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Help with math of this? Find T , 2T *sin45 = Mg

Physics
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wut?
\[T= \frac{ Mg }{ 2 \sin 45 }\]

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Other answers:

The answer says it = Mg/2 what happens to the root2 / 2
no idea what you're asking... tbh
hahha Just wanna know how they got from \[2T *\frac{ \sqrt{2} }{ 2 } = Mg \] to\[T = \]
mg/2
idk whats happening to the root 2 over 2
Yeah , but I don't see how they're getting rid of the square root?
\[\frac{ \sqrt{2}Mg }{ 2 }\]
right?
yes
so, I guess you're asking:\[\sqrt{2}T =Mg\] \[T = \frac{Mg }{ \sqrt{2}}\] \[T = \frac{Mg }{ \sqrt{2}} \frac{ \sqrt{2} }{ \sqrt{2} }\]
But they got T = Mg/2 in the book that I'm using?
I guess it's just a typo
possibly.
maybe the answer is actually Mg/2 ... I don't know.
Hahah thanks I know the questions was very vague but thanks :)
did you want me to look the actual problem?
I'll draw it, or attempt to
|dw:1350873308436:dw|
Find the tension in the two ropes
ok..
yep, you're right... should be sqrt(2) Mg /2

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