Новое решение

Valter правка от 5 авг. 2026 г., 18:07

правка #19855 позже →

en/5.5.4.md +47 −0

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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]