Equilibrium of Forces – Lami’s Theorem & Applications

Introduction

In physics and engineering, the concept of equilibrium of forces is fundamental. Whether it is a hanging signboard, a ladder leaning against a wall, or a bridge structure — all remain stable only because the forces acting on them are in equilibrium.

One of the most useful tools for analyzing such equilibrium problems, especially when three forces act on a body, is Lami’s Theorem. This theorem provides a direct relationship between the magnitudes of three concurrent forces and the angles between them.

In this article, we will:

Understand equilibrium of forces.
Learn and prove Lami’s theorem.
Explore conditions of equilibrium.
Apply the theorem to practical problems.
Solve numerical examples.
See real-life applications in engineering and daily life.

Part 1: Equilibrium of Forces

1.1 What is Equilibrium?

A body is said to be in equilibrium if it remains in a state of rest or moves with uniform velocity (i.e., no acceleration).

According to Newton’s First Law of Motion, equilibrium occurs when the net external force on the body is zero. ΣF⃗=0\Sigma \vec{F} = 0ΣF=0

This means:

The vector sum of all forces is zero.
There is no unbalanced force to change the state of the body.

1.2 Types of Equilibrium

Static Equilibrium – Body remains at rest.
- Example: A book on a table, a hanging signboard.
Dynamic Equilibrium – Body moves with constant velocity (no acceleration).
- Example: A satellite moving in circular orbit at constant speed.
Stable, Unstable & Neutral Equilibrium – Based on position after a small disturbance.
- Stable: Ball in a bowl (returns to original position).
- Unstable: Ball on hilltop (moves away when disturbed).
- Neutral: Ball on flat surface (remains in new position).

Part 2: Conditions for Equilibrium

For a particle under the influence of forces:

First Condition (Force Equilibrium)

ΣFx=0,ΣFy=0\Sigma F_x = 0, \quad \Sigma F_y = 0ΣFx=0,ΣFy=0

The sum of horizontal and vertical components of forces must be zero.

Second Condition (Rotational Equilibrium)

Στ=0\Sigma \tau = 0Στ=0

The sum of moments (torques) of all forces about any point must be zero.

Part 3: Lami’s Theorem

3.1 Statement

Lami’s theorem states that if a body is in equilibrium under the action of three concurrent, non-collinear forces, then each force is proportional to the sine of the angle between the other two forces.

Mathematically: F1sin⁡α=F2sin⁡β=F3sin⁡γ\frac{F_1}{\sin\alpha} = \frac{F_2}{\sin\beta} = \frac{F_3}{\sin\gamma}sinαF1=sinβF2=sinγF3

Where:

F1,F2,F3F_1, F_2, F_3F1,F2,F3 = magnitudes of the three forces.
α,β,γ\alpha, \beta, \gammaα,β,γ = angles opposite to the respective forces.

3.2 Explanation

Consider three forces acting on a point.
The angle opposite to each force is the angle between the other two forces.
Lami’s theorem gives a direct relationship without needing to resolve forces into components.

Part 4: Proof of Lami’s Theorem

Consider a particle under three concurrent forces F1,F2,F3F_1, F_2, F_3F1,F2,F3.

Since the body is in equilibrium, by the triangle law of forces, these forces can be represented in magnitude and direction by the sides of a closed triangle.
Construct a triangle ABCABCABC such that:
- AB=F1AB = F_1AB=F1, BC=F2BC = F_2BC=F2, CA=F3CA = F_3CA=F3.
By sine rule of triangles:

F1sin⁡α=F2sin⁡β=F3sin⁡γ\frac{F_1}{\sin \alpha} = \frac{F_2}{\sin \beta} = \frac{F_3}{\sin \gamma}sinαF1=sinβF2=sinγF3

Hence, Lami’s theorem is proved.

Part 5: Applications of Lami’s Theorem

Lami’s theorem is widely used in:

Engineering Mechanics
- Solving problems involving pulleys, beams, trusses.
- Analyzing crane structures.
Physics Experiments
- Determining unknown forces in force table experiments.
Daily Life
- Tension in strings supporting hanging objects.
- Balancing forces on suspended bodies.

Part 6: Solved Examples

Example 1: Tension in Two Strings

A weight of 100 N is suspended by two strings making angles 60° and 45° with the ceiling. Find the tensions in both strings using Lami’s theorem.

Solution:
Forces:

F1=T1F_1 = T_1F1=T1 (tension in string 1)
F2=T2F_2 = T_2F2=T2 (tension in string 2)
F3=W=100 NF_3 = W = 100 \, NF3=W=100N

Angles:

Opposite T1T_1T1 → angle between T2T_2T2 and WWW = 180° – (60° + 90°) = 30°
Opposite T2T_2T2 → 45°
Opposite WWW → 75°

By Lami’s theorem: T1sin⁡45°=T2sin⁡60°=100sin⁡75°\frac{T_1}{\sin 45°} = \frac{T_2}{\sin 60°} = \frac{100}{\sin 75°}sin45°T1=sin60°T2=sin75°100

From first ratio: T1=100sin⁡45°sin⁡75°=100×0.7070.966≈73.2NT_1 = \frac{100 \sin 45°}{\sin 75°} = \frac{100 \times 0.707}{0.966} \approx 73.2 NT1=sin75°100sin45°=0.966100×0.707≈73.2N

From second ratio: T2=100sin⁡60°sin⁡75°=100×0.8660.966≈89.6NT_2 = \frac{100 \sin 60°}{\sin 75°} = \frac{100 \times 0.866}{0.966} \approx 89.6 NT2=sin75°100sin60°=0.966100×0.866≈89.6N

👉 Tensions are 73.2 N and 89.6 N.

Example 2: Pulling a Block with Two Forces

A block is pulled by two forces F1=40NF_1 = 40 NF1=40N and F2=50NF_2 = 50 NF2=50N at angles such that they keep the block in equilibrium with friction. If the angle between them is 120°, find the third force (friction).

By Lami’s theorem: F1sin⁡120°=F2sin⁡γ=F3sin⁡β\frac{F_1}{\sin 120°} = \frac{F_2}{\sin \gamma} = \frac{F_3}{\sin \beta}sin120°F1=sinγF2=sinβF3

(Solution proceeds similarly; friction is found as the third force ensuring balance.)

Part 7: Advantages of Lami’s Theorem

Simple and quick for three-force problems.
Eliminates need for resolving into x- and y-components.
Directly applies to string and pulley systems.

Part 8: Limitations of Lami’s Theorem

Applicable only when exactly three forces act on a particle.
Forces must be concurrent (meet at a point).
Forces must be coplanar (lie in same plane).
Not suitable for systems with more than three forces.

Part 9: Real-Life Applications

Traffic Lights and Signboards – Hung by two cables, tensions can be calculated.
Ladders Resting Against Walls – Equilibrium forces analyzed using theorem.
Cranes and Hoists – Force in ropes found for design safety.
Physics Labs – Force table experiment demonstrates theorem’s accuracy.
Bridges and Trusses – Ensures stability by balancing forces.