How to Represent and Convert These Vectors?

[cary1234]

i think this is a very simple assignment, but I am confused and i don't know where to start.

and by the way do you think the question is corect?

here it is.

c(->) = 125m/s 75deg north of east

e(->) = 500 kg m/s (hey, what should i do in that kilogram meter/second? thatas what makes me confuse) 18 deg south of east.

s(->) = 707m 26 deg west of south

where do i start?
do you advice me to convert all of that to the same unit first?

im totally noob about this..

 

[tiny-tim]

Welcome to PF!

Hi cary1234! Welcome to PF!

What is the question?

(is it to repesent each vector in i and j? or as (x,y)? or graphically? or what?)

And are these three separate questions, or are c e and s connected in some way?

 

[cary1234]

Hi! thanks for your greetings.Im so sorry i forgot to type the question.
here it is.

find the resultant of
a) 800 Newton due South
and 600 Newton due East

b) 300 Newton due East and 450 Newton due north.


...................

by the way, were on just the basic part so i think we need to represent each vector in graphical...


all of them is connected..



im just starting to learn physics so plelase be patient with me. but i promise that i will try to follow whawt you will said.


by the way, I am so thankful for your response!

 

[wisvuze]

Those values are the "magnitudes" of the vectors, so you really only know the direction and "length" of each vector, since no points are given. If it says to the East, draw a vector to the right (positive right x-axis) and if it says North then just draw a vector of magnitude whatever to the positive Y-axis.. and otherwise, follow the angles given

 

[cary1234]

Those values are the "magnitudes" of the vectors, so you really only know the direction and "length" of each vector, since no points are given. If it says to the East, draw a vector to the right (positive right x-axis) and if it says North then just draw a vector of magnitude whatever to the positive Y-axis.. and otherwise, follow the angles given

hi! thanks for helping me.

is this correct?

i hope u understand.

im confused on putting the 75 deg. i don't know the rules, or other details regarding the degrees?

i need to know where to put the 75 deg.

heres the picture of what i did.

<a href="http://img134.imageshack.us/i/60341867.png/" target="_blank"><img src="http://img134.imageshack.us/img134/3150/60341867.th.png" border="0" alt="Free Image Hosting at www.ImageShack.us" /></a><br /><br /><a href="http://img604.imageshack.us/content.php?page=blogpost&files=img134/3150/60341867.png" title="QuickPost"><img src="http://imageshack.us/img/butansn.png" alt="QuickPost" border="0"></a> Quickpost this image to Myspace, Digg, Facebook, and others!

 

[wisvuze]

hi! thanks for helping me.

is this correct?

i hope u understand.

im confused on putting the 75 deg. i don't know the rules, or other details regarding the degrees?

i need to know where to put the 75 deg.

It says 75 degrees north of east, so that's telling you to use the East direction ( positive x-axis) as your base, and go in the north direction (positive angle), and make it 75 degrees.
In other words, imagine your x y axes to be a compass pointing North East South West. Now if it says 75 degrees northeast - from the east direction, draw an angle 75 degrees northeast

 

[charlestchan]

i thought this is like a momentum question, but i dun get it , why there's no mass given -_-

 

[cary1234]

i thought this is like a momentum question, but i dun get it , why there's no mass given -_-

find the resultant of
a) 800 Newton due South
and 600 Newton due East

b) 300 Newton due East and 450 Newton due north.


is this what you are looking for?

 

[modulus]

First of all, please clear up the relation between what you typed at the very beginning (the first post) and the question you stated later (the last post). How are they connected?

 

[modulus]

As for the latter question (if it is not related to the first one) you can solve it simply by adding the vectors using the vector addition equation.
Like wisvuze said, consider the positive x-axis to be the east direction, negative x-axis to be west, positive y-axis to be north, and negative y-axis to be south. In this way, the angle between the two forces in each of the questions will be 90 degrees.
If you have any doubts, post it, and please explain how the first and second are connected to each other.

 

[tiny-tim]

Hi cary1234!

(just got up :zzz: …)

hi! thanks for helping me.

is this correct?

i hope u understand.

im confused on putting the 75 deg. i don't know the rules, or other details regarding the degrees?

i need to know where to put the 75 deg.

heres the picture of what i did.
http://img134.imageshack.us/i/60341867.png/"

(hmm … I still don't understand what the questions (or answers?) in your post #1 have to do with the questions in your post #3. Anyway …)

No, your image is completely wrong …

(I'm referring to your imageshack image, not the picture in your post#8, which doesn't make it clear where the origin is).

The resultant should start at the origin.

From the origin, go along the first vector, then from the end of the first vector, add the second vector (so it's "half a parallelogram").

Then complete the triangle … the third side (starting at the origin) is the resultant (or sum).

