BloomLocke367
  • BloomLocke367
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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sweetburger
  • sweetburger
are these vectors?
BloomLocke367
  • BloomLocke367
Yes. I have to find the resultant vector using the component method
phi
  • phi
do the same as before: change to "standard angles" then you add up M*cos(x) of each vector

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sweetburger
  • sweetburger
I dont know how you approach these but i usually draw standard angles on a cartesain
phi
  • phi
for the x values and M*sin(x) for the y values
BloomLocke367
  • BloomLocke367
okay. hang on this may take a bit. Please be patient cx
BloomLocke367
  • BloomLocke367
would the first standard angle measure be 42 degrees? I just subtracted the original from 90
sweetburger
  • sweetburger
yes
BloomLocke367
  • BloomLocke367
and 117 for the second one?
BloomLocke367
  • BloomLocke367
wait.. no...
BloomLocke367
  • BloomLocke367
153
sweetburger
  • sweetburger
yes
BloomLocke367
  • BloomLocke367
the final is 94?
phi
  • phi
yes
BloomLocke367
  • BloomLocke367
now time to add.. one second.
BloomLocke367
  • BloomLocke367
I got Ax=-24.026 and Ay=39.991
BloomLocke367
  • BloomLocke367
so it will be in the 2nd quadrant, right?
sweetburger
  • sweetburger
yea
BloomLocke367
  • BloomLocke367
so \(A= \sqrt{(-24.026)^2+(39.991)^2}\)
sweetburger
  • sweetburger
I mean that looks correct but i havent done any of the corresponding math so im assuming your values are correct
BloomLocke367
  • BloomLocke367
alright.
BloomLocke367
  • BloomLocke367
so I got A=46.65
sweetburger
  • sweetburger
alright so now your looking for the direction via tan^-1(y/x)
BloomLocke367
  • BloomLocke367
mhm.. and I got -59.003°... I'm not sure in which direction though..
sweetburger
  • sweetburger
we know were in the second quadrant so we add 180
BloomLocke367
  • BloomLocke367
oh because right now we're in the 4th?
BloomLocke367
  • BloomLocke367
so 120.997°
sweetburger
  • sweetburger
wait is our x value positive and our y value negative?
phi
  • phi
** Ax=-24.026 and Ay=39.991*** yes that is good
BloomLocke367
  • BloomLocke367
x is negative and y is positive
sweetburger
  • sweetburger
i thought we were in the 2nd quadrant
BloomLocke367
  • BloomLocke367
that is the second quadrant
sweetburger
  • sweetburger
why did you say 4th then lol
phi
  • phi
sketch the vector |dw:1443215354419:dw|
BloomLocke367
  • 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
  • phi
now find the marked angle
phi
  • 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
  • BloomLocke367
I am so confused now >.<
sweetburger
  • sweetburger
|dw:1443215526042:dw| given you know where the vector you've drawn is on the cartesian you can find the correct direction
BloomLocke367
  • BloomLocke367
why do you disregard the negative sign?
sweetburger
  • sweetburger
in this case our vector is in the 2nd quadrant (generalized)|dw:1443215661242:dw|
sweetburger
  • 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
  • sweetburger
rip english
phi
  • 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
  • BloomLocke367
>.< this is so confusing. I
sweetburger
  • 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
  • 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
  • sweetburger
let me show you something |dw:1443216253502:dw|
phi
  • 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
  • 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
  • BloomLocke367
like what do you mean? >.< I don't know this is really confusing me. I was never good at trig.. :/
phi
  • phi
Let's say you have this vector |dw:1443216403123:dw|
phi
  • phi
I assume it is obvious how to find theta ?
BloomLocke367
  • BloomLocke367
yes lol
BloomLocke367
  • BloomLocke367
arctan(y/x)
phi
  • phi
now this problem |dw:1443216475660:dw|
phi
  • 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
  • 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
  • BloomLocke367
ohhhh I think I understand
phi
  • phi
now this problem |dw:1443216774766:dw|
BloomLocke367
  • BloomLocke367
I get it now.
BloomLocke367
  • BloomLocke367
I think lol
phi
  • phi
You should practice on the problems you just did (where you know the answer) just in case ...
BloomLocke367
  • BloomLocke367
I can't use the draw tool.. my mouse thing goes BEHIND google chrome.. I don't know why :/
sweetburger
  • sweetburger
could try alt+tabbing and tabbing back in has fixed it for me before
phi
  • 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
  • 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
  • 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
  • BloomLocke367
ohhh see that makes sense Phi.. I finally get why you said to ignore the sign XD
phi
  • phi
Of course, if you are writing a computer program, you have to work out the rules. (No pictures allowed)
sweetburger
  • sweetburger
phi are you computer software engineer?
BloomLocke367
  • 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
  • sweetburger
im in ap physics b too gl with your course
BloomLocke367
  • BloomLocke367
thanks :D so is this 59.003° S of W?... or is that completely wrong? >.<
sweetburger
  • sweetburger
south of west would be 3rd quadrant
BloomLocke367
  • 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
  • BloomLocke367
OH IT'S N OF W, RIGHT?
sweetburger
  • sweetburger
yes gj :P
BloomLocke367
  • BloomLocke367
sorry for the caps, kinda had an explosion of understanding XD
sweetburger
  • sweetburger
LOL
sweetburger
  • sweetburger
yea its great moment when stuff makes sense :P
BloomLocke367
  • BloomLocke367
okay now I have a word problem.. can I tag you in a new question?
sweetburger
  • sweetburger
sure

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