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Resolving vectors question

  1. Dec 14, 2007 #1
    1. The problem statement, all variables and given/known data
    Find the Magnitude and the direction of the resultant vector obtained by adding displacements of 5.0m W and 5.0m NW


    2. Relevant equations
    I think for these problems, pythagoras' theorem is needed (c^2=a^2+b^2) as well as the basic trigonometric formulae (sin = O/H etc)


    3. The attempt at a solution
    I've tried methods of representing the vectors as arrows and putting them "head-to-tail" but I'm finding it difficult to represent a vector in the north direction as well as the north-west direction. I'm not entirely sure on the proper way to start attempting this question
     
  2. jcsd
  3. Dec 14, 2007 #2

    Doc Al

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    Staff: Mentor

    Are you able to resolve the vectors into components and then add the components?
     
  4. Dec 14, 2007 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Draw a picture. As on a map, East is to the right, West to the left, North upward,and
    South downward. So your 5.0 m W vector is draw horizontally to the left. "North-West" is half way between N and W so it is drawn (from the tip of the W vector) 45 degrees upward and again 5.0 m long. Their sum is the vector from the base of the W vector to the tip of the NW vector. Look at that and you should see a triangle- but not a right triangle. Yes, you could resolve the two vectors into " xy-components" (i.e. East and North values)- for the W vector that is trivial. For the NW vector only slightly harder. Then add the component. However, the problem specifically asks for so then you would have to convert back to that form.

    Here, I think it is simplest to use the "cosine law". You know two sides of the triangle and the angle between them. If you call those sides a and b, and the angle between them [itex]theta[/itex], then the opposite side, the "magnitude" of the vector, c, is given by [itex]c^2= a^2+ b^2- 2ab cos(\theta)[/itex]. You could use the sine law to find the angle but here that is much easier. Since this is obviously an isosceles triangle (the two given sides are both 5.0 m) the two unknown angles must be the same; and the sum of those two and the 135 degree angle must be 180 degrees.

    Becareful how you write the direction. If you use [itex]\theta[/itex] itself, notice that the angle is from the horizontal (West) line up to the new vector- It must be given as that angle "North of West". Or you could use its complement- the angle from the new vector to the vertical (North) line, [itex]90- \theta[/itex] and call it "West of North".
     
  5. Dec 14, 2007 #4
    Thank you for your help...I found the magnitude using the cosine rule (which I have previously never learnt before)

    I'm still having trouble with finding the angle though. if i find the angle by doing (180-135=45 then 45/2 = 22.5 degrees), that's 2 degrees off the answer (20.5 degrees)

    I'm also unsure of the direction. The answer is apparently S 20.(5)degrees E

    I don't know how hey get South and East out of a question asking for North and North West

    By the way, what is the rule??? Like the cosine rule, I haven't learnt about it yet.
     
    Last edited: Dec 14, 2007
  6. Dec 14, 2007 #5
    Sorry about the last post

    The answer for the question is actually 9.2m N 68 degrees W...

    I'm not sure how they got 68 degrees if 135 degrees is already taken up in one angle

    Any ideas?
     
  7. Dec 15, 2007 #6
    in my opinion it would be easier to do the component method for this problem (add x components together, and then y components) but you can do it with the head-to-tail method as well.

    For the magnitude, you do the law of cosines, as you said (c^2 = 5^2 + 5^2 -50cos135)
    You get 9.2 m.

    The angle you found is correct, and if the answer in the book is 68 degrees W of N, it is the same as 22 degrees N of W (which is what you found).
     
  8. Dec 15, 2007 #7
    Ohh yes...Thank you everyone...I've solved the problem now:smile:
     
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