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Solving a system of equations!

  1. Feb 2, 2015 #1
    Hi,

    I have two equations and two unknowns which I need to get the relation T1/T2 from.

    The first one is ##T1-T2=(mg)/(cos(a))##
    and the second one: ##T1+T2=mw^2L##

    I need to solve for both T1 and T2. And if I add the bottom one to the upper I should eliminate T2.
    But what happens with the right hand side? Is T1 simply ##1/2(mg)/(cos(a))+mw^2L##?

    I can back off one step if I did anything wrong to this point.
    The two equations one step back were: ##T1cos(a)-T2cos(a)=mg## and ##T1sin(a)+T2sin(a)=mw^2Lsin(a)##

    Please help me, I need help with this!
     
    Last edited: Feb 2, 2015
  2. jcsd
  3. Feb 2, 2015 #2

    Simon Bridge

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    Put x=T1/T2 ... this is what you want to find.

    Sub for T1 in the first equation and for T2 in the second.

    See?


    Basically the answer to your first question is "yes", kinda, almost... check algebra. You can see why this is by doing the intermediate step.
     
  4. Feb 2, 2015 #3
    Sorry, but I don't. I've been sitting here for two hours+ now. I'm pretty bad at these things. :( Substituting T1 in the first equation for xT2? It made me even more confused!
     
  5. Feb 2, 2015 #4

    Simon Bridge

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    Oh I think I made a typo... Lets see...

    Write it out... you get T2 (x-1)=F ... where the F is everything on the rhs.
    Do the same with the second one... T2 (x+1)=G ...
    Do you still not see?

    Whever I have to solve for a ratio, I like to see if there is a way to divide the equations. It is often faster. Your approach works too... you just needed to be more careful with the algebra. When you divided by 2, you have to do it to all the rhs not just the first term.
     
  6. Feb 2, 2015 #5
    I'm pretty sure your method's a lot quicker and better but I'm very tired from doing a rough assignment all afternoon and all night so I cannot grasp the esscence of it, sorry! :(
    I made one more try on it with algebra, I'll attach a photo. Did I get it right this time?

    Edit: Oh dear, the downscaling of the picture. Here it is in a readable size. http://imgur.com/ngc8e9d
     

    Attached Files:

  7. Feb 2, 2015 #6

    Simon Bridge

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    No... the method looks good, but line 3 is wrong.
    How did you end up with two mg/cos when you only started with one?

    It should read, $$2T_1= \frac{mg}{\cos a}+ m\omega^2L $$ Those equations look odd to me... do the units match up?

    Looks like the 2am effect, get some rest.
     
    Last edited: Feb 2, 2015
  8. Feb 2, 2015 #7
    Yes I see now that I forgot to divide the whole thing with two. And I have no idea what happened with the m...
    But I tried to to it even more thoroughly again. I guess I can factor out the m and place before the parenthesis. Picture: http://imgur.com/2saPhqd

    It's part of my last answer on a thing that's due tomorrow, early. Like in 8 hours early.
     
  9. Feb 2, 2015 #8

    Simon Bridge

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    You subbed the result back into the 1st rquation?
    Consider...

    T1-T2=F
    T1+T2=G

    T1=( F+G)/2
    T2=( G-F)/2

    T1/T2 = ( G+F)/(G-F)

    But Id definitely check the units if I were you.

    Go splash some water on your face, then have a quick snack.
    Then try.

    You seem to have right idea but you are burning out.
     
  10. Feb 2, 2015 #9
    Thanks man! I'll try that on the train in the morning - gotta get some sleep before tomorrow's lecture. Thank you for your patience and help! :)
     
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