Physics Problem: Tension in a Flexible Chain
This solution details the calculation for the tension at the lowest point of a flexible chain based on the provided equilibrium conditions.
Chain Equilibrium Analysis
We analyze the forces acting on one half of the flexible chain hanging between two fixed points at the same level.
- Let the angle of inclination of the chain with the horizontal at the support be $\theta = 30^\circ$.
- We assume the mass of the considered half of the chain is $m$.
- The weight of this half chain is $W_{half} = mg$, acting vertically downwards.
- Let $T_0$ be the tension at the lowest point of the chain. This tension acts horizontally.
- Let $T_S$ be the tension at the point of support. This tension acts along the chain towards the support.
Calculating Tension at the Lowest Point
Applying the conditions for equilibrium to the half chain:
- Vertical Equilibrium: The upward vertical component of the tension at the support must balance the weight of the half chain. $ T_S \sin(\theta) = W_{half} $ Substituting the values $\theta = 30^\circ$ and $W_{half} = mg$: $ T_S \sin(30^\circ) = mg $ $ T_S \times \frac{1}{2} = mg $ $ T_S = 2mg $
- Horizontal Equilibrium: The tension at the lowest point ($T_0$) must balance the horizontal component of the tension at the support. $ T_0 = T_S \cos(\theta) $ Substituting the calculated $T_S = 2mg$ and $\theta = 30^\circ$: $ T_0 = (2mg) \cos(30^\circ) $ $ T_0 = 2mg \times \frac{\sqrt{3}}{2} $ $ T_0 = \sqrt{3}mg $
Therefore, the tension of the chain at the lowest point is $\sqrt{3}mg$.
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