Vector addition problem - component method

Oh, and don't forget to check your solution in the original equations.In summary, to find the unknown speed, Vg, and angle, theta, in the given system of equations, we can use the component method by isolating Vg in both equations and then using the trigonometric identities to solve for Vg. After finding Vg, we can plug it into either equation to solve for theta. However, it may be easier to solve the system by expanding the equations and using the quadratic formula to find Vg. Don't forget to check your solution in the original equations.
  • #1
Larrytsai
228
0
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 I am supposed to isolate Vg and then subsitute to find the unknown angle and then plug the angle into get the speed but I am not quite sure

I know i can use the cosine law or sine law but I have to learn both ways so that's why I am using the component method
 
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  • #2
Larrytsai said:
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 I am supposed to isolate Vg and then subsitute to find the unknown angle and then plug the angle into get the speed but I am not quite sure

I know i can use the cosine law or sine law but I have to learn both ways so that's why I am using the component method

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.
 
  • #3
hmmm, that's where i ran into trouble, isolating theta i don't understand how to isolate theta. like do i use Sin-1? and how would it look?
 
  • #4
Larrytsai said:
hmmm, that's where i ran into trouble, isolating theta i don't understand how to isolate theta. like do i use Sin-1? and how would it look?

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]?
 
  • #5
ohhh okay so now i got (Vg(Cos 165) - 47(Cos 211))/63 = Cos (theta)

i think that's right
 
  • #6
Larrytsai said:
ohhh okay so now i got (Vg(Cos 165) - 47(Cos 211))/63 = Cos (theta)

i think that's right

Yup, so what is [tex]cos^2(\theta)[/tex] then?
 
  • #7
by Cos^2 u mean Cos-1 right? and if so i don't see what will happen to my other unknown
 
  • #8
Larrytsai said:
by Cos^2 u mean Cos-1 right? and if so i don't see what will happen to my other unknown

Nope, I mean [tex](cos(\theta))^2 [/tex]...it will become clear why I want you to find this in a minute.
 
  • #9
oo I am just lost now sorrry =(
 
  • #10
Larrytsai said:
oo I am just lost now sorrry =(

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]
 
  • #11
okay so i got

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

well now I am wondering does that Cos^2 and Sin^2 isolate theta?
 
  • #12
Larrytsai said:
okay so i got

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

well now I am wondering does that Cos^2 and Sin^2 isolate theta?

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]?
 
  • #13
from what u said earlier it = 1
 
  • #14
Larrytsai said:
from what u said earlier it = 1

Yes, there is a trigonometric identity: http://en.wikipedia.org/wiki/Pythagorean_trigonometric_identity" [Broken] 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]?
 
Last edited by a moderator:
  • #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
 
  • #16
Larrytsai said:
3969=(Vg(Cos165)+40.287)^2 + (Vg(Sin165)+24.207)^2
square root both sides

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:

1. What is the component method for vector addition?

The component method for vector addition involves breaking down vectors into their horizontal and vertical components, then adding these components separately to find the resultant vector.

2. How do you find the horizontal and vertical components of a vector?

To find the horizontal and vertical components of a vector, you can use trigonometric functions. The horizontal component is found by multiplying the magnitude of the vector by the cosine of the angle it makes with the horizontal axis. Similarly, the vertical component is found by multiplying the magnitude of the vector by the sine of the angle.

3. Can the component method be applied to more than two vectors?

Yes, the component method can be applied to any number of vectors. Each vector's horizontal and vertical components can be added separately to find the resultant vector.

4. How do you represent vector addition using the component method?

Vector addition using the component method can be represented using a graphical method. Each vector's horizontal and vertical components can be drawn separately and then added to find the resultant vector. Alternatively, you can use mathematical equations to calculate the resultant vector.

5. What are the advantages of using the component method for vector addition?

The component method allows for easier visualization and calculation of vector addition, especially for complex vector systems. It also allows for the use of mathematical equations and trigonometric functions, making it a more efficient method for solving vector addition problems.

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