Updated greek laters @ latex compiling

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правка #10198 предыдущая #9024 GitHub dbb31ba ← раньше позже →
en/2.8.34.md +4 −4
@@ -82,14 +82,14 @@
<p>
Let's consider the Newton Second Law for both axis. Firstly, for $x$-axis,
$$N_1+f_{r2} = mg$$
− $$N_1+\mu_2 N_2 = mg \;(1)$$
+ $$N_1+\mu_2 N_2 = mg \quad(1)$$
For $y$-axis,
$$N_2 = f_{r1}$$
− $$N_2 = \mu_1 N_1 \;(2)$$
+ $$N_2 = \mu_1 N_1 \quad(2)$$
Substituting $(2)$ into $(1)$:
− $$N_1 (1+\mu_1 \mu_2) = mg \;(3)$$
+ $$N_1 (1+\mu_1 \mu_2) = mg \quad(3)$$
Applying Newton Second Law for the rotation about the pivot point,
$$\sum \vec{\tau} = \vec{0}$$
@@ -100,7 +100,7 @@
$$-mg\frac{l}{2}\sin{\frac{\pi}{2}-\alpha}+N_1 l\sin{\frac{\pi}{2}+\alpha}-\mu_1 N_1 l \sin{\pi-\alpha} \leq 0$$
Developing the above expression,
− $$mg \geq 2N_1(1-\mu_1 \tan{\alpha}) \;(4)$$
+ $$mg \geq 2N_1(1-\mu_1 \tan{\alpha}) \quad(4)$$
Putting $(3)$ into $(4)$, and separating $\tan{\alpha}$ term,
$$\boxed{\tan{\alpha} \geq \frac{1-\mu_1 \mu_2}{2 \mu_1}}$$
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