# Forces In Equilibrium

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.

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.

Suppose a force F1 is acted on a particle, then the state of motion of the particle is changed. But, if we apply a second force F2, equal in magnitude to F1 but opposite in direction, along the same line of action as F1, produces an equilibrium.

The forces F1 and F2 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 F1 and F2 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 F1, F2, …, Fn 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 F1, F2, …, Fn $X=X_1+X_2+…+X_n$ $X=\sum X_i$ and, Y is the algebraic sum of the Y-components of the forces F1, F2, …, Fn $Y=Y_1+Y_2+…+Y_n$ $Y=\sum Y_i$ 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 $\overrightarrow{P} \text{, } \overrightarrow{Q} \text{ and } \overrightarrow{R}$ 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 $\overrightarrow{AB} \text{ and } \overrightarrow{BC}$ 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 $\overrightarrow{P} \text{, } \overrightarrow{Q} \text{ and } \overrightarrow{R}$ be the forces acting at O be in equilibrium. Let $\overrightarrow{OA}=\overrightarrow{P}$ $\overrightarrow{OB}=\overrightarrow{Q}$ $\text{and, }\overrightarrow{OD}=\overrightarrow{R}$

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 $\overrightarrow{P} \text{, } \overrightarrow{Q} \text{ and } \overrightarrow{R}$ 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 P1, P2, P3, P4 and P5 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 P1, P2, P3 and P4 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. P1, P2, P3, P4 and P5 are in equilibrium.

Alternative Method

Let $\overrightarrow{P_1} \text{, } \overrightarrow{P_2} \text{, } \overrightarrow{P_3} \text{, } \overrightarrow{P_4} \text{ and } \overrightarrow{P_5}$ 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.

​$\text{Using Sine law,}$ $\frac{AB}{sinC}=\frac{BC}{sinA}=\frac{CA}{sinB}$ $\frac{\vec{P}}{sin(180°-α)}=\frac{\vec{Q}}{sin(180°-β)}=\frac{\vec{R}}{sin(180°-γ)}$ $\frac{\vec{P}}{sinα}=\frac{\vec{Q}}{sinβ}=\frac{\vec{R}}{sinγ}$ Alternative Method

Let $\overrightarrow{P} \text{, } \overrightarrow{Q} \text{ and } \overrightarrow{R}$ be the forces acting at O, be in equilibrium. Thus, $\overrightarrow{P}+\overrightarrow{Q}+\overrightarrow{R}=0$ $\implies \overrightarrow{P}×\overrightarrow{P}+\overrightarrow{P}×\overrightarrow{Q}+\overrightarrow{P}×\overrightarrow{R}=0$ $\implies \overrightarrow{P}×\overrightarrow{Q} = \overrightarrow{R}×\overrightarrow{P} \text{ } (\because \overrightarrow{P}×\overrightarrow{P}=0)$ $\implies |\overrightarrow{P}×\overrightarrow{Q}|=|\overrightarrow{R}×\overrightarrow{P}|$ $\implies PQsin(180-\gamma)=RPsin(180-\beta)$ $\implies Qsin\gamma=Rsin\beta$ $\therefore \frac{Q}{sin\beta}=\frac{R}{sin\gamma} \text{ ____(1)}$ Again, $\overrightarrow{P}+\overrightarrow{Q}+\overrightarrow{R}=0$ $\implies \overrightarrow{Q}×\overrightarrow{P}+\overrightarrow{Q}×\overrightarrow{Q}+\overrightarrow{Q}×\overrightarrow{R}=0$ $\implies \overrightarrow{Q}×\overrightarrow{R} = \overrightarrow{P}×\overrightarrow{Q} \text{ } (\because \overrightarrow{Q}×\overrightarrow{Q}=0)$ $\implies |\overrightarrow{Q}×\overrightarrow{R}|=|\overrightarrow{P}×\overrightarrow{Q}|$ $\implies QRsin(180-\alpha)=PQsin(180-\gamma)$ $\implies Rsin\alpha=Psin\gamma$ $\therefore \frac{R}{sin\gamma}=\frac{P}{sin\alpha} \text{ ____(2)}$ 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, $\lambda \overrightarrow{AO}+\mu \overrightarrow{BO}=(\lambda+\mu)\overrightarrow{CO}$