Define the moment of a force as moment = force × perpendicular distance from the pivot; recall and use this equation

1.5.2 Turning Effect of Forces

Learning Objectives

  • Define the moment (torque) of a force and write the equation M = F × d⊥.

  • State the *principle of moments* in the exact wording used in the Cambridge IGCSE 0625 syllabus.

  • State the *equilibrium condition* in the exact wording used in the syllabus.

  • Apply the principle of moments to a two‑force (balancing) situation.

  • Optional: Define “limit of proportionality”.

Key Definitions

  • Moment (torque) – the turning effect of a force about a pivot (axis of rotation). It is a vector quantity whose magnitude equals the product of the force magnitude and the perpendicular distance from the pivot to the line of action of the force.

  • Perpendicular distance (d⊥) – the shortest distance from the pivot to the line of action of the force; measured at right angles to the force.

  • Clockwise / anticlockwise moments – opposite directions are assigned opposite signs (commonly clockwise = negative, anticlockwise = positive).

Formula

M = F × d⊥

Units: force in newtons (N), distance in metres (m) → moment in newton‑metres (N·m).

Principle of Moments (exact syllabus wording)

For a body in rotational equilibrium, the sum of the clockwise moments equals the sum of the anticlockwise moments:

Σ Mclockwise = Σ Manticlockwise

Equivalently, the net moment about the pivot is zero.

Equilibrium Condition (exact syllabus wording)

An object is in equilibrium when both the resultant force and the resultant moment are zero.

Optional Supplement – Limit of Proportionality

Limit of proportionality – the point on a load‑extension (or stress‑strain) graph where the straight‑line (Hooke’s‑law) behaviour ceases. It is identified as the first noticeable deviation from a linear relationship between load and extension.

Application Example – Balancing a Beam (Two‑Force Situation)

Problem: A uniform beam is supported at its centre (pivot). A 50 N load acts 0.6 m to the left of the pivot and a 30 N load acts 1.2 m to the right of the pivot. Is the beam in rotational equilibrium?

  1. Choose a sign convention (anticlockwise = positive).

  2. Calculate each moment:

    • Left‑hand (clockwise) moment: Mcw = –(50 N × 0.6 m) = –30 N·m

    • Right‑hand (anticlockwise) moment: Macw = +(30 N × 1.2 m) = +36 N·m

  3. Sum the moments: ΣM = –30 + 36 = +6 N·m

  4. Because ΣM ≠ 0, the net moment is anticlockwise; the beam will rotate anticlockwise. The beam is not in rotational equilibrium.

Worked Example – Single Force on a Wrench

Problem: A force of 30 N is applied to a wrench at a distance of 0.25 m from the bolt (pivot). Calculate the moment and comment on equilibrium.

  1. Identify the quantities: F = 30 N, d⊥ = 0.25 m

  2. Apply the formula: M = F × d⊥ = 30 N × 0.25 m = 7.5 N·m (anticlockwise, therefore positive)

  3. Only one moment acts, so ΣM = +7.5 N·m ≠ 0. The bolt will rotate anticlockwise; the system is not in equilibrium.

Common Mistakes

  • Using the slant (hypotenuse) distance instead of the perpendicular distance.

  • Forgetting to assign a sign to the moment (direction of rotation).

  • Mixing units (e.g., centimetres with newtons) without converting to metres.

  • Assuming that the presence of a single moment automatically means equilibrium.

Practice Questions

  1. Door hinge: A door 0.9 m wide is pushed perpendicularly at the outer edge with a force of 15 N. What is the moment about the hinge?

    Solution: M = 15 N × 0.9 m = 13.5 N·m (anticlockwise).

  2. Wrench at an angle: A 0.20 m wrench is pulled with a 40 N force that makes an angle of 30° with the wrench. Find the effective perpendicular distance and the resulting moment.

    Solution: d⊥ = 0.20 m × sin 30° = 0.10 m;

    M = 40 N × 0.10 m = 4.0 N·m (anticlockwise).

  3. Seesaw equilibrium: Two forces act on a seesaw: 50 N clockwise at 0.6 m from the pivot and 30 N anticlockwise at 1.2 m from the pivot. Determine whether the seesaw is in rotational equilibrium.

    Solution: Clockwise moment = –50 × 0.6 = –30 N·m;

    Anticlockwise moment = +30 × 1.2 = +36 N·m;

    ΣM = –30 + 36 = +6 N·m ≠ 0 → not in equilibrium (net anticlockwise).

Variable Table

SymbolQuantityUnitTypical example
FForcenewton (N)30 N
d⊥Perpendicular distance from pivotmetre (m)0.25 m
MMoment (torque)newton‑metre (N·m)7.5 N·m

Suggested Diagram

Draw a simple lever: a horizontal bar with a pivot at its centre. Show a force vector applied at a point on the bar, the line of action of the force, and a right‑angled line from the pivot to this line of action to indicate the perpendicular distance d⊥. Use arrows to label clockwise and anticlockwise moments.

Summary

The turning effect of a force is quantified by its moment, calculated as M = F × d⊥. In rotational equilibrium the sum of clockwise moments equals the sum of anticlockwise moments, and both the resultant force and resultant moment must be zero. Remember to use the perpendicular distance, keep units consistent, and assign the correct sign to each moment. The optional concept of “limit of proportionality” marks where Hooke’s law ceases to apply on a load‑extension graph.