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Torque & Couples (A Level)

Key idea: Define moment (torque), apply the principle of moments, and solve equilibrium problems involving couples (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Apply moments, couples and force-and-torque equilibrium using free-body diagrams and vector triangles.
  • show an understanding that the weight of a body may be taken as acting at a single point known as its centre of gravity
  • apply the principle of moments to new situations or to solve related problems
  • Explain inertia and momentum, then apply Newton's laws using free-body diagrams.

1. Definitions (Must Know)

A. Moment (torque) of a force, τ

The moment (torque), τ, of a force about a point (pivot) is:

τ = Fd_⊥

where d_⊥ is the perpendicular distance from the pivot to the force’s line of action.

Unit: N m.

B. Couple

A couple is a pair of forces that are:

  • equal in magnitude,
  • opposite in direction,
  • parallel, but with different lines of action (separated by a distance).

A couple produces a turning effect with zero resultant force.

C. Equilibrium (force + moment)

A rigid body is in equilibrium when:

∑ vector F = 0 and ∑ τ = 0

2. Key Ideas (What Earns Marks)

  • Use perpendicular distance to the line of action (not the slanted length).
  • Choose a sign convention (e.g. anticlockwise +, clockwise −) and stick to it.
  • For equilibrium, use:
    • ∑ Fₓ = 0 and ∑ F_y = 0 (if needed),
    • ∑ τ = 0 about a convenient pivot.
  • The moment of a couple is: τ = Fs where s is the perpendicular separation of the forces; the torque of a couple does not depend on the pivot.
  • The weight of a body can be taken as acting at its centre of gravity: Centre of Gravity.
Exam pitfall: using non-perpendicular distance

Moment is force multiplied by the perpendicular distance to the line of action. Using the slanted length directly is one of the most common setup errors.

3. Detailed Explanations

A. Finding d_⊥ (the common trap)

d_⊥ is the shortest distance from the pivot to the force’s line of action.

If the force acts at an angle, you usually need:

  • d_⊥ = r sin θ (if r is the distance from pivot to point of application, and θ is the angle between vector r and vector F).

B. Choosing the pivot smartly

You can take moments about any point. Choose a pivot that removes unknown forces from the moment equation.

Example: if a beam is supported at a hinge, taking moments about the hinge eliminates the hinge reaction.

C. Why a couple gives “pure rotation”

For a couple:

  • resultant force is zero (forces cancel),
  • but there is still a net turning effect because the forces act at different lines of action.

That is why a steering wheel can be turned without translating.

D. Principle of moments (equilibrium condition)

If a body is in equilibrium, the sum of clockwise moments equals the sum of anticlockwise moments (about any point):

∑ τ_clockwise = ∑ τ_anticlockwise

This is just another way to say ∑ τ = 0.

4. Common Mistakes

  • Using the distance to the point of application instead of the perpendicular distance.
  • Swapping clockwise/anticlockwise signs halfway through.
  • Forgetting a force’s line of action (e.g. weight acts through the centre of gravity).
  • Assuming “balanced forces” implies “no rotation” (you also need ∑ τ = 0).

5. Exam Tips

  • Draw a clear diagram with the pivot marked and the line of action of each force.
  • State your sign convention explicitly.
  • In equilibrium problems, try moments about a point that eliminates an unknown.
  • If you get a negative value for a reaction force, your assumed direction is wrong.

6. Worked Examples

Modelled example 1

Simple torque calculation

Core

Problem

A 50 N force acts perpendicular to a spanner 0.20 m from the bolt. Find the torque magnitude about the bolt.
Study the worked solution

  1. Identify the lever arm

    Method

    d_⊥ = 0.20 m.

    Reason

    The force is perpendicular to the spanner, so the stated radius is already the perpendicular distance to its line of action.

    Working

    d_⊥ = 0.20 m

  2. Calculate the moment

    Method

    τ = 10 N m.

    Reason

    Moment magnitude is force times perpendicular distance.

    Working

    τ = Fd_⊥ = 50(0.20) = 10 N m

Guided practice 2

Beam in equilibrium (principle of moments)

About 4 min

Problem

A uniform 2.0 m beam of weight 40 N is held horizontally by a fixed support at its left end. A 60 N load hangs from the right end. Find the support’s total upward reaction. The fixed support also supplies the balancing moment, which is not required.

Try this before viewing the solution

Unit: N

Hints

Hint 1: balance vertical forces
The two downward forces total 40 + 60 N.
View solution step by step

  1. List the vertical forces

    Method

    The fixed support reaction acts upward; beam weight and load act downward.

    Reason

    The beam is in translational equilibrium.

