A flexible chain of mass m hangs between two fixed points at the same level. The inclination of the chain with the horizontal at the two points of support is $$30^{\circ}$$. Considering the equilibrium of each half of the chain, the tension of the chain at the lowest point is __________.

Take the chain and split it into two equal halves. Because both supports are at the same level, the situation is symmetric.

Now focus on only one half of the chain.

For this half:

• its weight acts downward and equals $$\frac{mg}{2}$$​
• the tension at the support is $$T_s$$ making an angle $$30^{\circ}$$ with the horizontal
• at the lowest point, the tension is horizontal (no vertical component there)

Now apply equilibrium.

Vertical direction:

Only the support tension provides an upward component, so it must balance the weight of half the chain:

$$T_s\sin30^{\circ}=\frac{mg}{2}$$

$$Since\ \sin30^{\circ}=\frac{1}{2}$$

$$T_s\cdot\frac{1}{2}=\frac{mg}{2}$$

$$\Rightarrow T_s=mg$$

Now look at horizontal direction.

The horizontal component of tension is the same everywhere in the chain. That means the tension at the lowest point is equal to the horizontal component of TsT_sTs​:

$$T_{lowest}=T\cos30^{\circ\ }$$

Substitute $$T_s=mg\ and\cos30^{\circ}=\frac{\sqrt{3}}{2}$$

$$T_{\text{lowest}}=mg\cdot\frac{\sqrt{3}}{2}$$

So the tension at the lowest point is

$$T_{\text{lowest}}=mg\cdot\frac{\sqrt{3}}{2}$$