Question

A flexible chain of mass M hangs between two fixed points A and B at the same level. The inclination of the chain with the horizontal at the Two points of support is theta. Find the tension at the mid point of the chain

Answer 1

See the freebody diagram with the tensions at one end (T1) and at the center of the chain (T2).

Balance the forces to get the answer.
T2 = T1* Cosθ
T1 = M/2 * g * Sinθ

So T2 = Mg/2 * Cotθ.

Answer 2

The mass of chain is M, A and B are the points between which the chain M hangs. Since the chain has certain weight it will sag between the points A and B and there will be a vertical tension as well as a horizontal tension so at the midpoint the mass will be taken as M/2 and the vertical force acting would be towards the ground that is due to gravity  

The tensions T1 is at one end and T2 at the center of the chain

Now to balance the forces to get the answer.  

T2 = T1 x Cos θ  

$\begin{array}{l}{\mathrm{T} 1=\frac{M}{2} \mathrm{x} \mathrm{g} \mathrm{x} \sin \theta} \\ {\text { So } \mathrm{T} 2=\frac{M g}{2} \mathrm{x} \cot \theta}\end{array}$