If you're just given the single vector "125m/s 75deg north of east", then again it starts from the origin … go right and up (mostly up) from the origin. You can mark "75º" as the angle from the East axis.

(Incidentally, I'm completely confused as to why you've marked the right-angle as "75º" )

 

[cary1234]

Hi tiny-tim I am sorry if I am confusing you..

i will type the question again and i promised to make this more clear.

represent the following vectors..

C = 125m/s 75 deg north of east.
E = 50 kg m/s 18 deg south of east
S = 707 m 26 deg west of south

they are all connected to each other..
thats the part one of my assignment.

II. part two.
find the resultant of ...
a) 800 N. due south and 600 N. due North.

b) 300 N. due east and 450 N. due North.

a and b is not connected to each other..

i will follow your instruction and i will try to upload the image.
(by the way, do you know any website where i can graph and upload easily?)

thanks for your kindness..

 

[cary1234]

Hi wisvuze and modulus thank you for your help. I am now less confuse and afraid of physics.. :)

As for the latter question (if it is not related to the first one) you can solve it simply by adding the vectors using the vector addition equation.
Like wisvuze said, consider the positive x-axis to be the east direction, negative x-axis to be west, positive y-axis to be north, and negative y-axis to be south. In this way, the angle between the two forces in each of the questions will be 90 degrees.
If you have any doubts, post it, and please explain how the first and second are connected to each other.

am i right about this? ( north west is -,+ ), ( north east is +,+), (south west is -,-) and (south east is +,-)..

so, with regard to what you say the question letter a) 800 N. due south and 600 N. due North is this correct?
please correct my English if it is wrong.

http://img134.imageshack.us/img134/6024/85620722.th.png

 

[wisvuze]

Hi! thanks for your greetings.Im so sorry i forgot to type the question.
here it is.

find the resultant of
a) 800 Newton due South
and 600 Newton due East

b) 300 Newton due East and 450 Newton due north....................

by the way, were on just the basic part so i think we need to represent each vector in graphical...all of them is connected..
im just starting to learn physics so plelase be patient with me. but i promise that i will try to follow whawt you will said.by the way, I am so thankful for your response!

Ah , I didn't see this before.

Okay well, use the method I told you about from my earlier post - but draw each vector as its own line/arrow. For example (if one vector is 50 N North and the other one is 100N West)^
|
|
| Vector 1: Length of 50 and pointing north
|
|

-------------------------------------->
Vector 2: length of 100 and pointing West.

What happens when you combine them?
?

Consider the direction of this ' / '

 

[cary1234]

in example# 1 on my notebook the question is find R.
150N (south)
90N (west)

i graph it.

R^2 = (x)square + (y)square
... = 174.93N


teta = tan-(-90N/-150N) = 30.96deg. W of S.


for my question..

how do i know the correct format. because i can make it wrong by making teta = tan-(150/90) = 59.09


i need to know what to consider in that case?
please correct my wrong grammars.. so that i will know and i will not repeat.

 

[cary1234]

Ah , I didn't see this before.

Okay well, use the method I told you about from my earlier post - but draw each vector as its own line/arrow. For example (if one vector is 50 N North and the other one is 100N West)


^
|
|
| Vector 1: Length of 50 and pointing north
|
|

-------------------------------------->
Vector 2: length of 100 and pointing West.

What happens when you combine them?



?

Consider the direction of this ' / '

if i apply what you say. the graph in C= 125 m/s 75 deg North of East
will be look like this?

http://img80.imageshack.us/img80/9152/23818104.th.png

 

[cary1234]

Ah , I didn't see this before.

Okay well, use the method I told you about from my earlier post - but draw each vector as its own line/arrow. For example (if one vector is 50 N North and the other one is 100N West)


^
|
|
| Vector 1: Length of 50 and pointing north
|
|

-------------------------------------->
Vector 2: length of 100 and pointing West.

What happens when you combine them?



?

Consider the direction of this ' / '

i think this is the answer to your question.

http://img145.imageshack.us/img145/866/assxc.th.png

r^2= x^2+y^2
r= square root of (50)^2 + (100)^2
r= 111.80

teta = tan-(-100/-50)
teta = 63.43

i don't know how to get the teta. i just folow the instruction on my notebook.

how do i know? because i can accidentally make it tan-(50/100)

 

[ideasrule]

First of all, why do you keep on labeling the right angle as 75 degrees? Does it look like 75 degrees to you? Second, displacement is a vector, so adding vectors is exactly like adding displacements. So instead of 50 N and 100 N, imagine if the quantities are 50 m and 100 m. You start at the origin. You go 50 m north. Then, you go 100 m west. How far have you gone? At what angle are you from the origin? Just draw out the relevant triangle and calculate an angle; any angle will do as long as you indicate clearly what angle it is.

