The key idea
A short topic with one equation and one principle, commonly tested through calculations. The marks are often lost in the distances: they must be perpendicular, and they must be measured from the pivot.
Understand it · 01
Find the force's line of action
Imagine extending a force arrow into a straight line. The moment arm is the shortest distance from the pivot to that line, measured at right angles. It is not automatically the distance from the pivot to the point where the force touches the object. A force whose line of action passes through the pivot has zero moment, however large the force is.
This explains why a door handle is far from the hinge and why pushing approximately perpendicular to the door is effective. The same force produces a larger moment with a larger perpendicular distance. Decide whether each force would turn the object clockwise or anticlockwise around the chosen pivot before calculating. Keep distances in metres if you report a moment in newton metres; a moment is not quoted in newtons alone.
Understand it · 02
Use one pivot consistently
Draw every force, including the beam's own weight when relevant. A uniform beam's centre of gravity is at its midpoint. Choose a pivot that simplifies the calculation, often a support whose reaction force is unknown. The reaction then has zero moment about that pivot, leaving fewer unknown moments. All remaining perpendicular distances must be measured from the same chosen point.
For rotational equilibrium, add all clockwise moments and equate their sum to all anticlockwise moments. After finding an unknown force, also check vertical force balance. A beam with zero resultant moment could still accelerate vertically if its upward and downward forces did not balance. In a lab investigation, account for the ruler's weight or place the pivot at its centre of gravity, and let the ruler settle before reading positions.
Understand it · 03
Find a centre of gravity by suspension
When a freely suspended flat object settles, its centre of gravity lies vertically below the suspension point. Otherwise its weight would provide a turning effect and it would rotate. Suspend a thin irregular lamina from one hole and hang a plumb line from the same point. Mark the vertical line on the lamina after it has stopped moving.
Repeat using a different suspension point. The crossing of the two marked lines locates the centre of gravity; a third line provides a useful consistency check. The method works for a freely hanging lamina, so excessive friction at the pin can prevent it from reaching the correct position. A centre of gravity can even lie outside the material, as it does for some curved or hollow shapes.
Understand it · 04
Test where the weight's vertical line falls
As an object tilts, compare the vertical line through its centre of gravity with the edge of its support base. While that line falls inside the base, weight tends to turn the object back towards its original position about the contact edge. Once the line falls outside the base, weight produces a moment that tips the object further over.
A wider base or lower centre of gravity generally allows a larger tilt before toppling. That is why stability is about the arrangement of mass and support, not simply an object being heavy. When comparing designs, state what changes and trace how it moves the line of action relative to the base. Remember that sliding is another possibility: the simple toppling argument assumes the object does not first slide away.
Another view of the physics
Choose the pivot, then mark each perpendicular distance
Key definitions
- Moment of a force
- The turning effect of a force: moment = force × perpendicular distance from the pivot.
- Principle of moments
- For an object in equilibrium, the sum of clockwise moments about any point equals the sum of anticlockwise moments about that point.
- Centre of gravity
- The point through which the whole weight of an object appears to act.
- Equilibrium
- The state in which there is no resultant force and no resultant moment on an object.
Equations, units & conditions
Moment
M = F × d
Units: N m
d = perpendicular distance from pivot
Balance condition
F₁ × d₁ = F₂ × d₂
Worked examples, step by step
Example 1
A uniform seesaw is pivoted at its centre. A child of weight 400 N sits 1.5 m from the pivot. Where must a 600 N adult sit to balance it?
Follow the method
- Balance: clockwise moment = anticlockwise moment.
- 400 × 1.5 = 600 × d.
- d = 600 ÷ 600 = 1.0 m.
d = 1.0 m from the pivot, on the other side
Include the beam's own weight
A uniform 2.0 m horizontal beam weighs 30 N and is hinged at its left end. A 50 N load hangs 1.6 m from the hinge. An upward support force acts at the right end. Find this support force.
Follow the method
- Take moments about the hinge so the unknown hinge force has zero moment.
- The beam's weight acts at its midpoint, 1.0 m from the hinge.
- Total clockwise moment = (30 × 1.0) + (50 × 1.6) = 110 N m.
