Tension at the lowest point of a hanging chain

[Joe8]

Homework Statement

Chain of mass M hangs between 2 walls with its ends at the same height. The chain makes an angle θ with each wall. Find the tension at the lowest point of the chain.

a) By considering the forces on half of the chain.

b) By using the fact that the height of the chain is given by y(x) = (1/α) cosh(αx), and considering the vertical forces on an infinitesimal piece at the bottom. This will give the tension in terms of α. Then find α in terms of the given angle θ.
Attempt:
[/B]

Homework Equations

The Attempt at a Solution

a) I called the tension force at the end of the chain attached to the wall T1 and the tension at the bottom To. Assuming that To is horizontal with no vertical component ( I am not sure that this is a fair assumption):

T1 Sinθ = To (ΣFx=0)
T1 Cosθ= Mg/2 (since we are working on half the chain and ΣFy=0)

Therefore To=Mg(tanθ)/2

b) Now here I have no idea. If I use the same assumption that allowed me to solve a (that the forces at the bottom segment of the chain are horizontal) then there is only the weight (dm * g) in the vertical direction...[/B]

 

[andrewkirk]

Consider a very small piece of chain of length dl, centred at the exact mid-point of the chain, at the bottom. You can calc its weight, and it is supported by the connections to the rest of the chain on either side. Take the derivative of the cosh formula at the points dl/2 from the centre on either side, to get the angle of those supporting pulls.

 

[hawkeye1029]

You could either calculate T1 and To separately using Newton's 2nd Law [involving tension and mass] and combine them after, or you could treat both as part of the same system.