Summary of Forces in Equilibrium and Lami's Theorem

Forces in Equilibrium & Lami's Theorem: A Student Guide

SummaryKnowledge testFlashcardsPodcastMindmap

Introduction

Lami's theorem provides a powerful and elegant relation between three non-collinear forces in equilibrium acting at a single point. It reduces a vector equilibrium statement into simple scalar relations involving magnitudes and the angles between the forces. This material explains the theorem, proves it, works through an example, and gives practice and applications.

Definition: Lami's theorem states that if three non-zero forces $\mathbf{A}$, $\mathbf{B}$ and $\mathbf{C}$ acting at a point are in equilibrium and none of the angles between them equals $0^\circ$ or $180^\circ$, then $$|\mathbf{A}|\sin\alpha = |\mathbf{B}|\sin\beta = |\mathbf{C}|\sin\gamma$$ where $\alpha$, $\beta$, $\gamma$ are the angles opposite to the forces $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$ respectively.

Background: Force equilibrium at a point

  • Equilibrium at a point means the vector sum of all forces acting at that point equals zero. For three forces this is $$\mathbf{A} + \mathbf{B} + \mathbf{C} = \mathbf{0}.$$
  • If three non-zero forces are in equilibrium they must be concurrent (act through the same point) and can be represented by the sides of a closed triangle when placed head-to-tail.

Proof of Lami's theorem (stepwise)

  1. Start with the equilibrium vector equation: $$\mathbf{A} + \mathbf{B} + \mathbf{C} = \mathbf{0}.$$
  2. Take the dot product of the equation with each force in turn. With $\mathbf{A}$: $$(\mathbf{A}+\mathbf{B}+\mathbf{C})\cdot\mathbf{A} = 0.$$ This gives $$|\mathbf{A}|^2 + |\mathbf{B}||\mathbf{A}|\cos\gamma + |\mathbf{C}||\mathbf{A}|\cos\beta = 0$$ Dividing by $|\mathbf{A}|$ (non-zero) yields $$|\mathbf{A}| + |\mathbf{B}|\cos\gamma + |\mathbf{C}|\cos\beta = 0.$$
  3. Analogously with $\mathbf{B}$ and $\mathbf{C}$ we obtain the system $$|\mathbf{A}| + |\mathbf{B}|\cos\gamma + |\mathbf{C}|\cos\beta = 0$$ $$|\mathbf{A}|\cos\gamma + |\mathbf{B}| + |\mathbf{C}|\cos\alpha = 0$$ $$|\mathbf{A}|\cos\beta + |\mathbf{B}|\cos\alpha + |\mathbf{C}| = 0.$$
  4. When none of $\alpha,\beta,\gamma$ equals $0^\circ$ or $180^\circ$ we can manipulate trigonometric relations (use sine and cosine addition identities or geometry of the force triangle) to obtain $$|\mathbf{A}|\sin\alpha = |\mathbf{B}|\sin\beta = |\mathbf{C}|\sin\gamma.$$

Note: If any angle equals $0^\circ$ or $180^\circ$ then the forces are collinear and Lami's theorem in this form does not apply; you must treat the problem as collinear equilibrium (scalar sum along a line).

Understanding the geometry

  • Represent forces $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$ as vectors from the common point O. Placing them head-to-tail forms a closed triangle. The side lengths of that triangle are proportional to the magnitudes of the forces, and the opposite angles in that triangle correspond to the angles between the other two forces.

Example: Determining a force magnitude

Problem: Forces $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$ act at point O and are in equilibrium. Given $|\mathbf{A}|=15\mathrm{~N}$, $|\mathbf{B}|=20\mathrm{~N}$ and the angle between $\mathbf{B}$ and $\mathbf{C}$ equals $90^\circ$, find $|\mathbf{C}|$.

Solution steps:

  1. By Lami's theorem $$15\sin\alpha = 20\sin\beta = |\mathbf{C}|\sin(90^\circ).$$ Since $\sin(90^\circ)=1$ the last equality is $$|\mathbf{C}| = 15\sin\alpha.$$
  2. Angles about the point sum to $360^\circ$, so $$\alpha+\beta+90^\circ=360^\circ \implies \alpha+\beta=270^\circ.$$ Thus $\beta=270^\circ-\alpha$.
  3. Use the equality between the first two terms: $$15\sin\alpha = 20\sin\beta = 20\sin(270^\circ-\alpha).$$
  4. Use the identity $\sin(270^\circ-\alpha)=\sin(270^\circ)\cos\alpha - \cos(270^\circ)\sin\alpha$ and values $\sin(270^\circ)=-1$, $\cos(270^\circ)=0$ to obtain $$15\sin\alpha = 20(-1)\cos\alpha - 20(0)\sin\alpha = -20\cos\alpha.$$ Rearranged: $$\tan\alpha = -\frac{15}{20} = -\frac{3}{4}.$$
Sign up for the full summary
FlashcardsKnowledge testSummaryPodcastMindmap
Start for free

