Edits to “Solution”, “Answer”

Valter edited Oct 27, 2026, 9:45 AM

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revision #22265

en/4.1.2.md +1 −1

			### Statement:

			$4.1.2.$ A rectangular prism is in a liquid; its dimensions are shown in the figure. Find the sum of the forces acting on the front and bottom faces of the prism if the pressure of the liquid is $2 \cdot 10^5$ Pa. What is the sum of the forces acting on the prism?

			![For problem $4.1.2$\|636x504, 40%](../../img/4.1.2/statement.svg)

			### Solution:

			![\|439x592, 25%](../../img/4.1.2/Picture1.svg)

			The pressure on the faces of the prism:

			\[
			P = \rho \cdot g \cdot h = 2 \cdot 10^5 ~ \text{Pa}.
			\]

			Since the dimensions of the prism are small compared with the depth, we consider the pressure constant over its entire height.

			The pressure forces on the front and bottom faces are equal and are determined by the hydrostatic pressure $P$ and the area of a face of the prism:

			\[
			F_1 = F_2 = P \cdot a^2.
			\]

			Since the two faces are perpendicular, we find the resultant force using the Pythagorean theorem:
			\[
			\|F_{\text{R}}\| = \sqrt{F_1^2 + F_2^2} = 2000 \sqrt{2} ~ \text{N}
			\]

			According to Pascal's law, the pressure is balanced, and the resultant force:

			\[
		@@ -33,4 +33,4 @@Solution:
			F_{\text{sum}} = 0.
			\]

		−	#### Answer: $F_1 = 2000 \sqrt{2} ~N, ~F_2 = 0$
		+	#### Answer: $F_1 = 2000 \sqrt{2} ~N, ~F_{\text{sum}} = 0$

	### Statement:		### Statement:

	$4.1.2.$ A rectangular prism is in a liquid; its dimensions are shown in the figure. Find the sum of the forces acting on the front and bottom faces of the prism if the pressure of the liquid is $2 \cdot 10^5$ Pa. What is the sum of the forces acting on the prism?		$4.1.2.$ A rectangular prism is in a liquid; its dimensions are shown in the figure. Find the sum of the forces acting on the front and bottom faces of the prism if the pressure of the liquid is $2 \cdot 10^5$ Pa. What is the sum of the forces acting on the prism?

	![For problem $4.1.2$\|636x504, 40%](../../img/4.1.2/statement.svg)		![For problem $4.1.2$\|636x504, 40%](../../img/4.1.2/statement.svg)

	### Solution:		### Solution:

	![\|439x592, 25%](../../img/4.1.2/Picture1.svg)		![\|439x592, 25%](../../img/4.1.2/Picture1.svg)

	The pressure on the faces of the prism:		The pressure on the faces of the prism:

	\[		\[
	P = \rho \cdot g \cdot h = 2 \cdot 10^5 ~ \text{Pa}.		P = \rho \cdot g \cdot h = 2 \cdot 10^5 ~ \text{Pa}.
	\]		\]

	Since the dimensions of the prism are small compared with the depth, we consider the pressure constant over its entire height.		Since the dimensions of the prism are small compared with the depth, we consider the pressure constant over its entire height.

	The pressure forces on the front and bottom faces are equal and are determined by the hydrostatic pressure $P$ and the area of a face of the prism:		The pressure forces on the front and bottom faces are equal and are determined by the hydrostatic pressure $P$ and the area of a face of the prism:

	\[		\[
	F_1 = F_2 = P \cdot a^2.		F_1 = F_2 = P \cdot a^2.
	\]		\]

	Since the two faces are perpendicular, we find the resultant force using the Pythagorean theorem:		Since the two faces are perpendicular, we find the resultant force using the Pythagorean theorem:
	\[		\[
	\|F_{\text{R}}\| = \sqrt{F_1^2 + F_2^2} = 2000 \sqrt{2} ~ \text{N}		\|F_{\text{R}}\| = \sqrt{F_1^2 + F_2^2} = 2000 \sqrt{2} ~ \text{N}
	\]		\]

	According to Pascal's law, the pressure is balanced, and the resultant force:		According to Pascal's law, the pressure is balanced, and the resultant force:

	\[		\[
	@@ -33,4 +33,4 @@Solution:
	F_{\text{sum}} = 0.		F_{\text{sum}} = 0.
	\]		\]

	#### Answer: $F_1 = 2000 \sqrt{2} ~N, ~F_2 = 0$		#### Answer: $F_1 = 2000 \sqrt{2} ~N, ~F_{\text{sum}} = 0$