Updated greek laters @ latex compiling

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revision #10055 parent #9402 GitHub dbb31ba ← older newer →
en/2.1.23.md +6 −6
@@ -72,34 +72,34 @@
the projection on the axis $OY$ of the applied force:
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<p style="text-align: center;">
−$N=mg-F \, sin \, \alpha$
+$N=mg-F \, \sin\alpha$
</p>
<p>
The friction force is determined as:
</p>
<p style="text-align: center;">
−$F_{тре} = (mg-F \,sin\, \alpha)$
+$F_{тре} = (mg-F \,\sin \alpha)$
</p>
<p>
2. The basic law of dynamics, therefore. will be written as follows:
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<p style="text-align: center;">
−$F \, cos \,\alpha = \mu (mg-F \,sin\, \alpha)$
+$F \, \cos \,\alpha = \mu (mg-F \,\sin \alpha)$
</p>
<p>
3. From the equation of Newton's second law it is easy to determine the desired acceleration
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<p style="text-align: center;">
−$a = \frac{1}{m}(F\,cos\,\alpha-\mu mg+F\,sin\,\alpha)$
+$a = \frac{1}{m}(F\cos\alpha-\mu mg+F\,\sin \alpha)$
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<p style="text-align: center;">
−$a = \frac{F}{m}(cos\,\alpha-\mu \,sin\,\alpha)$
+$a = \frac{F}{m}(\cos\alpha-\mu \,\sin \alpha)$
</p>
</p>
<h4>Answer</h4>
<p>
− $a = (F/m)(cos \,α + \mu \,sin \, \varphi)$$ − \mu g$ if this expression is greater than zero, otherwise $a = 0$.
+ $a = (F/m)(\cos \,\alpha + \mu \,\sin \varphi)$$ − \mu g$ if this expression is greater than zero, otherwise $a = 0$.
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