- For equilibrium, support force × 2.0 = 110, so the support force is 55 N.
Upward force at the right end = 55 N. The hinge provides the remaining 25 N upward.
Practice questions with worked solutions
Try each question on paper before opening the solution. These are original academy questions; the suggested marks are a self-checking guide, not an official exam-board mark scheme. Difficulty labels describe this practice set, not predicted grades.
Starter · 3 suggested marks
1. Convert the lever arm to metres
A 24 N force acts perpendicular to a door at a point 75 cm from its hinge. Calculate the moment. What perpendicular force at 30 cm would produce the same moment?
Show solution and self-check: question 1
- 75 cm = 0.75 m, so moment = 24 × 0.75 = 18 N m.
- At the shorter lever arm, F × 0.30 = 18.
- F = 18/0.30 = 60 N. A shorter perpendicular distance requires a larger force for the same moment.
Moment = 18 N m; required force at 30 cm = 60 N.
Self-check · one mark per point
- Converts the first distance to 0.75 m.
- Finds moment 18 N m.
- Finds force 60 N using the second distance in metres.
Build · 4 suggested marks
2. Balance a beam and its vertical forces
A light beam has a 12 N weight 0.30 m to the left of a pivot. A 9.0 N weight is placed on the right. Where should it be placed for balance, and what upward force must the pivot exert?
Show solution and self-check: question 2
- Anticlockwise moment = 12 × 0.30 = 3.6 N m.
- Clockwise moment must equal it: 9.0 × d = 3.6.
- d = 0.40 m to the right of the pivot.
- For vertical force equilibrium, the upward support force equals 12 + 9.0 = 21 N. A light beam's weight is neglected.
Place the 9.0 N weight 0.40 m to the right. Pivot force = 21 N upward.
Self-check · one mark per point
- Calculates the 3.6 N m anticlockwise moment.
- Sets clockwise and anticlockwise moments equal.
- Finds 0.40 m on the opposite side.
- Finds support force 21 N upward.
Build · 3 suggested marks
3. Explain when an object topples
A tall block is tilted slowly about one bottom edge. Explain why it returns to its starting position for a small tilt but topples after a sufficiently large tilt. Give one design change that makes it more stable.
Show solution and self-check: question 3
- For a small tilt, the vertical line through the centre of gravity is inside the base, and weight produces a restoring moment about the edge.
- Once that line passes outside the base, weight produces a moment that increases the tilt.
- A wider base or a lower centre of gravity requires a larger tilt before the line of action falls outside the base.
Toppling begins when the weight's line of action moves beyond the supporting edge. Widen the base or lower the centre of gravity.
Self-check · one mark per point
- Connects an inside-base line of action to a restoring moment.
- Connects an outside-base line of action to toppling.
- Gives a valid stability improvement.
Stretch · 4 suggested marks
4. Find two support reactions
A light horizontal beam 3.0 m long is supported at both ends. A 180 N load acts 1.0 m from the left support. Neglect the beam's weight. Calculate the upward force at each support.
Show solution and self-check: question 4
- Taking moments about the left support: right reaction × 3.0 = 180 × 1.0.
- Right reaction = 60 N.
- Upward forces total 180 N, so left reaction = 180 − 60 = 120 N.
Left support = 120 N upward; right support = 60 N upward.
Self-check · one mark per point
- Chooses a consistent pivot and correct distances.
- Balances moments to obtain 60 N at the right support.
- Uses vertical force balance.
- Finds 120 N at the left support.
Stretch · 4 suggested marks
5. Include the beam's own weight
A uniform 1.2 m beam weighs 20 N and is hinged at its left end. A 30 N load hangs at its right end. An upward force is applied at the right end to keep the beam horizontal. Find this upward force.
Show solution and self-check: question 5
- The beam's 20 N weight acts at its midpoint, 0.60 m from the hinge.
- Total clockwise moment = 20 × 0.60 + 30 × 1.2 = 48 N m.
- For balance, upward force × 1.2 = 48, so the force is 40 N.
Required upward force = 40 N.
Self-check · one mark per point
- Places the beam's weight 0.60 m from the hinge.
