## burhan101 Group Title Determine the resultant of this vector sum: 10 N at 045 degrees, and 8 N at 68 degrees one year ago one year ago

1. theEric Group Title

2. burhan101 Group Title

I can't do it with degrees :S

3. theEric Group Title

After a while, they'll become nearly routine!

4. burhan101 Group Title

Can you please go through the steps , i find these really confusing

5. theEric Group Title

Okay! So you've seen these? And have learned that any vector has separate components (x and y)?

6. theEric Group Title

We can draw a picture, if you like. Let's look at the first vector, 10N. Sound good?

7. burhan101 Group Title

Okay ! :)

8. theEric Group Title

Alright! |dw:1360640234142:dw| There's the vector in one piece.

9. burhan101 Group Title

How do we know that it's NE ?

10. theEric Group Title

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 Group Title

|dw:1360640574984:dw|

12. burhan101 Group Title

OH so that's the common starting point ?

13. theEric Group Title

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 Group Title

Yep! Angle measures start from there if no other information is given!

15. burhan101 Group Title

ohhh okay

16. theEric Group Title

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

17. theEric Group Title

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 Group Title

Now you know where your vector is! And you can split it up with trigonometric functions using the angle!

19. burhan101 Group Title

Oh alright

20. theEric Group Title

So you have seen that before? Using the trig functions?

21. theEric Group Title

@burhan101 so you've got it? Congrats and take care if you did!

22. burhan101 Group Title

No

23. burhan101 Group Title

i know how to do these with like normal vectors

24. theEric Group Title

Haha, okay!

25. burhan101 Group Title

|dw:1360641728957:dw|

26. burhan101 Group Title

i know thats not right -.-

27. burhan101 Group Title

Im so lost lol

28. theEric Group Title

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

29. theEric Group Title

Have you ever used sine or cosine for angles?

30. burhan101 Group Title

Yea

31. theEric Group Title

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 Group Title

|dw:1360642844365:dw|

33. theEric Group Title

If not, I can show you.

34. burhan101 Group Title

how do i add the other vector to this picture now

35. theEric Group Title

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 Group Title

Are you giving it a shot?

37. burhan101 Group Title

Yes im trying to draw it out :\$

38. theEric Group Title

I didn't see the picture you drew before! That's a good rough representation of the vectors! :) Just like mine! :)

39. theEric Group Title

|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 Group Title

So.. (wait for it...)

41. burhan101 Group Title

ohhh

42. theEric Group Title

|dw:1360644030205:dw|

43. theEric Group Title

aaannnndddd........

44. theEric Group Title

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 Group Title

Actually, nevermind! Retake!

46. theEric Group Title

|dw:1360644367333:dw|That one.

47. theEric Group Title

|dw:1360644483191:dw|and that one.

48. theEric Group Title

Do you see where this is going?

49. burhan101 Group Title

Not realll dont i add the other 8 N at 068 degrees ?

50. theEric Group Title

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 Group Title

Dont you just add then like the tail to tail method and then do tan inverse to find the angle ? :S

52. burhan101 Group Title

them *

53. theEric Group Title

You could, huh! But that's sort of what this is. Picture time! :)

54. burhan101 Group Title

okay so im not totally off track

55. theEric Group Title

|dw:1360645168898:dw|derfinately not!

56. theEric Group Title

|dw:1360645291825:dw|

57. theEric Group Title

That last one shows the resultant vector.

58. burhan101 Group Title

OH

59. theEric Group Title

|dw:1360645379901:dw|

60. burhan101 Group Title

makes so much sense !

61. theEric Group Title

$\huge :)$

62. burhan101 Group Title

so now i find the tan inverse right

63. theEric Group Title

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 Group Title

Does that make sense?

65. theEric Group Title

@burhan101 ?

66. burhan101 Group Title

i think so

67. theEric Group Title

68. burhan101 Group Title

tan-1 = 68/45

69. burhan101 Group Title

i think :S

70. theEric Group Title

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 Group Title

|dw:1360646989287:dw|

72. theEric Group Title

$\begin{matrix}x-component&=&10cos(45)&=&7.07\\y-component&=&10sin(45)&=&7.07\end{matrix}$

73. theEric Group Title

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 Group Title

thanks for your help ! :D

75. theEric Group Title

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 Group Title

$\sqrt{7.07^2+7.07^2}\approx 9.998\approx 10.00$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 Group Title

Switched back. :) You're very welcome! Take care! Tag me if you need any more help.