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Finding the resultant of three vectors

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  1. Jun 4, 2017 #1
    1. The problem statement, all variables and given/known data: Determine the resultant of three vectors of magnitude 1, 2 and 3 whose directions are those of sides of an equilateral triangle taken in order.


    2. Relevant equations:
    R2 = A2 + B2 + 2 A B cos θ
    tan α = B sin θ/(A+B cos θ)

    3. The attempt at a solution:
    https://drive.google.com/open?id=0B3IenBvKWb4PUG9kTUctcFJyLUk
    upload_2017-6-4_15-5-54.png
    The correct answer is: 1.732 units at an angle of 210 ° to the first vector. I have calculated the magnitude correctly but cannot match the direction.
     
    Last edited by a moderator: Jun 4, 2017
  2. jcsd
  3. Jun 4, 2017 #2

    jedishrfu

    Staff: Mentor

    The three vectors are:

    R1 = i + sqrt(3) j
    R2 = 2 i - 2 sqrt(3) j
    R3 = -3 i + 0 j

    The resultant vector is?

    ...

    And having that you can get the magnitude and the direction.
     
  4. Jun 4, 2017 #3
    @jedishrfu, How did you get R1, R2 and R3 in terms of i and j and what is R3?
     
    Last edited: Jun 4, 2017
  5. Jun 4, 2017 #4
    Draw a picture of an equatorial triangle. Add arrow heads to the triangle line segments so they all point clockwise or counter clockwise. These are your vectors. Add them together. What is the resultant?
     
  6. Jun 4, 2017 #5
    Basic trig.
     
  7. Jun 4, 2017 #6
    @Spinnor, I mean R1, R2 and R3 in terms in terms of i and j.
     
  8. Jun 4, 2017 #7
    Basic trigonometry.

    CNX_Precalc_Figure_08_08_020.jpg

    i and j are the x and y components of the three vectors found with trig.

    Read what I wrote, you don't need to do the work. ( possible snark removed)?
     
  9. Jun 4, 2017 #8
    I did not read carefully, sorry. I assumed the vectors were of equal magnitude. The above is incorrect.
     
  10. Jun 4, 2017 #9

    Merlin3189

    User Avatar
    Gold Member

    I don't see how R3 gets to be -3i whether by trig or any other method. I get -6i (by common sense rather than maths - so it could be wrong!)
    I can't check the trig, because I don't know what diagram you are using. My diagram (attached) doesn't give -3i, but then I've started by drawing it as -6i before I do the trig!

    Looking at your results,
    R1= i + sqrt(3)j so Magnitude[R1] = sqrt(1+3) = 2
    R2= 2i -2sqrt(3)j so Magnitude[R2] = sqrt(4+12) = 4
    R3= -3i +0j so Magnitude[R1] = sqrt(9 +0) = 3
    Seems to violate the condition that F1:F2:F3 = 1:2:3 though your directions seem OK.
     

    Attached Files:

  11. Jun 4, 2017 #10
    At the risk of being wrong again, R3 could be correct because it has magnitude 3? We are not told which vector has magnitude 3, it could be any of six directions?
     
  12. Jun 4, 2017 #11
    R2 and R1 need to be divided by the square root of 2?
     
  13. Jun 4, 2017 #12

    jedishrfu

    Staff: Mentor

    The problem says the directions come from the sides of an equilateral triangle so i wrote vectors following the perimeter and adhering to the lengths stated.

    I don't think R3 could go in the positive direction.
     
  14. Jun 4, 2017 #13

    jedishrfu

    Staff: Mentor

    Yes R1 and R2 should be divided by 2 to make all three have 1, 2 and 3 magnitudes respectively. That's my mistake as I was thinking they were unit vectors initially which is wrong.
     
  15. Jun 4, 2017 #14
    How do i express F1, F2 and F3 in terms of i and j?
     
  16. Jun 4, 2017 #15

    jedishrfu

    Staff: Mentor

    The R vectors are your F vectors.
     
  17. Jun 4, 2017 #16
    Then how do i express R vector in terms of i and j. BTW what is wrong with my approach?
     
  18. Jun 4, 2017 #17

    jedishrfu

    Staff: Mentor

    I didn't say your approach was wrong. I provided how I might do the problem without actually doing it in the hopes that it would help you find your mistake.
     
  19. Jun 4, 2017 #18
    Looks like there is more math involved with the way you did it, more places to go wrong?
     
  20. Jun 4, 2017 #19

    jedishrfu

    Staff: Mentor

    Okay I see now. You used F1, F2 and F3 whereas I used R1, R2, and R3 and the resultant is R = F1 + F2 +F3 or in my case R = R1 + R2 + R3.
     
  21. Jun 4, 2017 #20

    Merlin3189

    User Avatar
    Gold Member

    But if R3 has magnitude =3, then it must be R1:R2:R3 = 1:2:3 There is no permutation of 2, 4 and 3 which give a ratio of 1:2 :3 as required by the question. I simply took the offered R1 and R2 as being correct, then R3 follows from that to be -6i

    Now the 6 possible directions, if one of them is -3i, must be +i, -i, 0.5 i +0.5sqrt(3) j, -0.5 i -0.5sqrt(3) j, 0.5 i -0.5sqrt(3) j and -0.5 i +0.5sqrt(3)j

    Possible sequences of direction would be +i , -0.5 i +0.5sqrt(3) j , -0.5 i -0.5sqrt(3) j , +i , etc. or -i, +0.5 i + 0.5sqrt(3) j, +0.5 i -0.5sqrt(3) j , -i , etc.

    Wherever you start in a sequence, you then simply take each unit direction vector in turn ad multiply by the magnitudes 1,2 or 3.
    So if you want to finish with R3 in the -i direction, you need to choose the second sequence and work back from the -i term:
    R3 = -i x3 = -3 i magn= 3
    R2= +0.5 i -0.5sqrt(3) j x2 = +i - sqrt(3) j magn= 2
    R1= +0.5 i + 0.5sqrt(3) j x1 = +0.5 i + 0.5sqrt(3) j magn= 1

    In that case, your R3 would be correct, but both R1 and R2 wrong.
    Now, I think, whoever chose i and j made them half the magnitude of the unit force vector - for convenience, to avoid the fractions. So if you double all the above answers, R1 and R2 are correct and R3 becomes -6i. As far as I can see, you just take your pick.

    Then the resultant is either -1.5 i - 0.5sqrt(3) j when R1 is +0.5 i + 0.5sqrt(3) j or -3i -sqrt(3) j when R1 is i + sqrt(3) j

    The magnitude of the resultant is sqrt(3) x magnitude of R1 either way.
    I'm not sure how you calculate the angle between these without sketching a diagram and using trig or by calculating the angle of each wrt the unit vector i, but from the original diagram it is obvious that it must be 30 deg from the reverse of R1. How they can call this 210 deg when it could just as easily be 150 deg, since, as you say, no requirement is made as to following the sides of the triangle cw or acw, I can't see. I generally choose acw for angles with vectors, as this was the convention when I did electronics, but compass bearings use cw and I expect there are many groups of people using either convention.
     
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