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| ### Statement |
| ### Statement |
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| $4.2.10.$ [Insert the problem statement] |
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| ### Solution |
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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 $2 \text{ g}$. |
| $4.2.10.$ 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 |
| ### 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). |
| 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. |
| 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 $\alpha$: |
| 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. |
| * 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 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 $\frac{L}{6} \sin\alpha$. | | * 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 $O$: | | Let us write the torque equation about point $O$: |
| $$mg \cdot \frac{L}{6} \sin\alpha - T \cdot \frac{L}{3} \sin\alpha = 0$$ | | $$mg \cdot \frac{L}{6} \sin\alpha - T \cdot \frac{L}{3} \sin\alpha = 0$$ |
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| Canceling $L \sin\alpha$ from both sides of the equation: | | Canceling $L \sin\alpha$ from both sides of the equation: |
| $$mg \cdot \frac{1}{6} = T \cdot \frac{1}{3}$$ | | $$mg \cdot \frac{1}{6} = T \cdot \frac{1}{3}$$ |
| $$T = \frac{mg}{2}$$ | | $$T = \frac{mg}{2}$$ |
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| Substituting the numerical values ($m = 2 \text{ g} = 0.002 \text{ kg}$, $g \approx 10 \text{ m/s}^2$): | | Substituting the numerical values ($m = 2 \text{ g} = 0.002 \text{ kg}$, $g \approx 10 \text{ m/s}^2$): |
| $$T = \frac{0.002 \cdot 10}{2} = 0.01 \text{ N}$$ | | $$T = \frac{0.002 \cdot 10}{2} = 0.01 \text{ N}$$ |
