Equilibrium of Concurrent Forces and Lami's Theorem

When several forces act through a single point and the body neither accelerates nor rotates about that point, the body is in translational equilibrium. A particularly elegant statement covers the case of three coplanar concurrent forces — Lami's theorem.

Concept

Concurrent forces pass through a common point.

Equilibrium condition (translational):

$\sum \vec{F}_i = 0$

Equivalently, the vector polygon of the forces closes. For three forces, the triangle of forces closes.

Lami's theorem: If three coplanar concurrent forces $F_1$, $F_2$, $F_3$ keep a particle in equilibrium, then each force is proportional to the sine of the angle between the other two:

$\frac{F_1}{\sin \alpha} = \frac{F_2}{\sin \beta} = \frac{F_3}{\sin \gamma}$

where $\alpha$ is the angle between $F_2$ and $F_3$, $\beta$ between $F_1$ and $F_3$, $\gamma$ between $F_1$ and $F_2$, and $\alpha + \beta + \gamma = 360°$.

Derivation

Place the three forces tail-to-tail at a common point. Because they sum to zero, the head-to-tail arrangement forms a closed triangle (the "triangle of forces").

Let the interior angles of that triangle be $\pi - \alpha$, $\pi - \beta$, $\pi - \gamma$ (since each interior angle is the supplement of the angle between the two corresponding forces in tail-to-tail form). Then $\sin(\pi - x) = \sin x$.

By the sine rule applied to this triangle,

$\frac{F_1}{\sin(\pi - \alpha)} = \frac{F_2}{\sin(\pi - \beta)} = \frac{F_3}{\sin(\pi - \gamma)}$

i.e.

$\frac{F_1}{\sin \alpha} = \frac{F_2}{\sin \beta} = \frac{F_3}{\sin \gamma}$

This is Lami's theorem.

Worked Example

A street lamp of weight $W = 20\sqrt{3} \, \text{N}$ hangs from a horizontal strut and is tied to a wall by a rope making 60° with the strut. Find the tension in the rope and the compression in the strut.

Solution:

Three forces act at the joint: the rope tension $T$ along the rope (toward the wall), the strut compression $C$ along the strut (away from the wall, horizontal), and the weight $W$ pulling straight down.

Angles between forces (tail-to-tail at joint):

• Between $C$ (horizontal away from wall) and $W$ (vertically down): $90°$.
• Between $W$ (down) and $T$ (toward wall along rope at 60° above horizontal): $180° - 60° = 120°$.

Wait — let me set the angles cleanly. With strut horizontal and rope at 60° above horizontal toward the wall:

• Angle between rope and strut: $60°$ → opposite to $W$.
• Angle between strut and $W$: $90°$ → opposite to $T$.
• Angle between rope and $W$: $180° - 60° = 120°$ → opposite to $C$ but on the supplementary side; for Lami, the angle "between the other two" sweeping through the particle is $360° - 60° - 90° = 210°$? Cleaner to use the polygon angles.

Using tail-to-tail Lami angles with $W$ down, $T$ up-and-away-from-wall, $C$ toward-wall:

$\frac{W}{\sin 60°} = \frac{T}{\sin 90°} = \frac{C}{\sin (180° - 60°)} = \frac{C}{\sin 60°}$

Hmm — sign conventions can confuse. Easiest: resolve directly.

Horizontal: $T \cos 60° = C \Rightarrow C = T/2$.

Vertical: $T \sin 60° = W \Rightarrow T = W/\sin 60° = 20\sqrt{3}/(\sqrt{3}/2) = 40 \, \text{N}$.

So $T = 40 \, \text{N}$ and $C = 20 \, \text{N}$.

Common Confusions

• Lami's theorem applies only to three concurrent coplanar forces. Four-force problems need polygon-of-forces or component method.
• Angles in Lami are between the other two forces, not between a force and itself or the axes.
• Equilibrium $\neq$ rest. A body moving with constant velocity is also in equilibrium ($\sum \vec{F} = 0$).
• For rotational equilibrium, you additionally need $\sum \vec{\tau} = 0$, but for concurrent forces about the point of concurrence, $\sum\vec{\tau} = 0$ automatically.

Key Takeaways

• Equilibrium of a particle: $\sum \vec{F} = 0$.
• For three concurrent forces, the triangle of forces closes.
• Lami's theorem: $F_1/\sin\alpha = F_2/\sin\beta = F_3/\sin\gamma$, where each angle is between the other two forces.
• Lami follows from the sine rule applied to the closed triangle.
• Resolving into components is always a safe alternative.