Matter and forces are two inseparable twins. Even if there is no change in the state of motion of a material object, it does not mean that there is no force acting on the object. The forces acting on the object may be in **equilibrium** in such case.

No force does not mean zero force. When the forces acting on a material object mutually balance each other, then there is no effect in the state of motion of the object. This is when the zero force situation comes. Therefore, a zero force is a resultant of a number of forces acting on a material object (or particle or point), whose magnitude is zero. So, if the resultant is zero, then the forces acting on a particle or point are said to be in **equilibrium**. In this case, there is no change in the state of motion of particle i.e. at relative rest or uniform motion.

## First Condition for Equilibrium

Suppose a force $F_1$ is acted on a particle, then the state of motion of the particle is changed. But, if we apply a second force $F_2$, equal in magnitude to $F_1$ but opposite in direction, along the same line of action as $F_1$, produces an equilibrium.

The forces $F_1$ and $F_2$ are equal and opposite i.e. \[F_2=-F_1\]

If the resultant is $R$, then, \[R=F_1+F_2=F_1-F_2=0\]

Therefore, if the algebraic sum of the forces $F_1$ and $F_2$ acting on a particle along the same line is zero, then the forces are in equilibrium. This can be extended to any number of concurrent forces.

Suppose $F_1, F_2, …, F_n$ are $n$ coplanar forces acting on a particle are in equilibrium and let their resultant be $R$, then,

\[R^2=X^2+Y^2\] where, $X$ is the algebraic sum of the X-components of the forces $F_1, F_2, …, F_n$

\[X=X_1+X_2+…+X_n\] \[X=\sum X_i\]

And, $Y$ is the algebraic sum of the Y-components of the forces $F_1, F_2, …, F_n$

\[Y=Y_1+Y_2+…+Y_n\] \[Y=\sum Y_i\]

Since the forces are in equilibrium, \[R=0\] \[X^2+Y^2=0\] which implies, \[X=0 \text{ and, } Y=0\] \[\text{i.e. } \sum X_i=\sum Y_i=0\]

This is known as the **first condition for equilibrium**. In the case of non coplanar forces, the condition of equilibrium is, \[\sum X_i=\sum Y_i=\sum Z_i=0\]

*If each of a number of forces acting at a point is resolved into rectangular components, the forces will be in equilibrium if and only if the algebraic sum of the components of all the forces along each axis is zero.*

## Theorems on Equilibrium of Coplanar and Concurrent Forces

## Triangle of Forces

*If three forces, acting at a point, be represented in magnitude and direction by the sides of a triangle, taken in order, then the forces are in equilibrium.*

Suppose three forces $P$, $Q$ and $R$ are acting at a point $O$. Let the three sides $AB$, $BC$ and $CA$ of a triangle $ABC$ represent the forces $P$, $Q$ and $R$ respectively in magnitude and direction.

Now, complete the parallelogram $ABCD$. Since $AD$ is equal and parallel to $BC$, the force $Q$ is also represented in magnitude and direction by $AD$. According to parallelogram law of forces, the resultant of two forces $P$ and $Q$ represented in magnitude and direction by $AB$ and $AD$ respectively is represented in magnitude and direction by the diagonal $AC$. Here, $AC$ and $CA$ are equal in magnitude but opposite in direction, so they balance each other i.e. $R$ balances the resultant of $P$ and $Q$. Hence the three forces are in equilibrium.

