Updated spacing between @ latex expressions

astrosander edited
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revision #10471 parent #10074 GitHub c702ebe ← older newer →
en/2.1.45.md +8 −8
@@ -79,15 +79,15 @@
</center>
<p>
From the figure, we find the support reaction force:
−$$ N=mg \cos \alpha $$
+$$ N=mg \cos\alpha$$
According to the Amontons-Coulomb Law:
−$$ F_{тр}=\mu N=\mu mg \cos \alpha $$
−Since $\mu = tg \alpha$ (by condition):
−$$ F_{тр}=\mu mg \cos \alpha $$
+$$ F_{тр}=\mu N=\mu mg \cos\alpha$$
+Since $\mu = \tan\alpha$ (by condition):
+$$ F_{тр}=\mu mg \cos\alpha$$
−$$ F_{тр}=mg \cdot tg \alpha \cdot \cos \alpha=mg \sin \alpha $$
+$$ F_{тр}=mg \cdot \tan\alpha\cdot \cos\alpha =mg \sin\alpha$$
−$$ F=mg \sin \alpha $$
+$$ F=mg \sin\alpha$$
Redraw in the $XY$ plane
At the initial moment $(\varphi=90^{\circ})$:
</p>
@@ -138,8 +138,8 @@
\end{matrix}\right. $$
$$ \left\{\begin{matrix}
− \frac{d}{dt} (u \sin \varphi)= \frac{F \sin \varphi }{m}& \\
− \frac{d}{dt} (u \cos \varphi) = \frac{F (1- \cos \varphi) }{m} &
+ \frac{d}{dt} (u \sin\varphi)= \frac{F \sin \varphi }{m}& \\
+ \frac{d}{dt} (u \cos\varphi) = \frac{F (1- \cos \varphi) }{m} &
\end{matrix}\right. $$
We solve the system of differential equations:
$$ \left\{\begin{matrix}
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