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Adding Vectors:

  1. Jan 27, 2012 #1
    1. The problem statement, all variables and given/known data

    10.0km [N] + 5.0km [E15S]


    2. Relevant equations

    Cosine Law

    3. The attempt at a solution

    If you draw it out you see that 10km is 90 degrees north. From the 10km end, you do E15S for 5km.

    Therefore, a = 10km, b = 5km, c = ?, C = 90 - 15 = 75.

    Therefore:

    c^2 = 10^2 + 5^2 - 2(10)(5) Cos 75
    c^2 = 99.12
    c = 9.9

    Sin Law to find the angle of the resultant

    Sin A = Sin B
    ----- ------
    a = b

    Sin 75 Sin B
    ------- = -------
    9.9 5

    B = 29

    So final answer is 9.9km [N29E]


    Somehow the answer is 4.2[N35E]...
     
  2. jcsd
  3. Jan 27, 2012 #2

    PeterO

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    Your answer is the correct answer to the problem you presented.

    Either you miss read [and posted] the question, or you have compared your answer to the answer of a different question.

    Even if you went 10km North then 5km due South you would still be 5 km from you starting point. There is no way you finish within 4.2 km of your starting point!!
     
  4. Jan 27, 2012 #3

    Thanks. I also was confused how the displacement would be less than 5km.


    Appreciate it for clarifying!


    How would you do this one then?

    10km [N25E] + 5km [E]?
     
  5. Jan 27, 2012 #4

    SammyS

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    One vector is due north. The other is 15° South of due East. The angle made by the direction of the vectors is 105°.

    cos(105°) = -cos(75°)
     
  6. Jan 27, 2012 #5

    PeterO

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    When you add vectors you connect them head to to tail - that is why the angle is indeed 75° NOT 105°.

    Sure the vectors separately will make an angle of 105°, if they have a common start point, but that is not how you add vectors.
     
  7. Jan 27, 2012 #6

    PeterO

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    Do i the same way you did the previous question. You will be working with an angle bigger than 90° this time.
     
  8. Jan 27, 2012 #7

    PeterO

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    You can use components of each vector if you wish.

    10km [N25E] has a North component of about 9, and an East component about 4.2

    When you add components your final vector will have components about 9 North and 9,2 East.

    The final vector will be about 12.8 [N46E]

    All those figures I approximated - I even guessed 46 for the angle, knowing that of the components had been 9 N and 9 East the angle would have been 45, so I just went a little bit bigger.
    To get accurate components you use trig. and Pythagoras.
     
  9. Jan 29, 2012 #8
    How the hell do you find this angle? see attached pic

    says: determine the angle for which the resultant of the three forces is vertical.
     

    Attached Files:

  10. Jan 29, 2012 #9

    PeterO

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    When you read the question did you realise that the statement in read means: "there is zero horizontal component to the resultant of the three forces?"
     
  11. Jan 29, 2012 #10

    SammyS

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    Hello SUCRALOSE . Welcome to PF !

    You really should have posted this question in a new thread, rather than "hijacking" an existing link.
     
  12. Jan 29, 2012 #11
    was not my intention to take over this thread, I start a new one. and I still do not get it.
     
  13. Jan 29, 2012 #12

    PeterO

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    Can you answer my earlier question about what you saw in the wording please.
     
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