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Math - Trig algebra problem

  1. Jul 11, 2009 #1
    1. The problem statement, all variables and given/known data

    T₁cosθ₁+T₂cosθ₂=mg
    T₁sinθ₁=T₂sinθ₂

    Part A Isolating T₁ terms on one side,

    T₁cosθ₁=mg‐T₂cosθ₂
    T₁sinθ₁=T₂sinθ₂

    Part B Dividing the two equations, using tanθ=sinθ/cosθ,

    cosθ₁/sinθ₁=mg/(T₂sinθ₂)‐cosθ₂/sinθ₂

    1/tanθ₁=mg/(T₂sinθ₂)‐1/tanθ₂

    2. Relevant equations

    SOH CAH TOA

    3. The attempt at a solution

    I copy and paste the problem from my textbook. I will show you my attempt at this problem.
    My problem lies in Part B.

    It says that I should use tanθ=sinθ/cosθ, but it flips the equation.

    I want to set T₂ as 18.
    I want to set mg as 22.
    I want to set θ₂ as 64 degrees

    Accordingly tanθ=sinθ/cosθ, would then be (18sinθ₂)/(mg-18cosθ₂) which would result = 1.1466.

    (18sin64)/(22-18cos64)=

    But if I plug the numbers that I set above, it would give me 1.402

    mg/(T₂sinθ₂)‐cosθ₂/sinθ₂ = 22/(18sin64) - cos64/sin64 = 1.402
     
  2. jcsd
  3. Jul 11, 2009 #2
    Hi flyingpig,

    I would try typing this into your calculator again. For your very last step, I get 0.872 instead of 1.402 which if you take 1/0.872, you get 1.1466 as above.

    Hope this helps.
     
  4. Jul 11, 2009 #3
    I got it, but can you tell me why and how it is flipped?
     
  5. Jul 11, 2009 #4
    In the first case, tanθ₁ = (T₂sinθ₂)/(mg-T₂cosθ₂). Plugging in your numbers yields 1.1466. That is:

    [tex]\tan \theta _1 = 1.1466[/tex]

    In the second case (1/tanθ₁=mg/(T₂sinθ₂)‐1/tanθ₂), plugging in your numbers yields 0.872. That is:

    [tex]\frac{1}{\tan \theta _1} = 0.872[/tex]

    So you can see that you've actually calculated 1/tan in the second case instead of tan like in the first case.
     
  6. Jul 11, 2009 #5
    Why do we even flip it?
     
  7. Jul 11, 2009 #6
    Perhaps you're a bit confused with what you've done.

    Starting from your original post. In part B you have written some equations down. Here you took equation 1 from part A and divided it by equation 2 from part A to get a 1/tan on one side and the rest on the other. This is what was done in part B. Now it would just be a matter of plugging in your numbers to get a solution.

    However, in your "Attempt at the solution" section, for some reason, you re-did part B, but this time you divided equation 2 from Part A by equation 1 from Part A and hence you got a tan instead of a 1/tan, hence for the two solutions to compare you would have to do like I did in post #4.
     
  8. Jul 11, 2009 #7
    cosθ₁/sinθ₁=mg/(T₂sinθ₂)‐cosθ₂/sinθ₂

    mg/(T₂sinθ₂)‐cosθ₂/sinθ₂ <--- I want to focus on more on this part.

    It is a problem from my textbook, I got confused by the bracket.

    Originally the tan ratio is (T₂sinθ₂)/(mg - T₂cosθ₂), but when I flipped it, it should look like (mg - T₂cosθ₂)/(T₂sinθ₂).

    (mg - T₂cosθ₂)/(T₂sinθ₂).
    I know this is wrong, but how do I make the reciprocal of this and turn it into the verison my textbook had. Because I am really confused how my textbook seperate the "cosθ₂/sinθ₂" from the (mg - T₂cosθ₂)/(T₂sinθ₂) equation. Also, if you can don't use the math operators like I have.... that's probably where my question started. Thanks.
     
  9. Jul 11, 2009 #8
    Okay, well it seems like you're stuck on just some basic algebra:

    [tex]
    \tan \theta _1 = \frac{T_2 \sin \theta _2}{mg - T_2 \cos \theta _2}

    \implies \frac{1}{\tan \theta _1} = \frac{mg - T_2 \cos \theta _2}{T_2 \sin \theta _2}
    [/tex]

    Now for the part you seem to be stuck on:

    [tex]
    \frac{mg - T_2 \cos \theta _2}{T_2 \sin \theta _2}
    = \frac{mg}{T_2 \sin \theta _2} - \frac{T_2 \cos \theta _2}{T_2 \sin \theta _2}
    = \frac{mg}{T_2 \sin \theta _2} - \frac{1}{\tan \theta _2}
    [/tex]

    Let me know if you have any questions, but beyond this we are getting into some really basic algebra.
     
  10. Jul 11, 2009 #9
    You've cleared up ALOT! However....

    [tex]
    \frac{mg - T_2 \cos \theta _2}{T_2 \sin \theta _2}
    = \frac{mg}{T_2 \sin \theta _2} - \frac{T_2 \cos \theta _2}{T_2 \sin \theta _2}
    [/tex]

    This is troubling me

    [tex]
    \frac{T_2 \cos \theta _2}{T_2 \sin \theta _2}
    [/tex]

    I am not saying you are incorrect, because I believe you are correct. But the textbook canceled out the [tex] T_2 [/tex]s

    If they do they would it still make it equal? I am sorry for wasting your time on such a simple matter...I would've probably attracted more attention if I knew how to do nice operators.
     
  11. Jul 11, 2009 #10
    I neglected canceling out the [tex]T_2[/tex]s just to show you the step where I split the fraction. You could have done the cancellation then too.

    Notice in the last step, I make the cancellation, and the final answer is the same.
     
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