anonymous
  • anonymous
Determine the resultant of this vector sum: 10 N at 045 degrees, and 8 N at 68 degrees
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
theEric
  • theEric
This isn't so bad. Have you had practice with these?
anonymous
  • anonymous
I can't do it with degrees :S
theEric
  • theEric
After a while, they'll become nearly routine!

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Can you please go through the steps , i find these really confusing
theEric
  • theEric
Okay! So you've seen these? And have learned that any vector has separate components (x and y)?
theEric
  • theEric
We can draw a picture, if you like. Let's look at the first vector, 10N. Sound good?
anonymous
  • anonymous
Okay ! :)
theEric
  • theEric
Alright! |dw:1360640234142:dw| There's the vector in one piece.
anonymous
  • anonymous
How do we know that it's NE ?
theEric
  • 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
  • theEric
|dw:1360640574984:dw|
anonymous
  • anonymous
OH so that's the common starting point ?
theEric
  • 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
  • theEric
Yep! Angle measures start from there if no other information is given!
anonymous
  • anonymous
ohhh okay
theEric
  • theEric
|dw:1360640896289:dw|Now, a person might say, "10 degrees west of north" or something silly like that. That means....
theEric
  • 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
  • theEric
Now you know where your vector is! And you can split it up with trigonometric functions using the angle!
anonymous
  • anonymous
Oh alright
theEric
  • theEric
So you have seen that before? Using the trig functions?
theEric
  • theEric
@burhan101 so you've got it? Congrats and take care if you did!
anonymous
  • anonymous
No
anonymous
  • anonymous
i know how to do these with like normal vectors
theEric
  • theEric
Haha, okay!
anonymous
  • anonymous
|dw:1360641728957:dw|
anonymous
  • anonymous
i know thats not right -.-
anonymous
  • anonymous
Im so lost lol
theEric
  • theEric
You mean like ? Yeah, this is different. You aren't just told the components. However, you can find them!
theEric
  • theEric
Have you ever used sine or cosine for angles?
anonymous
  • anonymous
Yea
theEric
  • 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
  • theEric
|dw:1360642844365:dw|
theEric
  • theEric
If not, I can show you.
anonymous
  • anonymous
how do i add the other vector to this picture now
theEric
  • 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
  • theEric
Are you giving it a shot?
anonymous
  • anonymous
Yes im trying to draw it out :$
theEric
  • theEric
I didn't see the picture you drew before! That's a good rough representation of the vectors! :) Just like mine! :)
theEric
  • 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
  • theEric
So.. (wait for it...)
anonymous
  • anonymous
ohhh
theEric
  • theEric
|dw:1360644030205:dw|
theEric
  • theEric
aaannnndddd........
theEric
  • 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
  • theEric
Actually, nevermind! Retake!
theEric
  • theEric
|dw:1360644367333:dw|That one.
theEric
  • theEric
|dw:1360644483191:dw|and that one.
theEric
  • theEric
Do you see where this is going?
anonymous
  • anonymous
Not realll dont i add the other 8 N at 068 degrees ?
theEric
  • 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
  • anonymous
Dont you just add then like the tail to tail method and then do tan inverse to find the angle ? :S
anonymous
  • anonymous
them *
theEric
  • theEric
You could, huh! But that's sort of what this is. Picture time! :)
anonymous
  • anonymous
okay so im not totally off track
theEric
  • theEric
|dw:1360645168898:dw|derfinately not!
theEric
  • theEric
|dw:1360645291825:dw|
theEric
  • theEric
That last one shows the resultant vector.
anonymous
  • anonymous
OH
theEric
  • theEric
|dw:1360645379901:dw|
anonymous
  • anonymous
makes so much sense !
theEric
  • theEric
\[\huge :)\]
anonymous
  • anonymous
so now i find the tan inverse right
theEric
  • 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
  • theEric
Does that make sense?
theEric
  • theEric
@burhan101 ?
anonymous
  • anonymous
i think so
theEric
  • theEric
Okay.... Ready to try it?
anonymous
  • anonymous
tan-1 = 68/45
anonymous
  • anonymous
i think :S
theEric
  • 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
  • theEric
|dw:1360646989287:dw|
theEric
  • theEric
\[\begin{matrix}x-component&=&10cos(45)&=&7.07\\y-component&=&10sin(45)&=&7.07\end{matrix}\]
theEric
  • 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
  • anonymous
thanks for your help ! :D
theEric
  • 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
  • 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
  • theEric
Switched back. :) You're very welcome! Take care! Tag me if you need any more help.

Looking for something else?

Not the answer you are looking for? Search for more explanations.