Equilibrium of Concurrent Forces

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Equilibrium of Concurrent Forces

The equilibrium of concurrent forces is a fundamental concept in mechanics that deals with the condition where multiple forces acting at a single point balance each other so that the object remains at rest or moves with constant velocity. When forces are concurrent, they all intersect at a common point, and their vector sum determines the overall effect on the body. For the body to be in equilibrium, the resultant of these forces must be zero, meaning that the forces exactly counteract one another. This principle is essential in engineering and physics to analyze structures, machines, and objects under various force systems, ensuring stability and safety in design.

1.0Concurrent Forces

Concurrent forces are forces that all meet or pass through the same single point. They can pull or push in different directions, but they always come together at one spot.

The exact point where they all meet — called the point of concurrency — is very important.
These forces might lie in the same flat surface (called coplanar forces) or act in different planes.
To make calculations easier, these forces can be broken down into their horizontal (x-axis) and vertical (y-axis) parts.

2.0Equilibrium of Concurrent Forces

Forces that act on a body at the same point are known as concurrent forces. When multiple forces acting at that point result in a net force of zero, the body will either remain at rest or continue moving in a straight line at a constant speed. In this condition, the body is said to be in equilibrium.

Consider three concurrent forces $F_{1}, F_{2}, F_{3}$ acting at same point O of a body as shown in diagram,By the parallelogram Law
Resultant of $F_{1} an d F_{2} i s F_{1} + F_{2}$
If the third force $F_{3}$ acts on the body such that $F_{3} = - (F_{1} + F_{2})$ then the body will be in equilibrium.

$F_{3} = - (F_{1} + F_{2})$

$F_{1} + F_{2} + F_{3} = 0$

Note: As shown in the diagram these three forces in equilibrium can be represented by the sides of a triangle taken in the same order.

Condition for the equilibrium of a number of forces acting at the same point is that the vector sum of all these forces is equal to zero.

$F_{1} + F_{2} + F_{3} + F_{4} + \dots\dots\dots\dots F_{n} = 0$

A particle is in equilibrium under n forces if these forces, arranged in order, form a closed n-sided polygon.

3.0Types of Equilibrium

Type	Description	Example
Static Equilibrium	All forces acting on a body cancel out. Net force and acceleration are zero, and the body remains at rest.	A block resting on a floor with equal 6N forces acting in opposite directions.
Dynamic Equilibrium	Forces balance out, so net force and acceleration are zero, but the body moves with constant velocity.	A block in simple harmonic motion at its maximum speed, with net force = 0.
Coplanar Forces in Equilibrium	Forces acting in the same plane can be in equilibrium if: 1. Their vector sum is zero 2. Clockwise and counterclockwise moments are equal	Forces intersecting at a point on a 2D plane (e.g., cables holding a sign)

Type

Description

Example

Static Equilibrium

All forces acting on a body cancel out. Net force and acceleration are zero, and the body remains at rest.

A block resting on a floor with equal 6N forces acting in opposite directions.

Dynamic Equilibrium

Forces balance out, so net force and acceleration are zero, but the body moves with constant velocity.

A block in simple harmonic motion at its maximum speed, with net force = 0.

Coplanar Forces in Equilibrium

Forces acting in the same plane can be in equilibrium if:

1. Their vector sum is zero

2. Clockwise and counterclockwise moments are equal

Forces intersecting at a point on a 2D plane

(e.g., cables holding a sign)

4.0Translational Equilibrium

A body in state of rest or moving with constant velocity is said to be in translational equilibrium. Thus, if a body is in translational equilibrium in a particular inertial frame of reference, it must have no linear acceleration.
When it is at rest, it is in static equilibrium, whereas if it is moving at constant velocity it is in dynamic equilibrium.