Already have an account? Sign in

Lami's Theorem Overview

Klíčové pojmy: Three-force equilibrium: $\mathbf{A}+\mathbf{B}+\mathbf{C}=\mathbf{0}$, Lami's theorem: $\|\mathbf{A}\|\sin\alpha=\|\mathbf{B}\|\sin\beta=\|\mathbf{C}\|\sin\gamma$, Apply when no angle equals $0^\circ$ or $180^\circ$, Derive using dot products or force-triangle geometry, If collinear forces occur use scalar sum $\sum F=0$, Example: given $15\mathrm{~N}$ and $20\mathrm{~N}$ with right angle gives $\|\mathbf{C}\|=25\mathrm{~N}$, Angles around a point sum to $360^\circ$ when relating $\alpha,\beta,\gamma$, Use law of sines on the force triangle as alternative method

## Introduction Lami's theorem provides a powerful and elegant relation between three non-collinear forces in equilibrium acting at a single point. It reduces a vector equilibrium statement into simple scalar relations involving magnitudes and the angles between the forces. This material explains the theorem, proves it, works through an example, and gives practice and applications. > **Definition:** Lami's theorem states that if three non-zero forces $\mathbf{A}$, $\mathbf{B}$ and $\mathbf{C}$ acting at a point are in equilibrium and none of the angles between them equals $0^\circ$ or $180^\circ$, then > $$\|\mathbf{A}\|\sin\alpha = \|\mathbf{B}\|\sin\beta = \|\mathbf{C}\|\sin\gamma$$ > where $\alpha$, $\beta$, $\gamma$ are the angles opposite to the forces $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$ respectively. ## Background: Force equilibrium at a point - Equilibrium at a point means the vector sum of all forces acting at that point equals zero. For three forces this is $$\mathbf{A} + \mathbf{B} + \mathbf{C} = \mathbf{0}.$$ - If three non-zero forces are in equilibrium they must be concurrent (act through the same point) and can be represented by the sides of a closed triangle when placed head-to-tail. ## Proof of Lami's theorem (stepwise) 1. Start with the equilibrium vector equation: $$\mathbf{A} + \mathbf{B} + \mathbf{C} = \mathbf{0}.$$ 2. Take the dot product of the equation with each force in turn. With $\mathbf{A}$: $$(\mathbf{A}+\mathbf{B}+\mathbf{C})\cdot\mathbf{A} = 0.$$ This gives $$\|\mathbf{A}\|^2 + \|\mathbf{B}\|\|\mathbf{A}\|\cos\gamma + \|\mathbf{C}\|\|\mathbf{A}\|\cos\beta = 0$$ Dividing by $\|\mathbf{A}\|$ (non-zero) yields $$\|\mathbf{A}\| + \|\mathbf{B}\|\cos\gamma + \|\mathbf{C}\|\cos\beta = 0.$$ 3. Analogously with $\mathbf{B}$ and $\mathbf{C}$ we obtain the system $$\|\mathbf{A}\| + \|\mathbf{B}\|\cos\gamma + \|\mathbf{C}\|\cos\beta = 0$$ $$\|\mathbf{A}\|\cos\gamma + \|\mathbf{B}\| + \|\mathbf{C}\|\cos\alpha = 0$$ $$\|\mathbf{A}\|\cos\beta + \|\mathbf{B}\|\cos\alpha + \|\mathbf{C}\| = 0.$$ 4. When none of $\alpha,\beta,\gamma$ equals $0^\circ$ or $180^\circ$ we can manipulate trigonometric relations (use sine and cosine addition identities or geometry of the force triangle) to obtain $$\|\mathbf{A}\|\sin\alpha = \|\mathbf{B}\|\sin\beta = \|\mathbf{C}\|\sin\gamma.$$ > **Note:** If any angle equals $0^\circ$ or $180^\circ$ then the forces are collinear and Lami's theorem in this form does not apply; you must treat the problem as collinear equilibrium (scalar sum along a line). ## Understanding the geometry - Represent forces $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$ as vectors from the common point O. Placing them head-to-tail forms a closed triangle. The side lengths of that triangle are proportional to the magnitudes of the forces, and the opposite angles in that triangle correspond to the angles between the other two forces. ## Example: Determining a force magnitude Problem: Forces $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$ act at point O and are in equilibrium. Given $\|\mathbf{A}\|=15\mathrm{~N}$, $\|\mathbf{B}\|=20\mathrm{~N}$ and the angle between $\mathbf{B}$ and $\mathbf{C}$ equals $90^\circ$, find $\|\mathbf{C}\|$. Solution steps: 1. By Lami's theorem $$15\sin\alpha = 20\sin\beta = \|\mathbf{C}\|\sin(90^\circ).$$ Since $\sin(90^\circ)=1$ the last equality is $$\|\mathbf{C}\| = 15\sin\alpha.$$ 2. Angles about the point sum to $360^\circ$, so $$\alpha+\beta+90^\circ=360^\circ \implies \alpha+\beta=270^\circ.$$ Thus $\beta=270^\circ-\alpha$. 3. Use the equality between the first two terms: $$15\sin\alpha = 20\sin\beta = 20\sin(270^\circ-\alpha).$$ 4. Use the identity $\sin(270^\circ-\alpha)=\sin(270^\circ)\cos\alpha - \cos(270^\circ)\sin\alpha$ and values $\sin(270^\circ)=-1$, $\cos(270^\circ)=0$ to obtain $$15\sin\alpha = 20(-1)\cos\alpha - 20(0)\sin\alpha = -20\cos\alpha.$$ Rearranged: $$\tan\alpha = -\frac{15}{20} = -\frac{3}{4}.$$ 5.