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Addition of Vectors

  1. Sep 7, 2004 #1
    I think I have a trick question here. The section we are covering is addition of vectors by means of components. Here is my problem:

    You are on a treasure hunt and your map says, "Walk due east for 52 paces, then walk 26.6 degrees north of west for 42 paces, and finally walk due north for 25 paces."

    a) What is the magnitude of the component of your displacement in the direction due north?

    b) What is the magnitude of the component of your displacement in the direction of due east?

    I believe I figured out part (b) correctly getting an answer of 37.6 paces. But isn't the answer for part (a) already given? Wouldn't the answer to part (a) just be 25 paces??

    My drawing is attached.

    Attached Files:

  2. jcsd
  3. Sep 7, 2004 #2

    Doc Al

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    How did you get this answer? Did you add up all the east-west components?
    No. They want you to find the components of your final displacement from your starting point. You must take each leg of the journey, find the components (east-west & north-south), then add them to get the components of the final displacement.

    Start by writing out the components for each leg separately.
  4. Sep 7, 2004 #3
    The answer for part (b) is just the, well what would be the x component for that section. Is that not correct?

    Ok for your second part I'm confused. Refer to my drawing to correct me if I'm wrong. Are any of my drawings correct?

    So to develop further for B my x-component would be 52 paces and my y-component would look something like, (42 paces)sin26.6 which gives me 18.8 paces. Is that correct?

    Then I'm assuming I would need to find A then and get the results for the x and y components. Once those are found I would add the Ax and Bx components together? But my problem then would be how to find the A and Ax magnitudes? Because I'm thinking that my Ay component is 25 paces. So How do I go about finding Ax?

    I'm sorry this is all just soooo confusing for me. Am I on the right path here?

    Please take a look at my drawings, this is also a hard part for me. I'm not entirely sure how to draw them. Can u please tell me which one would be correct?


    Attached Files:

  5. Sep 7, 2004 #4

    Doc Al

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    I believe they want the total x component--not just the component for one section.
    Your first diagram shows the sections properly, but the labels make no sense to me. Be sure to mark the starting point and the end point. The arrow connecting those two points is the displacement.

    Your second diagram is not to scale (your 42 pace section seems longer than your 52 pace section). But at least it has the journey start at the origin.

    I have no idea what you are calling A and B. So let me start over and see if I can help you out. Let's start by labeling each section of the journey. I'll call them A, B, and C:

    A: 52 paces due east
    B: 42 paces 26.6 degrees north of west
    C: 25 paces due north

    Now I want you to find the x and y components of each section. I'll do the first one:
    [itex]A_x = + 52[/itex], [itex]A_y = 0[/itex]

    You do the next two. Then add up all the x-components and y-components: they will be the x and y components of your net displacement from the starting point. This is what I think they are looking for.
  6. Sep 7, 2004 #5
    Ok, so if I've done this correctly then.

    As you put it Ax = 52 paces Ay = 0

    Then this is what I found for the next two.

    Bx = 0 By = 18.8 paces

    Cx = 0 Cy = 25 paces

    Is what I've found correct? If so then I would just add all the x components and then y components? The x components would give me the magnitude of the component of my displacment in the direction of due east and the y components would give me the magnitude of my displacement in the direction of due north?
  7. Sep 7, 2004 #6

    Doc Al

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    By is correct, but not Bx. How did you figure out By?


  8. Sep 7, 2004 #7
    Ok, I see then. I know what Bx should be then. Bx would be 37.6 paces.

    To find By I used (42paces)sin26.6 to get 18.8 paces.

    To find Bx I used (42paces)cos26.6 to get 37.6 paces.

    I was confused because I assumed the 52 paces equaled the whole x component. But after looking at my diagram I realize that the 52 paces only accounts for one part of it.

