Question

Referring to the vectors in the figure, express the sum A⃗ +B⃗ +C⃗ in unit vector notation.
(Figure 1)
Express your answer using two significant figures.
Enter the x and y components of the vector separated by comma.

Answer 1

Alright, so you have to break each vector down. Unit vector notation is the\[\overrightarrow A=\text{(something)}\ \hat x+\text{(something)}\ \hat y\]right?

Answer 2

yeah I did all that, but its wrong

Answer 3

Di you make sure you got the signs right? Like By is negative?

Answer 4

prob not.

Answer 5

Cx is also negative.

Answer 6

And did you make sure to leave out \(\overrightarrow D\)?

Answer 7

that is my work

Answer 8

Alright, I'm checking it out!

Answer 9

ok

Answer 10

I think that's the problem.
Look where I put negatives in that long sum.

Answer 11

so everything else is right?

Answer 12

I verified the Ax, Ay, Bx, and so on. I think the only issue then, as long as you add them correctly, is that By and Cx are negative.

Answer 13

oh, nice, okay yea it was right, how did you know they were -?

Answer 14

So add up the \(\hat x\)'s and \(\hat y\)'s with the negatives. Your methodology was correct.

Answer 15

Remember that you need to adjust your angles if the angle doesn't point into the 1st quadrant. B is a negative angle, and C is 180-the angle

Answer 16

then y isnt bx - as well

Answer 17

No, because cos -x = cos x

Answer 18

But sin -x = - sin x

One is an even function, the other is odd, I don't recall which is which, just the behavior :-)

Answer 19

The Bx is positive because the vector is pointing in the postitive \(x\) direction!

Answer 20

hmm, alright

Answer 21

How's the problem going? Just curious.

Answer 22

i solved that, now i am stuck on another,

Answer 23

Okay, another screenshot?

Answer 24

haha yea

Answer 25

okay, so i find the angle, which is 11, now I am having trouble finding vector velocity pg:

Answer 26

\(\huge\color{blue}{\large\ \ o\ o\ \\\smile}\)

Answer 27

oops, i am trying to find pa sorry

Answer 28

which i though vector vpg = vpa + vag
so I plugged the values in 340= vpa + 65
so vps is 275? and the angle would be 90+11= 101

Answer 29

Can I see part A at all? I'm a little confused. And the vectors are blurry in the last screenshot. Are they \(\overrightarrow v_{ag}\) and \(\overrightarrow v_{pg'}\)?

Answer 30

okay, so I have at the top, the shortest vector is (65, 0 degrees) for Vag. Then the vector to the right is Vpg which is ( 340, 90 degrees) I know for sure those are right. Now i am suppose to find vector Vpa :)

Answer 31

so i thought i did it right, plugged them in i got (275, 101 degrees), but they are not connecting

Answer 32

Well, let's start from the beginning! You're trying to find \(\overrightarrow v_{pa}\), right? Well put the other two vectors into their components! And add the components like you have before. Did you try that?

Answer 33

what do you mean by that?

Answer 34

Well, \(\overrightarrow v_{ag}=\left(65\right)\hat x+\left(0\right)\hat y\), right?

Answer 35

yea and adding them it is (65)x^+(340)y^, right?

Answer 36

then i did sqrt(65^2 + 340^2) = 346.14 it touches :)

Answer 37

awesome :) all done! thanks!

Answer 38

You're welcome! Was that the very last one tonight?

Answer 39

no I just figured out the last one, I did A LOT today!!!!

Answer 40

Agreed! Congrats! It seems your professor likes assigning homework! Evil...

Answer 41

Just kidding. You learned, right? Homework goal accomplished.. Now you can relax. Take care!

Answer 42

thank you :)

Answer 43

:) Your welcome!