A flexible chain of weight W hangs between tw | vedclass

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(D) Let $T$ be the tension at the points of support $A$ and $B$. The vertical component of the tension at each support must balance half the weight of

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $T$ be the tension at the points of support $A$ and $B$. The vertical component of the tension at each support must balance half the weight of the chain. Thus,$2T \sin 	heta = W$,which gives $T \sin 	heta = \frac{W}{2}$.At the lowest point (mid-point) of the chain,the tension is purely horizontal,let this be $T'$.Considering the equilibrium of half of the chain,the horizontal forces must balance. The horizontal component of the tension at the support is $T \cos 	heta$,which must be equal to the tension $T'$ at the lowest point.Therefore,$T' = T \cos 	heta$.From $T \sin 	heta = \frac{W}{2}$,we have $T = \frac{W}{2 \sin 	heta}$.Substituting this into the expression for $T'$,we get $T' = \left( \frac{W}{2 \sin 	heta} ight) \cos 	heta = \frac{W}{2} \cot 	heta$.

$A$ 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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