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ksaimouli

  • one year ago

solve

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  1. ksaimouli
    • one year ago
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    \[T2 \cos \theta - \mu m1g=0\]

  2. ksaimouli
    • one year ago
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    \[T2 \sin \theta -m2g=0\]

  3. ksaimouli
    • one year ago
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    @zepdrix

  4. ksaimouli
    • one year ago
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    i subtracted those

  5. ksaimouli
    • one year ago
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    |dw:1360014922302:dw|

  6. ksaimouli
    • one year ago
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    =-m2g

  7. ksaimouli
    • one year ago
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    i solved for m2g is that rigjht

  8. zepdrix
    • one year ago
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    I don't really understand what this problem is in reference to D:

  9. zepdrix
    • one year ago
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    And your `Elimination` doesn't look right, how did the T's cancel out? <:o

  10. ksaimouli
    • one year ago
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    i subtracted the equation

  11. zepdrix
    • one year ago
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    |dw:1360015107658:dw|

  12. zepdrix
    • one year ago
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    See how they don't cancel out since they have the trig term attached to them? D: You could instead `divide` the equations I suppose.

  13. ksaimouli
    • one year ago
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    alright thx

  14. ksaimouli
    • one year ago
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    i solved to t2 and subed in to it

  15. zepdrix
    • one year ago
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    \[\large \frac{T_2\cos \theta}{T_2\sin \theta}=\frac{\mu m_1g}{m_2g}\]

  16. zepdrix
    • one year ago
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    Oh ok :)

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