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 ${\vec{F}}_1,{\vec{F}}_2,{\vec{F}}_3$ acting at same point O of a body as shown in diagram,By the parallelogram Law
• Resultant of ${\vec{F}}_{1}and{\vec{F}}_{2}is{\vec{F}}_1+{\vec{F}}_2$
• If the third force ${\vec{F}}_3$ acts on the body such that ${\vec{F}}_3=-\left({\vec{F}}_1+{\vec{F}}_2\right)$ then the body will be in equilibrium.

${\vec{F}}_3=-\left({\vec{F}}_1+{\vec{F}}_2\right)$

${\vec{F}}_1+{\vec{F}}_2+{\vec{F}}_3=0$

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.

${\vec{F}}_1+{\vec{F}}_2+{\vec{F}}_3+{\vec{F}}_4+\dots \dots \dots \dots {\vec{F}}_n=0$

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

3.0Types of Equilibrium

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 ${\vec{F}}_1,{\vec{F}}_2\dots \dots \dots \dots {\vec{F}}_i\text{ and }{\vec{F}}_n$ act simultaneously on a body and the body is in translational equilibrium, the resultant of these forces must be zero. $\sum {\vec{F}}_i=\vec{0}$

• If the forces ${\vec{F}}_1,{\vec{F}}_2\dots \dots \dots \dots ..{\vec{F}}_i\text{ and }{\vec{F}}_n$ are expressed in Cartesian components, we have

$\sum {\vec{F}}_{ix}=0\sum {\vec{F}}_{iy}=0\sum {\vec{F}}_{iz}=0$

• 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. ${\vec{F}}_1+{\vec{F}}_2+{\vec{F}}_3=0$

5.0Lami's Theorem

Diagram shows a particle O under the equilibrium of three concurrent forces ${\vec{F}}_1,{\vec{F}}_2\text{ and }{\vec{F}}_3$. Let $\alpha$ be angle between ${\vec{F}}_2\text{ and }{\vec{F}}_3$, $\beta$ between ${\vec{F}}_3\text{ and }{\vec{F}}_1$, $\beta$ and $\gamma$ between ${\vec{F}}_1\text{ and }{\vec{F}}_2$

The Forces ${\vec{F}}_1,{\vec{F}}_2\text{ and }{\vec{F}}_3$ can be represented by the sides of $△$ABC,taken in the same order.

Applying Law of Sines to $△$ABC,

$\frac{F_1}{\sin (\pi -\alpha )}=\frac{F_1}{\sin (\pi -\beta )}=\frac{F_1}{\sin (\pi -\gamma )}$

$\frac{F_1}{\sin \alpha }=\frac{F_2}{\sin \beta }=\frac{F_3}{\sin \gamma }$

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.

$\frac{F_1}{\sin A}=\frac{F_2}{\sin B}=\frac{F_3}{\sin C}$

• 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:

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

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

• 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 $\sum F_x=0(\text{ For Equilibrium })$

$F_1+10\sin 60^{\circ }=F_2\cos 60^{\circ }$

$F_1+10\left(\frac{\sqrt{3}}{2}\right)=F_2\left(\frac{1}{2}\right)$

$2F_1-F_2=-10\sqrt{3}....(1)$

In y-direction

$\sum F_y=0$

$F_2\sin 60^{\circ }+10\cos 60^{\circ }=10$

$F_2\times \left(\frac{\sqrt{3}}{2}\right)+10\left(\frac{1}{2}\right)=10\Rightarrow F_2(\sqrt{3})+10=20$

$F_2=\left(\frac{10}{\sqrt{3}}N\right)$

Now from Equation (1)

$2F_1-F_2=-10\sqrt{3}\Rightarrow 2F_1-\frac{10}{\sqrt{3}}=-10\sqrt{3}$

$2\sqrt{3}{F}_1-10=-30\Rightarrow 2\sqrt{3}{F}_1=-20\Rightarrow F_1=\frac{-10}{\sqrt{3}}N$