# Translational equilibrium: Definition, Formula, Conditions, Examples

As per the translational and rotational motion, the mechanical equilibrium is classified as translational equilibrium and rotational equilibrium. In this article, we are discussing the translational equilibrium in detail.

Contents:

## What is Translational Equilibrium?

The translational equilibrium indicates that the net force acting on the object is zero and thus the object has a zero linear acceleration. Therefore any object is said to be in translational equilibrium when the net force acting on the object is zero.

The condition can be expressed as,

\mathbf{\vec{F}}= 0

Here \vec{F} indicates the vector sum of all forces acting on an object.

In scalar form, the translational equilibrium conditions are as follows,

Fx = 0
Fy = 0
Fz = 0

Thus if all of these conditions are satisfied then the object is in translational equilibrium.

As the object under translational equilibrium has zero net force, thus according to Newton’s second law of motion (F = ma), the object has zero linear acceleration (a = 0). Therefore the object under the translational equilibrium may be steady or moves with a constant velocity (V = Constant).

## Explanation:

Following are the two different examples that will help to understand the translational equilibrium:-

1] Stationary object:-

In the case of the object shown in the above figure, there is no force is acting in a horizontal direction. Therefore the net force acting on an object in the x-direction is zero.

∑ Fx = 0 —– 

In the case of y-direction (vertical), the weight of the block (mg) is acting in the downward direction. In contrast, the same amount of normal reaction (N) developed by the ground opposes the downward movement of the block.

Therefore the net force acting on an object in the y-direction is given by,

∑ Fy = N – mg —- (Consider upward forces positive)

As the magnitudes of ‘N’ and ‘mg’ are the same,

∴ ∑ Fy = 0 —– 

The above equations  and  say that the net force acting in the x and y direction is zero. Hence the above block is in translational equilibrium.

2] Vehicle moving with constant velocity:-

As the above vehicle is moving with a constant velocity, thus it means that the vehicle has zero acceleration in the x-direction.

The zero acceleration says that the net force acting in the x-direction is zero.

∴ ∑ Fx = 0 —– 

In the case of y-direction, the weight of the car (mg) is acting in a downward direction. In contrast, the same magnitude of normal reaction (N) is developed by the ground that opposes the downward movement of the vehicle.

∑ Fy = N – mg —- (Consider upward forces positive)

As the magnitudes of ‘N’ and ‘mg’ are the same,

∴ ∑ Fy = 0 —– 

The above equation  states that the object has zero net force in the y-direction. Therefore from equation  and equation , it can be concluded that the above car moving with constant velocity is in translational equilibrium.

Therefore the object under the translational equilibrium may be stationary like the object shown in example-1 or it may move with a constant velocity like the vehicle shown in example-2.

## Translational equilibrium equation:

In vector form, the condition for the translational equilibrium is given by,

\mathbf{\vec{F}}= 0

In scalar form, the translational equilibrium conditions are given by,

Fx = 0
Fy = 0
Fz = 0

## Translational equilibrium solved examples:

1] A crate having a mass of 50 Kg is suspended with the help of rope A and rope B. If the crate is in translational equilibrium, find the tension in ropes A and B.

Given:
m = 50 Kg

Solution:-

Step: 1] Free body diagram:-

The free body diagram of the hanging crate is as follows,

The forces in the above free body diagram can be resolved in x and y direction as follows,

Step: 2] Apply tranlational equilibrium conditions:-

∑ Fx = 0

∑ Fx = – TAcos30 + TBcos40 = 0

TAcos30 – TBcos40 = 0 —– Equation 

∑ Fy = 0

∑ Fy = TAsin30 + TBsin40 – mg = 0

TAsin30 + TBsin40 = mg

TAsin30 + TBsin40 = 50 x 9.81

TAsin30 + TBsin40 = 490.5

T_{A} = \frac{1}{sin30}(490.5-T_{B}sin40)

Put this value of TA in equation ,

[\frac{1}{sin30}(490.5-T_{B}sin40)]cos30-T_{B}cos40= 0

TB = 452.047 N

Put the value of TB in equation ,

TAcos30 – 452.047 cos40 = 0

TA = 399.859 N

Therefore the tensions in ropes are,

TA = 399.859 N
TB = 452.047 N

2] The stationary wooden block on the inclined plane is held stationary by the frictional resistance between the wooden block and the ground.

If the weight of the block is 30 kg, thus find the frictional resistance offered by the ground.

Given:
m = 30 Kg

Solution:-

Step: 1] Free body diagram:-

The free body diagram of the wooden block is as follows,

Here,
N = Normal reaction
F = Frictional resistance

The forces in the above free body diagram can be resolved in x and y direction as follows,

Step: 2] Apply translational equation conditions:-

∑ Fx = 0

∑ Fx = Fcos30 – Nsin30 = 0

Fcos30 – Nsin30 = 0 —– eq 

∑ Fy = 0

∑ Fy = Fsin30 + Ncos30 – mg = 0

Fsin30 + Ncos30 = mg

Fsin30 + Ncos30 = 30 x 9.81

Fsin30 + Ncos30 = 294.3

F = \frac{1}{sin30}(294.3 – Ncos30)

Put the value of F in equation ,

{\frac{1}{sin30}(294.3-Ncos30)}cos30-Nsin30=0

N = 254.871 N

Put this value of N = 254.871 in equation ,

Fcos(30) – 254.871sin(30) = 0

F = 147.149 N

This is the frictional resistance offered by the ground to avoid the downward movement of the block.

## FAQs:

1. What indicates the translational equilibrium of a particle?

When the net force acting on the particle is zero, then the particle is in translational equilibrium.

2. How do you achieve translational equilibrium?

For achieving translational equilibrium, the object should have zero net force.

3. What is a real example of translational equilibrium?

Following are the examples of a translational equilibrium,
A person moving in a straight line with a constant velocity, houses, books kept on table, etc.

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