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Trig and Vector Components

  1. Sep 3, 2004 #1
    Hi -

    The problem is below and picture of diagram is attached. Could you tell me what is incorrect about my train of thought? I am suppose to express x and y.

    Find the components of the vector with length
    C and angle phi as shown, named C vector. Express your answer in terms of C and phi . Write the components in the form (x,y) .


    I expressed (x,y) as ... c*cos(phi),c*csc(phi) ... which is incorrect

    Thanks for your time.
     
  2. jcsd
  3. Sep 3, 2004 #2

    robphy

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    The component of any vector [itex]\vec V[/itex] along a given axis is
    [tex]V_{axis}=|V|\cos \theta_{\rm between\ \vec V\ and\ axis} [/tex]

    In your picture, what is [itex]\theta_{\rm between\ \vec V\ and\ x-axis} [/itex] and [itex]\theta_{\rm between\ \vec V\ and\ y-axis} [/itex]?
    With the trig identity for [itex]\cos(\phi+\alpha)[/itex], you can express all of your results in terms of [itex]\phi[/itex].
     
    Last edited: Sep 3, 2004
  4. Sep 3, 2004 #3
    I'm just wondering, can't you express the answer as
    c*sin(phi),c*cos(phi) ?

    correct me if im wrong...
     
    Last edited: Sep 3, 2004
  5. Sep 3, 2004 #4

    robphy

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    Almost... note that c_x < 0.
     
  6. Sep 3, 2004 #5
    thanks alot, i was just confused because iamhumble made it seem a lot harder...
     
  7. Sep 4, 2004 #6
    Vectors

    Hello Everyone,
    In school I had a test of Vector exercises. The problem is that I have to submit them online and I need the exact value for the questions. Anyway I don't want you to solve them for me. I just need some help where I am stuck.

    The first exercise: A boy runs 3.3 m North, 7.1 m NorthEast and 15.6 m West. a)Determine the length of the displacement vector tha goes from the starting point to his final position. b)Determine the direction of the displacement vector.

    I was confused by this exercise because there isn't any angle given. So I assumed that "NorthEast" means that he moves 7.1 m with an angle 45 Deg.
    Then part a) is easy. For part b) I found the angle formed by the North line and the Resultant vector. Am I right here?

    The next exercise is a little fuzzy I think: Let aA + bB + C=0, where
    A=(17,-99), B=(-99,28) and C=(4.3,78), where Ax=17 units, Ay=-99 units etc. What is the value of a and b? (All the Upper case letters A,B and C have the vector arrow above).
    What I did is: I opened aA + bB + C=0 using unit vectors i, j and k (k=0) and the numbers given above.
    Still I have a and b and it's an equation with 2 unknowns.
    The final expression I concluded is: (a*17i-b*99i+4.3i)+(b*28j-a*99j+78j)=0...........???

    Thank You for your time.
     
  8. Sep 4, 2004 #7

    Tide

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    Electro,

    Your final equation is really TWO equations since the unit vectors are independent of each other. You should be able to handle it from there.
     
  9. Sep 4, 2004 #8
    :smile: Thank You Tide,
    What I think now, is forming two simultaneous equations.
    1) 17a-99b= -4.3
    2) 28b-99a= -78
    Solving them, I think is the answer for a and b.
    Since the i units can't be added to the j units I formed two equations just neglecting i and j. So for i group it will give 0 and for j group too.

    Please I need an answer on the previous exercise I posted if I have concepted it right or not.

    Thank you
    Electro
     
  10. Sep 4, 2004 #9
    Thanks all for the much needed assistance. I understand what I did wrong.
     
  11. Sep 4, 2004 #10
    :uhh: Anyone can give the answer to my questions?
     
  12. Sep 4, 2004 #11
    i think you should start a new post of your own...
     
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