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en/4.2.10.md
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| + | ### Statement | ||
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| + | $4.2.10.$ [Insert the problem statement] | ||
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| + | ### Solution | ||
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| + | ### Statement | ||
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| + | $4.2.10.$ Determine the tension force of the lower fishing line of the float shown in the figure, if the float is submerged in water by two-thirds of its length. The mass of the float is $2 \text{ g}$. | ||
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| + | ### Solution | ||
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| + | Three forces act on the float: gravity $mg$, Archimedes' buoyancy force $F_A$, and the tension of the lower line $T$. (The upper line going to the fishing rod is assumed to be slack, so we neglect its tension). | ||
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| + | Since the float is in equilibrium, the sum of the torques about any point must be zero. It is most convenient to choose point $O$ — the **center of buoyancy** (the point where the Archimedes force is applied) — as the axis of rotation. In this case, the torque of the Archimedes force becomes zero. | ||
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| + | Let us determine the lever arms of the remaining forces. Let the total length of the uniform float be $L$, and its angle of deviation from the vertical be $\alpha$: | ||
| + | * Since the float is submerged by $2/3$ of its length, the center of the submerged part (point $O$) is at a distance of $\frac{L}{3}$ from the bottom end. | ||
| + | * The tension force $T$ is applied to the bottom end. Its lever arm relative to point $O$ is $\frac{L}{3} \sin\alpha$. | ||
| + | * The center of gravity of the uniform float is exactly in the middle, at a distance of $\frac{L}{2}$ from the bottom end. Therefore, the distance from point $O$ to the center of gravity is $\frac{L}{2} - \frac{L}{3} = \frac{L}{6}$. The lever arm of the gravity force is $\frac{L}{6} \sin\alpha$. | ||
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| + | Let us write the torque equation about point $O$: | ||
| + | $$mg \cdot \frac{L}{6} \sin\alpha - T \cdot \frac{L}{3} \sin\alpha = 0$$ | ||
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| + | Canceling $L \sin\alpha$ from both sides of the equation: | ||
| + | $$mg \cdot \frac{1}{6} = T \cdot \frac{1}{3}$$ | ||
| + | $$T = \frac{mg}{2}$$ | ||
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| + | Substituting the numerical values ($m = 2 \text{ g} = 0.002 \text{ kg}$, $g \approx 10 \text{ m/s}^2$): | ||
| + | $$T = \frac{0.002 \cdot 10}{2} = 0.01 \text{ N}$$ | ||
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| + | #### Answer | ||
| + | $T = 0.01 \text{ N}$ | ||
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| + | #### Answer | ||
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| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $4.2.10.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $4.2.10.$ Determine the tension force of the lower fishing line of the float shown in the figure, if the float is submerged in water by two-thirds of its length. The mass of the float is $2 \text{ g}$. | |||
|  | |||
| ### Solution | |||
| Three forces act on the float: gravity $mg$, Archimedes' buoyancy force $F_A$, and the tension of the lower line $T$. (The upper line going to the fishing rod is assumed to be slack, so we neglect its tension). | |||
| Since the float is in equilibrium, the sum of the torques about any point must be zero. It is most convenient to choose point $O$ — the **center of buoyancy** (the point where the Archimedes force is applied) — as the axis of rotation. In this case, the torque of the Archimedes force becomes zero. | |||
| Let us determine the lever arms of the remaining forces. Let the total length of the uniform float be $L$, and its angle of deviation from the vertical be $\alpha$: | |||
| * Since the float is submerged by $2/3$ of its length, the center of the submerged part (point $O$) is at a distance of $\frac{L}{3}$ from the bottom end. | |||
| * The tension force $T$ is applied to the bottom end. Its lever arm relative to point $O$ is $\frac{L}{3} \sin\alpha$. | |||
| * The center of gravity of the uniform float is exactly in the middle, at a distance of $\frac{L}{2}$ from the bottom end. Therefore, the distance from point $O$ to the center of gravity is $\frac{L}{2} - \frac{L}{3} = \frac{L}{6}$. The lever arm of the gravity force is $\frac{L}{6} \sin\alpha$. | |||
| Let us write the torque equation about point $O$: | |||
| $$mg \cdot \frac{L}{6} \sin\alpha - T \cdot \frac{L}{3} \sin\alpha = 0$$ | |||
| Canceling $L \sin\alpha$ from both sides of the equation: | |||
| $$mg \cdot \frac{1}{6} = T \cdot \frac{1}{3}$$ | |||
| $$T = \frac{mg}{2}$$ | |||
| Substituting the numerical values ($m = 2 \text{ g} = 0.002 \text{ kg}$, $g \approx 10 \text{ m/s}^2$): | |||
| $$T = \frac{0.002 \cdot 10}{2} = 0.01 \text{ N}$$ | |||
| #### Answer | |||
| $T = 0.01 \text{ N}$ | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||