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+ <meta name="description" content="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.">
+ <meta name="author" content="Aliaksandr Melnichenka">
+ <meta name="date" content="2023-10" scheme="YYYY-MM">
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+ <title>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.</title>
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+ <header style="text-align:center;">
+ <h2>Solutions of Savchenko Problems in Physics</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <h3 id="back-link"><a href="../../#2.4">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <p>
+ $2.4.9^*.$ 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.
+</p>
+<center>
+ <figure>
+ <img src="2.4.9.png"
+ loading="lazy" width="200" />
+ <figcaption>
+ For problem $2.4.9^*$
+ </figcaption>
+ </figure>
+</center>
+<p>
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+
+</p>
+<center>
+ <figure>
+ <img src="2.4.9_1.png"
+ loading="lazy" width="450" />
+ <figcaption>
+ Changing the position of weights
+ </figcaption>
+ </figure>
+</center>
+<p>
+
+From the geometry of the drawing, a trigonometric expression can be obtained by
+$$\tan\varphi=\frac{l}{h}\quad(1)$$
+$$\sin\varphi=\frac{l}{\sqrt{l^2+h^2}};\quad\cos\varphi=\frac{h}{\sqrt{l^2+h^2}}\quad(2)$$
+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
+$$\sqrt{l^2+h^2}=l+\Delta x $$
+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</p><p>
+Considering $h\gg l$,
+$$\sqrt{l^2+h^2}\approx l\Rightarrow \boxed{h-\Delta x=l}\quad(2)$$
+$L_1,~L_2$ — Length of threads at the initial moment</p><p>
+Conservation of the mechanical energy
+$$2mgL_1-2mgL_2=\frac{2mv_2^2}{2}-\frac{2mv_1^2}{2}$$
+$$v_1^2+v_2^2=2g\left(h-\Delta x\right)\quad(3)$$
+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
+$$\left|a\right|=\left|a^*\right|\Rightarrow v=v^*$$
+Given that the system is being run from a stationary state $(v_0=v_0^*=0)$, let's substitute into $(3)$
+$$2v^2=2g\left(h-\Delta x\right)$$
+Taking into consideration $(2)$
+$$v^2=gl\Rightarrow \boxed{v=\sqrt{gl}}$$
+
+
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$v=\sqrt{gl}$$
+ </p>
+
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