For simplicity, let us consider a system which is at rest at an equilibrium position, and which can move from that position of equilibrium in either direction by only one path.
Such a system is shown in Fig. 1, in which four different equilibrium configurations of a smooth cylinder are represented.

In (a) the cylinder rests on a concave surface; in (b) on a convex surface; in (c) on a point of inflection; and in (d) on a plane surface. Applying the principle of virtual displacements, it will be seen that all four of the positions are equilibrium positions, since the only active force is a vertical gravity force, and since the center of gravity of the cylinder moves horizontally for an infinitesimal rolling displacement of the cylinder. We can also relate these situations to the fact that the potential energy of the system must have a stationary value at the position of equilibrium. In (a), the potential energy of the system is seen to be a minimum, since the cylinder is in its lowest position, and work would have to be done on the cylinder to move it in either direction. In (b) the potential energy of the system is a maximum, since the cylinder is at the highest position, and work would be done by the system as it moved in either direction. The point of inflection of (c) corresponds to a stationary value, since the potential energy of the cylinder is neither a maximum nor a minimum.
It will also be recognized that the positions shown in Fig. 3-11 are physically somewhat different. In (a), we may say that we have stable equilibrium, since any movement away from the position of equilibrium sets up forces which will return the system to the equilibrium position, (b) and (c) are examples of unstable equilibrium, since any movement away from the position of equilibrium sets up forces which will move the system still further from the equilibrium position, and (d) illustrates neutral or indifferent equilibrium, in which motion of the system from the position of equilibrium does not affect the equilibrium of the system.
We can thus conclude that the maximum potential energy condition corresponds with the unstable condition, while the minimum potential energy condition corresponds with the stable condition.
The above considerations can be used to test an equilibrium position as to whether it is stable or unstable. It is only necessary to note whether the potential energy of the system increases or decreases as the system is moved from the position of equilibrium.
The conditions for equilibrium and the tests for stable and unstable equilibrium can be expressed analytically as follows. Suppose that for simplicity we consider a system whose displacement from a position of equilibrium can be described by one coordinate,
If we now write the Maclaurin series expansion of this function about the point
By the theorem of virtual displacements, we know that the work done by the forces upon an arbitrary infinitesimal displacement from the point
The change in the potential energy of the system, in the region near the equilibrium position, is then given by:
since the term
Thus by examining the sign of the second derivative, we determine whether the potential energy is increasing or decreasing as we move away from the equilibrium position, and hence whether the system is stable or unstable.
From the above considerations, we see that while the positions of equilibrium can be determined by the principle of virtual displacements by considering only the first-order terms of the small displacements, it is necessary to investigate the second-order terms in order to decide as to the stability or instability of the equilibrium position.
| Sign of | Potential Energy | Equilibrium Condition |
|---|---|---|
| minimum | stable | |
| maximum | unstable |
Example 1. A weight

Solution. We shall work this problem in two ways, first by the principle of virtual displacements, and second by means of potential energy considerations.

First Method
We take as a virtual displacement of the system a small lowering of the weight
This answer can be very simply checked by equating the moments about the point
In order to investigate the stability of this equilibrium position, we must see whether it requires work to be done upon the system to move it from the position of equilibrium, or whether the system itself can do work as it moves from the position of equilibrium.
The work done by the system during the displacement
Note that we have left only a second-order term
Since this work term is positive, we see that the system itself does work as it moves from the position of equilibrium, i.e., the potential energy would be decreasing, hence the position of equilibrium is unstable.
Second Method
We write down the expression for the potential energy of the system in a region close to the equilibrium position:
The equilibrium condition is:
The test for stability is:
Example 2. A uniform plank of thickness

Solution. Consider a virtual displacement

Originally, the points
From the figure, we have directly:
To find

For example, a
3.6.1 PROBLEMS
1. A door can swing through

2. Referring to the previous problem, determine whether the equilibrium position of the uniform bar is stable or unstable.
Answer
Unstable
3. A homogeneous body is composed of a semi-cylinder and a rectangular parallelepiped, as shown in the figure. Find the maximum value of

Answer
4. A rigid weightless bar
