Edit to “Solution”

astrosander edited Dec 29, 2024, 12:26 AM

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revision #12605 parent #12563 ← older

en/2.4.9.md +4 −4

			### Statement

			$2.4.9^*.$ At the ends of a long thread, weights of mass $m$ each are suspended. The thread is spanned over two light little blocks located at a distance of $2l$ from each other. A $2m$ weight is attached to it in the middle between the blocks, and the system starts moving. Find the speed of loads after a sufficiently long period of time has elapsed.

			![ For problem $2.4.9^*$ \|520x322, 34%](../../img/2.4.9/2.4.9.png)

			### Solution

			![ Changing the position of weights \|1613x520, 76%](../../img/2.4.9/2.4.9_1.png)

			From the geometry of the drawing, a trigonometric expression can be obtained by

			$$
			\tan\varphi=\frac{l}{h}\tag{1}
			$$

			$$
			\sin\varphi=\frac{l}{\sqrt{l^2+h^2}};\quad\cos\varphi=\frac{h}{\sqrt{l^2+h^2}}\tag{2}
			$$

			Over the long period of time the $2m$ weight would go down infinitely ($h\to \infty$) From $(1)$, we obtain that as $h$ increases $\tan\varphi$ decreases

			$$
			\tan\varphi \to 0;\quad\varphi \to 0
			$$

			For the small angle $\varphi$ we could use an approximation to the second order of magnitude

			$$
			\sin\varphi \approx\tan\varphi\approx\varphi;\quad\cos\varphi\approx 1
			$$

			From the geometry of the figure, the length of the thread

			$$
			\sqrt{l^2+h^2}=l+\Delta x
			$$

			by the Pythagorean theorem, the change in the bottom of the thread connecting $2m$

			$$
			\Delta x = \sqrt{l^2+h^2}-l
			$$

			Similarly, from the geometry of the figure

		@@ -47,7 +47,7 @@Solution
			Considering $\sin\varphi\approx\tan\varphi$ and $h\gg l$, we obtain

			$$
		−	\sqrt{l^2+h^2}\approx h\Rightarrow \boxed{l=h-\Delta x}\tag{2}
		+	\sqrt{l^2+h^2}\approx h\Rightarrow \boxed{l=h-\Delta x}\tag{3}
			$$

			$L_1,~L_2$ — Length of threads at the initial moment

			Conservation of the mechanical energy $(dE_p+dE_k=0)$

			Two weights with mass $m$ went up to $\Delta x$ and the $2m$ went down $h$, making change in potential energy

			$$
			dE_p=-2mgh+2\cdot mg\Delta x
			$$

			In the meantime, the velocities of $m$ and $2m$ bodies become $v_1$ and $v_2$, respectively, making total kinetic energy of the system

			$$
			dE_k = 2\frac{mv_1^2}{2}+\frac{2m\cdot v_2^2}{2}
			$$

			Substituing into equations of conservation of the mechanical energy

			$$
			dE_p+dE_k=0\Leftrightarrow -2mgh+2\cdot mg\Delta x+2\frac{mv_1^2}{2}+\frac{2m\cdot v_2^2}{2}=0
		@@ -73,7 +73,7 @@Solution
			$$

			$$
		−	v_1^2+v_2^2=2g\left(h-\Delta x\right)\tag{3}
		+	v_1^2+v_2^2=2g\left(h-\Delta x\right)\tag{4}
			$$

			By the Newton's second law for the $m$ weight

			$$
			ma=T-mg
			$$

			Similarly, for the $2m$ weight

			$$
			2ma^*=2mg-2T\cos\varphi\approx2mg-2T\Leftrightarrow
			$$

			$$
			\Leftrightarrow ma^*=mg-T
			$$

			As they are connected by the same inextensible thread, their acceleration will be equal

			$$
		@@ -98,13 +98,13 @@Solution
			\left\|a\right\|=\left\|a^\right\|\Rightarrow v=v^
			$$

		−	Given that the system is being run from a stationary state $(v_0=v_0^*=0)$, let's substitute into $(3)$
		+	Given that the system is being run from a stationary state $(v_0=v_0^*=0)$, let's substitute into $(4)$

			$$
			2v^2=2g\left(h-\Delta x\right)
			$$

		−	Taking into consideration $(2)$
		+	Taking into consideration $(3)$

			$$
			v^2=gl\Rightarrow \boxed{v=\sqrt{gl}}
			$$

			#### Answer

			$$
			v=\sqrt{gl}
			$$
		unchanged lines 7