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3psilon

  • 3 years ago

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

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  1. 3psilon
    • 3 years ago
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    @Algebraic!

  2. Algebraic!
    • 3 years ago
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    wut?

  3. Algebraic!
    • 3 years ago
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    \[T= \frac{ Mg }{ 2 \sin 45 }\]

  4. 3psilon
    • 3 years ago
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    The answer says it = Mg/2 what happens to the root2 / 2

  5. Algebraic!
    • 3 years ago
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    no idea what you're asking... tbh

  6. 3psilon
    • 3 years ago
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    hahha Just wanna know how they got from \[2T *\frac{ \sqrt{2} }{ 2 } = Mg \] to\[T = \]

  7. 3psilon
    • 3 years ago
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    mg/2

  8. 3psilon
    • 3 years ago
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    idk whats happening to the root 2 over 2

  9. 3psilon
    • 3 years ago
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    Yeah , but I don't see how they're getting rid of the square root?

  10. Algebraic!
    • 3 years ago
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    \[\frac{ \sqrt{2}Mg }{ 2 }\]

  11. Algebraic!
    • 3 years ago
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    right?

  12. 3psilon
    • 3 years ago
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    yes

  13. Algebraic!
    • 3 years ago
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    so, I guess you're asking:\[\sqrt{2}T =Mg\] \[T = \frac{Mg }{ \sqrt{2}}\] \[T = \frac{Mg }{ \sqrt{2}} \frac{ \sqrt{2} }{ \sqrt{2} }\]

  14. 3psilon
    • 3 years ago
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    But they got T = Mg/2 in the book that I'm using?

  15. 3psilon
    • 3 years ago
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    I guess it's just a typo

  16. Algebraic!
    • 3 years ago
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    possibly.

  17. Algebraic!
    • 3 years ago
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    maybe the answer is actually Mg/2 ... I don't know.

  18. 3psilon
    • 3 years ago
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    Hahah thanks I know the questions was very vague but thanks :)

  19. Algebraic!
    • 3 years ago
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    did you want me to look the actual problem?

  20. 3psilon
    • 3 years ago
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    I'll draw it, or attempt to

  21. 3psilon
    • 3 years ago
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    |dw:1350873308436:dw|

  22. 3psilon
    • 3 years ago
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    Find the tension in the two ropes

  23. Algebraic!
    • 3 years ago
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    ok..

  24. Algebraic!
    • 3 years ago
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    yep, you're right... should be sqrt(2) Mg /2

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