# How To Find Static Equilibrium: Problems and Examples

In the physical state, a system is referred to possess a static equilibrium condition when every particle involved in the system is a stationary state.

In rigid body dynamics, the large buildings, bridges, houses, and mountains are regarded as the physical systems which attain the static equilibrium condition as they do not move, tilt or rotate from their respective position. In this post, let us briefly discuss how to find static equilibrium acting on the physical system.

## How to find static equilibrium?

A physical system tends to possess static equilibrium if and only if it satisfies the following conditions:

• The Sum of all the forces acting in every direction must be equal to zero. i.e., ΣFX =ΣFY = 0
• The Sum of total torque acting in clockwise and counterclockwise directions must be zero. i.e., ΣτX=ΣτY=0
• The linear momentum of each particle in the physical system must be zero.

Strategy to find the static equilibrium.

• Calculate all the forces acting on the system. The force may have come from the other object in contact with the system like support, floor, weight, even gravity.
• Draw the free-body diagram, which helps you to resolve the forces acting on the system, including their magnitude and direction (if they are provided)
• Write the static equilibrium equation. Remember, while writing the torque equation, choose the axis that gives the simplest form of the equation to solve the equation.
• Then solve the resulting equation, which gives the result satisfying the static equilibrium condition.

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

To achieve the static equilibrium condition, we have to calculate the forces acting on the iron ball and the man using the above data. The man needs to hold his arm steady in order to make the iron ball stationary. The condition to achieve static equilibrium by his arm is

ΣFX=0

ΣFY=0

0=Fb-Fa-Fi

Where Fa is the force required by the upper arm pushing the elbow joint downward

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

Fi is the force pushing the iron ball down.

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

Στ=0

0=Fb.d+Fa.0-Fi.l

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

And l is distance from elbow joint to iron ball=30cm

0=Fb.d-Fi.l

Since the iron ball exerts the force downward, we take Fi as its weight according to Newton’s laws of motion.

0=Fb.d-5×9.8×0.30

Fb=Fi.l/d

Substituting the values

Fb= (5)(9.8)(0.30)/0.04

Fb=367.5 N

The force required for the upper arm to keep the ball steady is

0=Fb-Fa-Fi

Fa = Fb – Fi

Fa = 367.5-14.7

Fa=352.8 N

How to find static equilibrium in an inclined plane?

In order to find the static equilibrium in an inclined system, the angle of inclination must be included in the calculation. The further calculation will be similar to the static equation in the plane.

The inclined physical system is influenced by the forces. To find the static equilibrium in an inclined plane, the four forces must be resolved. The four forces influencing the static equilibrium in an inclined plane are given below.

For better understanding, let us consider an example of ladder slating on a wall. The four forces exerted on the ladder system can be resolved as

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

f-F=0

The net force along the y-axis is

N-w=0

The torque acting from the pivot point is

τwF=0; where τw and τF are the torque exerted on the ladder system due to weight and normal reaction on the wall respectively.

The torque acting on the ladder can be given as

τw=rwwsinθw

τF=rFFsinθF

On resolving the above equation, we can get

τwF=0

If any system attains the above condition then the system is said to be in static equilibrium.

## A truck of mass 850lb is parked in the middle of a rigid bridge. It is assumed that the bridge weighs around 100lb per foot, uniformly distributed along the length of 80ft. Calculate the resultant normal force acting on the two sides of the bridge to keep it steady.

Solution:

The total mass of the bridge mT= 100×80= 8000 lb

Let R1 and R2 be the two resultants.

Since there is no force acting along the x-direction, the equilibrium condition for x-direction is Fx=0

The equilibrium condition for y-direction is

ΣFy=0=850+8000-R1-R2

Or R1+R2=8850

The linear motion of the truck is zero, i.e., ΣM=0

If we consider the moment of force from the left, then the force exerted on the truck and bridge makes a clockwise moment, and the R2 will make the counterclockwise moment.

ΣM=0= 850*40 + 8000*40-R2*80

R2=(850*40)+(8000*40)/80

R2=4425lb

R1=8850-4425=4425lb

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

Solution:

Applying the equilibrium condition

ΣFX=0

ΣFX= f-F

ΣFY=0

ΣFY=N-W

The torque acting on the rod due to weight

τW=rwWsinθW

But the rw is given as half of the length of the rod

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

Similarly; toque due to normal reaction at the wall is given by

τF=rFFsinθF

But rF=L and sinθF=sin(180-β)=sinβ

τF=LFsinβ

To satisfy the equilibrium condition τW+τF=0

The weight of the rod is given by W=mg

W=8*9.8

W=78.4N

Substituting the known values in the equation, we get

Rearranging and solving the terms, we get the value for the normal reaction of the wall as

F=39.2*0.9325

F=36.55N.