monroe17
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 ;/
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Outkast3r09
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each vector has a y and x component
monroe17
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i and j :) ?
Outkast3r09
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well you're given a magnitude
Outkast3r09
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|dw:1359604762762:dw|
Outkast3r09
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you there?
Outkast3r09
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|dw:1359605259789:dw|
monroe17
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yeah, im here.. trying to understand
Outkast3r09
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|dw:1359605406061:dw|
these need direction due to vectors .. it says A direction is 16.6 degrees sooooo
Outkast3r09
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|dw:1359605476238:dw|
Outkast3r09
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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|
Outkast3r09
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now using trig functions you can find the components correct ? can you do that?
monroe17
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using sohcahtoa right?
monroe17
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A_x=sin(16.6)*A?
Outkast3r09
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yes
monroe17
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i got 3.16 as side A_y..
Outkast3r09
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so now we have
\[A_x=Acos(16.6)\]
\[A_y=Asin(16.6)\]
Outkast3r09
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you had them switched around =]
monroe17
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and 0.95 as a_y.. A=3.30 right?
Outkast3r09
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yep sounds correct
Outkast3r09
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atleast according to my diagram lol
Outkast3r09
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alright so the B is easy since there is only one component to it... which is the \[B_y\]
monroe17
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B=3.30 but theres no theta?
Outkast3r09
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yes but it's a simple line on the y axis so it only has one component an x is what i meant =/
Outkast3r09
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so you have
\[B_x=3.30\]
Outkast3r09
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|dw:1359606759899:dw|
Outkast3r09
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now you just add the like components to get 2 sides and the magnitude will be the square root of those squared total components
monroe17
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"like components" ? 8.1?
monroe17
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so B is 3.30 and for A do I add a_x and a_y and A?
Outkast3r09
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well lets do it like my book says
\[R_x=A_x+B_x\]
\[R_y=A_y+B_y\]
Outkast3r09
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\[|A+B|=\sqrt{R_x^2+R_y^2}\]
monroe17
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R_x=4.25
R_y= 3.16
Outkast3r09
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|dw:1359607102751:dw|
monroe17
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5.3 m?
Outkast3r09
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\[B_x=3.30\]
\[A_x=3.30cos(16.6)\]
\[A_y=3.30sin(16.6)\]
monroe17
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6.6 m?
Outkast3r09
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\[R_x=6.46\]
\[R_y=.943\]
\[|R|=\sqrt{6.46^2+.943^2}\]
Outkast3r09
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i'm getting 6.5
Outkast3r09
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6.53 to be more exact
monroe17
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yeah i got that too now. i rounded wrong earlier.
monroe17
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so for A-B would i just subtract?
Outkast3r09
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yes to do it easily i'm trying to remember what it looks like lol|dw:1359607683207:dw|
Outkast3r09
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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
Outkast3r09
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|dw:1359607791579:dw|
monroe17
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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 ;/
monroe17
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I'll go back later when i have more imte tomorrow and try to figure it out lol.
Outkast3r09
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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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