Translational equilibrium: Definition, Equation, Solved 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.

What is Translational Equilibrium?

The translational equilibrium indicates that the net force acting on the object is zero and the object has a zero linear acceleration.

Translational equilibrium

Thus any object is said to be in translational equilibrium when the net force acting on the object is zero.

It 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,

`\mathbf{F_{x}}` = 0
`\mathbf{F_{y}}` = 0
`\mathbf{F_{z}}` = 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).

Translational equilibrium explained:

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

1] Stationary object:-

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.

`\sum F_{x}` = 0 —–[1]

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,

`\sum F_{y}` = N – mg —- (Consider upward forces positive)

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

∴`\sum F_{y}` = 0 —–[2]

The above equations [1] and [2] 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:-

car 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.

∴`\sum F_{x}` = 0 —–[1]

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.

`\sum F_{y}` = N – mg —- (Consider upward forces positive)

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

∴`\sum F_{y}` = 0 —–[2]

The above equation [2] states that the object has zero net force in the y-direction. Therefore from equation [1] and equation [2], 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 formula:

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

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

In scalar form, the equilibrium conditions is given by,

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

Translational equilibrium vs Rotational equilibrium:

Sr. NoTranslational EquilibriumRotational Equilibrium
1In this type of equilibrium, an object experiences zero net force.In rotational equilibrium, an object experiences zero net torque.
2The object has zero linear acceleration.The object has zero angular acceleration.
3It is considered in the case of translational motion.It is considered in the case of rotational motion.
4Example: Stationary objects, car moving with constant velocity, etc.Example: Fan rotating with constant velocity, Balanced seesaw, etc.

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.
crate having a mass of 50 Kg

Given:-
m = 50 Kg

Solution:-

Step: 1] Free body diagram:-

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

free body diagram of the hanging crate

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

forces resolved in x and y direction
Step: 2] Apply tranlational equilibrium conditions:-

`\sum F_{x}` = 0

`\sum F_{x}` = – `T_{A}cos30` + `T_{B}cos40` = 0

`\mathbf{T_{A}cos30-T_{B}cos40}` = 0 —– Equation [1]

`\sum F_{y}` = 0

`\sum F_{y}` = `T_{A}sin30` + `T_{B}sin40` – mg = 0

`T_{A}sin30` + `T_{B}sin40` = mg

`T_{A}sin30` + `T_{B}sin40` = 50 x 9.81

`T_{A}sin30` + `T_{B}sin40` = 490.5

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

Put this value of `T_{A}` in equation [1]

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

`\mathbf{T_{B}}` = 452.047 N

Put the value of `T_{B}` in equation [1],

`T_{A}cos30` – 452.047 cos 40 = 0

`\mathbf{T_{A}}` = 399.859 N

Therefore the tensions in ropes are,

`T_{A}` = 399.859 N
`T_{B}` = 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.
stationary wooden block on the inclined plane

Given:-
m = 30 Kg

Solution:-

Step: 1] Free body diagram:-

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

free body diagram of the wooden block

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,

forces on wooden block resolved in x and y direction
Step: 2] Apply translational equation conditions:-

`\sum F_{x}` = 0

`\sum F_{x}` = Fcos30 – Nsin30 = 0

Fcos30 – Nsin30 = 0 —–eq [1]

`\sum F_{y}` = 0

Fsin30 + Ncos30 – mg = 0

Fsin30 + Ncos30 = mg

Fsin30 + Ncos30 = 30 x 9.81

Fsin30 + Ncos30 = 294.3

F = `\mathbf{\frac{1}{sin30}}`(294.3 – Ncos30)

Put the value of F in equation [1],

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

N = 254.871 N

Put this value of N = 254.871 in equation [1],

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 is meant by the translational equilibrium of a particle?

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

  2. How do you achieve translational equilibrium?

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

You might also be interested in this:

Pratik is a Graduated Mechanical engineer. He enjoys sharing the engineering knowledge learned by him with people.

Leave a Comment