Moment Of A Force
Key idea: Learn how to calculate the moment of a force about a pivot using moment = force × perpendicular distance (O Level Physics 6091).
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The core idea
On this page
Learning objectives
- Distinguish contact forces from non-contact forces
- State that mass measures the amount of matter in a body
- Describe a gravitational field as a region where a mass experiences gravitational force
- Define gravitational field strength as gravitational force per unit mass
- Apply weight = mass × gravitational field strength
- Distinguish mass from weight
- Describe the effect of balanced and unbalanced forces on a body
- Describe ways a force may change motion
- Identify action–reaction pairs on interacting bodies
- Draw free-body diagrams for force systems in at most two dimensions
- Solve three-force static equilibrium graphically
- Apply resultant force = mass × acceleration
- Relate mass to resistance to change in motion
- Explain the effects of friction on motion
- Describe falling with and without air resistance, including terminal velocity
- Describe a moment as a force's turning effect in everyday examples
- Apply moment = force × perpendicular distance from the pivot
- State the principle of moments for a body in equilibrium
- apply the principle of moments to new situations or to solve related problems
- show an understanding that the weight of a body may be taken as acting at a single point known as its centre of gravity
- Explain qualitatively how centre-of-gravity position affects stability
1. Definition
A. Moment of a force (turning effect)
The moment of a force about a pivot is:
M = F × d
where:
- F = force (N)
- d = perpendicular distance from the pivot to the line of action of the force (m)
2. Key Ideas
- Moment measures the turning effect of a force about a pivot.
- Unit: newton metre (N m).
- Bigger force or bigger perpendicular distance → bigger moment.
- If the line of action passes through the pivot, d = 0 so the moment is zero.
- Moments can be clockwise or anticlockwise (state the direction in answers).
Moment is proportional to perpendicular distance
Moment (turning effect) versus perpendicular distance from pivot for a fixed force, showing a straight line through the origin.
Scroll across the graph to read all labels.
View figure data
| Perpendicular distance from pivot, d (m) | F = 20 N |
|---|---|
| 0 | 0 |
| 0.1 | 2 |
| 0.2 | 4 |
| 0.3 | 6 |
| 0.4 | 8 |
3. Detailed Explanations
A. Everyday examples (why this matters)
Examples of turning effect:
- pushing a door to rotate about its hinges
- using a spanner to rotate a nut
- using a bottle opener or spoon as a lever
- turning a steering wheel by applying a force at the rim
B. The perpendicular distance is the key
The distance you multiply by is not “distance to where your hand is”. It is the perpendicular distance from the pivot to the line along which the force acts.
This is why pushing a door near its hinges is ineffective: d is small, so the moment is small.
C. Clockwise and anticlockwise moments
- A force can cause a clockwise turning effect about a pivot.
- A force can cause an anticlockwise turning effect about a pivot.
In many problems you will compare moments about the same pivot (the next lesson uses this idea for equilibrium).
4. Common Mistakes
- Using the distance along the bar instead of the perpendicular distance to the line of action.
- Using cm in the calculation and forgetting to convert to m (moment in N m).
- Taking moments about the wrong pivot.
- Using mass (kg) instead of weight (N) when the force is due to gravity.
- Giving a magnitude but not stating whether the moment is clockwise or anticlockwise.
5. Exam Tips
- Step 1: mark the pivot clearly.
- Step 2: draw the line of action of the force (extend it if needed).
- Step 3: measure/find the perpendicular distance, d.
- Step 4: use M = Fd and give the unit N m.
- If it’s an equilibrium question, you will often use “clockwise moment = anticlockwise moment” (see Rotational Equilibrium).
If you use signs, pick one convention and stick to it (e.g. clockwise positive, anticlockwise negative).
Many O Level answers avoid signs by writing:
- total clockwise moments = total anticlockwise moments
6. Worked Examples
- Distance is always the perpendicular distance to the force’s line of action (not “distance to your hand”).
- cm → m before calculating moments (unit is N m).
- “uniform rule/beam”: the weight acts at the centre.
Modelled example 1
Simple moment calculation
Problem
A force of 150 N acts at a perpendicular distance of 0.30 m from a pivot. Find the moment about the pivot.
Study the worked solution
Identify the relevant distance
Method
Use 0.30 m as the moment arm.Reason
The moment uses the shortest perpendicular distance from the pivot to the force’s line of action.Working
d_⊥ = 0.30 mApply the moment equation
Method
Multiply the force by the perpendicular distance.Reason
Both quantities required by M = Fd_⊥ are known in compatible SI units.Working
M = (150)(0.30) = 45 N m
Guided practice 2
Finding the required force (door handle idea)
Problem
A door requires a moment of 12 N m about its hinges to start opening. The handle is 0.80 m from the hinges. Find the minimum perpendicular force needed.
