A flexible chain of mass m hangs between two | vedclass

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Question

(A) Consider the free-body diagram of one half of the chain.The forces acting on this half are:$1$. The tension $T$ at the support point, acting at an

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{2} mg$

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(A) Consider the free-body diagram of one half of the chain.The forces acting on this half are:$1$. The tension $T$ at the support point, acting at an angle of $30^{\circ}$ with the horizontal.$2$. The tension $T_0$ at the lowest point, acting horizontally.$3$. The weight of the half-chain, which is $\frac{m}{2}g$, acting vertically downwards.For equilibrium in the vertical direction:$T \sin 30^{\circ} = \frac{m}{2}g$$T 	imes \frac{1}{2} = \frac{mg}{2} \implies T = mg$For equilibrium in the horizontal direction:$T \cos 30^{\circ} = T_0$Substituting $T = mg$:$T_0 = mg \cos 30^{\circ} = mg 	imes \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} mg$

$A$ 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 . . . . . . .

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