**Alternative Method**

Let $\overrightarrow{P}$, $\overrightarrow{Q}$ and $\overrightarrow{R}$ be the three forces acting at $O$. Let the three forces be represented by $∆ABC$ such that,

\[\overrightarrow{AB}=\overrightarrow{P}\] \[\overrightarrow{BC}=\overrightarrow{Q}\] \[\overrightarrow{CA}=\overrightarrow{R}\]

Complete the parallelogram $ABCD$. Since $BC$ and $AD$ are equal and parallel, \[\overrightarrow{AD}=\overrightarrow{BC}=\overrightarrow{Q}\]

By parallelogram law of forces, \[\overrightarrow{AB}+\overrightarrow{AD}=\overrightarrow{AC}\] \[\overrightarrow{P}+\overrightarrow{Q}=\overrightarrow{AC}\] Now, \[\overrightarrow{P}+\overrightarrow{Q}+\overrightarrow{R}=\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}\] \[=\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{CA}\] \[=\overrightarrow{AC}+\overrightarrow{CA}\] \[=0\]

Hence, the forces are in equilibrium.

**When applying triangle of forces, the forces $P$, $Q$ and $R$ do not act along the sides of the triangle $ABC$, they act at a same point.**

If forces $\overrightarrow{AB}$ and $\overrightarrow{BC}$ act along sides, then their resultant also acts at $B$ parallel to $AC$ and is equal to $AC$ in magnitude.

\[\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=0\] \[\overrightarrow{AB}+\overrightarrow{BC}-\overrightarrow{AC}=0\] \[\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}\]

Thus, if two forces be represented by two sides of a triangle, then the third side taken in opposite order represents their resultant.

For more: Triangle Law of Vector Addition

## Converse of the Triangle of Forces

*If three forces acting at a point are in equilibrium, then they can be represented in magnitude and direction by the three sides of a triangle taken in order.*

Let the three forces $P$, $Q$ and $R$ acting at a point $O$ be in equilibrium.

Let $OA$, $OB$ and $OD$ represent the forces $P$, $Q$ and $R$ respectively in magnitude and direction. Complete the parallelogram $OACB$. Since $AC$ and $OB$ are equal and parallel, $AC$ also represents the same force $Q$ in magnitude and direction. Join $OC$. Then, by parallelogram law of forces, the diagonal $OC$ represents the resultant of the forces $P$ and $Q$.

Since the forces $P$, $Q$ and $R$ are in equilibrium, the force $R$ balances the resultant of the forces $P$ and $Q$ i.e. the force represented by $OC$ in magnitude and direction is balanced by $R$. Hence, $R$ can be represented by $CO$ in magnitude and direction. Thus, the sides $OA$, $AC$ and $CO$ of triangle $OAC$, taken in order, represent the forces $P$, $Q$ and $R$ acting at $O$ respectively.

Draw any other triangle $LMN$ such that the sides $LM$, $MN$ and $NL$ are parallel to $OA$, $AC$ and $CO$ respectively. Then, the triangles $OAC$ and $LMN$ are similar. Hence, \[\frac{OA}{LM}=\frac{AC}{MN}=\frac{CO}{NL}\] \[\frac{P}{LM}=\frac{Q}{MN}=\frac{R}{NL}\]

i.e. the sides $LM$, $MN$ and $NL$ are proportional to the forces $P$, $Q$ and $R$ respectively and hence, the sides represent the forces in magnitude and direction on some scale.

**Alternative Method**

Let $\overrightarrow{P}$, $\overrightarrow{Q}$ and $\overrightarrow{R}$ be the forces acting at $O$ be in equilibrium.

Let \[\overrightarrow{OA}=\overrightarrow{P}\]

\[\overrightarrow{OB}=\overrightarrow{Q}\]

and, \[\overrightarrow{OD}=\overrightarrow{R}\]

Complete the parallelogram $OACB$. Since $AC$ and $OB$ are equal and parallel, \[\overrightarrow{AC}=\overrightarrow{Q}\]

By parallelogram law of forces, \[\overrightarrow{P}+\overrightarrow{Q}=\overrightarrow{OA}+\overrightarrow{OB}=\overrightarrow{OC}\]

Since the forces are in equilibrium, \[\overrightarrow{P}+\overrightarrow{Q}+\overrightarrow{R}=0\] \[\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{R}=0\] \[\overrightarrow{OC}+\overrightarrow{R}=0\] \[\overrightarrow{R}=-\overrightarrow{OC}\] \[\overrightarrow{R}=\overrightarrow{CO}\]

Hence, the forces $\overrightarrow{P}$, $\overrightarrow{Q}$ and $\overrightarrow{R}$ are represented by the sides of $OA$, $AC$ and $CO$ of the triangle $OAC$ taken in order.

