# Tension in Common Catenary

I was reading the intrinsic equation of common catenary , where the author assumes tension in the catenary and proceeds.

Why does a freely hanging rope with uniform mass per unit length has tension ? This is contradictory to the fact we use in solving problems in mechanics i.e. when the rope becomes slack, we take the tension in the rope to be zero . • catenary is the correct spelling. – John Alexiou Oct 20 '20 at 12:25
• corrected , thanks – Harsh Vardhan Singh Oct 20 '20 at 13:17

The catenary is the shape of a self-supporting cable with mass. Typically in mechanics ropes are massless and their shape when in not carrying tension is undefined, and when carrying tension are just straight lines.

But for a rope/chain/cable that has a defined mass per length, the shape must curve because the tension on the ends of each segment must balance out the weight of the segment in-between. Take any segment, and the weight $$W$$ must be counter-balanced by the vertical support loads $$V_1$$ and $$V_2$$. In addition, the combined tension on the left $$\sqrt{H^2+V_1^2}$$ must be tangent to the shape, and similarly for the combined tension on the right.

See this answer for the development of the equations.

Those are the necessary conditions to define the shape of the catenary.

In simple mechanics problems we assume that a string is massless. In the catenary the chain has mass.

• Let us say we have a rope hanging between two ends of a pole . Half of the rope has W mass per unit length and the other half is massless . Both the parts are symmetrical . Will there be tension in the massless part of the rope ? – Harsh Vardhan Singh Oct 20 '20 at 11:40
• Yes, and that massless part will be straight, with no sag. – Chemomechanics Nov 2 '20 at 21:54