A_1=16.2 N @ 48.0° E of N.
A_2=39.6 N @ 27.0° N of W.
A_3=11.2 N @ 4.00° W of N.
I have to do the same thing as the last problem @phi.

- BloomLocke367

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- sweetburger

are these vectors?

- BloomLocke367

Yes. I have to find the resultant vector using the component method

- phi

do the same as before: change to "standard angles"
then you add up M*cos(x) of each vector

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## More answers

- sweetburger

I dont know how you approach these but i usually draw standard angles on a cartesain

- phi

for the x values
and M*sin(x) for the y values

- BloomLocke367

okay. hang on this may take a bit. Please be patient cx

- BloomLocke367

would the first standard angle measure be 42 degrees? I just subtracted the original from 90

- sweetburger

yes

- BloomLocke367

and 117 for the second one?

- BloomLocke367

wait.. no...

- BloomLocke367

153

- sweetburger

yes

- BloomLocke367

the final is 94?

- phi

yes

- BloomLocke367

now time to add.. one second.

- BloomLocke367

I got Ax=-24.026 and Ay=39.991

- BloomLocke367

so it will be in the 2nd quadrant, right?

- sweetburger

yea

- BloomLocke367

so \(A=
\sqrt{(-24.026)^2+(39.991)^2}\)

- sweetburger

I mean that looks correct but i havent done any of the corresponding math so im assuming your values are correct

- BloomLocke367

alright.

- BloomLocke367

so I got A=46.65

- sweetburger

alright so now your looking for the direction via tan^-1(y/x)

- BloomLocke367

mhm.. and I got -59.003°... I'm not sure in which direction though..

- sweetburger

we know were in the second quadrant so we add 180

- BloomLocke367

oh because right now we're in the 4th?

- BloomLocke367

so 120.997°

- sweetburger

wait is our x value positive and our y value negative?

- phi

** Ax=-24.026 and Ay=39.991***
yes that is good

- BloomLocke367

x is negative and y is positive

- sweetburger

i thought we were in the 2nd quadrant

- BloomLocke367

that is the second quadrant

- sweetburger

why did you say 4th then lol

- phi

sketch the vector
|dw:1443215354419:dw|

- BloomLocke367

because -59.003.. don't we want to get in the 2nd quadrant? wouldn't -59.003 make us in the 4th? and we're trying to get to the second?

- phi

now find the marked angle

- phi

You can leave off the signs and just use y= 39.991 x= 24.026
and find the marked angle,
then looking at the picture, you see you found the angle between the neg x-axis and the vector
180 - that angle is the angle you want

- BloomLocke367

I am so confused now >.<

- sweetburger

|dw:1443215526042:dw| given you know where the vector you've drawn is on the cartesian you can find the correct direction

- BloomLocke367

why do you disregard the negative sign?

- sweetburger

in this case our vector is in the 2nd quadrant (generalized)|dw:1443215661242:dw|

- sweetburger

alright lets say the our y value in this case was also negative then our angle would be a positive 59 degrees would you then say its in the first quadrant??? no you would have to find the its actually angle in the 3rd quadrant being 180 + 59|dw:1443215872610:dw|

- sweetburger

rip english

- phi

I never remember the rules to figure out the angle algebraically
I draw the triangle, label the distances (but ignore the sign), and find the angle formed by the hypotenuse and x-axis.
once I have that angle, I can figure out what the true angle is in terms of 0 to 360

- BloomLocke367

>.< this is so confusing. I

- sweetburger

tan is y/x your angle is going to positive in the 1st and 3rd quadrant and negative in the 2nd and 4th this is why knowing what value to add to the found angle is useful. Or you could do what phi described which is also valid.

- BloomLocke367

I understand that they are negative in the 2nd and 4th.. and positive in the 1st and 3rd.. I guess I just don't understand when to add 180 and when not to and why.. if that makes sense.

- sweetburger

let me show you something |dw:1443216253502:dw|

- phi

The rules are a pain to memorize.
It should be clear if you treat the problem as a triangle.
then knowing what quadrant you are in, you can see whether to add 180, etc

- sweetburger

I agree that conceptually understanding it would prove to make you stronger with these problems in the long run. rather then memorizing the graph

- BloomLocke367

like what do you mean? >.< I don't know this is really confusing me. I was never good at trig.. :/

- phi

Let's say you have this vector
|dw:1443216403123:dw|

- phi

I assume it is obvious how to find theta ?

- BloomLocke367

yes lol

- BloomLocke367

arctan(y/x)

- phi

now this problem
|dw:1443216475660:dw|

- phi

if we treat the x and y as positive distances, then it is the same problem as in the first quadrant. You find theta the same way.
but we want this angle
|dw:1443216568948:dw|

- phi

Do you see if you know theta, you can find the angle to the vector starting from the positive x-axis ? i.e. find 180 - theta

- BloomLocke367

ohhhh I think I understand

- phi

now this problem
|dw:1443216774766:dw|

- BloomLocke367

I get it now.

- BloomLocke367

I think lol

- phi

You should practice on the problems you just did (where you know the answer)
just in case ...

- BloomLocke367

I can't use the draw tool.. my mouse thing goes BEHIND google chrome.. I don't know why :/

- sweetburger

could try alt+tabbing and tabbing back in has fixed it for me before

- phi

Notice, if you keep the signs on the x and y values, the tan returns negative angles in some of the quadrants, and that is confusing. It is simpler to always work with positive x and y values and find a positive theta. Then, knowing which quadrant the vector is in, you can find the angle using that positive theta

- BloomLocke367

\(\color{#0cbb34}{\text{Originally Posted by}}\) @sweetburger
could try alt+tabbing and tabbing back in has fixed it for me before
\(\color{#0cbb34}{\text{End of Quote}}\)
nope :/ it didn't work.

- sweetburger

I guess the only thing would to be force close your browser and then open it but wait till your done with the question.

- BloomLocke367

ohhh see that makes sense Phi.. I finally get why you said to ignore the sign XD

- phi

Of course, if you are writing a computer program, you have to work out the rules. (No pictures allowed)

- sweetburger

phi are you computer software engineer?

- BloomLocke367

\(\color{#0cbb34}{\text{Originally Posted by}}\) @phi
Of course, if you are writing a computer program, you have to work out the rules. (No pictures allowed)
\(\color{#0cbb34}{\text{End of Quote}}\)
I'm just in AP Physics cx but good to know :D

- sweetburger

im in ap physics b too gl with your course

- BloomLocke367

thanks :D
so is this 59.003° S of W?... or is that completely wrong? >.<

- sweetburger

south of west would be 3rd quadrant

- BloomLocke367

yes I know.. wait.. nevermind O.O I did a bad.. XD
I used the original vectors and my teacher said to not do that XD

- BloomLocke367

OH IT'S N OF W, RIGHT?

- sweetburger

yes gj :P

- BloomLocke367

sorry for the caps, kinda had an explosion of understanding XD

- sweetburger

LOL

- sweetburger

yea its great moment when stuff makes sense :P

- BloomLocke367

okay now I have a word problem.. can I tag you in a new question?

- sweetburger

sure

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