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burhan101

  • one year ago

Determine the resultant of this vector sum: 10 N at 045 degrees, and 8 N at 68 degrees

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  1. theEric
    • one year ago
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    This isn't so bad. Have you had practice with these?

  2. burhan101
    • one year ago
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    I can't do it with degrees :S

  3. theEric
    • one year ago
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    After a while, they'll become nearly routine!

  4. burhan101
    • one year ago
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    Can you please go through the steps , i find these really confusing

  5. theEric
    • one year ago
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    Okay! So you've seen these? And have learned that any vector has separate components (x and y)?

  6. theEric
    • one year ago
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    We can draw a picture, if you like. Let's look at the first vector, 10N. Sound good?

  7. burhan101
    • one year ago
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    Okay ! :)

  8. theEric
    • one year ago
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    Alright! |dw:1360640234142:dw| There's the vector in one piece.

  9. burhan101
    • one year ago
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    How do we know that it's NE ?

  10. theEric
    • one year ago
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    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.

  11. theEric
    • one year ago
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    |dw:1360640574984:dw|

  12. burhan101
    • one year ago
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    OH so that's the common starting point ?

  13. theEric
    • one year ago
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    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

  14. theEric
    • one year ago
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    Yep! Angle measures start from there if no other information is given!

  15. burhan101
    • one year ago
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    ohhh okay

  16. theEric
    • one year ago
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    |dw:1360640896289:dw|Now, a person might say, "10 degrees west of north" or something silly like that. That means....

  17. theEric
    • one year ago
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    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.

  18. theEric
    • one year ago
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    Now you know where your vector is! And you can split it up with trigonometric functions using the angle!

  19. burhan101
    • one year ago
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    Oh alright

  20. theEric
    • one year ago
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    So you have seen that before? Using the trig functions?

  21. theEric
    • one year ago
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    @burhan101 so you've got it? Congrats and take care if you did!

  22. burhan101
    • one year ago
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    No

  23. burhan101
    • one year ago
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    i know how to do these with like normal vectors

  24. theEric
    • one year ago
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    Haha, okay!

  25. burhan101
    • one year ago
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    |dw:1360641728957:dw|

  26. burhan101
    • one year ago
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    i know thats not right -.-

  27. burhan101
    • one year ago
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    Im so lost lol

  28. theEric
    • one year ago
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    You mean like <x,y>? Yeah, this is different. You aren't just told the components. However, you can find them!

  29. theEric
    • one year ago
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    Have you ever used sine or cosine for angles?

  30. burhan101
    • one year ago
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    Yea

  31. theEric
    • one year ago
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    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?

  32. theEric
    • one year ago
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    |dw:1360642844365:dw|

  33. theEric
    • one year ago
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    If not, I can show you.

  34. burhan101
    • one year ago
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    how do i add the other vector to this picture now

  35. theEric
    • one year ago
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    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).

  36. theEric
    • one year ago
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    Are you giving it a shot?

  37. burhan101
    • one year ago
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    Yes im trying to draw it out :$

  38. theEric
    • one year ago
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    I didn't see the picture you drew before! That's a good rough representation of the vectors! :) Just like mine! :)

  39. theEric
    • one year ago
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    |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.

  40. theEric
    • one year ago
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    So.. (wait for it...)

  41. burhan101
    • one year ago
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    ohhh

  42. theEric
    • one year ago
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    |dw:1360644030205:dw|

  43. theEric
    • one year ago
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    aaannnndddd........

  44. theEric
    • one year ago
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    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|

  45. theEric
    • one year ago
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    Actually, nevermind! Retake!

  46. theEric
    • one year ago
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    |dw:1360644367333:dw|That one.

  47. theEric
    • one year ago
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    |dw:1360644483191:dw|and that one.

  48. theEric
    • one year ago
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    Do you see where this is going?

  49. burhan101
    • one year ago
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    Not realll dont i add the other 8 N at 068 degrees ?

  50. theEric
    • one year ago
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    Well, yeah! You have to add its components to the other vectors components. So, first, you have to get each vector's components.

  51. burhan101
    • one year ago
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    Dont you just add then like the tail to tail method and then do tan inverse to find the angle ? :S

  52. burhan101
    • one year ago
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    them *

  53. theEric
    • one year ago
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    You could, huh! But that's sort of what this is. Picture time! :)

  54. burhan101
    • one year ago
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    okay so im not totally off track

  55. theEric
    • one year ago
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    |dw:1360645168898:dw|derfinately not!

  56. theEric
    • one year ago
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    |dw:1360645291825:dw|

  57. theEric
    • one year ago
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    That last one shows the resultant vector.

  58. burhan101
    • one year ago
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    OH

  59. theEric
    • one year ago
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    |dw:1360645379901:dw|

  60. burhan101
    • one year ago
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    makes so much sense !

  61. theEric
    • one year ago
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    \[\huge :)\]

  62. burhan101
    • one year ago
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    so now i find the tan inverse right

  63. theEric
    • one year ago
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    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.

  64. theEric
    • one year ago
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    Does that make sense?

  65. theEric
    • one year ago
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    @burhan101 ?

  66. burhan101
    • one year ago
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    i think so

  67. theEric
    • one year ago
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    Okay.... Ready to try it?

  68. burhan101
    • one year ago
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    tan-1 = 68/45

  69. burhan101
    • one year ago
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    i think :S

  70. theEric
    • one year ago
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    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!

  71. theEric
    • one year ago
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    |dw:1360646989287:dw|

  72. theEric
    • one year ago
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    \[\begin{matrix}x-component&=&10cos(45)&=&7.07\\y-component&=&10sin(45)&=&7.07\end{matrix}\]

  73. theEric
    • one year ago
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    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.

  74. burhan101
    • one year ago
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    thanks for your help ! :D

  75. theEric
    • one year ago
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    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.\[tan^{-1}{\LARGE (}\frac{y}{x}\LARGE )\]

  76. theEric
    • one year ago
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    \[\sqrt{7.07^2+7.07^2}\approx 9.998\approx 10.00\]There was some rounding error to get to 9.998. But if you round it to the nearest hundredth, you have 10! \[tan^{-1}{\LARGE (}\frac{7.07}{7.07}{\LARGE )}=tan^{-1}(1)=45^o\]

  77. theEric
    • one year ago
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    Switched back. :) You're very welcome! Take care! Tag me if you need any more help.

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