is in equilibrium when there is no resultant force and no resultant moment acting on it.

These are two separate conditions that must both be satisfied. The first condition — no resultant force — means the forces in every direction cancel out, so the object does not accelerate in any direction. The second condition — no resultant moment — means the total clockwise moment about any point equals the total anticlockwise moment about that point, so the object does not rotate.

A balanced seesaw with two children demonstrates equilibrium: their weights (downward forces) are balanced by the upward contact force at the pivot (no resultant force), and their clockwise and anticlockwise moments are equal (no resultant moment). A shelf bracket supporting a steady load is another example — it remains stationary because both conditions are met.

MisconceptionStudents sometimes believe that equilibrium means "nothing is happening" or "no forces act." Forces do act on an object in equilibrium — they simply cancel out in both magnitude/direction (forces) and turning effect (moments). Exam cue: always state both conditions — no resultant force AND no resultant moment.

Extended Principle of Moments with Multiple Forces

Extended The principle of moments extends to situations with more than one force on each side of the pivot.

The rule is the same: for equilibrium, the sum of all clockwise moments about the pivot equals the sum of all anticlockwise moments about the pivot. When multiple forces act on one side, calculate each individual moment (force × perpendicular distance) and then add them together for that side.

In equation form:

$${F}_{1}{d}_{1}+{F}_{2}{d}_{2}+…={F}_{3}{d}_{3}+{F}_{4}{d}_{4}+…$$

where the left side represents the sum of all anticlockwise moments and the right side represents the sum of all clockwise moments (or vice versa, depending on the arrangement).

The strategy for solving these problems is:

1. Draw and label all forces and their perpendicular distances from the pivot.

2. Identify which forces produce clockwise moments and which produce anticlockwise moments.

3. Calculate each moment separately.

4. Sum the clockwise moments and sum the anticlockwise moments.

5. Set them equal and solve for the unknown.

Worked Example

A beam is balanced on a pivot. On the left side, a 5.0 N weight hangs at 0.40 m from the pivot and a 3.0 N weight hangs at 0.20 m from the pivot. A single unknown weight $W$ hangs on the right side at 0.50 m from the pivot. Calculate $W$.

Finding the unknown weight

Applying the principle of moments:

$$\text{sum of anticlockwise moments}=\text{sum of clockwise moments}$$

The left-side weights produce anticlockwise moments. The right-side weight produces a clockwise moment.

Given

$${F}_{1}=5.0\text{ N}, {d}_{1}=0.40\text{ m}$$

$${F}_{2}=3.0\text{ N}, {d}_{2}=0.20\text{ m}$$

$${d}_{3}=0.50\text{ m}$$

Working

Calculating anticlockwise moments:

$${\text{Moment}}_{1}=5.0\times 0.40=2.0\text{ N m}$$

$${\text{Moment}}_{2}=3.0\times 0.20=0.60\text{ N m}$$

$$\text{Total anticlockwise}=2.0+0.60=2.6\text{ N m}$$

Setting equal to the clockwise moment:

$$2.6=W\times 0.50$$

Rearranging for $W$:

$$W=\frac{2.6}{0.50}$$

$$W=5.2\text{ N}$$

Answer

$$W=5.2\text{ N}$$

PRACTICAL: The Moment Balance Experiment

Aim & Principle: The experiment demonstrates that an object in equilibrium has no resultant moment, by showing that the sum of the clockwise moments about the pivot equals the sum of the anticlockwise moments.

Variables

• Independent variable (IDV): the positions (distances from the pivot) and the values of the hanging weights, in m and N
• Dependent variable (DV): whether the beam balances (remains horizontal), observed visually
• Control variables: the same beam is used throughout; the pivot remains at the centre of the beam; the same set of calibrated slotted masses is used

Method

1. Set up a uniform metre rule so that it balances horizontally on a pivot (knife edge or fulcrum) at the 50.0 cm mark. Check the beam balances with no loads — if not, adjust the pivot position until it does.

2. Hang a known weight (e.g. 2.0 N) on the left side of the pivot using a loop of thread at a measured perpendicular distance (e.g. 0.20 m) from the pivot. Record the weight and distance.

3. Hang a second known weight on the right side of the pivot. Adjust its position until the beam balances horizontally. Record this weight and its perpendicular distance from the pivot.

4. Calculate the anticlockwise moment (${F}_{1}\times {d}_{1}$) and the clockwise moment (${F}_{2}\times {d}_{2}$). Compare the two values.

5. Repeat with different combinations of weights and distances, including arrangements with more than one weight on each side. Each time, record all weights and distances, and calculate the total clockwise and anticlockwise moments.

Measurement technique

Measure each distance from the pivot to the point where the thread loop contacts the rule using a millimetre ruler, reading at eye level to avoid parallax error. The resolution of the ruler is 1 mm. Slotted masses of known weight provide the forces; check for zero error on any spring balance if used.

Key Observation & Explanation: In every case where the beam balances horizontally, the sum of the clockwise moments equals the sum of the anticlockwise moments. Small discrepancies arise from measurement uncertainty, but the values are very close. This confirms the principle of moments and demonstrates that no resultant moment acts on an object in equilibrium.

SafetyEnsure the pivot and beam are stable to prevent the beam and masses from falling. Place a soft surface (e.g. a foam mat) below the apparatus to cushion any falling masses.