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Homework Help: Help with algebra in a physics problem

  1. Oct 5, 2005 #1
    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?
    Last edited: Oct 5, 2005
  2. jcsd
  3. Oct 5, 2005 #2
    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.

  4. Oct 5, 2005 #3
    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...
  5. Oct 5, 2005 #4
    [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]
  6. Oct 5, 2005 #5
    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.

  7. Oct 5, 2005 #6
    alright just one moment....
  8. Oct 5, 2005 #7
    [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]
  9. Oct 5, 2005 #8
    I distributed, now one moment and I will attempt to do the second step you said. By the way this is REALLY helping.
  10. Oct 5, 2005 #9
    [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]
  11. Oct 5, 2005 #10
    [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]
  12. Oct 5, 2005 #11
    You are a godsend! Thanks alot friend.
  13. Oct 5, 2005 #12
    Glad I could help :smile:
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