A flexible chain of weight W hangs between two fixed points A and B which are at same horizontal lev

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4-1.Newton's Laws of Motion

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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?

A$\frac{W}{2} \cdot \operatorname{cosec} \theta$

B$\frac{W}{2} \cdot \tan \theta$

C$0$

D$\frac{W}{2} \cdot \cot \theta$

Solution

$T \cos = T ^{\prime}$
$T \sin \theta=\frac{ W }{2}$
$T ^{\prime}=\frac{ W }{2} \cot \theta$

Standard 11

Physics

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