Updated greek laters @ latex compiling

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revision #10141 parent #9557 GitHub dbb31ba ← older
en/2.8.1.md +4 −4
@@ -79,18 +79,18 @@
<p>
$a$. Applying Newton Second Law for $x$-direction
$$T_2~\cos{30^\circ}-T_1~\cos{30^\circ}=0$$
− $$T_1=T_2 \;(1)$$
+ $$T_1=T_2 \quad(1)$$
for $y$-direction
− $$(T_1+T_2)~\sin{30^\circ} = mg \;(2)$$
+ $$(T_1+T_2)~\sin{30^\circ} = mg \quad(2)$$
Let be $T_1=T_2=T$, so, back to $(2)$
$$2T~\sin{30^\circ}=mg$$
$$T = \frac{mg}{2~\sin{30^\circ}} = mg = \boxed{98~\rm{N}}$$
$b$. According figure,
$$F^2 = T^2+T^2 = 2T^2$$
We consider the same tension $T$ because we suppose the threads are unextensible.
− $$F = \sqrt{2}~T \;(3)$$
+ $$F = \sqrt{2}~T \quad(3)$$
Applying Newton Second Law
− $$T = mg \;(4)$$
+ $$T = mg \quad(4)$$
Putting $(4)$ into $(3)$
$$F = \sqrt{2}~mg = \boxed{138~\rm{N}}$$
Note: Calculations were made considering $g$ = 9.8 N/kg and $\sqrt{2}$ = 1.41.
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