Equilibrium of a Rigid Body and Center of Gravity
Equilibrium Of A Rigid Body
In statics, equilibrium refers to a state where an object is not undergoing any change in its motion. For a point particle, equilibrium means its velocity is constant (which can be zero, implying rest). For a rigid body, equilibrium is more complex because it can both translate and rotate.
A rigid body is said to be in equilibrium if it is not accelerating (its linear velocity is constant) and it is not undergoing angular acceleration (its angular velocity is constant). If the body is initially at rest and not rotating, and it remains in that state under the action of forces and torques, it is in static equilibrium.
For a rigid body to be in equilibrium, two conditions must be satisfied simultaneously:
First Condition for Equilibrium (Translational Equilibrium)
The first condition for equilibrium is that the net external force acting on the rigid body must be zero. This ensures that the centre of mass of the body has zero acceleration, meaning its linear velocity is constant.
$ \sum \vec{F}_{ext} = 0 $
In terms of components in a coordinate system:
$ \sum F_x = 0 $
$ \sum F_y = 0 $
$ \sum F_z = 0 $
This condition ensures translational equilibrium – the body is not changing its linear motion (either remaining at rest or moving with constant linear velocity).
Second Condition for Equilibrium (Rotational Equilibrium)
The second condition for equilibrium is that the net external torque acting on the rigid body about any point must be zero. This ensures that the body has zero angular acceleration, meaning its angular velocity is constant.
$ \sum \vec{\tau}_{ext} = 0 $
The reference point for calculating torques can be any point – the condition $\sum \vec{\tau}_{ext} = 0$ will hold true for any arbitrary point if it holds true for one point and the net force is zero. However, choosing a convenient pivot point (like where one of the forces acts) can simplify calculations as the torque due to that force will be zero ($\vec{r}=0$ for that force).
In terms of components about a chosen axis (e.g., z-axis if rotation is in the xy-plane):
$ \sum \tau_x = 0 $
$ \sum \tau_y = 0 $
$ \sum \tau_z = 0 $
This condition ensures rotational equilibrium – the body is not changing its rotational motion (either remaining without rotation or spinning with constant angular velocity).
For a rigid body to be in complete equilibrium (both translational and rotational), both conditions must be satisfied.
(Image Placeholder: A rectangular block with several forces applied to different points. Arrows indicating forces F1, F2, F3, F4. If F1+F2+F3+F4=0 and the sum of torques about any point (say, a corner or the center) is zero, the block is in equilibrium.)
Principle Of Moments
The second condition for equilibrium, $\sum \vec{\tau}_{ext} = 0$, is sometimes referred to as the Principle of Moments, especially when dealing with forces causing rotation in a plane about a fixed axis. In this context, moments usually refer to torques about that specific axis.
If a rigid body is in rotational equilibrium about a fixed axis, the sum of the moments (torques) tending to cause rotation in one direction (say, clockwise) about that axis is equal to the sum of the moments (torques) tending to cause rotation in the opposite direction (anti-clockwise) about the same axis.
Let's say we are considering rotation about the z-axis. Forces acting in the xy-plane will produce torques about the z-axis. We can assign a sign to these torques (e.g., positive for anti-clockwise rotation and negative for clockwise rotation, or vice-versa). The second condition for equilibrium becomes:
$ \sum \tau_{z, clockwise} = \sum \tau_{z, anti-clockwise} $
Or, taking direction into account:
$ \sum \tau_{z, i} = 0 $
where $\tau_{z, i}$ is the z-component of the torque due to the $i$-th force, with appropriate sign convention.
This principle is particularly relevant when dealing with levers, beams, or systems pivoted at a point or along an axis.
Example 1. A uniform meter scale is balanced at the 40 cm mark when a mass of 50 g is suspended at the 20 cm mark. Find the mass of the meter scale.
Answer:
Let the meter scale be a rigid body. It is uniform, so its centre of mass (and centre of gravity, assuming uniform gravity) is located at its geometrical centre, which is the 50 cm mark.
The scale is balanced (in static equilibrium) at the 40 cm mark. This is the pivot point.
