Question

A_1=16.2 N @ 48.0° E of N.
A_2=39.6 N @ 27.0° N of W.
A_3=11.2 N @ 4.00° W of N.

I have to do the same thing as the last problem @phi.

Answer 1

are these vectors?

Answer 2

Yes. I have to find the resultant vector using the component method

Answer 3

do the same as before: change to "standard angles"
then you add up M*cos(x) of each vector

Answer 4

I dont know how you approach these but i usually draw standard angles on a cartesain

Answer 5

for the x values
and M*sin(x) for the y values

Answer 6

okay. hang on this may take a bit. Please be patient cx

Answer 7

would the first standard angle measure be 42 degrees? I just subtracted the original from 90

Answer 8

yes

Answer 9

and 117 for the second one?

Answer 10

wait.. no...

Answer 11

153

Answer 12

yes

Answer 13

the final is 94?

Answer 14

yes

Answer 15

now time to add.. one second.

Answer 16

I got Ax=-24.026 and Ay=39.991

Answer 17

so it will be in the 2nd quadrant, right?

Answer 18

yea

Answer 19

so \(A=
\sqrt{(-24.026)^2+(39.991)^2}\)

Answer 20

I mean that looks correct but i havent done any of the corresponding math so im assuming your values are correct

Answer 21

alright.

Answer 22

so I got A=46.65

Answer 23

alright so now your looking for the direction via tan^-1(y/x)

Answer 24

mhm.. and I got -59.003°... I'm not sure in which direction though..

Answer 25

we know were in the second quadrant so we add 180

Answer 26

oh because right now we're in the 4th?

Answer 27

so 120.997°

Answer 28

wait is our x value positive and our y value negative?

Answer 29

** Ax=-24.026 and Ay=39.991***
yes that is good

Answer 30

x is negative and y is positive

Answer 31

i thought we were in the 2nd quadrant

Answer 32

that is the second quadrant

Answer 33

why did you say 4th then lol

Answer 34

sketch the vector