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Vector Addition scale diagram

  1. Jul 6, 2008 #1
    1. The problem statement, all variables and given/known data

    Determine the vector that must be added to the sum of A + B in the given attachment to give a resultant displacement of (a) O and (b) 4.0 km [W]. In case any one can't view the attachment, |A| is 5.1 km in length, 38 degrees N of E, and |B| is 6.8 km in length, 19 degrees E of S, both attached tail-to-tail.

    2. Relevant equations

    ∆d = d1 + d2 + ...
    c^2 = a^2 + b^2 -2abcosC

    3. The attempt at a solution

    So, in order to get a displacement of zero, your resultant has to be at the initial point, so I simply drew a vector from the end of |A| to the end of |B| and obtained the answer 7.0 km, which is the answer in the book. Where I'm a bit lost is where the angle's coming from? The answer in the book is 28 degrees N of W - I don't know where I'm supposed to indicate this angle from. I'm also a lost on part (b), where the resultant is 4.0 km [W]... I don't even know where to begin on that one. I've drawn my diagram - the 4.0 km is branching from the end of |B| in the direction of West, and I know the vector I'm looking for is from the end of |A| to the end of the 4.0 km one, I just don't know how to go about calculating it, other than drawing a scale diagram.

    Any insight is sincerely appreciated !

    Attached Files:

  2. jcsd
  3. Jul 6, 2008 #2


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    Hi pitaaa,

    When you draw the diagram to add the vectors, you need to draw the vectors head to tail. So redraw, with the tail of B at the head of A. That represents the motion going first along A, then along B. Now if the displacement is zero, that means you get back to your starting point. So the displacement you are looking for is from the head of B back to the tail of A. What do you get for the angle?
  4. Jul 6, 2008 #3
    Sweet! I get the 28 degrees, except its S of E, not N of W.

    And that's where my problem is, I was adding the vector straight from the diagram and not the head-to-tail. I'll give (b) another shot.
  5. Jul 6, 2008 #4


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    It sounds like you might be drawing the final vector in the wrong direction. Did you draw the final vector from the head of B back to the origin? If you draw the axes with north up the page and east to the right, vector A is to the right and upwards, and B is to the right and downwards. If the final vector is to take you back to the origin, it must have a leftwards (west) component.
  6. Jul 6, 2008 #5
    Hm, okay. When I do the head-to-tail addition exactly as you described, I no longer get 7.0 km as my answer. I understand where the 28 degrees N of W is coming from, but I still think I may be drawing my diagram wrong. :S
  7. Jul 6, 2008 #6


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    Can you post the mathematical steps you took when you got something besides 7km?
  8. Jul 6, 2008 #7
    Well I used the cosine law, so:

    c[tex]^{2}[/tex] = a[tex]^{2}[/tex] + b[tex]^{2}[/tex] - 2abcosC
    c[tex]^{2}[/tex] = 5.1[tex]^{2}[/tex] + 6.8[tex]^{2}[/tex] - 2(5.1)(6.8)cos123
    c= 10.48

    I attached the diagram I made... I'm guessing something's wrong with that, I just don't know how else to draw it.
    Last edited: Jul 6, 2008
  9. Jul 6, 2008 #8


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    I can't see the diagram at this point. (Is there any way you can post it at an image hosting site?)

    What I would say is try this: Go back to the original diagram from your book. Now those vectors A and B (both at the origin) can be two sides of a parallelogram. So draw the other two sides of the parallelogram.

    When you calculate this problem, use the original A vector; and the side of the parallelogram opposite the original B is your new B vector. The angle you want to find is the upper right corner of the parallelogram. What is it? (It should be less than ninety degrees.)
  10. Jul 6, 2008 #9
    Okaaay, that definitely explains it. It's 71 degrees :) And now I get 7.0 km. Thanks so much! Now I must get cracking on Part B.
  11. Jul 6, 2008 #10
    Woot! I managed to solve it using sine/cosine laws.
    The answer was 11 km [18 degrees N of W]
    Thanks for keeping me on the right track, alphysicist :)
  12. Jul 6, 2008 #11


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    Sure, glad to help!
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