Magnitude of vertical and horizonal componants

[kaydis]

Homework Statement

Find the magnitude of the vertical and horizontal components of this signal.

Relevant Equations

6∠(2pi/4) [W]
(∠ = angle)

I'm not too sure where to start, but 6 is the magnitude and (2pi/4) is the angle? So if I were to plot this on a graph I could get the value of the point on the horizontal and vertical axis. I've tried to plot is using graphing software but it wasn't working so I'm not sure if i was doing it correctly. Does anyone have any ideas?

 

[PeroK]

I'm not sure I understand your notation, but if the angle is $\frac{2\pi}{4}$ then that's $\frac{\pi}{2}$. Are you sure that's correct?

In any case, from what axis is the angle measured?

 

[kaydis]

sorry, i just double checked and its 3π/4

 

[PeroK]

sorry, i just double checked and its 3π/4

So, what do you know about components (of a vector)?

 

[kaydis]

i know that they're the parts of a vector and that you use trigonometry to find the componants?

 

[PeroK]

i know that they're the parts of a vector and that you use trigonometry to find the componants?

Exactly!

 

[kaydis]

so if:
horizontal: Vx = Vcos

verticle : Vy = Vsin

this means that:
Vx = Vcos

= (3π/4)cos6
= 2.262347939

Vy = Vsin

= (3π/4)sin6
= -0.658357257

is this correct?

 

[PeroK]

so if:
horizontal: Vx = VcosView attachment 248991
verticle : Vy = VsinView attachment 248992

this means that:
Vx = VcosView attachment 248991
= (3π/4)cos6
= 2.262347939

Vy = VsinView attachment 248992
= (3π/4)sin6
= -0.658357257

is this correct?

Are you sure you haven't calculated $v_x = \theta \cos(v)$?

 

[Delta2]

Btw to insert theta like this θ there is a toolbar above the text box we are writing and one of the options of the toolbar is Insert Symbol which you can use to insert various mathematical symbols.

Alternatively you got to learn to use Latex and type things like this

 ##\theta##

which will give $\theta$ as output text.

 

[kaydis]

Are you sure you haven't calculated $v_x = \theta \cos(v)$?

for Vx=Vcosθ
I got 2.262347939

if I do Vx=θcos(V)
i got -4.242640687

 

[PeroK]

if I do Vx=θcos(V)
i got -4.242640687

I've never seen anyone do that before. That's definitely a new approach.

 

[kaydis]

how would you do it? sorry I'm super confused

 

[PeroK]

for Vx=Vcosθ
I got 2.262347939

if I do Vx=θcos(V)
i got -4.242640687

If you are confused, so am I. I have no idea how you get those answers. I get:

$v_x = v \cos \theta = -4.24$

 

[kaydis]

no I think your right, I swapped them around