Moments and Equilibrium

Apply the principle of moments at A-Level. Solve equilibrium problems with beams, pivots, and multiple forces.

A moment is the turning effect of a force about a point. The principle of moments is used to solve problems involving beams, seesaws, and structures in equilibrium.

Core Concepts

Moment of a Force

Moment=F×d\text{Moment} = F \times d

where FF is the force and dd is the perpendicular distance from the pivot to the line of action of the force.

Units: Newton-metres (Nm).

Principle of Moments

For a body in equilibrium:

clockwise moments=anticlockwise moments\sum \text{clockwise moments} = \sum \text{anticlockwise moments}

Conditions for Equilibrium

  1. Resultant force = 0: F=0\sum F = 0 (translational equilibrium).
  2. Resultant moment = 0: M=0\sum M = 0 about any point (rotational equilibrium).

Uniform Beams

The weight of a uniform beam acts at its centre of mass (midpoint).

Tilting and Toppling

A beam is about to tilt when the reaction at one support becomes zero.

Worked Example: Simple Beam

Problem

A 4 m uniform beam (weight 20 N) rests on supports at A (left end) and B (right end). A 30 N load is placed 1 m from A. Find the reaction forces.

Taking moments about A: RB×4=30×1+20×2R_B \times 4 = 30 \times 1 + 20 \times 2 4RB=704R_B = 70RB=17.5R_B = 17.5 N.

Resolving vertically: RA+RB=50R_A + R_B = 50RA=32.5R_A = 32.5 N.

Solution

Worked Example: Non-Uniform Rod

Problem

A rod AB of length 5 m and weight 40 N has its centre of mass at 2 m from A. It's supported at A and a point C (3 m from A). A 60 N weight hangs from B.

Moments about A: RC×3=40×2+60×5R_C \times 3 = 40 \times 2 + 60 \times 5 3RC=3803R_C = 380RC=126.6R_C = 126.\overline{6} N.

RA=100126.6=26.6R_A = 100 - 126.\overline{6} = -26.\overline{6} N (acting downward — the support must pull down).

Solution

Worked Example: About to Tilt

Problem

A uniform plank (6 m, 80 N) overhangs a table. How far can it overhang before tipping?

When about to tilt, the reaction at the table edge supports all the weight. The pivot is the table edge.

Moments about the edge: weight acts at the midpoint of the plank. If overhang = xx, the centre of mass is at (3x)(3 - x) from the edge on the table side (or (x3)(x - 3) beyond, depending on setup).

The plank tips when the centre of mass reaches the edge: x=3x = 3 m.

Solution

Strategy

  1. Draw a clear diagram with all forces.
  2. Choose a pivot point (often where an unknown force acts — this eliminates it).
  3. Take moments about that point.
  4. Resolve vertically for remaining unknowns.

Common Mistakes

  • Forgetting the beam's own weight. It acts at the centre of mass.
  • Using incorrect perpendicular distances.
  • Inconsistent signs. Choose clockwise positive (or anticlockwise) and stick to it.
  • Not resolving when needed. If forces are at angles, resolve them before taking moments.

Practice Problems

  1. Problem 1

    A 3 m uniform beam (weight 60 N) has supports at each end. A 90 N weight is placed 1 m from the left end. Find the reactions at each support.

    Problem 2

    A non-uniform rod of length 4 m and weight 50 N has its centre of mass 1.5 m from end A. It's held horizontally by a string at B. Find the tension in the string and the reaction at A.

    Problem 3

    A uniform plank of mass 20 kg and length 8 m rests on two supports at 1 m and 6 m from one end. What is the maximum mass that can be placed at one end without the plank tipping?

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Key Takeaways

  • Moment = Force × perpendicular distance.

  • Equilibrium: F=0\sum F = 0 and M=0\sum M = 0.

  • Take moments about a point where an unknown force acts to eliminate it.

  • Uniform beam weight acts at the midpoint.

  • A beam is about to tilt when one reaction force becomes zero.

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