Conditions for Translational Equilibrium

For a body to be in translational equilibrium, no net force must act on it i.e. the vector sum of all the forces acting on it must be zero.
If several external forces $F_{1}, F_{2} \dots\dots\dots\dots F_{i} and F_{n}$ act simultaneously on a body and the body is in translational equilibrium, the resultant of these forces must be zero. $\sum F_{i} = 0$

If the forces $F_{1}, F_{2} \dots\dots\dots\dots .. F_{i} and F_{n}$ are expressed in Cartesian components, we have

$\sum F_{i x} = 0 \sum F_{i y} = 0 \sum F_{i z} = 0$

If a body is acted upon by a single external force, it cannot be in equilibrium. If a body is in equilibrium under the action of only two external forces, the forces must be equal and opposite.
If a body is in equilibrium under action of three forces, their resultant must be zero; therefore,according to the triangle law of vector addition they must be coplanar and make a closed triangle. $F_{1} + F_{2} + F_{3} = 0$

5.0Lami's Theorem

Diagram shows a particle O under the equilibrium of three concurrent forces $F_{1}, F_{2} and F_{3}$ . Let $α$ be angle between $F_{2} and F_{3}$ , $β$ between $F_{3} and F_{1}$ , $β$ and $γ$ between $F_{1} and F_{2}$

The Forces $F_{1}, F_{2} and F_{3}$ can be represented by the sides of $△$ ABC,taken in the same order.

Applying Law of Sines to $△$ ABC,

$\frac{F _{1}}{s i n ( π - α )} = \frac{F _{1}}{s i n ( π - β )} = \frac{F _{1}}{s i n ( π - γ )}$

$\frac{F _{1}}{s i n α} = \frac{F _{2}}{s i n β} = \frac{F _{3}}{s i n γ}$

If three forces acting on a particle keep it in equilibrium ,then each force is proportional to the sine of the angle between the other two forces.

$\frac{F _{1}}{s i n A} = \frac{F _{2}}{s i n B} = \frac{F _{3}}{s i n C}$

The analytical method uses Cartesian components to resolve forces.
Since the forces form a closed triangle, they lie in a plane—making a 2D Cartesian (x-y) system suitable.
Choose the x-y orientation so that force angles with axes are simple and convenient.
Apply equilibrium conditions:

$\sum F_{x} = 0 \Rightarrow F_{1 x} + F_{2 x} + F_{3 x} = 0$

$\sum F_{y} = 0 \Rightarrow F_{1 y} + F_{2 y} + F_{3 y} = 0$

Use the analytical method for systems with more than three forces.
If there are parallel or anti-parallel forces, combine them first to simplify the system.
Combining forces can sometimes reduce a multi-force problem to a simpler three-force system.

Illustration-1. In the given figure if body is in equilibrium then calculate $F_{1}$ and $F_{2}$ ?

Solution: Resolve the forces along x-direction & y-direction

In x-direction $\sum F_{x} = 0 (For Equilibrium)$

$F_{1} + 10 sin 6 0^{\circ} = F_{2} cos 6 0^{\circ}$

$F_{1} + 10 (\frac{3}{2}) = F_{2} (\frac{1}{2})$

$2 F_{1} - F_{2} = - 103 .... (1)$

In y-direction

$\sum F_{y} = 0$

$F_{2} sin 6 0^{\circ} + 10 cos 6 0^{\circ} = 10$

$F_{2} \times (\frac{3}{2}) + 10 (\frac{1}{2}) = 10 \Rightarrow F_{2} (3) + 10 = 20$

$F_{2} = (\frac{10}{3} N)$

Now from Equation (1)

$2 F_{1} - F_{2} = - 103 \Rightarrow 2 F_{1} - \frac{10}{3} = - 103$

$23 F_{1} - 10 = - 30 \Rightarrow 23 F_{1} = - 20 \Rightarrow F_{1} = \frac{- 10}{3} N$

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Equilibrium of Concurrent Forces

The equilibrium of concurrent forces is a fundamental concept in mechanics that deals with the condition where multiple forces acting at a single point balance each other so that the object remains at rest or moves with constant velocity. When forces are concurrent, they all intersect at a common point, and their vector sum determines the overall effect on the body. For the body to be in equilibrium, the resultant of these forces must be zero, meaning that the forces exactly counteract one another. This principle is essential in engineering and physics to analyze structures, machines, and objects under various force systems, ensuring stability and safety in design.

