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			<title>Determine the force exerted on the vertical wall by the falling dumbbell when the dumbbell axis is at an angle lpha with the horizon. The dumbbell starts moving from a vertical position without initial velocity. Mass of each dumbbell ball m.</title>
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			Solutions of Savchenko Problems in Physics <br>
			<i><b>knowledge must be free</b></i>
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			<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 $m$.

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			For problem $2.3.30$
			</figcaption>
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			<p>
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			<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":
		@@ -95,7 +95,7 @@
			$$N_y\cos\alpha=N_x\sin\alpha+mg\cos\alpha$$
			$$N_y=mg+N_x\tan\alpha$$
			$N_y$ found from the axis $OZ_\perp$ we substitute into $N_y$ found from the axis $OZ_\parallel$
		−	$$mg\sin\alpha\cos\alpha+N_x\sin^2\alpha+N_x\cos^2\alpha=2mg(2\sin\alpha-2)\cos\alpha$$
		+	$$mg\sin\alpha\cos\alpha+N_x\sin^2 \alpha+N_x\cos^2 \alpha=2mg(2\sin\alpha-2)\cos\alpha$$
			$$\frac{N_x}{mg}=4\sin\alpha\cos\alpha-2\cos\alpha-\sin\alpha\cos\alpha$$
			$$\boxed{N_x=(3\sin\alpha-2)\cos\alpha \cdot mg}$$
			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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			<p style="text-align: right; font-style: italic; font-size: 14;">
			Almaskhan Arsen<br>
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	Solutions of Savchenko Problems in Physics <br>		Solutions of Savchenko Problems in Physics <br>
	<i><b>knowledge must be free</b></i>		<i><b>knowledge must be free</b></i>
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	<h3 id="back-link"><a href="../../#2.3">$\leftarrow$Back</a></h3>		<h3 id="back-link"><a href="../../#2.3">$\leftarrow$Back</a></h3>

	<h3> Statement </h3>		<h3> Statement </h3>
	<p>		<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 $m$.		$2.3.30.$ 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 $m$.

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	<img src="statement.png"		<img src="statement.png"
	loading="lazy" width="130" />		loading="lazy" width="130" />
	<figcaption>		<figcaption>
	For problem $2.3.30$		For problem $2.3.30$
	</figcaption>		</figcaption>
	</figure>		</figure>
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	<p>		<p>
	</p>		</p>

	<h3>Solution</h3>		<h3>Solution</h3>
	<p>		<p>

	</p>		</p>
	<center>		<center>
	<figure>		<figure>
	<img src="draw.jpg"		<img src="draw.jpg"
	loading="lazy" width="280" />		loading="lazy" width="280" />
	<figcaption>		<figcaption>
	Forces acting on the system		Forces acting on the system
	</figcaption>		</figcaption>
	</figure>		</figure>
	</center>		</center>
	<p>		<p>

	Along the axis $OZ_\parallel$ "The Two-Ball System":		Along the axis $OZ_\parallel$ "The Two-Ball System":
	</p><p>		</p><p>
	Centripetal acceleration associated with circular rotation:		Centripetal acceleration associated with circular rotation:
	$$a=\frac{v^2}{R}$$		$$a=\frac{v^2}{R}$$
	From the Newton's second law:		From the Newton's second law:
	$$\frac{mv^2}{R}=2mg\sin\alpha-N_x\cos\alpha-N_y\sin\alpha$$		$$\frac{mv^2}{R}=2mg\sin\alpha-N_x\cos\alpha-N_y\sin\alpha$$
	Law of Conservation of Energy		Law of Conservation of Energy
	$$mgR=mgh+\frac{mv^2}{2}$$		$$mgR=mgh+\frac{mv^2}{2}$$
	$$h=R\sin\alpha$$		$$h=R\sin\alpha$$
	$$\frac{mv^2}{R}=2mg(1-\sin\alpha)$$		$$\frac{mv^2}{R}=2mg(1-\sin\alpha)$$
	$$N_y\sin\alpha+N_x\cos\alpha=(4\sin\alpha-2)mg$$		$$N_y\sin\alpha+N_x\cos\alpha=(4\sin\alpha-2)mg$$
	Along the axis $OZ_\perp$ "The Ball System":		Along the axis $OZ_\perp$ "The Ball System":
	@@ -95,7 +95,7 @@
	$$N_y\cos\alpha=N_x\sin\alpha+mg\cos\alpha$$		$$N_y\cos\alpha=N_x\sin\alpha+mg\cos\alpha$$
	$$N_y=mg+N_x\tan\alpha$$		$$N_y=mg+N_x\tan\alpha$$
	$N_y$ found from the axis $OZ_\perp$ we substitute into $N_y$ found from the axis $OZ_\parallel$		$N_y$ found from the axis $OZ_\perp$ we substitute into $N_y$ found from the axis $OZ_\parallel$
	$$mg\sin\alpha\cos\alpha+N_x\sin^2\alpha+N_x\cos^2\alpha=2mg(2\sin\alpha-2)\cos\alpha$$		$$mg\sin\alpha\cos\alpha+N_x\sin^2 \alpha+N_x\cos^2 \alpha=2mg(2\sin\alpha-2)\cos\alpha$$
	$$\frac{N_x}{mg}=4\sin\alpha\cos\alpha-2\cos\alpha-\sin\alpha\cos\alpha$$		$$\frac{N_x}{mg}=4\sin\alpha\cos\alpha-2\cos\alpha-\sin\alpha\cos\alpha$$
	$$\boxed{N_x=(3\sin\alpha-2)\cos\alpha \cdot mg}$$		$$\boxed{N_x=(3\sin\alpha-2)\cos\alpha \cdot mg}$$
	When $N_x \leq0$, the shoe will break away from the vertical wall and will no longer act on it:		When $N_x \leq0$, the shoe will break away from the vertical wall and will no longer act on it:
	$$N_x=0$$		$$N_x=0$$
	Condition of detachment:		Condition of detachment:
	$$(3\sin\alpha-2)\cos\alpha \leq0$$		$$(3\sin\alpha-2)\cos\alpha \leq0$$
	$$3\sin\alpha-2\leq0 \Leftrightarrow \boxed{\sin\alpha\leqslant2/3}$$		$$3\sin\alpha-2\leq0 \Leftrightarrow \boxed{\sin\alpha\leqslant2/3}$$
	</p>		</p>

	<h4>Answer</h4>		<h4>Answer</h4>
	<p>		<p>
	$$F=mg\cos\alpha(3\sin\alpha-2)\mathrm{~with~}\sin\alpha\geqslant2/3;$$		$$F=mg\cos\alpha(3\sin\alpha-2)\mathrm{~with~}\sin\alpha\geqslant2/3;$$
	$$F=0\mathrm{~with~}\sin\alpha\leqslant2/3.$$		$$F=0\mathrm{~with~}\sin\alpha\leqslant2/3.$$
	</p>		</p>
	<p style="text-align: right; font-style: italic; font-size: 14;">		<p style="text-align: right; font-style: italic; font-size: 14;">
	Almaskhan Arsen<br>		Almaskhan Arsen<br>
	</p>		</p>


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