## anonymous 3 years ago The displacement vectors A and B both have magnitudes of 3.30 m. The direction of vector A is 16.6 degrees. Find the magnitude of |A+B|, |A-B|, |B-A|, and |A-2B| graphically. PLEASE HELP ;/

1. anonymous

each vector has a y and x component

2. anonymous

i and j :) ?

3. anonymous

well you're given a magnitude

4. anonymous

|dw:1359604762762:dw|

5. anonymous

you there?

6. anonymous

|dw:1359605259789:dw|

7. anonymous

yeah, im here.. trying to understand

8. anonymous

|dw:1359605406061:dw| these need direction due to vectors .. it says A direction is 16.6 degrees sooooo

9. anonymous

|dw:1359605476238:dw|

10. anonymous

so in order to make sense of each vector you need to break them down into right triangles that show components of x and y i'll start with A|dw:1359605592468:dw|

11. anonymous

now using trig functions you can find the components correct ? can you do that?

12. anonymous

using sohcahtoa right?

13. anonymous

A_x=sin(16.6)*A?

14. anonymous

yes

15. anonymous

i got 3.16 as side A_y..

16. anonymous

so now we have $A_x=Acos(16.6)$ $A_y=Asin(16.6)$

17. anonymous

you had them switched around =]

18. anonymous

and 0.95 as a_y.. A=3.30 right?

19. anonymous

yep sounds correct

20. anonymous

atleast according to my diagram lol

21. anonymous

alright so the B is easy since there is only one component to it... which is the $B_y$

22. anonymous

B=3.30 but theres no theta?

23. anonymous

yes but it's a simple line on the y axis so it only has one component an x is what i meant =/

24. anonymous

so you have $B_x=3.30$

25. anonymous

|dw:1359606759899:dw|

26. anonymous

now you just add the like components to get 2 sides and the magnitude will be the square root of those squared total components

27. anonymous

"like components" ? 8.1?

28. anonymous

so B is 3.30 and for A do I add a_x and a_y and A?

29. anonymous

well lets do it like my book says $R_x=A_x+B_x$ $R_y=A_y+B_y$

30. anonymous

$|A+B|=\sqrt{R_x^2+R_y^2}$

31. anonymous

R_x=4.25 R_y= 3.16

32. anonymous

|dw:1359607102751:dw|

33. anonymous

5.3 m?

34. anonymous

$B_x=3.30$ $A_x=3.30cos(16.6)$ $A_y=3.30sin(16.6)$

35. anonymous

6.6 m?

36. anonymous

$R_x=6.46$ $R_y=.943$ $|R|=\sqrt{6.46^2+.943^2}$

37. anonymous

i'm getting 6.5

38. anonymous

6.53 to be more exact

39. anonymous

yeah i got that too now. i rounded wrong earlier.

40. anonymous

so for A-B would i just subtract?

41. anonymous

yes to do it easily i'm trying to remember what it looks like lol|dw:1359607683207:dw|

42. anonymous

a-b is the diagonal from left to right if i remember right so you have to take A and move it infront of b like this

43. anonymous

|dw:1359607791579:dw|

44. anonymous

oh my lordy... i give up... ;/ my grade is good enough.. 96.3% on this assignment. I'm done with this question.. i still have to do a lab report and study for calc 2... thanks for the help though ;/

45. anonymous

I'll go back later when i have more imte tomorrow and try to figure it out lol.

46. anonymous

i looked at my book you have to face the vectors in certain ways so a-b says to face the vecotrs inward to eachother \