Editing Mechanical equilibrium (section)

== Stability ==
An important property of systems at mechanical equilibrium is their [[Stability theory|stability]].

===Potential energy stability test===
In a function which describes the system's potential energy, the system's equilibria can be determined using [[calculus]]. A system is in mechanical equilibrium at the [[critical point (mathematics)|critical point]]s of the function describing the system's [[potential energy]]. These points can be located using the fact that the [[First derivative test|derivative]] of the function is zero at these points. To determine whether or not the system is stable or unstable, the [[second derivative test]] is applied. With <math>V </math> denoting the static [[Equations of motion|equation of motion]] of a system with a single [[Degrees of freedom (mechanics)|degree of freedom]] the following calculations can be performed: 

[[File:Diagram of a ball placed in an unstable equilibrium.svg|thumb|Diagram of a ball placed in an unstable equilibrium.]]

;Second derivative < 0: The potential energy is at a local maximum, which means that the system is in an unstable equilibrium state. If the system is displaced an arbitrarily small distance from the equilibrium state, the forces of the system cause it to move even farther away.

[[File:Diagram of a ball placed in a stable equilibrium.svg|thumb|Diagram of a ball placed in a stable equilibrium.]]

;Second derivative > 0: The potential energy is at a local minimum. This is a stable equilibrium. The response to a small perturbation is forces that tend to restore the equilibrium. If more than one stable equilibrium state is possible for a system, any equilibria whose potential energy is higher than the absolute minimum represent metastable states.

;Second derivative = 0: The state is neutral to the lowest order and nearly remains in equilibrium if displaced a small amount. To investigate the precise stability of the system, [[Taylor expansion|higher order derivatives]] can be examined. The state is unstable if the lowest nonzero derivative is of odd order or has a negative value, stable if the lowest nonzero derivative is both of even order and has a positive value. If all derivatives are zero then it is impossible to derive any conclusions from the derivatives alone. For example, the function <math>e^{-1/x^2}</math> (defined as 0 in x=0) has all derivatives equal to zero. At the same time, this function has a local minimum in x=0, so it is a stable equilibrium. If this function is multiplied by the [[Sign function]], all derivatives will still be zero but it will become an unstable equilibrium.

[[File:Diagram of a ball placed in a neutral equilibrium.svg|thumb|Diagram of a ball placed in a neutral equilibrium.]]

;Function is locally constant: In a truly neutral state the energy does not vary and the state of equilibrium has a finite width. This is sometimes referred to as a state that is marginally stable, or in a state of indifference, or astable equilibrium.

When considering more than one dimension, it is possible to get different results in different directions, for example stability with respect to displacements in the ''x''-direction but instability in the ''y''-direction, a case known as a [[saddle point]]. Generally an equilibrium is only referred to as stable if it is stable in all directions.

===Statically indeterminate system===
{{main|Statically indeterminate}}
Sometimes the [[Mechanical equilibrium|equilibrium]] equations{{snd}} force and moment equilibrium conditions{{snd}} are insufficient to determine the forces and [[Reaction (physics)|reactions]]. Such a situation is described as ''statically indeterminate''.

Statically indeterminate situations can often be solved by using information from outside the standard equilibrium equations.

[[File:Ship stability.svg|thumb|[[Ship stability]] illustration explaining the stable and unstable dynamics of buoyancy (B), center of buoyancy (CB), center of gravity (CG), and weight (W)]]