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BloomLocke367

  • one year ago

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.

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  1. sweetburger
    • one year ago
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    are these vectors?

  2. BloomLocke367
    • one year ago
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    Yes. I have to find the resultant vector using the component method

  3. phi
    • one year ago
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    do the same as before: change to "standard angles" then you add up M*cos(x) of each vector

  4. sweetburger
    • one year ago
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    I dont know how you approach these but i usually draw standard angles on a cartesain

  5. phi
    • one year ago
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    for the x values and M*sin(x) for the y values

  6. BloomLocke367
    • one year ago
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    okay. hang on this may take a bit. Please be patient cx

  7. BloomLocke367
    • one year ago
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    would the first standard angle measure be 42 degrees? I just subtracted the original from 90

  8. sweetburger
    • one year ago
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    yes

  9. BloomLocke367
    • one year ago
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    and 117 for the second one?

  10. BloomLocke367
    • one year ago
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    wait.. no...

  11. BloomLocke367
    • one year ago
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    153

  12. sweetburger
    • one year ago
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    yes

  13. BloomLocke367
    • one year ago
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    the final is 94?

  14. phi
    • one year ago
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    yes

  15. BloomLocke367
    • one year ago
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    now time to add.. one second.

  16. BloomLocke367
    • one year ago
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    I got Ax=-24.026 and Ay=39.991

  17. BloomLocke367
    • one year ago
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    so it will be in the 2nd quadrant, right?

  18. sweetburger
    • one year ago
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    yea

  19. BloomLocke367
    • one year ago
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    so \(A= \sqrt{(-24.026)^2+(39.991)^2}\)

  20. sweetburger
    • one year ago
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    I mean that looks correct but i havent done any of the corresponding math so im assuming your values are correct

  21. BloomLocke367
    • one year ago
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    alright.

  22. BloomLocke367
    • one year ago
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    so I got A=46.65

  23. sweetburger
    • one year ago
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    alright so now your looking for the direction via tan^-1(y/x)

  24. BloomLocke367
    • one year ago
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    mhm.. and I got -59.003°... I'm not sure in which direction though..

  25. sweetburger
    • one year ago
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    we know were in the second quadrant so we add 180

  26. BloomLocke367
    • one year ago
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    oh because right now we're in the 4th?

  27. BloomLocke367
    • one year ago
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    so 120.997°

  28. sweetburger
    • one year ago
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    wait is our x value positive and our y value negative?

  29. phi
    • one year ago
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    ** Ax=-24.026 and Ay=39.991*** yes that is good

  30. BloomLocke367
    • one year ago
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    x is negative and y is positive

  31. sweetburger
    • one year ago
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    i thought we were in the 2nd quadrant

  32. BloomLocke367
    • one year ago
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    that is the second quadrant

  33. sweetburger
    • one year ago
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    why did you say 4th then lol

  34. phi
    • one year ago
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    sketch the vector |dw:1443215354419:dw|

  35. BloomLocke367
    • one year ago
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    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?

  36. phi
    • one year ago
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    now find the marked angle

  37. phi
    • one year ago
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    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

  38. BloomLocke367
    • one year ago
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    I am so confused now >.<

  39. sweetburger
    • one year ago
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    |dw:1443215526042:dw| given you know where the vector you've drawn is on the cartesian you can find the correct direction

  40. BloomLocke367
    • one year ago
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    why do you disregard the negative sign?

  41. sweetburger
    • one year ago
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    in this case our vector is in the 2nd quadrant (generalized)|dw:1443215661242:dw|

  42. sweetburger
    • one year ago
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    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|

  43. sweetburger
    • one year ago
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    rip english

  44. phi
    • one year ago
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    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

  45. BloomLocke367
    • one year ago
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    >.< this is so confusing. I

  46. sweetburger
    • one year ago
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    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.

  47. BloomLocke367
    • one year ago
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    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.

  48. sweetburger
    • one year ago
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    let me show you something |dw:1443216253502:dw|

  49. phi
    • one year ago
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    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

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

  51. BloomLocke367
    • one year ago
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    like what do you mean? >.< I don't know this is really confusing me. I was never good at trig.. :/

  52. phi
    • one year ago
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    Let's say you have this vector |dw:1443216403123:dw|

  53. phi
    • one year ago
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    I assume it is obvious how to find theta ?

  54. BloomLocke367
    • one year ago
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    yes lol

  55. BloomLocke367
    • one year ago
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    arctan(y/x)

  56. phi
    • one year ago
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    now this problem |dw:1443216475660:dw|

  57. phi
    • one year ago
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    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|

  58. phi
    • one year ago
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    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

  59. BloomLocke367
    • one year ago
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    ohhhh I think I understand

  60. phi
    • one year ago
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    now this problem |dw:1443216774766:dw|

  61. BloomLocke367
    • one year ago
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    I get it now.

  62. BloomLocke367
    • one year ago
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    I think lol

  63. phi
    • one year ago
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    You should practice on the problems you just did (where you know the answer) just in case ...

  64. BloomLocke367
    • one year ago
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    I can't use the draw tool.. my mouse thing goes BEHIND google chrome.. I don't know why :/

  65. sweetburger
    • one year ago
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    could try alt+tabbing and tabbing back in has fixed it for me before

  66. phi
    • one year ago
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    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

  67. BloomLocke367
    • one year ago
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    \(\color{#0cbb34}{\text{Originally Posted by}}\) @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.

  68. sweetburger
    • one year ago
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    I guess the only thing would to be force close your browser and then open it but wait till your done with the question.

  69. BloomLocke367
    • one year ago
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    ohhh see that makes sense Phi.. I finally get why you said to ignore the sign XD

  70. phi
    • one year ago
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    Of course, if you are writing a computer program, you have to work out the rules. (No pictures allowed)

  71. sweetburger
    • one year ago
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    phi are you computer software engineer?

  72. BloomLocke367
    • one year ago
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    \(\color{#0cbb34}{\text{Originally Posted by}}\) @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

  73. sweetburger
    • one year ago
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    im in ap physics b too gl with your course

  74. BloomLocke367
    • one year ago
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    thanks :D so is this 59.003° S of W?... or is that completely wrong? >.<

  75. sweetburger
    • one year ago
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    south of west would be 3rd quadrant

  76. BloomLocke367
    • one year ago
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    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

  77. BloomLocke367
    • one year ago
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    OH IT'S N OF W, RIGHT?

  78. sweetburger
    • one year ago
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    yes gj :P

  79. BloomLocke367
    • one year ago
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    sorry for the caps, kinda had an explosion of understanding XD

  80. sweetburger
    • one year ago
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    LOL

  81. sweetburger
    • one year ago
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    yea its great moment when stuff makes sense :P

  82. BloomLocke367
    • one year ago
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    okay now I have a word problem.. can I tag you in a new question?

  83. sweetburger
    • one year ago
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    sure

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