
#1
Oct505, 04:19 PM

P: 222

I don't need help with the physics, it is simply the algebra that I can't figure out in this problem. Here am where I am at:
[tex] m_{1} a + \mu ( m_{1} g \cos \theta ) + m_{1} g \sin \theta = \frac{\m_{2} a +m_{2} g \cos \theta }{ \cos \theta } [/tex] I need to solve for a. How the heck do I do this? How can I factor the a's into just one a? 



#2
Oct505, 04:34 PM

P: 669

Alex 



#3
Oct505, 04:35 PM

P: 222

in the fraction it is supposed to say m2, not just a subscript two. I don't know what I did wrong.
And I did that. Here I will show you how far I have gotten past that. Just a few minutes... 



#4
Oct505, 04:37 PM

P: 222

help with algebra in a physics problem
[tex] \cos \theta [m_{1} a + \mu ( m_{1} g \cos \theta ) + m_{1} g \sin \theta] = m_{2} a +m_{2} g \cos \theta [/tex]




#5
Oct505, 04:39 PM

P: 669

Alex 



#6
Oct505, 04:40 PM

P: 222

alright just one moment....




#7
Oct505, 04:42 PM

P: 222

[tex] m_{1} a \cos \theta + \mu ( m_{1} g \cos ^2 \theta ) + m_{1} g \sin \theta \cos \theta = m_{2} a +m_{2} g \cos \theta [/tex]




#8
Oct505, 04:43 PM

P: 222

I distributed, now one moment and I will attempt to do the second step you said. By the way this is REALLY helping.




#9
Oct505, 04:47 PM

P: 222

[tex] \mu ( m_{1} g \cos ^2 \theta ) + m_{1} g \sin \theta \cos \theta  m_{2} g \cos \theta = m_{1} a \cos \theta + m_{2} a [/tex]




#10
Oct505, 04:49 PM

P: 222

[tex] \mu ( m_{1} g \cos ^2 \theta ) + m_{1} g \sin \theta \cos \theta  m_{2} g \cos \theta = a (m_{1} \cos \theta + m_{2}) [/tex]




#11
Oct505, 04:49 PM

P: 222

You are a godsend! Thanks alot friend.




#12
Oct505, 04:54 PM

P: 669




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