Translated 2.1.34-2.1.47

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+ <meta name="description" content="A coin is placed on a plane whose slope tangent is equal to the coefficient of friction. In the horizontal direction along the plane, the coin was given a velocity v. Find the steady speed of the coin.">
+ <meta name="author" content="Aliaksandr Melnichenka">
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+ <header style="text-align:center;">
+ <div id = "logo">
+ <span><img src = "../../img/book.png"><span><span>Savchenko Solutions</span>
+ </div>
+ <p class="author">
+ Solutions of Savchenko Problems in Physics <br>
+ <i><b>knowledge must be free</b></i>
+ </p>
+ </header>
+
+ <h3 id="back-link"><a href="../../#2.1">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <p>
+ $2.1.45^*.$ A coin is placed on a plane whose slope tangent is equal to the coefficient of friction. In the horizontal direction along the plane, the coin was given a velocity $v$. Find the steady speed of the coin.
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+ <center>
+<figure>
+<img src="https://savchenkosolutions.com/2/2.1.45/sol.jpg"
+loading="lazy" width="250" />
+<figcaption>
+Forces acting on the body
+</figcaption>
+</figure>
+</center>
+<p>
+Projecting $m\vec{g}$ on $OY$:
+</p>
+<br>
+<center>
+<figure>
+<img src="https://savchenkosolutions.com/2/2.1.45/sol1.jpg"
+loading="lazy" width="250" />
+<figcaption>
+Projecting on $OY$
+</figcaption>
+</figure>
+</center>
+<p>
+From the figure, we find the support reaction force:
+$$ N=mg \cos \alpha $$
+According to the Amontons-Coulomb Law:
+$$ F_{тр}=\mu N=\mu mg \cos \alpha $$
+Since $\mu = tg \alpha$ (by condition):
+$$ F_{тр}=\mu mg \cos \alpha $$
+
+$$ F_{тр}=mg \cdot tg \alpha \cdot \cos \alpha=mg \sin \alpha $$
+
+$$ F=mg \sin \alpha $$
+Redraw in the $XY$ plane
+At the initial moment $(\varphi=90^{\circ})$:
+</p>
+<br>
+<center>
+<figure>
+<img src="https://savchenkosolutions.com/2/2.1.45/sol3.jpg"
+loading="lazy" width="250" />
+<figcaption>
+Drawing in the $xOy$ plane
+</figcaption>
+</figure>
+</center>
+<p>
+At the final moment $(\varphi=0^{\circ})$:
+</p>
+<br>
+<center>
+<figure>
+<img src="https://savchenkosolutions.com/2/2.1.45/sol4.jpg"
+loading="lazy" width="250" />
+<figcaption>
+Forces acting on the body at the final moment
+</figcaption>
+</figure>
+</center>
+<p>
+Let's consider an arbitrary moment:
+</p>
+<br>
+<center>
+<figure>
+<img src="https://savchenkosolutions.com/2/2.1.45/sol2.jpg"
+loading="lazy" width="250" />
+<figcaption>
+Forces acting on a body at an arbitrary moment
+</figcaption>
+</figure>
+</center>
+<p>
+Note that $\varphi$ changes from $0^{\circ}$ to $90^{\circ}$:
+</p>
+<p>
+Let's write Newton's second law:
+$$ \left\{\begin{matrix}
+ \frac{du_x}{dt} = \frac{F \sin \varphi }{m}& \\
+ \frac{du_y}{dt} = \frac{F (1- \cos \varphi) }{m} &
+ \end{matrix}\right. $$
+
+$$ \left\{\begin{matrix}
+ \frac{d}{dt} (u \sin \varphi)= \frac{F \sin \varphi }{m}& \\
+ \frac{d}{dt} (u \cos \varphi) = \frac{F (1- \cos \varphi) }{m} &
+ \end{matrix}\right. $$
+We solve the system of differential equations:
+$$ \left\{\begin{matrix}
+ \frac{d\varphi}{dt} = \frac{F \sin \varphi }{mu}& (a)\\
+ \frac{du}{dt} = \frac{F (1- \cos \varphi) }{m} & (b)
+ \end{matrix}\right. $$
+We divide $(b)$ by $(a)$:
+$$ \frac{du}{d\varphi} = u\frac{1- \cos \varphi}{\sin \varphi} $$
+
+$$ \frac{du}{u} = \frac{1- \cos \varphi}{\sin \varphi} d\varphi $$
+
+$$ \frac{du}{u} = tg (\frac{\varphi}{2}) d\varphi $$
+We integrate both parts of the equation:
+<!-- $\int_{v_0}^{v}\frac{du}{u} = \int_{90^{\circ}}^{0^{\circ}}tg (\frac{\varphi}{2}) d\varphi \; (c)$ -->
+$$ \int \frac{du}{u} = \int tg (\frac{\varphi}{2}) d\varphi \; (c) $$
+
+$$ \int\frac{du}{u}=ln|u| \; (d) $$
+
+$$ \int tg (\frac{\varphi}{2}) d\varphi=-2 ln(\cos(\frac{x}{2})) \; (e) $$
+Substitute $(e)$ and $(d)$ into $(c)$:
+<div class="scroll-wrapper">
+$$ ln(v)+C=ln(\sin(\varphi)) -(ln(\sin(\frac{\varphi}{2})-\cos(\frac{\varphi}{2})))\; (e) $$
+</div>
+$$ \boxed{v_x=\frac{v}{2}} $$
+</p>
+ <h4>Answer</h4>
+ <p>
+ $$ v_x=\frac{v}{2} $$
+ </p>
+
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