Forces acting on the scale causing torque about the pivot (40 cm mark):
- Weight of the 50 g mass, suspended at the 20 cm mark. This force acts downwards. Let the mass be $m_1 = 50$ g = 0.05 kg. The distance from the pivot (40 cm) to the point of application (20 cm) is $r_1 = 40 \text{ cm} - 20 \text{ cm} = 20$ cm = 0.2 m. The force is $F_1 = m_1 g$. This force will tend to rotate the scale clockwise about the pivot.
- Weight of the meter scale itself, acting at its centre of gravity (50 cm mark). This force acts downwards. Let the mass of the scale be $m_s$. The distance from the pivot (40 cm) to the point of application (50 cm) is $r_2 = 50 \text{ cm} - 40 \text{ cm} = 10$ cm = 0.1 m. The force is $F_2 = m_s g$. This force will tend to rotate the scale anti-clockwise about the pivot.
- The normal force from the support at the 40 cm mark. This force acts upwards at the pivot point itself. Since the distance from the pivot is zero ($\vec{r}=0$), the torque due to this force is zero ($\tau_{Normal} = 0$).
For rotational equilibrium, the net torque about the pivot must be zero. Applying the Principle of Moments:
Torque due to $m_1$: $\tau_1 = r_1 F_1 \sin(90^\circ) = r_1 (m_1 g)$ (Assuming the force and position vector relative to pivot are perpendicular, valid for horizontal scale and vertical forces). Direction: Clockwise.
Torque due to $m_s$: $\tau_2 = r_2 F_2 \sin(90^\circ) = r_2 (m_s g)$. Direction: Anti-clockwise.
For equilibrium, sum of clockwise torques equals sum of anti-clockwise torques:
$ \tau_1 = \tau_2 $
$ r_1 (m_1 g) = r_2 (m_s g) $
The factor $g$ (acceleration due to gravity) cancels out:
$ r_1 m_1 = r_2 m_s $
Substitute the known values:
$ (0.2 \text{ m}) \times (0.05 \text{ kg}) = (0.1 \text{ m}) \times m_s $
$ 0.01 \text{ kg} \cdot \text{m} = 0.1 m_s \text{ m} $
$ m_s = \frac{0.01}{0.1} $ kg
$ m_s = 0.1 $ kg
The mass of the meter scale is 0.1 kg (or 100 g).
(For translational equilibrium, the upward normal force at 40 cm would balance the total downward forces: $N = m_1 g + m_s g$.)
Centre Of Gravity
Every object is composed of a vast number of particles, and each of these particles is attracted towards the Earth by the force of gravity. The centre of gravity (CG) of a body is the point through which the resultant of all the gravitational forces acting on the individual particles of the body passes, regardless of the orientation of the body.
In a uniform gravitational field (where the acceleration due to gravity $g$ is the same in magnitude and direction for all parts of the body), the centre of gravity coincides with the centre of mass. For objects near the Earth's surface, the gravitational field is approximately uniform, so the centre of gravity and centre of mass are practically at the same location.
However, if the gravitational field is not uniform (e.g., for a very large object like a mountain range, or in a region with significant density variations, or when considering objects at very large distances from the Earth), the centre of gravity might be slightly different from the centre of mass. In such cases, the CG is located closer to the region where the gravitational field is stronger.
Locating the Centre of Gravity
The concept of CG is particularly useful when considering the effect of gravity on the rotational equilibrium of a body. The total effect of gravity on a rigid body can be represented by a single downward force equal to the total weight of the body, acting vertically downwards through the centre of gravity.
For a body suspended from a point, it will come to rest in a position where its centre of gravity is vertically below the point of suspension. If a body is suspended from two different points, the intersection of the two vertical lines drawn from the points of suspension will locate the centre of gravity.
(Image Placeholder: An irregular flat shape. It is suspended from one point, and a plumb line is shown hanging down from the suspension point. The shape is then suspended from a different point, and another plumb line is shown. The intersection of the lines where the plumb lines would fall through the object marks the CG.)
For stable equilibrium when a body is placed on a surface, the vertical line passing through its centre of gravity must fall within its base of support. If this line falls outside the base, the gravitational torque will cause the body to topple.