Moments and Equilibrium

Moments and Equilibrium

A moment measures the turning effect of a force about a point. The study of moments is essential for analysing beams, levers, and any system where rotation (or the prevention of rotation) is important.

Definition of a Moment

The moment of a force about a point is:

Moment = Force × Perpendicular distance from the point to the line of action of the force

M = Fd where F is in newtons and d is in metres. The moment is measured in newton-metres (Nm).

The direction of the moment is either clockwise or anticlockwise about the chosen point.

Worked Example: A horizontal beam is 4 m long, supported at point A (left end). A downward force of 50 N acts at the right end. What is the moment about A?

Moment = 50 × 4 = 200 Nm clockwise

When the Force Is at an Angle

If the force acts at angle θ to the beam, only the perpendicular component creates a moment:

M = Fsinθ × d

Alternatively, find the perpendicular distance from the pivot to the line of action of the force, and multiply by the full force.

Principle of Moments

For a body in rotational equilibrium (not rotating), the sum of clockwise moments about any point equals the sum of anticlockwise moments:

Σ clockwise moments = Σ anticlockwise moments

Equivalently: the net moment about any point is zero.

Worked Example: A uniform beam of length 6 m and weight 200 N is supported at its centre. A child of weight 300 N sits 2 m from the left end. Where should a child of weight W sit to balance the beam?

Taking moments about the centre (at 3 m):

The beam's weight acts at the centre, so its moment about the centre is zero.

Clockwise: 300 × (3 - 2) = 300 × 1 = 300 Nm

Anticlockwise: W × d (where d is the distance from the centre to where W sits)

For equilibrium: W × d = 300

If W = 400 N: d = 300/400 = 0.75 m from the centre (i.e., 3.75 m from the left end).

Uniform Beams

A uniform beam has its weight evenly distributed, which means the weight acts at the geometric centre (midpoint) of the beam.

A non-uniform beam has its centre of mass at a specified point, which may not be the midpoint.

Supports, Reactions and Equilibrium

For a beam in complete equilibrium (no linear acceleration and no rotation):

1. Vertical equilibrium: Σ upward forces = Σ downward forces

2. Horizontal equilibrium: Σ forces right = Σ forces left

3. Rotational equilibrium: Σ moments about any point = 0

Use conditions 1 and 3 together (choosing the moment point wisely eliminates unknowns).

Worked Example: A uniform plank AB of length 5 m and mass 20 kg rests horizontally on supports at A and B. A box of mass 30 kg is placed 1 m from A. Find the reactions at A and B.

Weight of plank = 20g = 196 N (acts at midpoint, 2.5 m from A)

Weight of box = 30g = 294 N (acts 1 m from A)

Taking moments about A:

R_B × 5 = 196 × 2.5 + 294 × 1

5R_B = 490 + 294 = 784

R_B = 156.8 N

Vertical equilibrium: R_A + R_B = 196 + 294 = 490

R_A = 490 - 156.8 = 333.2 N

Tilting and Limiting Equilibrium

A beam begins to tilt about a support when one of the reactions becomes zero. At the point of tilting:

• One support loses contact (its reaction = 0)
• The beam rotates about the other support
• Take moments about the tilting point to find the critical condition

Worked Example: The same plank (5 m, 20 kg) rests on supports at A and at a point 4 m from A. How far from A can a 50 kg person walk before the plank tilts about the support at 4 m?

At the point of tilting, R_A = 0. Taking moments about the support at 4 m:

Anticlockwise from plank weight: 196 × (4 - 2.5) = 196 × 1.5 = 294 Nm

Clockwise from person: 490 × (d - 4)

For tilting: 490(d - 4) = 294

d - 4 = 0.6

d = 4.6 m from A

Ladder Problems

A common application: a ladder leaning against a wall. Forces acting:

• Weight of the ladder (at its midpoint if uniform)
• Normal reaction from the ground (vertical, at the base)
• Normal reaction from the wall (horizontal, at the top)
• Friction at the ground (horizontal, opposing potential sliding)
• Sometimes friction at the wall (vertical)

Set up three equations (horizontal, vertical, moments) and solve.

Exam Tips

• Choose your moment point wisely — pick a point where an unknown force acts. Its moment is zero, eliminating it from the equation.
• Always state the direction (clockwise or anticlockwise) of each moment.
• For uniform objects, the weight acts at the centre. This is a common source of error — do not place it at the end.
• In tilting problems, clearly identify which support the beam is about to tilt around, and set the other reaction to zero.
• Draw a clear diagram with all forces labelled and their distances from the pivot marked.