- Includes both clockwise moments.
- Obtains total clockwise moment 48 N m.
- Divides by 1.2 m to find 40 N.
Stretch · 5 suggested marks
6. Find the reactions at two supports
A uniform 3.0 m plank weighs 60 N and rests on supports at both ends. A 180 N student stands 0.50 m from the left end. Find the upward force at each support.
Show solution and self-check: question 6
- The plank's weight acts at 1.5 m from the left end.
- Taking moments about the left support gives Rright × 3.0 = (60 × 1.5) + (180 × 0.50).
- Rright = 180/3.0 = 60 N.
- Total upward force = total weight = 60 + 180 = 240 N.
- Rleft = 240 − 60 = 180 N. The nearer support carries more of the student's load.
Left support = 180 N upward; right support = 60 N upward.
Self-check · one mark per point
- Places plank weight at the midpoint.
- Writes a moment balance about an end.
- Finds right support force 60 N.
- Uses total upward force 240 N.
- Finds left support force 180 N.
Common mistakes & how to avoid them
- Measuring distances from the wrong point. Every distance in a moments calculation is measured from the pivot — not from the end of the beam, not from the other force.
- Forgetting the weight of the beam itself in Extended questions. A uniform beam's weight acts at its centre — include it as a force at the midpoint.
- Defining equilibrium with only 'no resultant force'. Full marks require both: no resultant force and no resultant moment.
- Mixing N cm and N m. You may balance moments using centimetres on both sides, but convert distances to metres when reporting a moment in N m.
Understand the why
Questions, explained
Start with the short answer, then follow the reasoning. Each explanation has a permanent link you can share with a classmate or return to when revising.
How do you find the centre of gravity of a lamina?
Suspend the lamina freely from a hole and mark the vertical line shown by a plumb line hanging from the same suspension point. Repeat from a different point. The lines intersect at the centre of gravity; a third suspension provides a useful check on the result.
A freely hanging object rotates until its centre of gravity is below the suspension point. In that position, weight's line of action passes through the pivot and gives no turning effect. The first suspension therefore identifies a line containing the centre of gravity, not a unique point on its own.
Wait for both the lamina and plumb line to settle before marking. Avoid friction that prevents free rotation and choose well-separated suspension positions so the lines cross clearly. If repeated lines form a small triangle instead of meeting exactly, use their near-intersection to estimate the position and acknowledge the experimental uncertainty rather than choosing one line as perfect.
What makes an object stable?
A wider support base and a lower centre of gravity generally increase resistance to toppling. As an object tilts, follow the vertical line through its centre of gravity. Weight tends to restore it while that line is inside the base, but tips it farther once the line passes outside.
The edge touching the ground acts as the possible pivot. What matters is which side of that edge weight's line of action passes, because that determines the direction of the moment. The same object can move from a restoring situation to a toppling situation as its tilt increases, without its weight changing.
A lower centre of gravity requires a larger tilt before the vertical line crosses the support edge; a wider base has a similar benefit. Simply making an object heavier does not necessarily improve this geometry. This comparison assumes the object tips about the edge rather than sliding first and that its mass does not shift significantly as it tilts.
What are the two conditions for equilibrium?
The resultant force must be zero and the resultant moment must be zero. The first prevents acceleration of the object's centre of mass; the second prevents angular acceleration. For a stationary beam to remain at rest, both its force balance and its turning-effect balance must hold.
Equal upward and downward forces alone are insufficient if their lines of action are different. They can form a turning pair that rotates the object despite zero resultant force. Likewise, a collection of forces can have zero net moment about a chosen point while still pushing the object bodily in one direction.
In a beam problem, use moments to find an unknown support force and then use vertical force balance to find the other. Do not equate the two support forces unless the loading is symmetric. More generally, equilibrium does not automatically mean zero velocity: an object can translate at constant velocity when the resultant force is zero.
Which distance do I use when calculating a moment?
Use the perpendicular distance from the pivot to the force's line of action. Imagine extending the force arrow into a straight line, then draw the shortest distance to it from the pivot. The distance to the point where the force is applied is not always the correct moment arm.