    Thank you so much for your help. I'm sure I'll be back in a bit with another question for another problem. LOL
  9. Sep 7, 2004 #8

    Doc Al

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    Good. But since B is 26.6 degrees north of west, the x component will be negative. So Bx = -37.6 paces.
  10. Sep 7, 2004 #9

    Ok, right, I overlooked that part. But when I add the x components the negative sign would not come in to play right? Since it's just showing direction?

    Because after adding, my total x component was 89.6 paces. Which makes sense. But if I added using the negative sign my total x component would be 14.4 paces. That couldn't be right. The negative sign is only there to show direction, right?

    Btw, I really appreciate this help. I was wondering if it would be possible to look at this other thread I started. I have most of the problem figured out except for a small bit of it. And I still cannot seem to get it.

    Here is the url:
    My other thread

    I would really appreciate it. It's just a small bit. Or at least I believe it to only be something small I am missing.
    Last edited: Sep 7, 2004
  11. Sep 7, 2004 #10

    Doc Al

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    The sign does show direction, but direction is vital to understanding vectors. You must include the sign in adding components!

    Displacement is a vector. It shows where you ended up, compared to where you started. So direction matters! Here's a silly example: you walk 5 steps east, then 5 steps west. What's your displacement? Zero! You are right back where you started.

    They aren't asking how many total paces you made in your journey. That would be easy: just add them up = 52 + 42 + 25 = 119 paces. But that's got nothing to with your displacement.
  12. Sep 7, 2004 #11
    Ah, I see that does make sense then. The question does ask for the magnitude of the displacement.

    So even though the walker starts out with 52 paces he comes back 37.6 paces. Thus the 37.6 paces would need to be subtracted from the total of 52 paces, leaving the total displacement of 14.4 paces. Ok, I undertand that.

    This whole physics thing is a whole lot easier to understand when someone explains it properly.
  13. Sep 7, 2004 #12
    I have another question, but this one is relating to a different problem.

    My diagram is attached below. It's another vector addition problem.

    I have to find the resultant plus theta.

    So this is what I have currently to find the components.

    Ax = 0 (but I think I have to find this?) Ay = 5m

    Bx = 16.0m By = 0

    Cx = 0 Cy = 16.4m

    I know from here I would add all x's and y's and then use the forumula

    R = (the square root of) x^2 + y^2

    Then find theta accordingly. However, my problem lies in finding theta. I believe that I should have an amount for Ax. Is that correct?

    Attached Files:

  14. Sep 7, 2004 #13

    Doc Al

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    Both right.

    Both are wrong!
  15. Sep 7, 2004 #14
    Hmm, ok let me go back to the drawing board on this one.
  16. Sep 7, 2004 #15
    Alright how about this?

    Cx = - 11.5m Cy = - 14m
  17. Sep 7, 2004 #16
    Ok never mind on finding the resultant. I've got that. But now I'm facing a different problem. I am using tan to find theta by using:

    theta = tan^-1 (27.5/14) but when I enter the answer I get from this it says it's wrong. (By it, I'm referring to webassign). Anyway my problem it would seem is that either I'm not using the right function, i.e tan, or I'm not using the correct distance for y.

    You can look at the above diagram in post # 12 for referrence.
  18. Sep 8, 2004 #17

    Doc Al

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    Here's what I get:
    Cx = 20 cos(35) = 16.4
    Cy = -20 sin(35) = -11.5
  19. Sep 8, 2004 #18

    Doc Al

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    Your distances (components of the resultant vector) for both x and y are wrong.

    There's nothing wrong with your method for finding the angle, but since you have calculated the components wrong you will of course get a wrong answer. Add up the components correctly, then redo the calculation to find the angle.
  20. Sep 8, 2004 #19
    That's interesting considering when I submited my resultant answer for my x and y components, it was correct. The only difference was that my resultant answer was 33.4 and when I tried it using the Cx and Cy components you gave the resultant was 33.04.

    In any case, I found theta to be 11.3.
    Last edited: Sep 8, 2004
  21. Sep 8, 2004 #20

    Doc Al

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    What did you get for your x and y components of the resultant?
    Sounds right.
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