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burhan101
Determine the resultant of this vector sum: 10 N at 045 degrees, and 8 N at 68 degrees
This isn't so bad. Have you had practice with these?
I can't do it with degrees :S
After a while, they'll become nearly routine!
Can you please go through the steps , i find these really confusing
Okay! So you've seen these? And have learned that any vector has separate components (x and y)?
We can draw a picture, if you like. Let's look at the first vector, 10N. Sound good?
Alright! |dw:1360640234142:dw| There's the vector in one piece.
How do we know that it's NE ?
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.
OH so that's the common starting point ?
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
Yep! Angle measures start from there if no other information is given!
|dw:1360640896289:dw|Now, a person might say, "10 degrees west of north" or something silly like that. That means....
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.
Now you know where your vector is! And you can split it up with trigonometric functions using the angle!
So you have seen that before? Using the trig functions?
@burhan101 so you've got it? Congrats and take care if you did!
i know how to do these with like normal vectors
|dw:1360641728957:dw|
i know thats not right -.-
You mean like <x,y>? Yeah, this is different. You aren't just told the components. However, you can find them!
Have you ever used sine or cosine for angles?
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?
If not, I can show you.
how do i add the other vector to this picture now
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 <x, y> form. You can, however, get it back to how it was, with a given magnitude and given direction (as an angle).
Are you giving it a shot?
Yes im trying to draw it out :$
I didn't see the picture you drew before! That's a good rough representation of the vectors! :) Just like mine! :)
|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.
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|
Actually, nevermind! Retake!
|dw:1360644367333:dw|That one.
|dw:1360644483191:dw|and that one.
Do you see where this is going?
Not realll dont i add the other 8 N at 068 degrees ?
Well, yeah! You have to add its components to the other vectors components. So, first, you have to get each vector's components.
Dont you just add then like the tail to tail method and then do tan inverse to find the angle ? :S
You could, huh! But that's sort of what this is. Picture time! :)
okay so im not totally off track
|dw:1360645168898:dw|derfinately not!
That last one shows the resultant vector.
makes so much sense !
so now i find the tan inverse right
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.
Okay.... Ready to try it?
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!
\[\begin{matrix}x-component&=&10cos(45)&=&7.07\\y-component&=&10sin(45)&=&7.07\end{matrix}\]
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.
thanks for your help ! :D
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.\[tan^{-1}{\LARGE (}\frac{y}{x}\LARGE )\]
\[\sqrt{7.07^2+7.07^2}\approx 9.998\approx 10.00\]There was some rounding error to get to 9.998. But if you round it to the nearest hundredth, you have 10! \[tan^{-1}{\LARGE (}\frac{7.07}{7.07}{\LARGE )}=tan^{-1}(1)=45^o\]
Switched back. :) You're very welcome! Take care! Tag me if you need any more help.