## Polygon of Forces

*If any number of forces acting at a point be represented in magnitude and direction by the sides of a polygon, taken in order, then the forces are in equilibrium.*

Let the forces $P_1$, $P_2$, $P_3$, $P_4$ and $P_5$ acting at a point $O$ be represented in magnitude and direction by the sides $AB$, $BC$, $CD$, $DE$ and $EA$ respectively taken in the order of the polygon $ABCDE$. Join $AC$ and $AD$.

Now, in triangle $ABC$, the resultant of the forces $AB$ and $BC$ is represented by $AC$. Similarly, in triangle $ACD$, the resultant of the forces $AC$ and $CD$ is represented by $AD$ and that of the forces $AD$ and $DE$ is represented by $AE$.

Hence, the resultant of the forces $P_1$, $P_2$, $P_3$ and $P_4$ which are represented respectively by $AB$, $BC$, $CD$ and $DE$ is represented by $AE$. Since $AE$ and $EA$ are equal and opposite, their resultant is zero. Thus, the resultant of all the forces is zero i.e. $P_1$, $P_2$, $P_3$, $P_4$ and $P_5$ are in equilibrium.

**Alternative Method**

Let $\overrightarrow{P_1}$, $\overrightarrow{P_2}$, $\overrightarrow{P_3}$, $\overrightarrow{P_4}$ and $\overrightarrow{P_5}$ be the forces acting at $O$ and are represented in magnitude and direction by the sides $AB$, $BC$, $CD$, $DE$ and $EA$ of the polygon $ABCDE$ respectively.

Now, \[\overrightarrow{P_1}+\overrightarrow{P_2}+\overrightarrow{P_3}+\overrightarrow{P_4}+\overrightarrow{P_5}\] \[=\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DE}+\overrightarrow{EA}\] \[=\overrightarrow{AC}\overrightarrow{CD}+\overrightarrow{DE}+\overrightarrow{EA}\] \[=\overrightarrow{AD}+\overrightarrow{DE}+\overrightarrow{EA}\] \[=\overrightarrow{AE}+\overrightarrow{EA}\] \[=0\] Hence, the forces are in equilibrium.

A similar method of proof can be extended whatever be the number of forces. The converse of the polygon of the forces is not true.

## Lami’s Theorem

*If a body is in equilibrium under the action of three forces, then each of the forces is proportional to the sine of the angle between the other two forces.*

Let three forces $P$, $Q$ and $R$ be acting on a body. These three forces are represented by three lines as shown in figure. Let $α$ be the angle between $Q$ and $R$, $β$ be the angle between $P$ and $R$, and $γ$ be the angle between $P$ and $Q$.

If the body is in equilibrium under the action of these forces, then these forces **can be represented as the sides of a triangle as shown in figure.**

Using Sine law, \[\frac{AB}{\sin C}=\frac{BC}{\sin A}=\frac{CA}{\sin B}\] \[\frac{\vec{P}}{\sin(180°-α)}=\frac{\vec{Q}}{\sin(180°-β)}=\frac{\vec{R}}{\sin(180°-γ)}\] \[\frac{\vec{P}}{\sinα}=\frac{\vec{Q}}{\sinβ}=\frac{\vec{R}}{\sinγ}\]

#### Al**ternative Method**

Let $\overrightarrow{P}$, $\overrightarrow{Q}$ and $\overrightarrow{R}$ be the forces acting at $O$, be in equilibrium.

