System of Particles and Rotational Dynamics

Equilibrium of a Rigid Body

A rigid body is a system of many particles. It is not essential that each of the particles of a rigid body behaves in a similar manner like the other particle. Depending on the type of motion every particle behaves in a specific way. This is where the equilibrium of rigid bodies comes into play. How this equilibrium affects the whole system of particles is what we shall learn now.

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Equilibrium of Rigid Bodies

Rigid bodies are those bodies in which the distance between particles is constant despite any kind of external force. So while studying the equilibrium of rigid bodies, we mainly aim to define the behavior of these constituting particles in changed conditions of force or torque.Since we are concentrating on the equilibrium of rigid bodies under motion, therefore we need to take both translational and rotational motion into consideration.

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Equilibrium is a defined as any point where the total amount of external force or torque is zero. This point may be anywhere near the center of mass. External force in translational motion of the rigid body changes the linear momentum of that body. While the external torque in rotational motion can change the angular momentum of the rigid body.

In the mechanical equilibrium of a rigid body, the linear momentum and angular momentum remain unchanged with time. This implies that the body under the influence of external force neither has a linear acceleration nor an angular acceleration. We, therefore, can say that:

  • If the total force on a rigid body is zero then the body shows translational equilibrium as the linear momentum remains unchanged despite the change in time:
  • If the total torque on a rigid body is zero then the body shows rotational equilibrium as the angular momentum does not change with time.

Mechanical Equilibrium

When we sum up the above findings of translational and rotational equilibrium we get the following assumptions:

F1+F2+F3+F4+…..Fn= Fi = 0  (For translational equilibrium)

τ1234+……τn = τi = 0 (For rotational equilibrium)

These equations are the vector in nature. As scalars the force and torque in their x,y, and z components are:

Fix = 0 ,  Fiy = 0, and Fiz= 0 and τix = 0 ,  τiy = 0, and τiz= 0

The independent condition of force and torque helps in reaching the rigid bodies to a state of mechanical equilibrium. Generally, the forces acting on the rigid body are coplanar. The three conditions if satisfied, help the rigid body attain equilibrium. The condition of translational equilibrium arises when any of the two components along any perpendicular axis sum up to be zero.

For rotational equilibrium, it is necessary that all the three components result in a zero. Moreover, as the translational equilibrium is a condition that depends on a particle’s behaviour, therefore, the vector sum of forces on all the particles must be a zero.

Partial Equilibrium

Equilibrium in a rigid body may also be partial in nature. Partial equilibrium of a body is that state where a rigid body shows only one kind of equilibrium. For example, consider the following figure:


The figure shows an instance of rotational equilibrium. Q being the center with sides PQ=QR=a. The forces F at points P and Q are equal in magnitude but opposite in direction.The system is in rotational equilibrium as the net moment on the rod is zero. The translational equilibrium can only be seen if the forces from points P and Q are opposite in perpendicular direction to the rod.

The figure below shows translational equilibrium. In the figure below, the moments from forces at points P and Q are equal. The forces F are not opposite rather they act in the same sense thus causing anti-clockwise rotation of the rod. The total force hence is zero. Here, since the rod without any translation shows rotation. The kinds of force acting on the rod are termed as couple or torque.

A pair of equal and opposite forces with different lines of action that produces rotation of the body is called the torque on that body.

The Principle of Moments

For understanding mechanical equilibrium we need to understand the working of a fulcrum and lever. These pose as the best examples of mechanical equilibrium. A See-saw in parks best explains the principle of lever and fulcrum. See-saw is the lever while the point at which the rod is pivoted is the fulcrum.

The lever here shows mechanical equilibrium. R, the reaction of the support from the fulcrum, is directed opposite the forces, F1 and F2 and at R – F– F2 = 0, we see that the rigid body attains translational equilibrium. Rotational equilibrium is attained when d1F– d2F2 = 0. For rotational equilibrium, the sum of moments about the fulcrum is zero.

Here, F1 = load. F2 = effort needed to lift the load, d1 = load arm and d2 = effort arm. At rotational equilibrium,  d1F1 = d2 F2 or F1/F2 = d2/d1The ratio F1/F2 is also called the Mechanical advantage. Now, if d2 > d1 then mechanical advantage is greater than 1, which means that a small effort can lift the load.

Centre of Gravity

Center of gravity is the point of balance of a rigid body. This situation is the result of the mechanical equilibrium between the two rigid bodies. For example, when you hold a book on the tip of your finger the center at which the book is balanced is called the center of gravity. The mechanical equilibrium between the finger and book has made the balance possible.

The reaction of your fingertip at the center is equal and opposite to Mg (the force of gravity). This balance is an example of translational and rotational equilibrium. The center of gravity of the book is located at the point where the total torque due to force mg is zero. Center of Gravity is, therefore, that point where the total gravitational torque on the rigid body is zero.

Sample Question For You

Q: Consider the following two statements A and B and identify the correct choice:
A) The torques produced by two forces of the couple are opposite to each other
B) The direction of torque is always perpendicular to the plane of rotation of the body

a) A is true and B is false.      b) B is true and A is false.       c) Both are true.        d) A and B are false.

Solution: b) The direction of torque is always perpendicular to the plane of rotation of body as a cross product is in the perpendicular plane to r and F vectors and the torques produced by two forces of the couple are in the same direction to each other.

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