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en/5.5.4.md
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| + | ### Statement | ||
| + | |||
| + | $5.5.4.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | ### Problem | ||
| + | |||
| + | $5.5.4^*.$ A slab of mass $m$ is placed on two long cylindrical bags of radius $r$ and length $L \gg r$, made of an inextensible material and filled with gas. As a result, they are flattened to a thickness of $h \ll r$. The external pressure is $P_0$. Determine the initial pressure in the bags if the temperature of the gas in them did not change. | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | Since the material of the bags is inextensible, the perimeter of their cross-section does not change during deformation. Initially, the perimeter of one bag is $C = 2\pi r$. | ||
| + | |||
| + | After the slab is placed on them, the bags are flattened to a thickness of $h \ll r$. In this flattened state, the cross-section of the bag can be approximated as a rectangle with width $w$ and height $h$. The perimeter of this cross-section is $2w + 2h$. Using the condition $h \ll r$, we can neglect $2h$, so the perimeter is approximately $2w$. | ||
| + | Equating the initial and final perimeters: | ||
| + | $$ 2w \approx 2\pi r \implies w = \pi r $$ | ||
| + | |||
| + | Now we can write down the initial and final volumes for one bag: | ||
| + | Initial volume: $V_1 = \pi r^2 L$ | ||
| + | Final volume: $V_2 = w \cdot h \cdot L = \pi r h L$ | ||
| + | |||
| + | The slab of mass $m$ rests on two bags. The total horizontal contact area of the slab with both bags is: | ||
| + | $$ S = 2 \cdot w \cdot L = 2\pi r L $$ | ||
| + | |||
| + | In a state of mechanical equilibrium, the force of gravity acting on the slab is balanced by the difference between the force from the gas pressure inside the bags ($P_2$) and the external atmospheric pressure ($P_0$) acting on this contact area: | ||
| + | $$ mg = (P_2 - P_0)S $$ | ||
| + | From this, we express the final pressure in the bags: | ||
| + | $$ P_2 = P_0 + \frac{mg}{2\pi r L} $$ | ||
| + | |||
| + | According to the problem statement, the gas temperature does not change ($T = \text{const}$). Applying Boyle's Law for the initial and final states of the gas inside one bag: | ||
| + | $$ P_1 V_1 = P_2 V_2 $$ | ||
| + | |||
| + | Substituting the expressions for the volumes: | ||
| + | $$ P_1 (\pi r^2 L) = P_2 (\pi r h L) $$ | ||
| + | $$ P_1 \cdot r = P_2 \cdot h \implies P_1 = P_2 \frac{h}{r} $$ | ||
| + | |||
| + | Substituting the expression for $P_2$ into this equation and expanding the brackets: | ||
| + | $$ P_1 = \left( P_0 + \frac{mg}{2\pi r L} \right) \frac{h}{r} = P_0 \frac{h}{r} + \frac{mgh}{2\pi r^2 L} $$ | ||
| + | |||
| + | #### Answer: | ||
| + | |||
| + | $$ P_1 = P_0 \frac{h}{r} + \frac{mgh}{2\pi r^2 L} $$ | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $5.5.4.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Problem | |||
| $5.5.4^*.$ A slab of mass $m$ is placed on two long cylindrical bags of radius $r$ and length $L \gg r$, made of an inextensible material and filled with gas. As a result, they are flattened to a thickness of $h \ll r$. The external pressure is $P_0$. Determine the initial pressure in the bags if the temperature of the gas in them did not change. | |||
| ### Solution | |||
| Since the material of the bags is inextensible, the perimeter of their cross-section does not change during deformation. Initially, the perimeter of one bag is $C = 2\pi r$. | |||
| After the slab is placed on them, the bags are flattened to a thickness of $h \ll r$. In this flattened state, the cross-section of the bag can be approximated as a rectangle with width $w$ and height $h$. The perimeter of this cross-section is $2w + 2h$. Using the condition $h \ll r$, we can neglect $2h$, so the perimeter is approximately $2w$. | |||
| Equating the initial and final perimeters: | |||
| $$ 2w \approx 2\pi r \implies w = \pi r $$ | |||
| Now we can write down the initial and final volumes for one bag: | |||
| Initial volume: $V_1 = \pi r^2 L$ | |||
| Final volume: $V_2 = w \cdot h \cdot L = \pi r h L$ | |||
| The slab of mass $m$ rests on two bags. The total horizontal contact area of the slab with both bags is: | |||
| $$ S = 2 \cdot w \cdot L = 2\pi r L $$ | |||
| In a state of mechanical equilibrium, the force of gravity acting on the slab is balanced by the difference between the force from the gas pressure inside the bags ($P_2$) and the external atmospheric pressure ($P_0$) acting on this contact area: | |||
| $$ mg = (P_2 - P_0)S $$ | |||
| From this, we express the final pressure in the bags: | |||
| $$ P_2 = P_0 + \frac{mg}{2\pi r L} $$ | |||
| According to the problem statement, the gas temperature does not change ($T = \text{const}$). Applying Boyle's Law for the initial and final states of the gas inside one bag: | |||
| $$ P_1 V_1 = P_2 V_2 $$ | |||
| Substituting the expressions for the volumes: | |||
| $$ P_1 (\pi r^2 L) = P_2 (\pi r h L) $$ | |||
| $$ P_1 \cdot r = P_2 \cdot h \implies P_1 = P_2 \frac{h}{r} $$ | |||
| Substituting the expression for $P_2$ into this equation and expanding the brackets: | |||
| $$ P_1 = \left( P_0 + \frac{mg}{2\pi r L} \right) \frac{h}{r} = P_0 \frac{h}{r} + \frac{mgh}{2\pi r^2 L} $$ | |||
| #### Answer: | |||
| $$ P_1 = P_0 \frac{h}{r} + \frac{mgh}{2\pi r^2 L} $$ | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||