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ksaimouli Group Title

solve

  • one year ago
  • one year ago

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  1. ksaimouli Group Title
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    \[T2 \cos \theta - \mu m1g=0\]

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

    • one year ago
  3. ksaimouli Group Title
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    @zepdrix

    • one year ago
  4. ksaimouli Group Title
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    i subtracted those

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

    • one year ago
  6. ksaimouli Group Title
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    =-m2g

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

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

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

    • one year ago
  10. ksaimouli Group Title
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    i subtracted the equation

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

    • one year ago
  12. zepdrix Group Title
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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.

    • one year ago
  13. ksaimouli Group Title
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    alright thx

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

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

    • one year ago
  16. zepdrix Group Title
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    Oh ok :)

    • one year ago
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