Question

Can someone please help me figure this out. It's die tomorrow.Add each of the following sets of vectors together using the
component method. F1 = 140 N [W], F2 = 200 N [E30°N], F3 = 100 N [S35°W]

Answer 1

First we need a convention... let's set north and east to be positive directions while setting west and south to be negative, ok?

Answer 2

ok

Answer 3

It's all a matter of adding components... Let's start with the first, shall we?
\[\Large \left.\begin{matrix}FORCE & Horizontal & Vertical\\ F_1 & \color{red}?&\color{red}? \\ F_2 & & \\ F_3 & & \\ \color{red}{total} \end{matrix}\right.\]

Answer 4

Now, if it was 140 newtons northwest, how much is its horizontal (in this case, westward) component?

Answer 5

oh okay. So I just have to find the components then add them up.

Answer 6

Yup.

Answer 7

Would the horizontal component would be 0? And vertical component be 140sin30

Answer 8

For which?
oh wait... F1 is just west, right?
sorry... it's the vertical component that has to be zero if F1 is only going west -_-

Answer 9

F1 is just West yup

Answer 10

oh okay. sorry, physics is not my strong subject

Answer 11

okay, let's key that in
\[\Large \left.\begin{matrix}FORCE & Horizontal & Vertical\\ F_1 & \color{red}{140}&\color{red}0 \\ F_2 & & \\ F_3 & & \\ \color{red}{total} \end{matrix}\right.\]

Answer 12

Is that right?

Answer 13

Yeah. I do believe it would be

Answer 14

LOL
No it isn't :P
Be very careful with signs... remember that west is negative (as is south)
\[\Large \left.\begin{matrix}FORCE & Horizontal & Vertical\\ F_1 & \color{blue}{-140}&\color{blue}0 \\ F_2 &\color{red}? &\color{red}? \\ F_3 & & \\ \color{red}{total} \end{matrix}\right.\]
Now what about F2?

Answer 15

okay so F2 would equal to horizontal 200cos30=173.2 and vertical 200sin30=100

Answer 16

Okay, yes :D
\[\Large \left.\begin{matrix}FORCE & Horizontal & Vertical\\ F_1 & \color{blue}{-140}&\color{blue}0 \\ F_2 &\color{blue}{173.2} &\color{blue}{100} \\ F_3 &\color{red}? &\color{red}? \\ \color{red}{total} \end{matrix}\right.\]

Answer 17

Is that right tho?

Answer 18

Yup :)

Answer 19

oh good okay. So F3 would have horizontal 100cos30=86.60 and vertical 100sin30=50

Answer 20

Not right :D

Answer 21

for one thing, it's 35 degrees in F3

Answer 22

oops :p

Answer 23

oh aahh

Answer 24

okay so would it be horizontal 100cos35=81.9 and vertical 100sin35=57.36

Answer 25

or do I have cosine and sine mixed up :p

Answer 26

the horizontal doesn't always get the cosine...
the one to the left of the angle measure (in this case, S)
It's S35E
So the one to the left is south, a vertical component...

So, vertical gets cosine, horizontal gets sine.
try again :P

Answer 27

whoops, I meant S35W
sorry

Answer 28

Oh now I get it. :P Okay so it would be vertical 100cos35=81.9 and horizontal would be 100sine35=57.35. Is that right?

Answer 29

Almost, but not quite. Do remember that these are south and west that we're dealing with...

Answer 30

oh yes so it would have to be negative

Answer 31

so it would be vertical -100cos35=-81.9 and horizontal would be -00sine35=-57.35.

Answer 32

that's better :P

Answer 33

-100sin35=-57.35

Answer 34

\[\Large \left.\begin{matrix}FORCE & Horizontal & Vertical\\ F_1 & \color{blue}{-140}&\color{blue}0 \\ F_2 &\color{blue}{173.2} &\color{blue}{100} \\ F_3 &\color{blue}{-57.35} &\color{blue}{-81.9} \\ \color{red}{total} \end{matrix}\right.\]

Answer 35

Now just add them up.

Answer 36

alright so horizontal = to -24.15 and vertical adds up to 18.1.

Answer 37

Is that all you need?

Answer 38

I got it from here. Thank you so much. :)

Answer 39

No problem :)

Answer 40

Be very careful with signs... VERY CAREFUL
LOL
signing off

---------------------------------
Terence out