Laws of MotionMedium
Question
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 $\_\_\_\_$ .
Options
A.$\frac{\sqrt{3}}{2}mg$
B.$\frac{1}{2}mg$
C.$mg$
D.$\sqrt{3}mg$
Solution
$$\begin{matrix} & Tsin30^{\circ} = \frac{m}{2}\text{ }g \\ & \text{ }Tcos30^{\circ} = T_{0} \\ & tan30^{\circ} = \frac{mg}{2{\text{ }T}_{0}} \\ & {\text{ }T}_{0} = \frac{\sqrt{3}}{2}mg \end{matrix}$$
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