### Equilibrium of a rigid body-conditions and physics

##### December 20, 2017

# Equilibrium of a Rigid body

A rigid body is said to be in mechanical **equilibrium**, if both its **linear momentum** and **angular momentum** are not changing with time. In other words, the body is in mechanical equilibrium when it has neither linear acceleration nor angular acceleration.This means the **conditions for equilibrium** can be expressed as the following:

**conditions for equilibrium**

**(1) The total or net force i.e. the vector sum of all the forces, on the rigid body is zero.**

i.e., **ΣF _{i} = 0 when i=1 to n …………………… (1 a)**

**or
**

**dp**

_{tot}/dt = 0**where p**

_{tot}=net linear momentum………………… (1b)**(2) The total Torque i.e. the vector sum of the torques on the rigid body is zero.**

**ΣΤ _{i} = 0 when i=1 to n………………….. (2a)**

or

**dl _{tot}/dt = 0 where l_{tot}=net angular momentum…………………(2b)**

## Conditions for mechanical equilibrium

Expanding equation 1a and 2a:

**ΣF _{ix} = 0 ΣF_{iy} = 0 ΣF_{iz} = 0 when i=1 to n**

**ΣΤ _{ix} = 0 ΣΤ_{iy} = 0 ΣΤ_{iz} = 0 when i=1 to n**

For m.e. these 6 independent conditions are to be satisfied.

### Conditions when forces are coplanar

We need only 3 conditions to be satisfied for this when the forces working on the rigid body are coplanar.

* **Two of these conditions **correspond to translational equilibrium: The sum of forces along any 2 perpendicular axes in the plane must be zero. so if we take x and y axes, then **ΣF _{x} = 0 ΣF_{y} = 0**

* The third **condition** is about Rotational Equilibrium: The sum of the torques along any axis perpendicular to the plane of the above said forces, must be zero. Here it will be: **ΣΤz = 0**

### Partial Equilibrium

A body may be in this condition. It means that the body may be in translational equilibrium but not in rotational equilibrium and vice versa.

**Case 1:**

AB is a light rod and C is its midpoint. So CA = CB = r (say). At point A a force is applied downwards and at B same amount of force f is applied but in opposite direction.

Here from the diagram it’s clear that the net force on the rod is zero. [ F + (-F) =0 ]

But what is the net Torque?

Say at point A the torque is T1 and at point B it’s T2. T1= F r .

Similarly T2 = F r. (both torques trying to move the rod anticlockwise).

So certainly net Torque on the rod is non zero.

So in this case, the rod is in translational equilibrium (net force zero) but it’s not in rotational equilibrium. (net torque nonzero)

**Case 2:**

in this case, both the forces at point A and B are equal (F) like case 1 but this time both are working in the same direction (downwards here). As a result, the net force is nonzero.

Let’s consider the torques now.

At point A and B magnitudes of the torque are same, i.e. r F. but in this case their directions are opposite to each other.

At point A the torque tries to move the rod anticlockwise, at point B the torque plays clockwise.

Therefore, in this case net torque is zero.

So in this case, the rod is in rotational equilibrium (net torque zero) but it’s not in translational equilibrium. (net force nonzero)

#### Couple-Rotation without translation

A pair of equal and opposite forces with different lines of action is known as a **couple**.

A couple produces rotation without translation.

**Example of couple**; When we open the lid of a bottle by turning it, we apply a couple to the lid through our fingers.

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