1.0Concurrent Forces

Concurrent forces are forces that all meet or pass through the same single point. They can pull or push in different directions, but they always come together at one spot.

The exact point where they all meet — called the point of concurrency — is very important.
These forces might lie in the same flat surface (called coplanar forces) or act in different planes.
To make calculations easier, these forces can be broken down into their horizontal (x-axis) and vertical (y-axis) parts.

2.0Equilibrium of Concurrent Forces

Forces that act on a body at the same point are known as concurrent forces. When multiple forces acting at that point result in a net force of zero, the body will either remain at rest or continue moving in a straight line at a constant speed. In this condition, the body is said to be in equilibrium.

Consider three concurrent forces $F_{1}, F_{2}, F_{3}$ acting at same point O of a body as shown in diagram,By the parallelogram Law
Resultant of $F_{1} an d F_{2} i s F_{1} + F_{2}$
If the third force $F_{3}$ acts on the body such that $F_{3} = - (F_{1} + F_{2})$ then the body will be in equilibrium.

$F_{3} = - (F_{1} + F_{2})$

$F_{1} + F_{2} + F_{3} = 0$

Note: As shown in the diagram these three forces in equilibrium can be represented by the sides of a triangle taken in the same order.

Condition for the equilibrium of a number of forces acting at the same point is that the vector sum of all these forces is equal to zero.

$F_{1} + F_{2} + F_{3} + F_{4} + \dots\dots\dots\dots F_{n} = 0$

A particle is in equilibrium under n forces if these forces, arranged in order, form a closed n-sided polygon.

3.0Types of Equilibrium

Type	Description	Example
Static Equilibrium	All forces acting on a body cancel out. Net force and acceleration are zero, and the body remains at rest.	A block resting on a floor with equal 6N forces acting in opposite directions.
Dynamic Equilibrium	Forces balance out, so net force and acceleration are zero, but the body moves with constant velocity.	A block in simple harmonic motion at its maximum speed, with net force = 0.
Coplanar Forces in Equilibrium	Forces acting in the same plane can be in equilibrium if: 1. Their vector sum is zero 2. Clockwise and counterclockwise moments are equal	Forces intersecting at a point on a 2D plane (e.g., cables holding a sign)

Type

Description

Example

Static Equilibrium

All forces acting on a body cancel out. Net force and acceleration are zero, and the body remains at rest.

A block resting on a floor with equal 6N forces acting in opposite directions.

Dynamic Equilibrium

Forces balance out, so net force and acceleration are zero, but the body moves with constant velocity.

A block in simple harmonic motion at its maximum speed, with net force = 0.

Coplanar Forces in Equilibrium

Forces acting in the same plane can be in equilibrium if:

1. Their vector sum is zero

2. Clockwise and counterclockwise moments are equal

Forces intersecting at a point on a 2D plane

(e.g., cables holding a sign)

4.0Translational Equilibrium

A body in state of rest or moving with constant velocity is said to be in translational equilibrium. Thus, if a body is in translational equilibrium in a particular inertial frame of reference, it must have no linear acceleration.
When it is at rest, it is in static equilibrium, whereas if it is moving at constant velocity it is in dynamic equilibrium.

Conditions for Translational Equilibrium

For a body to be in translational equilibrium, no net force must act on it i.e. the vector sum of all the forces acting on it must be zero.
If several external forces $F_{1}, F_{2} \dots\dots\dots\dots F_{i} and F_{n}$ act simultaneously on a body and the body is in translational equilibrium, the resultant of these forces must be zero. $\sum F_{i} = 0$

If the forces $F_{1}, F_{2} \dots\dots\dots\dots .. F_{i} and F_{n}$ are expressed in Cartesian components, we have

$\sum F_{i x} = 0 \sum F_{i y} = 0 \sum F_{i z} = 0$

If a body is acted upon by a single external force, it cannot be in equilibrium. If a body is in equilibrium under the action of only two external forces, the forces must be equal and opposite.
If a body is in equilibrium under action of three forces, their resultant must be zero; therefore,according to the triangle law of vector addition they must be coplanar and make a closed triangle. $F_{1} + F_{2} + F_{3} = 0$

5.0Lami's Theorem

Diagram shows a particle O under the equilibrium of three concurrent forces $F_{1}, F_{2} and F_{3}$ . Let $α$ be angle between $F_{2} and F_{3}$ , $β$ between $F_{3} and F_{1}$ , $β$ and $γ$ between $F_{1} and F_{2}$

The Forces $F_{1}, F_{2} and F_{3}$ can be represented by the sides of $△$ ABC,taken in the same order.