 

[cary1234]

First of all, why do you keep on labeling the right angle as 75 degrees? Does it look like 75 degrees to you? Second, displacement is a vector, so adding vectors is exactly like adding displacements. So instead of 50 N and 100 N, imagine if the quantities are 50 m and 100 m. You start at the origin. You go 50 m north. Then, you go 100 m west. How far have you gone? At what angle are you from the origin? Just draw out the relevant triangle and calculate an angle; any angle will do as long as you indicate clearly what angle it is.

ow. sorry for that.
after researching on how to get angle for right triangle i m pretty sure this is correct.


C= 125 m/s 75 deg North of East
http://img228.imageshack.us/img228/8565/69844695.th.png

is this correct?


am i wrong in my answer 111.80?
i though that i will need to use the R^2 formula that my teacher teach to me.

 

[modulus]

Am i right about this? ( north west is -,+ ), ( north east is +,+), (south west is -,-) and (south east is +,-)..

Oh yeah, you're completely right about that! But, your imageshack image is wrong. That’s why you still seem to be confused about representing the cardinals on the axes. Look at this, I hope it will help:

North
^
l
l
l
l
l
l
l
l
l___________________________> East

And, for adding the vectors, we do it like this:

800N North (coinciding with y-axis)
^
l
l
l
l
l Resultant (in North-east direction)
l /
l /
l /
l /
l /
l /angle=theta
l/_______________________________>600N East (coinciding with x-axis)

By doing it this way, you can add them using the parallelogram rule of addition. This is the same thing as:

800N North (coinciding with y-axis)
^
l
l
l 800N North
l ^
l l
l l
l l
l l
l l
l l
l_________________________________________________________>l600N East (on x-axis)

Here, we have done nothing but, 'shifted' the 800N vector to the right so that its tail coincides with the 600N vector's head. In this way, we can use the triangle rule of vector addition.

 

[tiny-tim]

D'oh!

You mean …​

North
^
l
l
l
l
l
l
l
l
l___________________________> East

And, for adding the vectors, we do it like this:

800N North (coinciding with y-axis)

^
 l
 l
 l 
 l    
 l              Resultant (in North-east direction)        
 l            /
 l          /
 l        /
 l      /
 l    /
 l  /angle=theta
 l/_______________________________>600N East (coinciding with x-axis)

By doing it this way, you can add them using the parallelogram rule of addition. This is the same thing as:

800N North (coinciding with y-axis)

^ 
 l                                                                                              
 l                          
 l                                                                         800N North 
 l                                                                            ^
 l                                                                            l                
 l                                                                            l
 l                                                                            l 
 l                                                                            l
 l                                                                            l
 l                                                                            l
 l___________________________________________________________________________>l
                                                                            600N East (on x-axis)

 

[cary1234]

thanks modulus, thanks tiny-tim. I am more sure of my answer by now.
for a) 800N. due south and 600N. due north. i think i don't need to find the theta.
because it will be like these.

...800N due South.
..^...l
..l...l
..l...l
..l...l
..l...V
600N due North

the answer will be 200N south right?


for b) 300N. due East and 450N. due north.

http://img209.imageshack.us/img209/5395/75863050.png http://g.imageshack.us/img209/75863050.png/1/

the answer for that will be.
R= square root of 292500
R= 540.83N is this correct?


my problem is how to get the theta?

 

[tiny-tim]

Hi cary1234!

(have a theta: θ )

my problem is how to get the theta?

You need to learn these:

cos = adj/hyp

sin = opp/hyp

tan = opp/adj

In this case, use tanθ = opp/adj (opposite/adjacent) ​

 

[cary1234]

so it mean that iam correct?


by the way.. what is the easiest clue for me to know what is the adjacent and opposite.

its easy to know the hypotenuse but i don't have any clue on how to know the adjacent and opposite.

 

[cary1234]

is this correct? 0.03??


in the example of my teacher. she use tan-(opposite/adjacent)

 

[tiny-tim]

so it mean that iam correct?

To be honest, I've got a bit lost as to exactly what the question was.

by the way.. what is the easiest clue for me to know what is the adjacent and opposite.

its easy to know the hypotenuse but i don't have any clue on how to know the adjacent and opposite.

They're "exactly what it says on the tin" …

draw the triangle with θ marked in one corner …

then the side next to θ is "adjacent", and the side furthest is "opposite".

(rather like the sine rule, sina/A = sinb/B, where A is the side opposite angle a, and B is the side opposite angle b)

 

[cary1234]

thanks now i know..

heres the question:


thanks modulus, thanks tiny-tim. I am more sure of my answer by now.
for a) 800N. due south and 600N. due north. i think i don't need to find the theta.
because it will be like these.

...800N due South.
..^...l
..l...l
..l...l
..l...l
..l...V
600N due North

the answer will be 200N south right?


for b) 300N. due East and 450N. due north.

and my answer is in page


and my answer for letter b can be found in my post #22.

please tell me if i am correct.

