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., ΣF**_{X}=ΣF_{Y}= 0**The Sum of total torque acting in clockwise and counterclockwise directions must be zero.****i.e., Στ**_{X}=Στ_{Y}=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

ΣF_{X}=0

ΣF_{Y}=0

0=F_{b}-F_{a}-F_{i}

Where F_{a} is the force required by the upper arm pushing the elbow joint downward

F_{b} is the force pulling the forearm at a distance 4cm from the elbow

F_{i} 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=F_{b}.d+F_{a}.0-F_{i}.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=F_{b}.d-F_{i}.l

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

0=F_{b}.d-5×9.8×0.30

F_{b}=F_{i}.l/d

Substituting the values

F_{b}= (5)(9.8)(0.30)/0.04

F_{b}=367.5 N

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

0=F_{b}-F_{a}-F_{i}

F_{a} = F_{b} – F_{i}

F_{a} = 367.5-14.7

F_{a}=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.**

- The first force is normal force exerted on the floor pointing vertically upward.
- The second force is the static friction force exerted horizontally on the floor, which restrict the motion.
- The third force is due to gravity acting vertically downward acting from the center of mass position.
- The last force is the normal reaction force acting horizontally at the top of the inclined surface.

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

τ_{w}-τ_{F}=0; where τ_{w} and τ_{F} 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

τ_{w}=r_{w}wsinθ_{w}

τ_{F}=r_{F}Fsinθ_{F}

On resolving the above equation, we can get

τ_{w}+τ_{F}=0

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

**Solved problems on how to find 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 m_{T}= 100×80= 8000 lb

Let R_{1} and R_{2} be the two resultants.

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

The equilibrium condition for y-direction is

ΣF_{y}=0=850+8000-R_{1}-R_{2}

Or R_{1}+R_{2}=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 R_{2} will make the counterclockwise moment.

ΣM=0= 850*40 + 8000*40-R_{2}*80

R_{2}=(850*40)+(8000*40)/80

R_{2}=4425lb

R_{1}=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

ΣF_{X}=0

ΣF_{X}= f-F

ΣF_{Y}=0

ΣF_{Y}=N-W

The torque acting on the rod due to weight

τ_{W}=r_{w}Wsinθ_{W}

But the r_{w} 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

τ_{F}=r_{F}Fsinθ_{F}

But r_{F}=L and sinθ_{F}=sin(180-β)=sinβ

τ_{F}=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.

**Read more on Dynamic Equilibrium**