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Represent the following vector

  1. Nov 26, 2009 #1
    i think this is a very simple assignment, but im confused and i dont 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..
     
  2. jcsd
  3. Nov 26, 2009 #2

    tiny-tim

    User Avatar
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    Welcome to PF!

    Hi cary1234! Welcome to PF! :smile:

    What is the question? :confused:

    (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?
     
  4. Nov 27, 2009 #3
    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, im so thankful for your response!
     
  5. Nov 27, 2009 #4
    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
     
  6. Nov 27, 2009 #5
    hi! thanks for helping me.

    is this correct?

    i hope u understand.

    im confused on putting the 75 deg. i dont 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/" [Broken] target="_blank"><img src="http://img134.imageshack.us/img134/3150/60341867.th.png" [Broken] 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" [Broken] title="QuickPost"><img src="http://imageshack.us/img/butansn.png" alt="QuickPost" border="0"></a> Quickpost this image to Myspace, Digg, Facebook, and others!
     
    Last edited by a moderator: May 4, 2017
  7. Nov 27, 2009 #6
    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
     
  8. Nov 27, 2009 #7
    i thought this is like a momentum question, but i dun get it , why there's no mass given -_-
     
  9. Nov 27, 2009 #8

    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?
     
  10. Nov 27, 2009 #9
    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?
     
  11. Nov 27, 2009 #10
    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 eachother.
     
  12. Nov 27, 2009 #11

    tiny-tim

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    Hi cary1234! :smile:

    (just got up :zzz: …)
    (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" :confused:)
     
    Last edited by a moderator: May 4, 2017
  13. Nov 27, 2009 #12
    Hi tiny-tim im sorry if im 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..
     
  14. Nov 27, 2009 #13
    Hi wisvuze and modulus thank you for your help. im now less confuse and afraid of physics.. :)

    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 [Broken]
     
    Last edited by a moderator: May 4, 2017
  15. Nov 27, 2009 #14
    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 ' / '
     
  16. Nov 27, 2009 #15
    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.
     
  17. Nov 27, 2009 #16

    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 [Broken]
     
    Last edited by a moderator: May 4, 2017
  18. Nov 28, 2009 #17

    i think this is the answer to your question.

    http://img145.imageshack.us/img145/866/assxc.th.png [Broken]

    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 dont 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)
     
    Last edited by a moderator: May 4, 2017
  19. Nov 28, 2009 #18

    ideasrule

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    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.
     
  20. Nov 28, 2009 #19

    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 [Broken]

    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.
     
    Last edited by a moderator: May 4, 2017
  21. Nov 29, 2009 #20
    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.
     
    Last edited: Nov 29, 2009
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