Updated greek laters @ latex compiling

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revision #10076 parent #9331 GitHub dbb31ba ← older newer →
en/2.1.48.md +13 −13
@@ -93,32 +93,32 @@
<p>
From geometric considerations:
−$$ dy = dx \cdot ctg\,\beta $$
+$$ dy = dx \cdot c\tan \beta $$
Let us differentiate both parts of the expression twice:
−$$ \frac{dy}{d^2t} = \frac{dx}{d^2t} \cdot ctg\,\beta $$
+$$ \frac{dy}{d^2t} = \frac{dx}{d^2t} \cdot c\tan \beta $$
−$$ a_1 = a_2 \cdot tg\,\beta\;(1) $$
+$$ a_1 = a_2 \cdot \tan \beta\quad(1) $$
Since the Author did not say anything about friction, example $F_{tr}=0$
Next, we write Newton's 2nd law for the bar $m_1$ on the $OY$ axis:
−$$ m_1a_1=N_1\,cos\,\beta $$
+$$ m_1a_1=N_1\cos \beta $$
−$$ N_1=\frac{m_1a_1}{cos\,\beta}\;(a) $$
+$$ N_1=\frac{m_1a_1}{\cos \beta}\quad(a) $$
Similarly for $m_2$:
−$$ m_2a_2=m_2g-2N_1\,sin\,\beta\;(b) $$
+$$ m_2a_2=m_2g-2N_1\,\sin \beta\quad(b) $$
Substitute $(a)$ into $(b)$:
−$$ m_2a_2=m_2g-2m_1a_1\cdot tg\,\beta $$
+$$ m_2a_2=m_2g-2m_1a_1\cdot \tan \beta $$
We substitute $(1)$ into $(c)$:
−$$ m_2a_2=m_2g-2m_1a_2\cdot tg^2\,\beta $$
+$$ m_2a_2=m_2g-2m_1a_2\cdot \tan^2\beta $$
We express $a_2$:
−$$ \boxed{a_2=\frac{m_2g}{m_2+2m_1\,tg^2\,\beta}}\;(d) $$
−Given $(1)$, we multiply $(d)$ by $tg\,\beta$:
−$$ \boxed{a_1=\frac{m_2g\,tg\,\beta}{m_2+2m_1\,tg^2\,\beta}} $$
+$$ \boxed{a_2=\frac{m_2g}{m_2+2m_1\,\tan^2\beta}}\quad(d) $$
+Given $(1)$, we multiply $(d)$ by $\tan \beta$:
+$$ \boxed{a_1=\frac{m_2g\,\tan \beta}{m_2+2m_1\,\tan^2\beta}} $$
</p>
<h4>Answer</h4>
<p>
− $$a_1=\frac{m_2g\,tg\,\beta}{m_2+2m_1\,tg^2\,\beta}$$
− $$a_2=\frac{m_2g}{m_2+2m_1\,tg^2\,\beta}$$
+ $$a_1=\frac{m_2g\,\tan \beta}{m_2+2m_1\,\tan^2\beta}$$
+ $$a_2=\frac{m_2g}{m_2+2m_1\,\tan^2\beta}$$
</p>
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