anonymous
  • anonymous
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 ;/
Physics
chestercat
  • chestercat
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anonymous
  • anonymous
each vector has a y and x component
anonymous
  • anonymous
i and j :) ?
anonymous
  • anonymous
well you're given a magnitude

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anonymous
  • anonymous
|dw:1359604762762:dw|
anonymous
  • anonymous
you there?
anonymous
  • anonymous
|dw:1359605259789:dw|
anonymous
  • anonymous
yeah, im here.. trying to understand
anonymous
  • anonymous
|dw:1359605406061:dw| these need direction due to vectors .. it says A direction is 16.6 degrees sooooo
anonymous
  • anonymous
|dw:1359605476238:dw|
anonymous
  • 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|
anonymous
  • anonymous
now using trig functions you can find the components correct ? can you do that?
anonymous
  • anonymous
using sohcahtoa right?
anonymous
  • anonymous
A_x=sin(16.6)*A?
anonymous
  • anonymous
yes
anonymous
  • anonymous
i got 3.16 as side A_y..
anonymous
  • anonymous
so now we have \[A_x=Acos(16.6)\] \[A_y=Asin(16.6)\]
anonymous
  • anonymous
you had them switched around =]
anonymous
  • anonymous
and 0.95 as a_y.. A=3.30 right?
anonymous
  • anonymous
yep sounds correct
anonymous
  • anonymous
atleast according to my diagram lol
anonymous
  • anonymous
alright so the B is easy since there is only one component to it... which is the \[B_y\]
anonymous
  • anonymous
B=3.30 but theres no theta?
anonymous
  • anonymous
yes but it's a simple line on the y axis so it only has one component an x is what i meant =/
anonymous
  • anonymous
so you have \[B_x=3.30\]
anonymous
  • anonymous
|dw:1359606759899:dw|
anonymous
  • 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
anonymous
  • anonymous
"like components" ? 8.1?
anonymous
  • anonymous
so B is 3.30 and for A do I add a_x and a_y and A?
anonymous
  • anonymous
well lets do it like my book says \[R_x=A_x+B_x\] \[R_y=A_y+B_y\]
anonymous
  • anonymous
\[|A+B|=\sqrt{R_x^2+R_y^2}\]
anonymous
  • anonymous
R_x=4.25 R_y= 3.16
anonymous
  • anonymous
|dw:1359607102751:dw|
anonymous
  • anonymous
5.3 m?
anonymous
  • anonymous
\[B_x=3.30\] \[A_x=3.30cos(16.6)\] \[A_y=3.30sin(16.6)\]
anonymous
  • anonymous
6.6 m?
anonymous
  • anonymous
\[R_x=6.46\] \[R_y=.943\] \[|R|=\sqrt{6.46^2+.943^2}\]
anonymous
  • anonymous
i'm getting 6.5
anonymous
  • anonymous
6.53 to be more exact
anonymous
  • anonymous
yeah i got that too now. i rounded wrong earlier.
anonymous
  • anonymous
so for A-B would i just subtract?
anonymous
  • anonymous
yes to do it easily i'm trying to remember what it looks like lol|dw:1359607683207:dw|
anonymous
  • 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
anonymous
  • anonymous
|dw:1359607791579:dw|
anonymous
  • 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 ;/
anonymous
  • anonymous
I'll go back later when i have more imte tomorrow and try to figure it out lol.
anonymous
  • 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 \

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