    Working

    ∑ F_y = R-40-60

  2. Apply vertical equilibrium

    Method

    R = 100 N upward.

    Reason

    Zero vertical resultant requires the support force to balance both downward loads.

    Working

    R-40-60 = 0 ⇒ R = 100 N

  3. Keep rotational equilibrium distinct

    Method

    The fixed support must also provide a reaction moment.

    Reason

    A single upward force at the left end cannot by itself balance the clockwise moments of the two loads.

    Working

    Mₛᵤₚₚₒᵣₜ = 40(1.0) + 60(2.0) = 160 N m anticlockwise.

Common misconception 3

Torque of a couple

Find and correct the mistake

Learner claim

Two equal and opposite 12 N forces form a couple with perpendicular line-of-action separation 0.30 m. A learner calculates 2Fs = 7.2 N m because there are two forces. Diagnose the method.

Try this before viewing the solution

Unit: N m

View solution step by step

  1. Interpret the separation

    Method

    s = 0.30 m is the full perpendicular distance between the forces.

    Reason

    Each force’s moment about a midpoint would use half this distance.

    Working

    s/2 = 0.15 m

  2. Add the two force moments

    Method

    The moments reinforce to 3.6 N m.

    Reason

    Using the midpoint gives two terms F(s/2), equivalent to one Fs.

    Working

    2(12)(0.15) = 12(0.30) = 3.6 N m

Examiner practice 4

Beam supported at both ends (find reactions)

5 marks

Examination question

A uniform 3.0 m beam of weight 30 N is supported at both ends. A 60 N load hangs 2.0 m from the left end. Find the upward reactions R_A and R_B. [5 marks]

Try this before viewing the solution

Unit: N
Unit: N

View solution step by step

  1. Write vertical force equilibrium

    1 mark

    Method

    R_A + R_B = 90 N.

    Reason

    The reactions balance the beam and added-load weights.

    Working

    R_A + R_B = 30 + 60

  2. Take moments about the left support

    2 marks

    Method

    3.0R_B = 165 N m.

    Reason

    The left reaction has zero lever arm; the uniform beam’s weight acts at 1.5 m.

    Working

    R_B(3.0) = 30(1.5) + 60(2.0) = 165

  3. Calculate the right reaction

    1 mark

    Method

    R_B = 55 N.

    Reason

    Divide the balancing moment by the 3.0 m support separation.

    Working

    R_B = 165/3.0 = 55 N

  4. Calculate the left reaction

    1 mark

    Method

    R_A = 35 N.

    Reason

    The two reactions must sum to 90 N.

    Working

    R_A = 90-55 = 35 N

Challenge 5

Torque when the force is at an angle

Minimal support

Independent transfer

A 40 N force is applied 0.25 m from a bolt at 30° to the spanner. Find the torque magnitude about the bolt.

Try this before viewing the solution

Unit: N m

Hints

Hint 1: extract the perpendicular contribution
Only F sin 30° is perpendicular to the spanner.
View solution step by step

  1. Find the perpendicular factor

    Method

    The moment uses rF sin θ.

    Reason

    θ is the angle between the position vector along the spanner and the force.

    Working

    d_⊥ = r sin θ

  2. Calculate torque

    Method

    τ = 5.0 N m.

    Reason

    The parallel component of force produces no moment about the bolt.

    Working

    τ = (0.25)(40) sin 30° = 5.0 N m

7. Mind Stretchers

Mind stretcher 1: “Net force zero” does not guarantee equilibriumExtension

Give an example where the resultant force is zero but the object still rotates. Explain using moments.

Click here to show/hide answer

A couple: two equal and opposite forces separated by a distance.

Resultant force is zero, but there is a non-zero turning effect (τ = Fs), so it rotates unless an opposing moment balances it.

Mind stretcher 2: Pivot choice checkExtension

In a beam problem, explain why you can take moments about a point where an unknown reaction acts, and why that helps.

Click here to show/hide answer

If you take moments about the point where the unknown reaction acts, its moment is zero because d_⊥ = 0.

This removes that unknown from the moment equation, making the algebra simpler.

Mind stretcher 3: Optional (Enrichment)Extension

A. Torque as a vector (right-hand rule)

At a more advanced level, torque is treated as a vector vector τ = vector r × vector F (direction by right-hand rule).

For most A Level exam questions, clockwise/anticlockwise moments about a pivot are sufficient.

B. Three-force equilibrium (concurrency idea)

For a rigid body in equilibrium under three non-parallel forces, the lines of action are concurrent (they meet at one point). This is useful but not always required.

O Level refresher on moments: Moment Of A Force.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027