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Vector addition problem - component method

  1. Sep 13, 2008 #1
    Vplane = 63km/hr at x degrees and direction
    Vwind = 47km/hr at 211 degrees
    Vg = unknown speed at 165 degrees

    find the unknowns


    my attempt using component method:

    (Cos 165)Vg = (Cos 211)47 + (Cos theta)63

    (Sin 165)Vg = (Sin 211)47 + (Sin theta)63

    from here i think im supposed to isolate Vg and then subsitute to find the unknown angle and then plug the angle in to get the speed but im not quite sure

    I know i can use the cosine law or sine law but I have to learn both ways so thats why im using the component method
     
  2. jcsd
  3. Sep 13, 2008 #2

    gabbagabbahey

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    I think the easiest way for you to solve this system of equations is to isolate [tex]cos(\theta)[/tex] in the first equation, [tex]sin(\theta)[/tex] in the second one, and then square both equations, add them together and take advantage of the fact that [tex]sin^2(\theta)+cos^2(\theta)=1[/tex] to sollve for [tex]V_g[/tex]. After that, plug your solution into either equation and solve for [tex]\theta[/tex]. If you post your solution, I'll be happy to check it.
     
  4. Sep 13, 2008 #3
    hmmm, thats where i ran in to trouble, isolating theta i dont understand how to isolate theta. like do i use Sin-1? and how would it look?
     
  5. Sep 13, 2008 #4

    gabbagabbahey

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    Start with the first equation:
    [tex]V_gcos(165^{\circ}) = 47cos(211^{\circ}) + 63cos(\theta)[/tex]

    Can you solve this for [tex]cos(\theta)[/tex]?
     
  6. Sep 13, 2008 #5
    ohhh okay so now i got (Vg(Cos 165) - 47(Cos 211))/63 = Cos (theta)

    i think thats right
     
  7. Sep 13, 2008 #6

    gabbagabbahey

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    Yup, so what is [tex]cos^2(\theta)[/tex] then?
     
  8. Sep 13, 2008 #7
    by Cos^2 u mean Cos-1 right? and if so i dont see what will happen to my other unknown
     
  9. Sep 13, 2008 #8

    gabbagabbahey

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    Nope, I mean [tex](cos(\theta))^2 [/tex]...it will become clear why I want you to find this in a minute.
     
  10. Sep 13, 2008 #9
    oo im just lost now sorrry =(
     
  11. Sep 13, 2008 #10

    gabbagabbahey

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    If
    [tex]cos(\theta)=\frac{V_gcos(165^{\circ})-47cos(211^{\circ})}{63}[/tex]
    then
    [tex]cos^2(\theta)=\frac{[V_gcos(165^{\circ})-47cos(211^{\circ})]^2}{63^2}[/tex]

    ......Now, solve the second equation for [tex]sin(\theta)[/tex] and then find [tex]sin^2(\theta)[/tex]
     
  12. Sep 13, 2008 #11
    okay so i got

    Sin^2(theta) = ((Vg(Sin165)-47(Sin211))^2)/(63)^2

    well now im wondering does that Cos^2 and Sin^2 isolate theta?
     
  13. Sep 13, 2008 #12

    gabbagabbahey

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    Okay, so now you have:
    [tex]cos^2(\theta)=\frac{[V_gcos(165^{\circ})-47cos(211^{\circ})]^2}{63^2}[/tex]
    and
    [tex]sin^2(\theta)=\frac{[V_gsin(165^{\circ})-47sin(211^{\circ})]^2}{63^2}[/tex]

    so what is [tex]sin^2(\theta)+cos^2(\theta)[/tex]?
     
  14. Sep 13, 2008 #13
    from what u said earlier it = 1
     
  15. Sep 13, 2008 #14

    gabbagabbahey

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    Yes, there is a trigonometric identity: http://en.wikipedia.org/wiki/Pythagorean_trigonometric_identity that it will equal 1 for any theta.

    so you get:
    [tex]sin^2(\theta)+cos^2(\theta)=\frac{[V_gsin(165^{\circ})-47sin(211^{\circ})]^2}{63^2} + \frac{[V_gcos(165^{\circ})-47cos(211^{\circ})]^2}{63^2}=1[/tex]
    And so,
    [tex]\frac{[V_gsin(165^{\circ})-47sin(211^{\circ})]^2}{63^2} + \frac{[V_gcos(165^{\circ})-47cos(211^{\circ})]^2}{63^2}=1[/tex]

    Now, can you expand this equation to find a quadratic equation for [tex]V_g[/tex]?
     
  16. Sep 13, 2008 #15
    okay so i tried to find wat Vg is equal to and i get 2.11 which seems wrong to me

    my work is shown here

    1=((Vg(cos165)+40.287)^2 + (Vg(Sin165)+24.207)^2)/(3969)
    multiply both sides by 3969

    3969=(Vg(Cos165)+40.287)^2 + (Vg(Sin165)+24.207)^2
    square root both sides

    63=-0.7071Vg+64.494
    subtract 64 to both sides

    -1.494= -0.7071Vg

    0.7072 came from cos165+sin165

    Vg = -1.494/-0.7071
    Vg= 2.11
     
  17. Sep 13, 2008 #16

    gabbagabbahey

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    you can't just take the square root of the right hand side here, because it is a sum of 2 squares not just a perfect square.....you have to expand each term first like this:


    [tex]\frac{[V_gsin(165^{\circ})-47sin(211^{\circ})]^2}{63^2} + \frac{[V_gcos(165^{\circ})-47cos(211^{\circ})]^2}{63^2}=1[/tex]


    [tex]\Rightarrow [V_gsin(165^{\circ})-47sin(211^{\circ})]^2 +[V_gcos(165^{\circ})-47cos(211^{\circ})]^2=63^2[/tex]



    [tex]\Rightarrow [V_g^2sin^2(165^{\circ})-94V_gsin(165^{\circ})sin(211^{\circ})+47^2sin^2(211^{\circ})] +[V_g^2cos^2(165^{\circ})-94V_gcos(165^{\circ})cos(211^{\circ})+47^2cos^2(211^{\circ})]=63^2[/tex]



    [tex]\Rightarrow [sin^2(165^{\circ})+cos^2(165^{\circ})]V_g^2-94V_g[sin(165^{\circ})sin(211^{\circ})+cos(165^{\circ})cos(211^{\circ})]+47^2[sin^2(211^{\circ})] +cos^2(211^{\circ})]=63^2[/tex]

    Now, there are two useful Trig Identities: (1) [tex]sin^2(x)+cos^2(x)=1[/tex] and (2) [tex]sin(x)sin(y)+cos(x)cos(y)=cos(y-x)[/tex].
    Using these you get:

    [tex](1)V_g^2-94cos(211^{\circ}-165^{\circ})V_g+(1)47^2=63^2[/tex]


    [tex]\Rightarrow V_g^2-94cos(46^{\circ})V_g+(47^2-63^2)=0[/tex]

    Now, can you solve for [tex]V_g[/tex]?

    You should get 2 solutions!
     
    Last edited: Sep 13, 2008
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