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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>
+ $2.1.23.$ A body of mass $m$ lying on a horizontal plane is affected by a force $F$ at an angle $\alpha$ to the horizon. Coefficient of friction $\mu$. Find the acceleration of the body if it does not detach from the plane.
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+ <center>
+<figure>
+<img src="https://savchenkosolutions.com/2/2.1.23/sol.jpg"
+loading="lazy" width="250" />
+<figcaption>
+Forces acting on the body
+</figcaption>
+</figure>
+</center>
+
+<div class="solutions">
+<p>
+1. The normal reaction of the connection in this case
+will be determined both by the force of gravity $mg$ and
+the projection on the axis $OY$ of the applied force:
+</p>
+<p style="text-align: center;">
+$N=mg-F \, sin \, \alpha$
+</p>
+<p>
+The friction force is determined as:
+</p>
+<p style="text-align: center;">
+$F_{тре} = (mg-F \,sin\, \alpha)$
+</p>
+<p>
+2. The basic law of dynamics, therefore. will be written as follows:
+</p>
+<p style="text-align: center;">
+$F \, cos \,\alpha = \mu (mg-F \,sin\, \alpha)$
+</p>
+<p>
+3. From the equation of Newton's second law it is easy to determine the desired acceleration
+</p>
+<p style="text-align: center;">
+$a = \frac{1}{m}(F\,cos\,\alpha-\mu mg+F\,sin\,\alpha)$
+</p>
+<p style="text-align: center;">
+$a = \frac{F}{m}(cos\,\alpha-\mu \,sin\,\alpha)$
+</p>
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $a = (F/m)(cos \,α + \mu \,sin \, \varphi)$$ − \mu g$ if this expression is greater than zero, otherwise $a = 0$.
+ </p>
+
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