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+ <meta name="date" content="2023-10" scheme="YYYY-MM">
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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="../../#2.1">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <p>
+ $2.1.21.$ The belt lift forms an angle $\alpha$ with the horizon. With what maximum acceleration can a box be lifted on such a lift, if the coefficient of friction is equal to $\mu$? The tape doesn't bend.
+</p>
+<center>
+ <figure>
+ <img src="https://savchenkosolutions.com/2/2.1.21/statement.png"
+ loading="lazy" width="180" />
+ <figcaption>
+ For problem $2.1.21$
+ </figcaption>
+ </figure>
+</center>
+<p>
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+ <center>
+<figure>
+<img src="https://savchenkosolutions.com/2/2.1.21/sol.png"
+loading="lazy" width="180" />
+<figcaption>
+Forces acting on the box
+</figcaption>
+</figure>
+</center>
+
+<p>
+1. By discarding the constraints imposed on the box and replacing them with reactions, we can consider it as a free body capable of moving along the $OX$ axis. The friction force in this case is directed toward acceleration, i.e. against the possible movement of the box.
+</p>
+<p>
+2. The equation of Newton's second law allows us to determine the maximum value of acceleration
+$$ \mu mg \cos \alpha-mg \sin \alpha \leq ma $$
+
+$$ \boxed{a \leq g (\mu \cos \alpha-\sin \alpha)} $$
+</p>
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$a_{max} = g(\mu \cos \alpha − \sin \alpha)$$
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