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ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
\[T2 \cos \theta  \mu m1g=0\]
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
\[T2 \sin \theta m2g=0\]
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
@zepdrix
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
i subtracted those
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
dw:1360014922302:dw
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
i solved for m2g is that rigjht
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
I don't really understand what this problem is in reference to D:
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
And your `Elimination` doesn't look right, how did the T's cancel out? <:o
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
i subtracted the equation
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
dw:1360015107658:dw
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
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

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
alright thx
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
i solved to t2 and subed in to it
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
\[\large \frac{T_2\cos \theta}{T_2\sin \theta}=\frac{\mu m_1g}{m_2g}\]
 one year ago
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