Equilibrium of forces

Two conditions for rigid body equilibrium:

1. Translational. $\sum F = 0$ (resolved in perpendicular directions: $\sum F_x = 0$ and $\sum F_y = 0$).
2. Rotational. $\sum M = 0$ about ANY point.

For a particle, only condition 1 applies (no rotation possible).

Method.

1. Sketch free-body diagram.
2. Resolve forces into perpendicular axes.
3. Apply $\sum F_x = 0$, $\sum F_y = 0$.
4. Take moments about chosen pivot, $\sum M = 0$.
5. Solve system.

Cambridge tip. Choose pivot through an unknown force to eliminate it from moment equation — simplifies algebra.

Three concurrent forces in equilibrium form a CLOSED TRIANGLE when drawn head-to-tail.

Approach 1: components. Resolve into perpendicular directions; apply $\sum F = 0$ in each.

Approach 2: triangle of forces. Draw triangle scale-and-direction; apply trigonometry / sine rule.

Lami's theorem. For three concurrent forces in equilibrium: $\frac{F_1}{\sin\alpha} = \frac{F_2}{\sin\beta} = \frac{F_3}{\sin\gamma}$ where $\alpha$ is opposite $F_1$, etc.

Example. Object 20 N hanging from two strings at $30°$ and $45°$ to horizontal.

• Resolve: $T_1\cos 30° = T_2\cos 45°$; $T_1\sin 30° + T_2\sin 45° = 20$.
• Solve: $T_1 \approx 14.6$ N; $T_2 \approx 17.9$ N.

Cambridge tip. Both methods OK; pick whichever gives cleaner algebra.

Stable equilibrium. Body returns to original position after small displacement.

• Centre of mass at MINIMUM height locally.
• Example: ball at bottom of bowl.

Unstable equilibrium. Body moves further from equilibrium.

• Centre of mass at MAXIMUM height locally.
• Example: ball balanced on dome.

Neutral equilibrium. Body remains in displaced position.

• Centre of mass at CONSTANT height.
• Example: ball on horizontal surface.

Toppling. A body topples when its vertical line through centre of mass falls OUTSIDE its base of support.

Cambridge tip. Sketch + label centre of mass for top marks.