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burhan101
Group Title
Determine the resultant of this vector sum: 10 N at 045 degrees, and 8 N at 68 degrees
 one year ago
 one year ago
burhan101 Group Title
Determine the resultant of this vector sum: 10 N at 045 degrees, and 8 N at 68 degrees
 one year ago
 one year ago

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theEric Group TitleBest ResponseYou've already chosen the best response.1
This isn't so bad. Have you had practice with these?
 one year ago

burhan101 Group TitleBest ResponseYou've already chosen the best response.1
I can't do it with degrees :S
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
After a while, they'll become nearly routine!
 one year ago

burhan101 Group TitleBest ResponseYou've already chosen the best response.1
Can you please go through the steps , i find these really confusing
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Okay! So you've seen these? And have learned that any vector has separate components (x and y)?
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
We can draw a picture, if you like. Let's look at the first vector, 10N. Sound good?
 one year ago

burhan101 Group TitleBest ResponseYou've already chosen the best response.1
Okay ! :)
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Alright! dw:1360640234142:dw There's the vector in one piece.
 one year ago

burhan101 Group TitleBest ResponseYou've already chosen the best response.1
How do we know that it's NE ?
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
dw:1360640574984:dw
 one year ago

burhan101 Group TitleBest ResponseYou've already chosen the best response.1
OH so that's the common starting point ?
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Yep! Angle measures start from there if no other information is given!
 one year ago

burhan101 Group TitleBest ResponseYou've already chosen the best response.1
ohhh okay
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
dw:1360640896289:dwNow, a person might say, "10 degrees west of north" or something silly like that. That means....
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
But as a starting point, positive xaxis is what you'll use. Hopefully your teacher agrees. It's a farreaching convention.. Goes into lots of maths.
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Now you know where your vector is! And you can split it up with trigonometric functions using the angle!
 one year ago

burhan101 Group TitleBest ResponseYou've already chosen the best response.1
Oh alright
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
So you have seen that before? Using the trig functions?
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
@burhan101 so you've got it? Congrats and take care if you did!
 one year ago

burhan101 Group TitleBest ResponseYou've already chosen the best response.1
i know how to do these with like normal vectors
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Haha, okay!
 one year ago

burhan101 Group TitleBest ResponseYou've already chosen the best response.1
dw:1360641728957:dw
 one year ago

burhan101 Group TitleBest ResponseYou've already chosen the best response.1
i know thats not right .
 one year ago

burhan101 Group TitleBest ResponseYou've already chosen the best response.1
Im so lost lol
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
You mean like <x,y>? Yeah, this is different. You aren't just told the components. However, you can find them!
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Have you ever used sine or cosine for angles?
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Well that's what you'll be doing here. Are you well practiced with them? Going back to this picture, dw:1360642770812:dwDo you see how to get the x and y components?
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
dw:1360642844365:dw
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
If not, I can show you.
 one year ago

burhan101 Group TitleBest ResponseYou've already chosen the best response.1
how do i add the other vector to this picture now
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
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).
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Are you giving it a shot?
 one year ago

burhan101 Group TitleBest ResponseYou've already chosen the best response.1
Yes im trying to draw it out :$
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
I didn't see the picture you drew before! That's a good rough representation of the vectors! :) Just like mine! :)
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
So.. (wait for it...)
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
dw:1360644030205:dw
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
aaannnndddd........
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Actually, nevermind! Retake!
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
dw:1360644367333:dwThat one.
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
dw:1360644483191:dwand that one.
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Do you see where this is going?
 one year ago

burhan101 Group TitleBest ResponseYou've already chosen the best response.1
Not realll dont i add the other 8 N at 068 degrees ?
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Well, yeah! You have to add its components to the other vectors components. So, first, you have to get each vector's components.
 one year ago

burhan101 Group TitleBest ResponseYou've already chosen the best response.1
Dont you just add then like the tail to tail method and then do tan inverse to find the angle ? :S
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
You could, huh! But that's sort of what this is. Picture time! :)
 one year ago

burhan101 Group TitleBest ResponseYou've already chosen the best response.1
okay so im not totally off track
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
dw:1360645168898:dwderfinately not!
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
dw:1360645291825:dw
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
That last one shows the resultant vector.
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
dw:1360645379901:dw
 one year ago

burhan101 Group TitleBest ResponseYou've already chosen the best response.1
makes so much sense !
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
\[\huge :)\]
 one year ago

burhan101 Group TitleBest ResponseYou've already chosen the best response.1
so now i find the tan inverse right
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
After you get the components of the resultant vector, you use them to find the magnitude (however many N) and that inverse tangent (ycomponent divided by xcomponent) to find the angle.
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Does that make sense?
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
@burhan101 ?
 one year ago

burhan101 Group TitleBest ResponseYou've already chosen the best response.1
i think so
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Okay.... Ready to try it?
 one year ago

burhan101 Group TitleBest ResponseYou've already chosen the best response.1
tan1 = 68/45
 one year ago

burhan101 Group TitleBest ResponseYou've already chosen the best response.1
i think :S
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
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 ycomponent divided by its xcomponent. 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 refind the angle!
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
dw:1360646989287:dw
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
\[\begin{matrix}xcomponent&=&10cos(45)&=&7.07\\ycomponent&=&10sin(45)&=&7.07\end{matrix}\]
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
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 xcomponents together. Then add the ycomponents together. Then you have the resultant vectors components. That would be how you knew vectors before.
 one year ago

burhan101 Group TitleBest ResponseYou've already chosen the best response.1
thanks for your help ! :D
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
And to rediscover that magnitude and angle of the 10N vector, using it's x and ycomponents, Magnitude: Use Pythagorean Theorem.\[\sqrt{x^2+y^2}\] For the angle: Use inverse tangent.\[tan^{1}{\LARGE (}\frac{y}{x}\LARGE )\]
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
\[\sqrt{7.07^2+7.07^2}\approx 9.998\approx 10.00\]There was some rounding error to get to 9.998. But if you round it to the nearest hundredth, you have 10! \[tan^{1}{\LARGE (}\frac{7.07}{7.07}{\LARGE )}=tan^{1}(1)=45^o\]
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Switched back. :) You're very welcome! Take care! Tag me if you need any more help.
 one year ago
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