A flexible chain of weight W hangs between tw | vedclass

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(C) Consider half of the chain. The vertical component of the tension at the support must balance half the weight of the chain.$T \sin 	heta = \frac{

Vedclass · Accepted Answer

$\frac{W}{2} \cot 	heta$

Vedclass · Answer

(C) Consider half of the chain. The vertical component of the tension at the support must balance half the weight of the chain.$T \sin 	heta = \frac{W}{2}$At the lowest point (mid-point),the tension is purely horizontal,let this be $T_0$.For the equilibrium of the half-chain,the horizontal component of the tension at the support must be equal to the tension at the lowest point.$T \cos 	heta = T_0$From the first equation,$T = \frac{W}{2 \sin 	heta}$.Substituting this into the second equation:$T_0 = \left( \frac{W}{2 \sin 	heta} ight) \cos 	heta$$T_0 = \frac{W}{2} \cot 	heta$

$A$ flexible chain of weight $W$ hangs between two fixed points $A$ and $B$ which are at the same horizontal level. The inclination of the chain with the horizontal at both the points of support is $\theta$. What is the tension of the chain at the mid-point?

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