Updated spacing between @ latex expressions

astrosander edited
revision #10506 parent #9982 GitHub c702ebe ← older newer →
@@ -6,14 +6,14 @@
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<meta http-equiv="content-language" content="en">
<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
− <meta name="description" content="At what angle to the horizon will the surface of the liquid in the vessel slide on an inclined plane that makes an angle \alpha to the horizon if the friction coefficient is \mu?">
+ <meta name="description" content="At what angle to the horizon will the surface of the liquid in the vessel slide on an inclined plane that makes an angle \alpha to the horizon if the friction coefficient is \mu ?">
<meta name="author" content="Aliaksandr Melnichenka">
<meta name="date" content="2023-10" scheme="YYYY-MM">
− <meta property="og:title" content="At what angle to the horizon will the surface of the liquid in the vessel slide on an inclined plane that makes an angle \alpha to the horizon if the friction coefficient is \mu?">
+ <meta property="og:title" content="At what angle to the horizon will the surface of the liquid in the vessel slide on an inclined plane that makes an angle \alpha to the horizon if the friction coefficient is \mu ?">
<meta property="og:image" content="img/logo.png">
− <meta property="og:description" content="At what angle to the horizon will the surface of the liquid in the vessel slide on an inclined plane that makes an angle \alpha to the horizon if the friction coefficient is \mu?">
+ <meta property="og:description" content="At what angle to the horizon will the surface of the liquid in the vessel slide on an inclined plane that makes an angle \alpha to the horizon if the friction coefficient is \mu ?">
<meta name="yandex-verification" content="6cfda41f74038368">
− <title>At what angle to the horizon will the surface of the liquid in the vessel slide on an inclined plane that makes an angle \alpha to the horizon if the friction coefficient is \mu?</title>
+ <title>At what angle to the horizon will the surface of the liquid in the vessel slide on an inclined plane that makes an angle \alpha to the horizon if the friction coefficient is \mu ?</title>
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@@ -71,8 +71,8 @@
Given the reaction force $N=mg\cos\alpha$, the sliding friction force is found as
$$F_\text{fr} = \mu N = \mu mg\cos\alpha$$
Newton's second law for the vessel
−$$ma = mg\sin\alpha-F_\text{fr}$$
−$$\boxed{a=g(\sin\alpha-\mu\cos\alpha)}\quad(1)$$
+$$ma = mg\sin\alpha -F_\text{fr}$$
+$$\boxed{a=g(\sin\alpha -\mu\cos\alpha )}\quad(1)$$
Let us introduce the effective acceleration of free fall $\vec{g^*}$ in a non-inertial frame of reference associated with the vessel, then vectorially, it is found as
$$\vec{g^*}=\vec{g}-\vec{a}$$
@@ -91,12 +91,12 @@
From the drawing it could be found that
$$\tan\beta = \frac{a - g\sin\alpha}{g\cos\alpha} = -\mu$$
The angle between the water level and the horizon will be
−$$\boxed{\gamma= \alpha-\operatorname{\arctan }\mu}$$
+$$\boxed{\gamma = \alpha -\operatorname{\arctan }\mu}$$
</p>
<h4>Answer</h4>
<p>
− $$\gamma=\alpha-\operatorname{\arctan }\mu $$
+ $$\gamma =\alpha -\operatorname{\arctan }\mu$$
</p>
<p style="text-align: right; font-style: italic; font-size: 14;">
Aliaksandr Melnichenka<br>
unchanged lines 15