Правка раздела «Statement»

Valter правка от 26 авг. 2026 г., 15:07

правка #20341 предыдущая #10161 ← раньше

en/4.2.14.md +1 −1

		@@ -1,6 +1,6 @@
			### Statement

		−	$4.2.14$ Determine the pressure force of the logs of mass $m$ on the walls of the channel. The upper log is half submerged in water, and the lower log touches the upper part of the water surface. The logs are equal.
		+	$4.2.14.$ Determine the pressure force of the logs of mass $m$ on the walls of the channel. The upper log is half submerged in water, and the lower log touches the upper part of the water surface. The logs are equal.

			![ For problem 4.2.14 \|325x312, 51%](../../img/4.2.14/statement.png)

			### Solution

			Let's consider the following figure

			![ Force analysis \|1057x447, 92%](../../img/4.2.14/draw.png)

			From right figure

			$$
			\cos{\theta} = \frac{R}{2R} = 0.5
			$$

			so, $\theta = 60^\circ$. Applying Newton Second Law for submerged trunk on $x$-axis,

			$$
			N = N_t~\sin{\theta} \tag{1}
			$$

			on $y$-axis

			$$
			F_{A1} = mg + N_t~\cos{\theta}
			$$

			$$
			\rho_w~g~V = mg + N_t~\cos{\theta} \tag{2}
			$$

			For half-submerged trunk, on $x$-axis,

			$$
			N = N_t~\sin{\theta}
			$$

			on $y$-axis

			$$
			F_{A2} + N_t~\cos{\theta} = mg
			$$

			$$
			\rho_w~g~\frac{V}{2} + N_t~\cos{\theta} = mg \tag{3}
			$$

			Substituting $(2)$ into $(3)$, it is obtained

			$$
			N_t = \frac{mg}{3~\cos{\theta}} \tag{4}
			$$

			Putting $(4)$ into $(1)$

			$$
			N = \frac{mg}{3}~\tan{\theta}
			$$

			$$
			\boxed{N = \frac{mg}{\sqrt{3}}}
			$$

			Note: You can prove that both pressure forces generated by trunks are equals.






		ещё строк без изменений 67

	@@ -1,6 +1,6 @@
	### Statement		### Statement

	$4.2.14$ Determine the pressure force of the logs of mass $m$ on the walls of the channel. The upper log is half submerged in water, and the lower log touches the upper part of the water surface. The logs are equal.		$4.2.14.$ Determine the pressure force of the logs of mass $m$ on the walls of the channel. The upper log is half submerged in water, and the lower log touches the upper part of the water surface. The logs are equal.

	![ For problem 4.2.14 \|325x312, 51%](../../img/4.2.14/statement.png)		![ For problem 4.2.14 \|325x312, 51%](../../img/4.2.14/statement.png)

	### Solution		### Solution

	Let's consider the following figure		Let's consider the following figure

	![ Force analysis \|1057x447, 92%](../../img/4.2.14/draw.png)		![ Force analysis \|1057x447, 92%](../../img/4.2.14/draw.png)

	From right figure		From right figure

	$$		$$
	\cos{\theta} = \frac{R}{2R} = 0.5		\cos{\theta} = \frac{R}{2R} = 0.5
	$$		$$

	so, $\theta = 60^\circ$. Applying Newton Second Law for submerged trunk on $x$-axis,		so, $\theta = 60^\circ$. Applying Newton Second Law for submerged trunk on $x$-axis,

	$$		$$
	N = N_t~\sin{\theta} \tag{1}		N = N_t~\sin{\theta} \tag{1}
	$$		$$

	on $y$-axis		on $y$-axis

	$$		$$
	F_{A1} = mg + N_t~\cos{\theta}		F_{A1} = mg + N_t~\cos{\theta}
	$$		$$

	$$		$$
	\rho_w~g~V = mg + N_t~\cos{\theta} \tag{2}		\rho_w~g~V = mg + N_t~\cos{\theta} \tag{2}
	$$		$$

	For half-submerged trunk, on $x$-axis,		For half-submerged trunk, on $x$-axis,

	$$		$$
	N = N_t~\sin{\theta}		N = N_t~\sin{\theta}
	$$		$$

	on $y$-axis		on $y$-axis

	$$		$$
	F_{A2} + N_t~\cos{\theta} = mg		F_{A2} + N_t~\cos{\theta} = mg
	$$		$$

	$$		$$
	\rho_w~g~\frac{V}{2} + N_t~\cos{\theta} = mg \tag{3}		\rho_w~g~\frac{V}{2} + N_t~\cos{\theta} = mg \tag{3}
	$$		$$

	Substituting $(2)$ into $(3)$, it is obtained		Substituting $(2)$ into $(3)$, it is obtained

	$$		$$
	N_t = \frac{mg}{3~\cos{\theta}} \tag{4}		N_t = \frac{mg}{3~\cos{\theta}} \tag{4}
	$$		$$

	Putting $(4)$ into $(1)$		Putting $(4)$ into $(1)$

	$$		$$
	N = \frac{mg}{3}~\tan{\theta}		N = \frac{mg}{3}~\tan{\theta}
	$$		$$

	$$		$$
	\boxed{N = \frac{mg}{\sqrt{3}}}		\boxed{N = \frac{mg}{\sqrt{3}}}
	$$		$$

	Note: You can prove that both pressure forces generated by trunks are equals.		Note: You can prove that both pressure forces generated by trunks are equals.






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