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Adding Vectors Geometrically

  1. Jan 19, 2013 #1
    1. The problem statement, all variables and given/known data
    I am a new physics student am having trouble with a sample problem from the book.
    I have the answer (below) but I don't not understand how they are finding that number.
    I have been trying to make sense of it using the pythagorean theorem and Law of Cosines but I just cannot break it. I have copied and pasted the problem below and included the graphic from the book.
    Please help.

    In an orienteering class, you have the goal of moving as far (straight-line distance) from base camp as possible by making three straight-line moves. You may use the following displacements in any order: (a) , 2.0 km due east (directly toward the east); (b), 2.0 km 30° north of east (at an angle of 30° toward the north from due east); (c), 1.0 km due west. Alternatively, you may substitute either -b for b or -c for c. What is the greatest distance you can be from base camp at the end of the third displacement?

    Reasoning:

    Using a convenient scale, we draw vectors a, b, c, -b, and -c as in Fig. 3-7a. We then mentally slide the vectors over the page, connecting three of them at a time in head-to-tail arrangements to find their vector sum d. The tail of the first vector represents base camp. The head of the third vector represents the point at which you stop. The vector sum d extends from the tail of the first vector to the head of the third vector. Its magnitude d is your distance from base camp.


    2. Relevant equations
    Also, I don't understand what happens to c for it to become -c.
    Does that mean instead of 1, it is -1?
    I have been fumbling around with the Law of Cosines because i think it is related to finding the answer to this problem, but I'm not clear on how to use it.
    c2=a2+b2-2abcosC
    b2=a2+c2-2accosC
    a2=b2+c2-2bccosC

    3. The attempt at a solution
    D=4.8, but I have not been successful at reverse engineering.
     

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    Last edited: Jan 19, 2013
  2. jcsd
  3. Jan 19, 2013 #2

    haruspex

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    It's pretty easy to see what combinations of signs produce the max distance, and you appear to have solved that. So I assume the remaining problem is to work out what that distance is.
    Can you work out what the northerly and easterly displacements are from vector b?
     
  4. Jan 19, 2013 #3
    no, i don't.
    i don't know how to get to that point.
    it's not like regular algebraic addition.
     
  5. Jan 19, 2013 #4

    haruspex

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    If you travel at 30 degrees north of east for 2km, how far north have you gone? It's simple trigonometry.
     
  6. Jan 19, 2013 #5
    I seriously do not know how to setup what you are asking me to answer that question.
     
  7. Jan 19, 2013 #6

    Mark44

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    Your book should have some examples where they decompose a vector into components, with one component parallel to the x-axis, and another component parallel to the y-axis. If the vector makes an angle of 30° to the pos. x-axis and has a length of |r|, the x component is |r|cos 30°, and the y component is |r| sin 30°. This is right triangle trigonometry. Using the Law of Cosines isn't very helpful here, and the Pythagorean theorem (a special case of the Law of Cosines) isn't any use either.
     
  8. Jan 19, 2013 #7
    Oh, i see. There is a section on that, but it is later in the book. The section I'm in is doing it geometrically. I was thinking that I should know how to do this example w/o reading ahead. I was thinking this meant the book was assuming I know some concept or technique that I don't and therefore I was asking here what I'm missing.
     
  9. Jan 19, 2013 #8

    haruspex

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    That'll work here, by recognising that a 30 degree right angled triangle is half an equilateral triangle. That's enough to figure out how far north b goes, then you can use Pythagoras for the easting.
     
  10. Jan 19, 2013 #9
    try breaking each vector into X and Y vectors or or north/south/east/west and figure out the sum of X, and sum of Y. that will tell you how far away from the origin you actually are, and if you go the exact opposite of that you will be back at your origin.

    like if you go 45 degrees northeast, that is the resultant of the two component vectors (which can be broken down as the X and Y directions and the resultant vector is 45 degrees north. just by knowing that its 45 degrees is helpful because that means the sin45=Y/hyp and cos45=X/hyp (as you can see you solve for X and Y by knowing a) the angle, and b) what HYP is (the hypoentuse aka the resultant vector)

    for example, a classic 3,4,5 triangle. if the hypotenuse is 5cm (in this example we will call it 45 degrees northeast, aka you went 5 cm to the northeast. that is the same as you going 3 cm east and 4 cm north)
     
    Last edited: Jan 19, 2013
  11. Jan 20, 2013 #10
    but what happens to c. It goes west 1km, so i switch it to go east 1km which makes it -c.
    When i connect -c to the head of a (2km east) does that length now become 1, because 2 + -1 = 1?
     
  12. Jan 20, 2013 #11
    anyone?
     
  13. Jan 20, 2013 #12

    SammyS

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    Patience ... patience .

    c is 1 km west, so -c is 1 km east.

    Therefore, a + -c is (2 km east) + (1 km east) =  ? 
     
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