Updated spacing between @ latex expressions

astrosander правка от 6 окт. 2024 г., 00:48

правка #10454 предыдущая #9388 GitHub c702ebe ← раньше позже →

en/2.1.21.md +7 −7

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		−	<title>A bar of mass m is placed on the horizontal board. The board is slowly tilted. Determine the dependence of the friction force acting on the bar on the angle of inclination of the board \alpha. Coefficient of friction \mu.</title>
		+	<title>A bar of mass m is placed on the horizontal board. The board is slowly tilted. Determine the dependence of the friction force acting on the bar on the angle of inclination of the board \alpha . Coefficient of friction \mu .</title>
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			<span><img src="../../img/book.png"></span><span>Savchenko Solutions</span>
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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="../../#2.1">$\leftarrow$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>
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			<figcaption>
			For problem $2.1.21$
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			<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.
		@@ -81,15 +81,15 @@
			</p>
			<p>
			2. The equation of Newton's second law allows us to determine the maximum value of acceleration
		−	$$ \mu mg \cos \alpha-mg \sin \alpha \leq ma $$
		+	$$ \mu mg \cos\alpha -mg \sin\alpha\leq ma $$

		−	$$ \boxed{a \leq g (\mu \cos \alpha-\sin \alpha)} $$
		+	$$ \boxed{a \leq g (\mu\cos\alpha -\sin\alpha )} $$
			</p>
			</p>

			<h4>Answer</h4>
			<p>
		−	$$a_{max} = g(\mu \cos \alpha − \sin \alpha)$$
		+	$$a_{max} = g(\mu\cos\alpha − \sin\alpha )$$
			</p>


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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="../../#2.1">$\leftarrow$Back</a></h3>		<h3 id="back-link"><a href="../../#2.1">$\leftarrow$Back</a></h3>

	<h3> Statement </h3>		<h3> Statement </h3>
	<p>		<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.		$2.1.21.$ 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>		</p>
	<center>		<center>
	<figure>		<figure>
	<img src="https://savchenkosolutions.com/2/2.1.21/statement.png"		<img src="https://savchenkosolutions.com/2/2.1.21/statement.png"
	loading="lazy" width="180" />		loading="lazy" width="180" />
	<figcaption>		<figcaption>
	For problem $2.1.21$		For problem $2.1.21$
	</figcaption>		</figcaption>
	</figure>		</figure>
	</center>		</center>
	<p>		<p>
	</p>		</p>

	<h3>Solution</h3>		<h3>Solution</h3>
	<p>		<p>
	<center>		<center>
	<figure>		<figure>
	<img src="https://savchenkosolutions.com/2/2.1.21/sol.png"		<img src="https://savchenkosolutions.com/2/2.1.21/sol.png"
	loading="lazy" width="180" />		loading="lazy" width="180" />
	<figcaption>		<figcaption>
	Forces acting on the box		Forces acting on the box
	</figcaption>		</figcaption>
	</figure>		</figure>
	</center>		</center>

	<p>		<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.		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.
	@@ -81,15 +81,15 @@
	</p>		</p>
	<p>		<p>
	2. The equation of Newton's second law allows us to determine the maximum value of acceleration		2. The equation of Newton's second law allows us to determine the maximum value of acceleration
	$$ \mu mg \cos \alpha-mg \sin \alpha \leq ma $$		$$ \mu mg \cos\alpha -mg \sin\alpha\leq ma $$

	$$ \boxed{a \leq g (\mu \cos \alpha-\sin \alpha)} $$		$$ \boxed{a \leq g (\mu\cos\alpha -\sin\alpha )} $$
	</p>		</p>
	</p>		</p>

	<h4>Answer</h4>		<h4>Answer</h4>
	<p>		<p>
	$$a_{max} = g(\mu \cos \alpha − \sin \alpha)$$		$$a_{max} = g(\mu\cos\alpha − \sin\alpha )$$
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