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burhan101

  • 2 years 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
    • 2 years ago
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    This isn't so bad. Have you had practice with these?

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

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

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

  5. theEric
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    Okay ! :)

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

  9. burhan101
    • 2 years ago
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    How do we know that it's NE ?

  10. theEric
    • 2 years 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
    • 2 years ago
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    |dw:1360640574984:dw|

  12. burhan101
    • 2 years ago
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    OH so that's the common starting point ?

  13. theEric
    • 2 years 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
    • 2 years ago
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    Yep! Angle measures start from there if no other information is given!

  15. burhan101
    • 2 years ago
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    ohhh okay

  16. theEric
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    Oh alright

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

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

  22. burhan101
    • 2 years ago
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    No

  23. burhan101
    • 2 years ago
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    i know how to do these with like normal vectors

  24. theEric
    • 2 years ago
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    Haha, okay!

  25. burhan101
    • 2 years ago
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    |dw:1360641728957:dw|

  26. burhan101
    • 2 years ago
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    i know thats not right -.-

  27. burhan101
    • 2 years ago
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    Im so lost lol

  28. theEric
    • 2 years 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
    • 2 years ago
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    Have you ever used sine or cosine for angles?

  30. burhan101
    • 2 years ago
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    Yea

  31. theEric
    • 2 years 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
    • 2 years ago
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    |dw:1360642844365:dw|

  33. theEric
    • 2 years ago
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    If not, I can show you.

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

  35. theEric
    • 2 years 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
    • 2 years ago
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    Are you giving it a shot?

  37. burhan101
    • 2 years ago
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    Yes im trying to draw it out :$

  38. theEric
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    So.. (wait for it...)

  41. burhan101
    • 2 years ago
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    ohhh

  42. theEric
    • 2 years ago
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    |dw:1360644030205:dw|

  43. theEric
    • 2 years ago
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    aaannnndddd........

  44. theEric
    • 2 years 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
    • 2 years ago
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    Actually, nevermind! Retake!

  46. theEric
    • 2 years ago
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    |dw:1360644367333:dw|That one.

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

  48. theEric
    • 2 years ago
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    Do you see where this is going?

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

  50. theEric
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    them *

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

  54. burhan101
    • 2 years ago
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    okay so im not totally off track

  55. theEric
    • 2 years ago
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    |dw:1360645168898:dw|derfinately not!

  56. theEric
    • 2 years ago
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    |dw:1360645291825:dw|

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

  58. burhan101
    • 2 years ago
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    OH

  59. theEric
    • 2 years ago
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    |dw:1360645379901:dw|

  60. burhan101
    • 2 years ago
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    makes so much sense !

  61. theEric
    • 2 years ago
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    \[\huge :)\]

  62. burhan101
    • 2 years ago
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    so now i find the tan inverse right

  63. theEric
    • 2 years 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
    • 2 years ago
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    Does that make sense?

  65. theEric
    • 2 years ago
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    @burhan101 ?

  66. burhan101
    • 2 years ago
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    i think so

  67. theEric
    • 2 years ago
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    Okay.... Ready to try it?

  68. burhan101
    • 2 years ago
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    tan-1 = 68/45

  69. burhan101
    • 2 years ago
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    i think :S

  70. theEric
    • 2 years 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
    • 2 years ago
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    |dw:1360646989287:dw|

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

  73. theEric
    • 2 years 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
    • 2 years ago
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    thanks for your help ! :D

  75. theEric
    • 2 years 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
    • 2 years 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
    • 2 years 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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