Updated spacing between @ latex expressions

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правка #10582 предыдущая #9535 GitHub c702ebe ← раньше позже →
en/2.8.14.md +8 −8
@@ -79,23 +79,23 @@
<p>
State of forces equilibrium for the wedge for the horizontal axis:
−$$N_3+\mu_1N_1=\left(\mu_2\cos\alpha+\sin\alpha\right)N_2$$
+$$N_3+\mu_1N_1=\left(\mu_2\cos\alpha +\sin\alpha\right)N_2$$
In the set mode, the force that will act on the left wall will be equal to zero ($N_3=0$)
−$$\frac{N_1}{N_2}=\frac{\mu_2\cos\alpha+\sin\alpha}{\mu_1}\quad(1)$$
+$$\frac{N_1}{N_2}=\frac{\mu_2\cos\alpha +\sin\alpha}{\mu_1}\quad(1)$$
Similarly, for the vertical axis
−$$N_1+\mu_2N_2\sin\alpha=N_2\cos\alpha$$
−$$\frac{N_1}{N_2}=\cos\alpha-\mu_2\sin\alpha\quad(2)$$
+$$N_1+\mu_2N_2\sin\alpha =N_2\cos\alpha$$
+$$\frac{N_1}{N_2}=\cos\alpha -\mu_2\sin\alpha\quad(2)$$
Substituing $(2)$ into $(1)$
−$$\mu_1\cos\alpha-\mu_1\mu_2\sin\alpha=\mu_2\cos\alpha+\sin\alpha$$
+$$\mu_1\cos\alpha -\mu_1\mu_2\sin\alpha =\mu_2\cos\alpha +\sin\alpha$$
Divide the left and right parts by $\cos\alpha$
−$$\left(1+\mu_1\mu_2\right)\tan\alpha=\mu_1-\mu_2$$
+$$\left(1+\mu_1\mu_2\right)\tan\alpha =\mu_1-\mu_2$$
Whence the required angle
−$$\boxed{\tan\alpha=\frac{{\mu}_{1}-\mu_{2}}{1+\mu_{1}\mu_{2}}}$$
+$$\boxed{\tan\alpha =\frac{{\mu}_{1}-\mu_{2}}{1+\mu_{1}\mu_{2}}}$$
</p>
<h4>Answer</h4>
<p>
− $$\operatorname{tg}\alpha=({\mu}_{1}-\mu_{2})/(1+\mu_{1}\mu_{2})$$
+ $$\operatorname{tg}\alpha =({\mu}_{1}-\mu_{2})/(1+\mu_{1}\mu_{2})$$
</p>
<p style="text-align: right; font-style: italic; font-size: 14;">
Almaskhan Arsen<br>
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