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+ <h2>Solutions of Savchenko Physics Textbook</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <h3 id="back-link"><a href="../#2.3">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <p>
+ $2.3.30.$ Determine the force exerted on the vertical wall by the falling dumbbell when the dumbbell axis is at an angle $\alpha$ with the horizon. The dumbbell starts moving from a vertical position without initial velocity. Mass of each dumbbell ball $m$.
+
+</p>
+<center>
+ <figure>
+ <img src="statement.png"
+ loading="lazy" width="130" />
+ <figcaption>
+ For problem $2.3.30$
+ </figcaption>
+ </figure>
+</center>
+<p>
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+
+</p>
+<center>
+ <figure>
+ <img src="draw.jpg"
+ loading="lazy" width="280" />
+ <figcaption>
+ Forces acting on the system
+ </figcaption>
+ </figure>
+</center>
+<p>
+
+Along the axis $OZ_\parallel$ "The Two-Ball System":
+</p><p>
+Centripetal acceleration associated with circular rotation:
+$$a=\frac{v^2}{R}$$
+From the Newton's second law:
+$$\frac{mv^2}{R}=2mg\sin\alpha-N_x\cos\alpha-N_y\sin\alpha$$
+Law of Conservation of Energy
+$$mgR=mgh+\frac{mv^2}{2}$$
+$$h=R\sin\alpha$$
+$$\frac{mv^2}{R}=2mg(1-\sin\alpha)$$
+$$N_y\sin\alpha+N_x\cos\alpha=(4\sin\alpha-2)mg$$
+Along the axis $OZ_\perp$ "The Ball System":
+$$N_y\cos\alpha=N_x\sin\alpha+mg\cos\alpha$$
+$$N_y=mg+N_x\tan\alpha$$
+$$mg\sin\alpha\cos\alpha+N_x\sin^2\alpha+N_x\cos^2\alpha=2mg(2\sin\alpha-2)\cos\alpha$$
+$$\frac{N_x}{mg}=4\sin\alpha\cos\alpha-2\cos\alpha-\sin\alpha\cos\alpha$$
+$$\boxed{N_x=(3\sin\alpha-2)\cos\alpha \cdot mg}$$
+When $N_x \leq0$, the shoe will break away from the vertical wall and will no longer act on it:
+$$N_x=0$$
+Condition of detachment:
+$$(3\sin\alpha-2)\cos\alpha \leq0$$
+$$3\sin\alpha-2\leq0 \Leftrightarrow \boxed{\sin\alpha\leqslant2/3}$$
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$F=mg\cos\alpha(3\sin\alpha-2)\mathrm{~with~}\sin\alpha\geqslant2/3;$$
+$$F=0\mathrm{~with~}\sin\alpha\leqslant2/3.$$
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