# Static equilibrium (Physics): Definition, Equation, Solved examples

Static equilibrium is an important term in physics that indicates the stability of the objects at rest. Thus, in this article, we will discuss static equilibrium, which is a type of mechanical equilibrium.

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## What is Static equilibrium?

The static equilibrium is a physical state in which an object is at rest with no net force and no net torque acting on it. Thus for static equilibrium, the object must be in translational equilibrium as well as in rotational equilibrium and another vital condition is that the object must be at rest.

The conditions required for the static equilibrium are as follows:-

1] The object must be in translational equilibrium:-

It means that the net force acting on an object must be equal to zero.

\sum \vec{F} = 0

Where \sum \vec{F} is the vector summation of all the forces acting onto the object.

In scalar form, the condition becomes,

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

Each of the above equations indicates that the sum of the forces in each of the three directions must be equal to zero.

2] The object must be in rotational equilibrium:-

It means that the net torque acting on the object must be equal to zero.

\sum \vec{T} = 0

Where \sum \vec{T} is the vector summation of the torques acting onto the object.

In scalar form, the condition becomes,

\sum T_{x} = 0
\sum T_{y} = 0
\sum T_{z} = 0

Each of the above equations indicates that the sum of the torques in each of the three directions must be equal to zero.

3] The object must be at rest:-

The object under static equilibrium must be in a resting condition and it should not move or tilt.

Thus the object must have zero angular velocity and zero translational velocity.

v = 0, \omega = 0

## Static equilibrium equation:

Following are the equations for checking the static equilibrium of the object:-

a]

\mathbf{V_{\text{object}}} = 0, \mathbf{\omega_{\text{object}}} = 0

b]

\mathbf{\sum \vec{F}} = 0

OR

\mathbf{\sum F_{x}} = 0
\mathbf{\sum F_{y}} = 0
\mathbf{\sum F_{z}} = 0

c]

\mathbf{\sum \vec{T}} = 0

OR

\mathbf{\sum T_{x}} = 0
\mathbf{\sum T_{y}} = 0
\mathbf{\sum T_{z}} = 0

## Static equilibrium explained:

The below example of a cantilever beam will help to understand the static equilibrium easily:-

The free body diagram of the above beam can be drawn as follows,

Now check the cantilever beam for conditions of static equilibrium,

1] The object must be in translational equilibrium:-

In the above figure, no force is acting in the horizontal direction, thus the net force acting on the beam in the x-direction is zero.

\sum F_{x} = 0 –

In the vertical direction, the weight of the beam (mg) acts in a downward direction. In contrast, the fixed support applies equal and opposite reactions (R_{\text{support}}). Thus the net force on the beam in the y-direction is given by,

\sum F_{y} = R_{\text{support}} – mg —-[Consider positive]

As magnitude of ‘R_{\text{support}}’ and ‘mg’ are same,

∴ \sum F_{y} = 0 –

Therefore equation- and equation- says that the above cantilever beam is in translational equilibrium.

2] The object must be in rotational equilibrium:-

The torque developed by the weight of the beam T_{\text{weight}} tries to rotate the beam in a clockwise direction.
In contrast, the support also develops equal and opposite torque T_{\text{support}} to avoid the tilting of the beam.

Thus the net torque acting on the beam is given by,

\sum T = T_{\text{weight}} – T_{\text{support}} [Consider ↻ +ve]

As T_{\text{weight}} and T_{\text{support}} has same magnitude

∴ \sumT = 0

Thus the above cantilever beam is in rotational equilibrium.

3] The object must be at rest:-

The above rigid cantilever beam is not moving as well as it is not rotating.

∴ V_{\text{beam}} = 0, \omega_{\text{beam}} = 0

The above cantilever beam fulfills all conditions, Thus the above rigid cantilever beam is in static equilibrium.

## Static equilibrium solved examples:

1] For the below cantilever beam, find the support reactions developed by fixed support at A. (Reaction force R_{A} & Reaction torque T_{A})

Solution:-

1] Free body diagram:-

To draw the FBD, replace the fixed support at A with R_{A} and T_{A}.

Where, R_{A} is the reaction force and T_{A} is the reaction moment developed by the support.

Step 2] Apply static equilibrium conditions:-

Condition 1] Net force acting on the beam is zero.

\sum F_{y} = 0 [Take ↑ +ve]

R_{A} – 20 – 40 = 0

\mathbf{R_{A}} = 60 N

Condition 2] Net torque acting on beam is zero.

Apply this condition at point A.

\sum T_{\text{@}A} = 0 [Consider ↻ +ve]

[20 x AB] + [40 x AC] + [R_{A} x 0] – T_{A} = 0

[20 x 0.5] + [40 x 1] – T_{A} = 0

\mathbf{T_{A}} = 50 N.m

Therefore the support reactions are,

\mathbf{R_{A}} = 60 N
\mathbf{T_{A}} = 50 N.m

2] The beam shown in the below figure is in static equilibrium. Find the support reaction at points A & D.

Solution:-

Step 1] Free body diagram:-

To draw the free body diagram, replace the support at A and D with the support reaction R_{A} and R_{D} respectively.

Step 2] Apply static equilibrium condition:-

Consider net torque at hinge support A is zero.

\sum T_{\text{@}A} = 0 [Consider ↻ +ve]

[100 x AB] + [200 x AC] – [R_{D} x AD] + [R_{A} x 0] = 0

[100 x 0.5] + [200 x 1.5] – [R_{D} x 2] = 0

\mathbf{R_{D}} = 175 N

Consider net force acting on the beam is zero.

\sum F_{y} = 0 [Take ↑ +ve]

R_{A} – 100 – 200 + R_{D} = 0

R_{A} – 100 – 200 + 175 = 0

\mathbf{R_{A}} = 125 N

Therefore the support reaction are,

\mathbf{R_{A}} = 125 N
\mathbf{R_{D}} = 175 N

FAQs:

1. Why is Static equilibrium important?

The static equilibrium is important during the construction of structures like buildings and bridges. To stand properly on the ground it is necessary that structures must be in static equilibrium so that they can stand without falling or tilting.

2. Which are the Static equilibrium examples in real life?

The things around us that are in resting condition without showing translational or rotational motion are examples of static equilibrium. For example, Books kept on tables, Houses, Buildings, Cell towers, Lighthouse, etc. Pratik is a Graduated Mechanical engineer. He enjoys sharing the engineering knowledge learned by him with people.