Thus,

\[\overrightarrow{P}+\overrightarrow{Q}+\overrightarrow{R}=0\]

\[\implies \overrightarrow{P}×\overrightarrow{P}+\overrightarrow{P}×\overrightarrow{Q}+\overrightarrow{P}×\overrightarrow{R}=0\]

\[\implies \overrightarrow{P}×\overrightarrow{Q} = \overrightarrow{R}×\overrightarrow{P} \text{ } (\because \overrightarrow{P}×\overrightarrow{P}=0)\]

\[\implies |\overrightarrow{P}×\overrightarrow{Q}|=|\overrightarrow{R}×\overrightarrow{P}|\] \[\implies PQ\sin(180-\gamma)=RP\sin(180-\beta)\]

\[\implies Q\sin\gamma=R\sin\beta\]

\[\therefore \frac{Q}{\sin\beta}=\frac{R}{\sin\gamma} \text{ ____(1)}\]

Again,

\[\overrightarrow{P}+\overrightarrow{Q}+\overrightarrow{R}=0\]

\[\implies \overrightarrow{Q}×\overrightarrow{P}+\overrightarrow{Q}×\overrightarrow{Q}+\overrightarrow{Q}×\overrightarrow{R}=0\]

\[\implies \overrightarrow{Q}×\overrightarrow{R} = \overrightarrow{P}×\overrightarrow{Q} \text{ } (\because \overrightarrow{Q}×\overrightarrow{Q}=0)\]

\[\implies |\overrightarrow{Q}×\overrightarrow{R}|=|\overrightarrow{P}×\overrightarrow{Q}|\]

\[\implies QR\sin(180-\alpha)=PQ\sin(180-\gamma)\] \[\implies R\sin\alpha=P\sin\gamma\]

\[\therefore \frac{R}{\sin\gamma}=\frac{P}{\sin\alpha} \text{ ____(2)}\]

From $(1)$ and $(2)$, \[\frac{P}{\sin\alpha}=\frac{Q}{\sin\beta}=\frac{R}{\sin\gamma}\]

## λ-μ Theorem

*The resultant of two forces, acting at a point $O$ in direction $OA$ and $OB$ in magnitude by $λOA$ and $μOB$, is represented by $(λ+μ)OC$, where $C$ divides the line segment $AB$ in the ratio $μ:λ$.*

\[\text{i.e. } \lambda AC=\mu BC\]

From $ΔOAC$, \[\overrightarrow{OA}+\overrightarrow{AC}=\overrightarrow{OC}\] \[\lambda \overrightarrow{OA}+\lambda \overrightarrow{AC}=\lambda \overrightarrow{OC}\text{ ___(i)}\]

From $ΔOBC$, \[\overrightarrow{OB}+\overrightarrow{BC}=\overrightarrow{OC}\] \[\mu \overrightarrow{OB}+\mu \overrightarrow{BC}=\mu \overrightarrow{OC} \text{ ___(ii)}\]

Adding $(i)$ and $(ii)$, we get, \[\lambda \overrightarrow{OA}+\lambda \overrightarrow{AC}+\mu \overrightarrow{OB}+\mu \overrightarrow{BC}=\lambda \overrightarrow{OC}+\mu \overrightarrow{OC}\]

Since $λAC=μBC$ and $λAC$ and $μBC$ are acting in opposite direction along the same line, they cancel out. Therefore, \[\lambda \overrightarrow{OA}+\mu \overrightarrow{OB}=(\lambda+\mu)\overrightarrow{OC}\]

This is $λ-μ$ theorem.

**Case I:** If $λ=μ$, then $AC=BC$ i.e. $C$ is the midpoint of $AB$, then, \[\overrightarrow{OA}+\overrightarrow{OB}=2\overrightarrow{OC}\]

**Case II:** If the direction of the forces are reversed, then, \[\lambda \overrightarrow{AO}+\mu \overrightarrow{BO}=(\lambda+\mu)\overrightarrow{CO}\]