Updated spacing between @ latex expressions

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@@ -6,14 +6,14 @@
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− <meta name="description" content="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.">
+ <meta name="description" content="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.">
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+ <meta property="og:title" content="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.">
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+ <meta property="og:description" content="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.">
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− <title>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.</title>
+ <title>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.</title>
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@@ -50,7 +50,7 @@
<h3> Statement </h3>
<p>
− $2.1.22.$ After what time will the velocity of a body that has been sent up an inclined plane with velocity $v$ be equal to $v$ again? Coefficient of friction $\mu$, the angle between the plane and the horizon $\alpha$, $tg \alpha > \mu$.
+ $2.1.22.$ After what time will the velocity of a body that has been sent up an inclined plane with velocity $v$ be equal to $v$ again? Coefficient of friction $\mu$, the angle between the plane and the horizon $\alpha$, $\tan\alpha > \mu$.
</p>
<h3>Solution</h3>
@@ -72,31 +72,31 @@
The equation of Newton's second law in projection onto the OX axis when the body moves upward in vector form is written as follows
$$ \vec{F}_{1} + \vec{F}_{2} = m \vec{a} $$
Let us determine the projections of the acting forces on the OX axis and write down the equation of Newton's second law
−$$ F_{1x} = mg \sin \alpha, F_{2x} = \mu mg \cos \alpha $$
+$$ F_{1x} = mg \sin\alpha , F_{2x} = \mu mg \cos\alpha$$
−$$ \mu mg \cos \alpha + mg \sin \alpha = ma $$
+$$ \mu mg \cos\alpha + mg \sin\alpha = ma $$
We divide both parts of the last equation by the mass $m$ and express the magnitude of the acceleration
−$$ a = g( \sin \alpha + \mu \cos \alpha ) $$
+$$ a = g( \sin\alpha + \mu\cos\alpha ) $$
The time of the body's upward movement $t_1$ is determined from the condition that the speed at the end of the ascent is equal to zero
−$$ 0 = v_{0} - at \Rightarrow t_{1} = \frac{v_{0} }{a} = \frac{v_{0} }{g ( \sin \alpha + \mu \cos \alpha ) } $$
+$$ 0 = v_{0} - at \Rightarrow t_{1} = \frac{v_{0} }{a} = \frac{v_{0} }{g ( \sin\alpha + \mu\cos\alpha ) } $$
The downward motion of the body corresponds to the equation
−$$ mg \sin \alpha - \mu mg \cos \alpha = ma $$
+$$ mg \sin\alpha - \mu mg \cos\alpha = ma $$
−$$ a = g( \sin \alpha - \mu \cos \alpha ) $$
+$$ a = g( \sin\alpha - \mu\cos\alpha ) $$
The speed will become equal to $v_0$ only at the end of the descent, because the law of conservation of energy has not been repealed, therefore
−$$ \frac{dv}{dt} = g( \sin \alpha + \mu \cos \alpha ) $$
+$$ \frac{dv}{dt} = g( \sin\alpha + \mu\cos\alpha ) $$
−$$ \int_{0}^{v_{0} } dv = g( \sin \alpha + \mu \cos \alpha ) = \int_{0}^{t_{2} } dt $$
+$$ \int_{0}^{v_{0} } dv = g( \sin\alpha + \mu\cos\alpha ) = \int_{0}^{t_{2} } dt $$
−$$ t_{2} = \frac{v_{0} }{g ( \sin \alpha + \mu \cos \alpha ) } $$
+$$ t_{2} = \frac{v_{0} }{g ( \sin\alpha + \mu\cos\alpha ) } $$
The required time will be determined as the sum $t = t_{1} + t_{2}$
−$$ \boxed{t = \frac{2v_{0} \sin \alpha }{g( \sin^{2} \alpha - \mu^{2} \cos^{2} \alpha )}} $$
+$$ \boxed{t = \frac{2v_{0} \sin\alpha}{g( \sin^{2} \alpha - \mu^{2} \cos^{2} \alpha )}} $$
</p>
</p>
<h4>Answer</h4>
<p>
− $$t = \frac{2v_{0} \sin \alpha }{g( \sin^{2} \alpha - \mu^{2} \cos^{2} \alpha )}$$
+ $$t = \frac{2v_{0} \sin\alpha}{g( \sin^{2} \alpha - \mu^{2} \cos^{2} \alpha )}$$
</p>
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