Apply the conditions for translational and rotational equilibrium, using the principle of moments and the resolution of forces, to extended rigid bodies

What this dot point is asking

SEAB wants you to apply the two conditions for the equilibrium of an extended rigid body: the resultant force is zero and the resultant moment is zero. This requires the principle of moments, the resolution of forces, and an understanding of moments and couples. Beam and ladder problems are standard structured-question material.

The answer

Common types of force

In mechanics problems you meet weight (gravity acting at the centre of gravity), the normal contact force (perpendicular to a surface), friction (along a surface, opposing relative motion), tension (along a string or rod) and the upthrust on a body in a fluid. Identifying every force is the first step in any equilibrium problem.

The moment of a force

The moment (or torque) of a force about a point measures its turning effect:

where $d$ is the perpendicular distance from the pivot to the line of action of the force. The unit is the newton metre ($\text{N m}$). Moments are taken as clockwise or anticlockwise about a chosen pivot.

Couples

A couple is a pair of equal, opposite, parallel forces whose lines of action do not coincide. A couple produces a turning effect with no resultant force. Its torque is:

where $s$ is the perpendicular separation of the two forces. A steering wheel turned with both hands is a couple.

The two conditions for equilibrium

An extended rigid body is in equilibrium when both conditions hold:

1. Translational equilibrium: the resultant force is zero in every direction, so $\sum F_x = 0$ and $\sum F_y = 0$.
2. Rotational equilibrium: the resultant moment about any point is zero, $\sum \text{(clockwise moments)} = \sum \text{(anticlockwise moments)}$.

The second condition is the principle of moments. A point particle needs only the first condition; an extended body needs both.

Strategy: choosing the pivot

You may take moments about any point. Choosing the pivot at the line of action of an unknown force eliminates that force from the moment equation, because its moment arm is zero. This is the single most useful trick for beam problems.

Examples in context

Example 1. A ladder against a wall. A ladder leaning on a smooth wall and a rough floor is held in equilibrium by its weight, the wall's normal force, the floor's normal force and friction. Resolving forces horizontally and vertically and taking moments about the base gives three equations for the unknown forces, which determine whether the ladder slips.

Example 2. A balanced seesaw. Two children on a seesaw balance when their moments about the pivot are equal: $W_1 d_1 = W_2 d_2$. A lighter child sits further from the pivot to compensate, a direct everyday demonstration of the principle of moments.

Try this

Q1. State the two conditions for the equilibrium of an extended body. [2 marks]

• Cue. The resultant force is zero in all directions, and the resultant moment about any point is zero.

Q2. A $30\ \text{N}$ force acts at the end of a spanner $0.25\ \text{m}$ long, perpendicular to it. Find the moment about the nut. [1 mark]

• Cue. Moment $= F \times d = 30 \times 0.25 = 7.5\ \text{N m}$.

Q3. A uniform bar of weight $40\ \text{N}$ and length $2.0\ \text{m}$ is pivoted at its centre. A $10\ \text{N}$ weight hangs $0.80\ \text{m}$ to the left of the pivot. Where must a $16\ \text{N}$ weight hang to balance it? [3 marks]

• Cue. Clockwise must equal anticlockwise about the pivot: $10 \times 0.80 = 16 \times d \Rightarrow d = 0.50\ \text{m}$ to the right.