Determine the resultant of this vector sum: 10 N at 045 degrees, and 8 N at 68 degrees

- anonymous

Determine the resultant of this vector sum: 10 N at 045 degrees, and 8 N at 68 degrees

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- theEric

This isn't so bad. Have you had practice with these?

- anonymous

I can't do it with degrees :S

- theEric

After a while, they'll become nearly routine!

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## More answers

- anonymous

Can you please go through the steps , i find these really confusing

- theEric

Okay! So you've seen these? And have learned that any vector has separate components (x and y)?

- theEric

We can draw a picture, if you like. Let's look at the first vector, 10N. Sound good?

- anonymous

Okay ! :)

- theEric

Alright!
|dw:1360640234142:dw|
There's the vector in one piece.

- anonymous

How do we know that it's NE ?

- theEric

That is the trick with degrees, and all you have to remember is this - and it'll be in the next picture. You're working with two dimensional space, here. A great way to draw this is the x and y axes that I bet you're familiar with. Everything has an x and y component, like the vectors. And people like to use angles! So they decided on a common place to start the angle, and where to go from there.
I'll show you.

- theEric

|dw:1360640574984:dw|

- anonymous

OH so that's the common starting point ?

- theEric

That's the convention. Convention just means it's a proper way to do it in some organization of math rules. So pretty much everybody does this.
You could make up your own convention, but it'd be hard to explain it to the rest of us :P

- theEric

Yep! Angle measures start from there if no other information is given!

- anonymous

ohhh okay

- theEric

|dw:1360640896289:dw|Now, a person might say, "10 degrees west of north" or something silly like that. That means....

- theEric

But as a starting point, positive x-axis is what you'll use. Hopefully your teacher agrees. It's a far-reaching convention.. Goes into lots of maths.

- theEric

Now you know where your vector is! And you can split it up with trigonometric functions using the angle!

- anonymous

Oh alright

- theEric

So you have seen that before? Using the trig functions?

- theEric

@burhan101 so you've got it? Congrats and take care if you did!

- anonymous

No

- anonymous

i know how to do these with like normal vectors

- theEric

Haha, okay!

- anonymous

|dw:1360641728957:dw|

- anonymous

i know thats not right -.-

- anonymous

Im so lost lol

- theEric

You mean like ? Yeah, this is different. You aren't just told the components. However, you can find them!

- theEric

Have you ever used sine or cosine for angles?

- anonymous

Yea

- theEric

Well that's what you'll be doing here.
Are you well practiced with them? Going back to this picture, |dw:1360642770812:dw|Do you see how to get the x and y components?

- theEric

|dw:1360642844365:dw|

- theEric

If not, I can show you.

- anonymous

how do i add the other vector to this picture now

- theEric

Well, vectors are easy to split up into components with sine and cosine. Then you use "vector addition," which really means you just add all like components! Then you have one vector as a result. And it is called the "resultant vector." And it'll be in form. You can, however, get it back to how it was, with a given magnitude and given direction (as an angle).

- theEric

Are you giving it a shot?

- anonymous

Yes im trying to draw it out :$

- theEric

I didn't see the picture you drew before! That's a good rough representation of the vectors! :) Just like mine! :)

- theEric

|dw:1360643739640:dw|
funy thing with vectors is that they are a magnitude (like 10N) and direction (like 45 degrees). They aren't a location. You can the same arrowed line anywheres, and they're still equal vectors.

- theEric

So.. (wait for it...)

- anonymous

ohhh

- theEric

|dw:1360644030205:dw|

- theEric

aaannnndddd........

- theEric

The first drawing of the components is common, where the vectors start at the origin. But looking at it this next way will help see sine and cosine's roles.|dw:1360644139377:dw|

- theEric

Actually, nevermind! Retake!

- theEric

|dw:1360644367333:dw|That one.

- theEric

|dw:1360644483191:dw|and that one.

- theEric

Do you see where this is going?

- anonymous

Not realll dont i add the other 8 N at 068 degrees ?

- theEric

Well, yeah! You have to add its components to the other vectors components. So, first, you have to get each vector's components.

- anonymous

Dont you just add then like the tail to tail method and then do tan inverse to find the angle ? :S

- anonymous

them *

- theEric

You could, huh! But that's sort of what this is. Picture time! :)

- anonymous

okay so im not totally off track

- theEric

|dw:1360645168898:dw|derfinately not!

- theEric

|dw:1360645291825:dw|

- theEric

That last one shows the resultant vector.

- anonymous

OH

- theEric

|dw:1360645379901:dw|

- anonymous

makes so much sense !

- theEric

\[\huge :)\]

- anonymous

so now i find the tan inverse right

- theEric

After you get the components of the resultant vector, you use them to find the magnitude (however many N) and that inverse tangent (y-component divided by x-component) to find the angle.

- theEric

Does that make sense?

- theEric

@burhan101 ?

- anonymous

i think so

- theEric

Okay.... Ready to try it?

- anonymous

tan-1 = 68/45

- anonymous

i think :S

- theEric

Not quite! Inverse tangent will give you an angle, but you have to give it a y/x amount. Since you want to find the angle of the really big resultant vector, you'll use its y-component divided by its x-component. You'll have those components after you do the adding.
Let's break your 10N vector into components for practice. We can even then use those components to re-find the angle!

- theEric

|dw:1360646989287:dw|

- theEric

\[\begin{matrix}x-component&=&10cos(45)&=&7.07\\y-component&=&10sin(45)&=&7.07\end{matrix}\]

- theEric

And units are N.
I'm gonna head to bed. Good luck!
If you want to continue on your own, do the same thing to the 8N vector! Then, add the x-components together. Then add the y-components together. Then you have the resultant vectors components. That would be how you knew vectors before.

- anonymous

thanks for your help ! :D

- theEric

And to rediscover that magnitude and angle of the 10N vector, using it's x- and y-components,
Magnitude:
Use Pythagorean Theorem.\[\sqrt{x^2+y^2}\]
For the angle:
Use inverse tangent.\[tan^{-1}{\LARGE (}\frac{y}{x}\LARGE )\]

- theEric

\[\sqrt{7.07^2+7.07^2}\approx 9.998\approx 10.00\]There was some rounding error to get to 9.998. But if you round it to the nearest hundredth, you have 10!
\[tan^{-1}{\LARGE (}\frac{7.07}{7.07}{\LARGE )}=tan^{-1}(1)=45^o\]

- theEric

Switched back. :)
You're very welcome! Take care! Tag me if you need any more help.

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