Describe the moment of a force as a measure of its turning effect and give everyday examples

1.5.2 Turning Effect of Forces – Moment (Torque)

Quick‑scan of the Cambridge IGCSE Physics (0625) syllabus

What is a Moment?

• A moment (or torque) measures a force’s ability to cause rotation about a pivot (fulcrum).
• It depends on:
  • Force magnitude (F) – newtons (N)
  • Perpendicular distance from the pivot to the line of action (lever arm, d) – metres (m)
  • Angle (θ) between the force direction and the lever arm. Only the component of the force acting ⟂ to the lever arm contributes.

τ = F d sin θ

When the force is applied at right angles (θ = 90°), sin θ = 1 and the formula reduces to τ = F d.

Units

SI unit: newton‑metre (N·m). Although dimensionally equivalent to the joule, N·m is used for rotational effects, while J is used for energy.

Sign convention

Any consistent convention may be used; apply it throughout a problem.

Principle of Moments (Rotational Equilibrium)

An object will not rotate when the algebraic sum of all moments about any pivot is zero:

∑τ = 0 ⟹ ∑τ_clockwise = ∑τ_{anticlockwise}

Everyday Examples

1. Opening a door – The handle is far from the hinges, giving a large lever arm and a large moment for a modest push.
2. Using a wrench – Doubling the length of the wrench halves the required force for the same torque.
3. Seesaw (teeter‑totter) – Balance is achieved when the clockwise and anticlockwise moments are equal.
4. Steering wheel – Hands on the rim apply force at a large radius, producing a sizeable turning moment.
5. Crowbar – A longer bar provides a greater lever arm, making it easier to lift or pry heavy objects.
6. Cork‑screw – The spiral acts as a lever; a longer handle reduces the force needed to twist the cork.

Worked Example – Seesaw Balance

Problem: A seesaw is pivoted at its centre. Child A (weight = 300 N) sits 1.2 m to the left of the pivot. Child B (weight = 200 N) sits on the right. How far from the pivot must Child B sit to keep the seesaw level?

1. Choose a sign convention (clockwise = –, anticlockwise = +).
2. Write the equilibrium condition: ∑τ = 0.
3. Calculate each moment:
   • τ_A = –(300 N)(1.2 m) = –360 N·m
   • τ_B = +(200 N)(d) (where d is the unknown distance)
4. Set the sum to zero and solve:

   –360 N·m + 200 N·d = 0 ⇒ d = 360 / 200 = 1.8 m

Child B must sit 1.8 m to the right of the pivot.

Sample Calculations

Practical Activity – Verifying τ = F d

Objective: Demonstrate that the moment produced by a force applied at right angles equals the product of the force and its perpendicular distance.

1. Set up a rigid board (≈1 m) on a low fulcrum (e.g., a wooden block).
2. Attach a spring scale at one end; this will apply a known force F.
3. Measure the distance d from the fulcrum to the point where the scale is attached.
4. Place a known weight W on the opposite side of the fulcrum. Adjust its distance d′ until the board stays horizontal (rotational equilibrium).
5. Using the principle of moments, calculate the experimental torque: τ_exp = W d′.
6. Compare τ_exp with the calculated torque τ = F d. Discuss sources of error (friction at the fulcrum, non‑perpendicular force, scale calibration).

Safety: Secure the board, keep fingers away from the moving scale, and avoid over‑loading the fulcrum.

Quick‑Check Questions

1. If you double the lever arm d while keeping the force unchanged, what happens to the moment?
2. A 10 N force is applied at the end of a 0.30 m wrench at an angle of 60°. Calculate the moment.
3. Explain why a longer crowbar makes it easier to lift a heavy object.
4. In the seesaw example above, if Child A moves 0.30 m farther from the pivot, how must Child B’s distance change to keep balance?

Model Answers

1. The moment doubles because τ = F d (τ ∝ d).
2. τ = 10 N × 0.30 m × sin 60° = 10 × 0.30 × 0.866 ≈ 2.6 N·m.
3. A longer crowbar increases the lever arm d, so for the same applied force the torque τ = F d is larger, reducing the effort needed.
4. Original balance: 300 N × 1.2 m = 200 N × d → d = 1.8 m.
   New left‑hand moment: 300 N × (1.2 + 0.3) = 300 × 1.5 = 450 N·m.
   To balance: 200 N × d′ = 450 N·m → d′ = 2.25 m.
   Child B must move 0.45 m farther from the pivot.

Summary of Key Points

• Moment (torque) quantifies a force’s turning effect: τ = F d sin θ.
• Both the magnitude of the force and its perpendicular distance from the pivot are crucial; the angle determines the effective component.
• Assign opposite signs to clockwise and anticlockwise moments; rotational equilibrium requires ∑τ = 0.
• Everyday tools (doors, wrenches, seesaws, steering wheels, crowbars, cork‑screws) illustrate how increasing the lever arm reduces the required effort.
• Practical investigations reinforce the concept and develop essential AO3 skills for the IGCSE exam.

Note to Teachers & Learners

These notes cover sub‑topic 1.5.2 in detail. They should be used alongside the full IGCSE Physics syllabus, which also includes motion, thermal physics, waves, electricity & magnetism, nuclear physics, space physics, and practical skills. Cross‑reference the relevant sections for a complete revision programme.