Stable Equilibrium

In the physical state, a system is referred to as having a stable state of equilibrium when every particle involved in the system is in a stable state.

In rigid body dynamics, large buildings, bridges, houses and mountains are regarded as physical systems that achieve a state of stable equilibrium as they do not move, bend or rotate from their respective positions. In this post, let us briefly discuss how to find the stable equilibrium acting on a physical system.

How to find stable equilibrium?

A physical system tends to have a stable equilibrium if and only if it satisfies the following conditions:

  • The sum of all the forces acting in each direction must be equal to zero. That is, F X =ΣF Y = 0
  • The sum of the total torques acting in the clockwise and counterclockwise directions must be zero. i.e., X =Στ Y =0
  • The linear momentum of every particle in a physical system must be zero.

Strategies for finding a stable equilibrium .

  • Calculate all the forces acting on the system. The force can also come from other objects in contact with the system, such as supports, floors, weights, even gravity.
  • Draw a free-body diagram, which helps you solve the forces acting on the system including their magnitude and direction (if they are provided)
  • Write the stable equilibrium equation. Remember, when writing the torque equation, choose the axis that gives the simplest form of the equation to solve for.
  • Then solve the resulting equation, which gives a result satisfying the stable equilibrium condition.

Let us consider an example to help you understand how to find stable equilibrium in a stationary system . A man is holding an iron ball of mass 5kg. The distance between the ball and its elbow joint is 30 cm, and the distance between the elbow joint and the forearm is 4 cm. Then find the force required to keep the ball stationary?

To obtain the stable equilibrium position, we have to calculate the forces acting on the iron ball and the one using the above data. To make the iron ball stable, the man has to keep his hand steady. The condition for attaining a stable balance by his arm is

X =0

Y =0

0 = F b -F a -F i

where F a is the force required by the upper arm to push the elbow joint down

b is the force pulling the forearm at a distance of 4 cm from the elbow

i is the force pushing down the iron ball.

All three forces are acting in the vertical direction. The torque acting on the man and the iron ball is given by


0 = f b .d+f a .0-f i .l

where d is the distance from the forearm to the elbow joint=4cm

and l is the distance of the iron ball from the elbow joint=30cm

0 = F b .dF i .l

Since the iron ball exerts a downward force, we take F’ i as its weight in accordance with Newton’s laws of motion.

0 = f b .d-5×9.8×0.30


replace values


FB = 367.5 N

The force required by the upper arm to keep the ball stationary is

0 = F b -F a -F i

a = F b – F i

a = 367.5-14.7

a =352.8 N

How to find stable equilibrium in an inclined plane?

To find stable equilibrium in a tilting system, the angle of inclination must be included in the calculation. Further calculations will be similar to the equation stationary in the plane.

An inclined physical system is affected by forces. To find stable equilibrium in an inclined plane, four forces must be resolved. Given below are four forces affecting a stable equilibrium in an inclined plane.

  • The first force is the normal force pointing vertically upward on the floor.
  • The second force is the static friction force applied horizontally on the floor, which restricts the motion.
  • The third force is due to gravity acting vertically downward from the center of mass’s position.
  • The final force is the normal reaction force acting horizontally on the top of the inclined surface.

To understand better, let us consider the example of climbing a ladder on a wall. The four forces exerted on the ladder system can be solved by

In the x-axis, the net force is given by

ff = 0

The net force along the y-axis is

nw = 0

The torque acting from the pivot point is

w -τ F = 0; where w and F are the torque exerted on the ladder system due to the load and normal reactance on the wall respectively.

The torque acting on the ladder can be given as

w = r w Wsinθ w

F = r F Fsinθ F

On solving the above equation, we can get

w +τ F = 0

If a system achieves the above condition then the system is said to be in stable equilibrium.

Solved problems on how to find stable equilibrium

A truck of mass 850lb is standing in the middle of a rigid bridge. It is believed that the bridge weighs about 100lb per foot, which is evenly distributed along the length of 80 feet. Calculate the resultant normal force acting on both sides of the bridge to keep the bridge stationary.


Total mass of the bridge m T = 100 × 80 = 8000 lb

Let R 1 and R 2 be the two resultant.

Since there is no force acting along the x-direction, the equilibrium position for the x-direction is F. is x =0

The RSI equilibrium position is for the y-direction

y =0=850+8000-R 1 -R 2

or R 1 +R 2 = 8850

The linear speed of the truck is zero, i.e. M=0

If we consider the moment of force from the left, then the force exerted on the truck and the bridge creates a clockwise moment, and R 2 will create a counterclockwise moment.

m = 0 = 850 * 40 + 8000 * 40-r 2 * 80


2 =4425lb

1 =8850-4425=4425lb

A rod of mass 8kg and length 12m rests on a wall. The angle of inclination of the rod is 47°. Find the normal reaction force of the wall.


Apply balance condition

X =0

X = FF

Y =0

Y = NW

torque on the rod due to the load

W = r w Wsinθ W

but r w is given as half the length of the rod


sinθW can be written as sin(180+90-β)=sin(90-β)=-cosβ


Similarly; The torque due to normal reaction on the wall is given by

F = r F Fsinθ F

But Su F =L and sinθ F =sin(180-β)=sinβ

F = lfsinβ

To satisfy the equilibrium condition W+τF=0


Weight of rod W=mg. is given by

W = 8 * 9.8

W = 78.4N

Substituting the known values ​​in the equation, we get


By rearranging and solving the terms, we get the value of the normal response of the wall:



F = 36.55 N.