Complete the guided steps
Hints
Hint 1: choose the relationship
Hint 2: make force the subject
View solution step by step
Choose and rearrange the relationship
Method
Make F the subject of M = Fd_⊥.Reason
The required turning effect and moment arm are known, while the force is unknown.Working
F = M/d_⊥Substitute and state the unit
Reason
The force is perpendicular, so the handle distance is already the correct moment arm.Working
F = 12/0.80 = 15 N
Common misconception 3
When the moment is zero
Learner response
A force acts on a 0.50 m rod, but its line of action passes through the pivot. A student writes M = F(0.50) because the rod is 0.50 m long. Locate and correct the first error.
Diagnose before viewing the correction
View solution step by step
Locate the first error
Method
Reject 0.50 m as the moment arm.Reason
The rod length is not automatically the perpendicular distance to the force’s line of action.Working
Here the line of action passes through the pivot, so d_⊥ = 0.Correct the moment
Reason
A force has no turning leverage about a point lying on its line of action.Working
M = F(0) = 0 N m
Examiner practice 4
Resultant moment (clockwise vs anticlockwise)
Examination question
Two forces act on a door. Force A produces a clockwise moment of 6.0 N m and Force B produces an anticlockwise moment of 2.5 N m. Calculate the resultant moment and state its direction. [3 marks]
Write your answer before viewing the mark scheme
View solution step by step
Combine the opposing moments
1 markMethod
Treat clockwise as positive.Reason
Opposite turning effects must have opposite signs before they are combined.Working
M_clockwise = +6.0 N m and M_anticlockwise = -2.5 N m.Calculate the magnitude
1 markReason
The moments act in opposite rotational directions, so their magnitudes subtract.Working
Mᵣₑₛᵤₗₜₐₙₜ = 6.0-2.5 = 3.5 N mState the direction
1 markWorking
The resultant moment is 3.5 N m clockwise.
Self-mark with the mark scheme
Compare your response with each mark point. Select a point only when your response contains that evidence.
Self-mark the calculation, unit and direction separately.
Challenge 5
Spoon vs coin (why a longer lever helps)
New context
It takes a moment of 0.60 N m to pry open a lid. A coin gives d_⊥ = 0.02 m while a spoon gives d_⊥ = 0.10 m. Find the force needed with each tool and use the results to explain why the spoon is more effective.
Try this without the worked method
Hints
Hint 1: identify what stays constant
Hint 2: rearrange for force
View solution step by step
Calculate the force for each tool
Reason
The lid requires a fixed turning effect, so changing the lever distance changes the required force.Working
F_coin = 0.60/0.02 = 30 N, Fₛₚₒₒₙ = 0.60/0.10 = 6.0 NExplain the comparison
Working
The spoon provides five times the moment arm, so it needs one fifth of the force.
7. Mind Stretchers
Mind stretcher 1: Pivot choice matters (but the physics doesn’t change)Extension
A uniform metre rule is pivoted at the 20 cm mark. A downward force of 4.0 N acts at the 80 cm mark.
Find the moment of the force about the pivot and state its direction.
Show Answer
Perpendicular distance from pivot:
d = (80-20) cm = 60 cm = 0.60 m
Moment:
moment = Fd = (4.0)(0.60) = 2.4 N m
The force acts on the right of the pivot and downwards, so it produces a clockwise moment.
Mind stretcher 2: Link to equilibrium (preview)Extension
A seesaw is balanced about a central pivot. A child of weight 300 N sits 0.40 m from the pivot on one side.
How far from the pivot must a child of weight 240 N sit on the other side to balance the seesaw?
Show Answer
For balance, clockwise moment = anticlockwise moment:
(300)(0.40) = (240)d
d = (300 × 0.40)/240 = 0.50 m
8. Practice and next step
Calculate the moment of a 25 N force acting 0.18 m perpendicularly from a pivot, then explain how your answer would change if the same force acted closer to the pivot. Check your moments, equilibrium, centre-of-gravity and stability reasoning with the Turning Effects of Forces check.
Continue to Rotational Equilibrium, where clockwise and anticlockwise moments are combined in one system. Study Centre of Gravity and Stability afterwards rather than trying to learn all four ideas on this page.
Continue with the next resource in this course.
Course and syllabus information
- Course
- SEC G3 Physics
- Edition
- SEC G3 Physics 2027