^^

 

[tiny-tim]

for a) 800N. due south and 600N. due north. i think i don't need to find the theta.
because it will be like these.

...800N due South.
..^...l
..l...l
..l...l
..l...l
..l...V
600N due North

the answer will be 200N south right?

Yes, for "800N. due south and 600N. due north", that's correct (though you don't really need a diagram for that, do you?) …

but what happened to "800N. due south and 600N. due East" …

are you confusing Newtons with north?

for b) 300N. due East and 450N. due north.

and my answer is in page

page … ?

and my answer for letter b can be found in my post #22.

Yes, your diagram is correct.

(and have you worked out what that θ is?)

 

[cary1234]

for letter a. i just subtract the 800N. with 600N. so i the answer will be 200N south. :)


thanks for clarifying my answer. the theta for letter b. will be 0.03 right?
450/300 = 1.5
tan(1.5) = 0.03..

im pretty sure that I am correct.

 

[tiny-tim]

the theta for letter b. will be 0.03 right?
450/300 = 1.5
tan(1.5) = 0.03..

No!

(0.03 what? does it look like 0.03º ??)

1.5 is opp/adj …

1.5 is the tan …

to find the degrees, you need tan^-1(1.5), inother words the angle whose tangent is 1.5.

 

[cary1234]

ow I am sorry.haha!

i forgot to tan-1 it..

the answer = 56.31 deg.

:)

for this question

c(->) = 125m/s 75deg north of east

e(->) = 500 kg m/s (hey, what should i do in that kilogram meter/second? thatas what makes me confuse) 18 deg south of east.

s(->) = 707m 26 deg west of south


do u think they are connected to each other sir?


i create a graph of them individual here it is.
please correct me if I am wrong.

http://img509.imageshack.us/img509/251/78228931.png http://g.imageshack.us/img509/78228931.png/1/

 

[tiny-tim]

Hi cary1234 !

(been out all afternoon)

c(->) = 125m/s 75deg north of east

Yes, except that that angle looks nothing like 75º …

75º should be nearly vertical.

And (this goes for all three answers) you should either put an arrow on your vector, or mark which end the origin is.​

e(->) = 500 kg m/s (hey, what should i do in that kilogram meter/second? thatas what makes me confuse) 18 deg south of east.

(kg m/s is a momentum.

Momentum is a vector, just like velocity.

The drawing is exactly the same.)​

No, you've drawn 72º south of east.

Read it … "18 deg south of east" means you start at east, and go 18º south.

s(->) = 707m 26 deg west of south

Yes, the hypotenuse is in the correct direction, but that's a very strange way of drawing it.

You've drawn the triangle as if the question said "64º south of west".

I agree they're the same thing, but it would be more natural to draw the triangle under the hypotenuse, rather than on top of it, so we can see the 26º being south-and-west-y instead of north-and-east-y, as you've drawn it.

do u think they are connected to each other sir?

i can't see any connection.

and don't call me sir! …

i'm only a little goldfish ​

 

[cary1234]

you mean like this?
the truth is i only know how to get the other angles but i don't know where toplace them.
thanks for the information now i know where to place them.
is this correct?

http://img341.imageshack.us/img341/1923/41728774.png http://g.imageshack.us/img341/41728774.png/1/

And (this goes for all three answers) you should either put an arrow on your vector, or mark which end the origin is.

do you mean like this?

http://img215.imageshack.us/img215/9232/43404868.png http://g.imageshack.us/img215/43404868.png/1/

and don't call me sir! …

I'm only a little goldfish

ow youre not a little goldfish. :)
youre a cute little goldfish.. hehe.. :)
or maybe a shark? :)

 

[tiny-tim]

HI cary1234!

(just got up :zzz: …)

you mean like this?
the truth is i only know how to get the other angles but i don't know where toplace them.

That's why those nice professors give you plenty of practice, so that you get used to it before the exam!

thanks for the information now i know where to place them.
is this correct?

(1. was ok)
2. yes, that's now correct.
3. no, you've swapped the angles (as in 2), but the angles were in the right place … i was just saying that the triangle would look better if the right-angle was at bottom right, instead of top left.

do you mean like this?

Sort-of.

You don't need a big red dot at the origin, provided it's clear where the origin is …

drawing the two coordinate axes (even if they're not marked as such) makes it clear that the origin is where they cross.

And you don't need arrows on the axes …

(in fact, if you do put arrow on them, you should only put the arrow on one end …

an axis only points in one direction, doesn't it? )

What I meant about the arrow was that, if you don't draw coordinate axes (so the origin isn't marked), then you must put an arrow on your answer vector.

 

[cary1234]

ok!

now i know..

and now i understand those basic parts. thanks!

im waiting for more than hours.. haha!

thank you very much.