Applying Law of Sines to $△$ ABC,

$\frac{F _{1}}{s i n ( π - α )} = \frac{F _{1}}{s i n ( π - β )} = \frac{F _{1}}{s i n ( π - γ )}$

$\frac{F _{1}}{s i n α} = \frac{F _{2}}{s i n β} = \frac{F _{3}}{s i n γ}$

If three forces acting on a particle keep it in equilibrium ,then each force is proportional to the sine of the angle between the other two forces.

$\frac{F _{1}}{s i n A} = \frac{F _{2}}{s i n B} = \frac{F _{3}}{s i n C}$

The analytical method uses Cartesian components to resolve forces.
Since the forces form a closed triangle, they lie in a plane—making a 2D Cartesian (x-y) system suitable.
Choose the x-y orientation so that force angles with axes are simple and convenient.
Apply equilibrium conditions:

$\sum F_{x} = 0 \Rightarrow F_{1 x} + F_{2 x} + F_{3 x} = 0$

$\sum F_{y} = 0 \Rightarrow F_{1 y} + F_{2 y} + F_{3 y} = 0$

Use the analytical method for systems with more than three forces.
If there are parallel or anti-parallel forces, combine them first to simplify the system.
Combining forces can sometimes reduce a multi-force problem to a simpler three-force system.

Illustration-1. In the given figure if body is in equilibrium then calculate $F_{1}$ and $F_{2}$ ?

Solution: Resolve the forces along x-direction & y-direction

In x-direction $\sum F_{x} = 0 (For Equilibrium)$

$F_{1} + 10 sin 6 0^{\circ} = F_{2} cos 6 0^{\circ}$

$F_{1} + 10 (\frac{3}{2}) = F_{2} (\frac{1}{2})$

$2 F_{1} - F_{2} = - 103 .... (1)$

In y-direction

$\sum F_{y} = 0$

$F_{2} sin 6 0^{\circ} + 10 cos 6 0^{\circ} = 10$

$F_{2} \times (\frac{3}{2}) + 10 (\frac{1}{2}) = 10 \Rightarrow F_{2} (3) + 10 = 20$

$F_{2} = (\frac{10}{3} N)$

Now from Equation (1)

$2 F_{1} - F_{2} = - 103 \Rightarrow 2 F_{1} - \frac{10}{3} = - 103$

$23 F_{1} - 10 = - 30 \Rightarrow 23 F_{1} = - 20 \Rightarrow F_{1} = \frac{- 10}{3} N$

Frequently Asked Questions

What are concurrent forces?

Can three non-parallel concurrent forces be in equilibrium? Under what condition?

What happens if the vector sum of concurrent forces is not zero?

Why must both the horizontal and vertical components of forces be zero for equilibrium?

Is torque or moment considered in the equilibrium of concurrent forces?

Join ALLEN!

Equilibrium of Concurrent Forces

1.0Concurrent Forces

2.0Equilibrium of Concurrent Forces

3.0Types of Equilibrium

4.0Translational Equilibrium

5.0Lami's Theorem

Table of Contents

Frequently Asked Questions

What are concurrent forces?

Can three non-parallel concurrent forces be in equilibrium? Under what condition?

What happens if the vector sum of concurrent forces is not zero?

Why must both the horizontal and vertical components of forces be zero for equilibrium?

Is torque or moment considered in the equilibrium of concurrent forces?

Join ALLEN!

Equilibrium of Concurrent Forces

1.0Concurrent Forces

2.0Equilibrium of Concurrent Forces

3.0Types of Equilibrium

4.0Translational Equilibrium

5.0Lami's Theorem

Table of Contents