Editing Catenary (section)

===Chain under a general force===
With no assumptions being made regarding the force {{math|'''G'''}} acting on the chain, the following analysis can be made.<ref>Follows [[#Routh|Routh]] Art. 455</ref>

First, let {{math|1='''T''' = '''T'''(''s'')}} be the force of tension as a function of {{mvar|s}}. The chain is flexible so it can only exert a force parallel to itself. Since tension is defined as the force that the chain exerts on itself, {{math|'''T'''}} must be parallel to the chain. In other words,

<math display=block>\mathbf{T} = T \mathbf{u}\,,</math>

where {{mvar|T}} is the magnitude of {{math|'''T'''}} and {{math|'''u'''}} is the unit tangent vector.

Second, let {{math|1='''G''' = '''G'''(''s'')}} be the external force per unit length acting on a small segment of a chain as a function of {{mvar|s}}. The forces acting on the segment of the chain between {{mvar|s}} and {{math|''s'' + Δ''s''}} are the force of tension {{math|'''T'''(''s'' + Δ''s'')}} at one end of the segment, the nearly opposite force {{math|−'''T'''(''s'')}} at the other end, and the external force acting on the segment which is approximately {{math|'''G'''Δ''s''}}. These forces must balance so

<math display=block>\mathbf{T}(s+\Delta s)-\mathbf{T}(s)+\mathbf{G}\Delta s \approx \mathbf{0}\,.</math>

Divide by {{math|Δ''s''}} and take the limit as {{math|Δ''s'' → 0}} to obtain

<math display=block>\frac{d\mathbf{T}}{ds} + \mathbf{G} = \mathbf{0}\,.</math>

These equations can be used as the starting point in the analysis of a flexible chain acting under any external force. In the case of the standard catenary, {{math|1='''G''' = (0, −''w'')}} where the chain has weight {{mvar|w}} per unit length.