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Help with algebra in a physics problem

by JoshHolloway
Tags: algebra, physics
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JoshHolloway
#1
Oct5-05, 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?
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amcavoy
#2
Oct5-05, 04:34 PM
P: 668
Quote Quote by JoshHolloway
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?
My suggestion is to multiply both sides by cosθ, thereby eliminating any fractions. From there, rearrange terms so that any term with an a in it is on one side, and everything else is on another. Now you can factor an a out of this and solve.

Alex
JoshHolloway
#3
Oct5-05, 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...

JoshHolloway
#4
Oct5-05, 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]
amcavoy
#5
Oct5-05, 04:39 PM
P: 668
Quote Quote by JoshHolloway
[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]
That's step 1. Now distribute the cosine and put all of the terms containing an a in them on one side. Tell me what you get.

Alex
JoshHolloway
#6
Oct5-05, 04:40 PM
P: 222
alright just one moment....
JoshHolloway
#7
Oct5-05, 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]
JoshHolloway
#8
Oct5-05, 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.
JoshHolloway
#9
Oct5-05, 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]
JoshHolloway
#10
Oct5-05, 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]
JoshHolloway
#11
Oct5-05, 04:49 PM
P: 222
You are a godsend! Thanks alot friend.
amcavoy
#12
Oct5-05, 04:54 PM
P: 668
Quote Quote by JoshHolloway
You are a godsend! Thanks alot friend.
Glad I could help


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