When a force acts at right angles to a straight lever, the lever length from the pivot can equal the required perpendicular distance. If the force tilts, that convenient equality may no longer hold. If its line of action passes through the pivot, the moment is zero even when the force is applied far away along the lever.
Draw the pivot, the extended force line and a right-angle marker before substituting. This makes it easier to see why pushing a door near its handle is effective and why pushing along a spanner's length produces little turning effect. Keep distances in metres when reporting a moment in N m, and state the turning direction separately.
A force perpendicular to a lever
A 40 N force acts at right angles to a handle, 0.25 m from its pivot. Calculate the moment. What happens if the same force acts along a line through the pivot?
- For the perpendicular force, moment = 40 × 0.25 = 10 N m.
- For a force through the pivot, perpendicular distance = 0, so moment = 0.
Moment = 10 N m in the first case and zero in the second.
When do I include a beam's weight in a moments calculation?
Include the beam's weight whenever it is relevant and not explicitly neglected. For a uniform beam in a uniform gravitational field, weight acts through the midpoint. Its moment is zero only if that line of action passes through the chosen pivot; a non-uniform beam may have a different centre of gravity.
A diagram can show hanging loads without drawing the beam's own weight. Read the written information as well: a stated beam mass is often there because its weight contributes a moment. Convert mass to weight using the supplied gravitational field strength before adding it to other forces already given in newtons.
Choosing the beam's centre as the pivot removes its weight's moment, but it does not remove the force itself from a vertical force balance. Choose the pivot that simplifies the actual unknowns, not one that avoids noticing a load. The word uniform supports using the midpoint; geometric length alone does not establish the centre of gravity of every object.
Include a uniform beam's moment
A uniform 2.0 m beam weighing 30 N is hinged at its left end. A 20 N load hangs at the right end. What upward force at the right end balances the moments?
- Beam weight acts 1.0 m from the hinge.
- Clockwise moment = 30 × 1.0 + 20 × 2.0 = 70 N m.
- Upward force × 2.0 = 70, so force = 35 N.
An upward force of 35 N is required at the right end.
Why does the support nearer a load carry more of its weight?
The two support forces must balance both the load's weight and its moment. On a light beam supported at its ends, moving a load towards one support reduces the reaction needed at the far support. The nearer support therefore provides a larger share of the upward force.
Taking moments about the nearer support is a quick way to see this. The load has a small moment arm, while the far support has the full beam length. Only a relatively small far-support reaction is needed to balance the load's moment. The remaining upward force must come from the nearer support.
This conclusion assumes a downward load positioned between the supports and an arrangement that can maintain contact. If a question includes the beam's own weight, include its moment too. Loads beyond a support can require different reactions or cause loss of contact, so the usual two-positive-reactions picture should not be applied without checking the physical arrangement.
Share a load between supports
A light 4.0 m beam has a 240 N load 1.0 m from its left support. Find the two support reactions.
- About the left support: right reaction × 4.0 = 240 × 1.0.
- Right reaction = 60 N.
- Left reaction = 240 − 60 = 180 N from vertical force balance.
Left support: 180 N upward. Right support: 60 N upward.
Your learning checklist
Use these goals to check your understanding. Requirements vary by specification, tier and exam year; the official references below define your full course.
- Define the moment of a force and calculate it
- Apply the principle of moments to balanced beams and levers
- Describe an experiment to verify the principle of moments
- Define centre of gravity and describe how to find it for a plane lamina
- Relate the position of the centre of gravity to stability
Sources & how to cite these notes
Original explanations, illustrations and practice by Dr Desouky Physics Academy. Official specifications guide the course scope; questions and suggested marks on this page are our own revision material.
- Cambridge IGCSE Physics 0625 — 2026–2028 syllabus
- Pearson Edexcel International GCSE Physics 4PH1 — specification
Reference this page
Dr Desouky Physics Academy. IGCSE Physics: Moments and turning effects. Updated 2026-09-25. https://www.drdesouky.com/revision/igcse-physics/moments-and-turning-effects
Use the arrow beside an explanation heading to link straight to that section. Spotted a problem? Contact the academy with the topic and the step you want us to check.