Equilibrium of a rigid body

Two conditions

A rigid body stays at rest (or moves steadily without changing its spin) only if both of these hold:

The first stops the centre of mass accelerating. The second stops the turning from changing. Neither is enough alone: two equal and opposite forces that are not along one line, a couple, give zero net force, but they still turn a body.

Choosing the pivot

The torque condition holds about every point. So choose a point where an unknown force acts: that force has no torque about it and drops out.

Slide a load along a plank and watch the supports share it.

You should find the two pushes always adding up to the total weight, and the support nearer the load always carrying more.

A ladder

A ladder against a smooth wall has three forces from outside besides its weight: the wall's push (horizontal, since the wall is smooth), and the ground's push and friction at its foot.

Take torques about the foot to find the wall's push; then friction must balance it.

Watch out. The weight of a uniform body acts at its middle, not at an end. Its moment arm is measured to that point.

A worked example

A $12\ \text{kg}$ uniform board $2.4\ \text{m}$ long rests on two trestles at its ends. A $30\ \text{kg}$ tin of paint sits $0.60\ \text{m}$ from the left end.

1. Torques about the left end. $R_B \times 2.4\ \text{m} = 117.6\ \text{N} \times 1.2\ \text{m} + 294\ \text{N} \times 0.60\ \text{m}$, so $R_B = \dfrac{141.1\ \text{N\,m} + 176.4\ \text{N\,m}}{2.4\ \text{m}} = 132.3\ \text{N}$.
2. Forces. $R_A = 117.6\ \text{N} + 294\ \text{N} - 132.3\ \text{N} = 279.3\ \text{N}$.
3. Check about the right end. $279.3\ \text{N} \times 2.4\ \text{m} = 117.6\ \text{N} \times 1.2\ \text{m} + 294\ \text{N} \times 1.8\ \text{m}$: $670\ \text{N\,m} = 670\ \text{N\,m}$.

Think first. Draw every force, mark where each acts, then pick the pivot that removes the most unknowns.

Practise with new numbers

Try it in Python

The program moves the tin of paint along the board and finds how the trestles share the load.

Remember these

• Equilibrium: $\sum \vec{F} = 0$ and $\sum \vec{\tau} = 0$ about any point.
• Take torques about a point where an unknown force acts.
• A couple: zero net force, torque = force × distance between the lines.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A $30\ \text{kg}$ child sits $2.0\ \text{m}$ from a seesaw's pivot. Where must a $40\ \text{kg}$ child sit to balance?

2. Two forces of $10\ \text{N}$, equal and opposite, act on a ruler at points $0.30\ \text{m}$ apart, at right angles to it. What is their net force? Their torque?

3. Why is a ladder more likely to slip when it leans at a shallow angle?

4. A uniform $2.0\ \text{m}$ shelf of $6.0\ \text{kg}$ rests on brackets at its two ends. What does each bracket carry? ($g = 9.8\ \text{m/s}^2$.)