Moments, couples and centre of gravity
Key idea: A complete H2 Physics lesson on field and contact forces, Hooke’s law, moments, couples, centre of gravity and equilibrium.
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The core idea
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Big question: How can forces produce turning even when their resultant is zero?
The moment of a force about a point is force multiplied by the perpendicular distance from that point to the force's line of action. A couple is two equal, opposite, parallel forces with separated lines of action: its resultant force is zero but its torque is one force times their perpendicular separation. A body's weight may be treated as acting through its centre of gravity.
The line of action controls turning
Moment is not determined by the distance to the point where the force arrow begins. It uses the shortest distance from the pivot to the force’s complete line of action. If that line passes through the pivot, the moment is zero even when the force acts far from the pivot on the drawing.
Check your understanding: Why does pushing directly towards a door hinge produce almost no turning?
The force’s line of action passes close to the hinge, so its perpendicular moment arm is nearly zero.
A couple turns without translating
The equal and opposite forces of a couple cancel as a resultant force, yet both moments have the same rotational sense. Its torque Fd uses the separation between the two parallel lines of action and does not depend on the chosen pivot.
Centre of gravity lets a distributed weight be replaced by one resultant force for force and moment calculations. This is why the weight of a uniform beam can be drawn at its midpoint.
Check your understanding: If the separation of a couple doubles at the same force, what happens to its torque?
It doubles because couple torque is Fd.
Key ideas to keep
- Use perpendicular distance, not simply the length of a lever.
- A couple's torque is independent of the chosen origin.
- For a uniform symmetrical body, the centre of gravity lies at its geometrical centre.
See the reasoning
Worked example
Moment of an angled force
Question: A 30 N force acts at the end of a 0.50 m handle and makes 40° with the handle. Find its moment about the pivot.
Step 1: Identify the turning component
Why: Only the component perpendicular to the handle contributes to the moment.
Working: F⊥ = 30 sin40° = 19.3 N.
Step 2: Multiply by the radius
Why: The perpendicular component acts 0.50 m from the pivot.
Working: M = F⊥r = 19.3(0.50) = 9.64 N m.
Step 3: State the rotational sense
Why: Moment is a directed turning effect in a plane.
Working: Use the diagram to label the result clockwise or anticlockwise.
Answer: The moment magnitude is 9.6 N m, with the sense determined by the force arrow in the diagram.
Check: The value is below the maximum 30(0.50) = 15 N m because the force is not perpendicular.
Another worked model
Question
A 30 N force acts at 40° to a 0.50 m spanner measured from its pivot. Find the moment.
Check the worked solution
The perpendicular component is 30 sin 40°, so moment = rF sin 40° = 0.50 × 30 × sin 40° = 9.64 N m.
Use a hint if needed
Practise with support
Try this
A 60 N load is 0.80 m left of a pivot. Where should a 40 N load act on the right for rotational equilibrium?
Hint: Equate clockwise and anticlockwise moments.
Check your answer
60 × 0.80 = 40d, so d = 1.20 m to the right. Translational equilibrium would still require the pivot reaction.
Now work without the hint
Practise independently
Your turn
Two opposite parallel 20 N forces are 0.18 m apart. Find the couple torque and explain why the couple has no resultant force.
Check your answer
Torque = 20 × 0.18 = 3.6 N m. The forces have equal magnitudes and opposite directions, so their vector sum is zero while their separated lines of action produce rotation.
Avoid these traps
Common mistakes
Common mistake
Any distance from a pivot can be used in a moment.
What is wrong with this reasoning?
Show better thinking
Use the perpendicular distance from the pivot to the force’s line of action. Couple torque uses the perpendicular separation of the two lines of action.
Write for the examiner
Exam guidance
Mark the line of action and moment arm on the diagram before calculating the turning effect.
Exam-style practice [6 marks]
A uniform 4.0 m beam of weight 120 N is pivoted 1.0 m from its left end. An 80 N load acts at the left end and an upward force F acts at the right end. Find F and the vertical pivot force for equilibrium.
Plan before you answer
- Place the beam’s weight at its midpoint.
- Take moments about the pivot.
- Use vertical force balance after finding F.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
About the pivot, the 80 N load and F act anticlockwise while the beam’s weight acts clockwise: 80(1.0) + F(3.0) = 120(1.0). Thus F = 13.3 N upward. Vertical balance gives Rpivot + 13.3 = 120 + 80, so Rpivot = 186.7 N upward.
Come back in three days
Check what stayed with you
Recall question 1
Define the moment of a force about a point.
Check the answer
Force multiplied by the perpendicular distance from the point to the force’s line of action.
Recall question 2
Why does a couple have zero resultant force?
Check the answer
Its two forces are equal in magnitude and opposite in direction.
Recall question 3
Where does the weight of a uniform beam act in the model?
Check the answer
Through its centre of gravity at the midpoint.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. For friction and viscous force, qualitative treatment is required; coefficients of friction and viscosity are excluded.
- GCE A-Level H2 PhysicsTopic 2(d) / Topic 2(e) / Topic 2(f) / Topic 2(g) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027