help with algebra in a physics problem

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?

amcavoy

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

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

[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

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

alright just one moment....

JoshHolloway

[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

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

JoshHolloway

[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

[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

You are a godsend! Thanks alot friend.

amcavoy

Glad I could help