Translated 2.1.48

astrosander edited
View solution History
revision #6392 GitHub 6f04e54 newer →
en/2.1.48.md +132 −0
@@ -0,0 +1,132 @@
+<!DOCTYPE html>
+<html lang="en">
+
+<head>
+ <meta charset="utf-8">
+ <meta name="viewport" content="width=device-width, initial-scale=1.0">
+ <meta http-equiv="content-language" content="en">
+ <meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
+ <meta name="description" content="A wedge of mass m_2 with an angle \alpha is inserted between two identical smooth bars of mass m_1. Determine the acceleration of bodies.">
+ <meta name="author" content="Aliaksandr Melnichenka">
+ <meta name="date" content="2023-10" scheme="YYYY-MM">
+ <meta property="og:title" content="A wedge of mass m_2 with an angle \alpha is inserted between two identical smooth bars of mass m_1. Determine the acceleration of bodies.">
+ <meta property="og:image" content="img/logo.png">
+ <meta property="og:description" content="A wedge of mass m_2 with an angle \alpha is inserted between two identical smooth bars of mass m_1. Determine the acceleration of bodies.">
+ <meta name="yandex-verification" content="6cfda41f74038368">
+ <title>A wedge of mass m_2 with an angle \alpha is inserted between two identical smooth bars of mass m_1. Determine the acceleration of bodies.</title>
+ <link rel="stylesheet" href="https://savchenkosolutions.com/css/css-latex/style.css">
+ <link rel="icon" href="https://savchenkosolutions.com/img/logo.png" type="image/png">
+ <script src="https://savchenkosolutions.com/js/jquery-1.10.1.min.js"></script>
+ <script async src="https://cdnjs.cloudflare.com/ajax/libs/mathjax/2.7.7/MathJax.js?config=TeX-MML-AM_CHTML"></script>
+ <script type="text/x-mathjax-config">
+ MathJax.Hub.Config({
+ extensions: ['tex2jax.js'],
+ jax: ['input/TeX', 'output/HTML-CSS'],
+ tex2jax: {
+ inlineMath: [['$', '$'], ['$', '$']],
+ processEscapes: true,
+ processClass: 'tex2jax',
+ ignoreClass: 'html'
+ },
+ showProcessingMessages: false,
+ messageStyle: 'none'
+ });
+ </script>
+</head>
+ <body style="">
+ <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="../../#2.1">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <p>
+ $2.1.48.$ A wedge of mass $m_2$ with an angle $\alpha$ is inserted between two identical smooth bars of mass $m_1$. Determine the acceleration of bodies.
+</p>
+<center>
+ <figure>
+ <img src="https://savchenkosolutions.com/2/2.1.48/statement.png"
+ loading="lazy" width="230" />
+ <figcaption>
+ For problem $2.1.48$
+ </figcaption>
+ </figure>
+</center>
+<p>
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+
+</p>
+<center>
+ <figure>
+ <img src="https://savchenkosolutions.com/2/2.1.48/sol.png"
+ loading="lazy" width="400" />
+ <figcaption>
+ Forces acting on the system
+ </figcaption>
+ </figure>
+</center>
+<p>
+Let us consider a small displacement over time $dt$ from the point of view of kinematics:
+
+</p>
+<center>
+ <figure>
+ <img src="https://savchenkosolutions.com/2/2.1.48/sol1.png"
+ loading="lazy" width="280" />
+ <figcaption>
+ Small transition in time $dt$
+ </figcaption>
+ </figure>
+</center>
+<p>
+
+From geometric considerations:
+$$ dy = dx \cdot ctg\,\beta $$
+Let us differentiate both parts of the expression twice:
+$$ \frac{dy}{d^2t} = \frac{dx}{d^2t} \cdot ctg\,\beta $$
+
+$$ a_1 = a_2 \cdot tg\,\beta\;(1) $$
+Since the Author did not say anything about friction, example $F_{tr}=0$
+Next, we write Newton's 2nd law for the bar $m_1$ on the $OY$ axis:
+$$ m_1a_1=N_1\,cos\,\beta $$
+
+$$ N_1=\frac{m_1a_1}{cos\,\beta}\;(a) $$
+Similarly for $m_2$:
+$$ m_2a_2=m_2g-2N_1\,sin\,\beta\;(b) $$
+Substitute $(a)$ into $(b)$:
+$$ m_2a_2=m_2g-2m_1a_1\cdot tg\,\beta $$
+We substitute $(1)$ into $(c)$:
+$$ m_2a_2=m_2g-2m_1a_2\cdot tg^2\,\beta $$
+We express $a_2$:
+$$ \boxed{a_2=\frac{m_2g}{m_2+2m_1\,tg^2\,\beta}}\;(d) $$
+Given $(1)$, we multiply $(d)$ by $tg\,\beta$:
+$$ \boxed{a_1=\frac{m_2g\,tg\,\beta}{m_2+2m_1\,tg^2\,\beta}} $$
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$a_1=\frac{m_2g\,tg\,\beta}{m_2+2m_1\,tg^2\,\beta}$$
+ $$a_2=\frac{m_2g}{m_2+2m_1\,tg^2\,\beta}$$
+ </p>
+
+
+ <footer class="row container">
+ <br>
+ <p>
+ <small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small>
+ </p>
+ <p>
+ <small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
+ </p>
+ </footer>
+ </body>
+
+</html>