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Vectors (x/y-components)

  1. Sep 13, 2013 #1
    I just want to clarify. I have a problem that says:
    Draw each of the following vectors, then find its x and y components.

    d=(100m, 45° below the +x-axis.

    So this would be a line drawn in the fourth quadrant, that bisects the quadrant in half. But does it mean the bisecting line is 100m? I first assumed that it was the y-axis, but now I'm not so sure, and am thinking it's more the bisecting line is 100m instead of the y-axis.
     
  2. jcsd
  3. Sep 13, 2013 #2

    SteamKing

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    I would take the magnitude of the vector to be 100m, and its direction to be -45 degrees.
     
  4. Sep 13, 2013 #3
    I also have another one part c) "a=(5m/s/s, -y-direction)". I'm not sure if I'm on the right track with this one as there is no degree given, but I'm figuring that in this example acceleration is constant at 5/m/s/s (otherwise how would you figure it out, right?). So I have an angle of 90 degrees that the line of acceleration makes with the y-axis. What would be the best way to solve this?
     
  5. Sep 13, 2013 #4

    SteamKing

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    The -y direction indicates that the acceleration is along the -y axis, not 90 degrees to the y-axis.
     
  6. Sep 13, 2013 #5
    But wouldn't the acceleration graph have a line going through y=-5? I mean if the acceleration ran the length of y axis wouldn't that mean acceleration is zero? Or maybe I'm not understanding what your saying...
     
  7. Sep 13, 2013 #6

    SteamKing

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    Are you trying to make a plot of acceleration or are you trying to express the magnitude of the acceleration and indicate its direction? These are two different things.

    P.S.: It would help if you would follow the HW template and post your complete problem statement. Answering questions about snippets of a HW problem is counterproductive and is frankly not fair to others on these forums. You have the advantage of having the complete problem, while we can only see what portions you give us.
     
  8. Sep 13, 2013 #7
    b) v=(300m/s, 20° above +x-axis.)
    c) a=(5m/s/s, -y-direction)

    That is the whole question. There is no other information that what I've already given. I skipped part b, as I know how to do it, regardless of whether the value is bound to the x-axis or to the line the dissects the first quadrant.

    I gave the question, and part a in the original post as that was originally the part I was concerned about and hadn't read the question all the way through. I apologize if this caused an inconvenience.

    Perhaps the answer to my question about part c is that it is a line starting at the origin and having a negative slope?
     
  9. Sep 13, 2013 #8

    SteamKing

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    No, according to the problem, these are vectors you are drawing. Each vector consists of a magnitude and a direction.
     
  10. Sep 13, 2013 #9
    So it would just be 5m/s/s below the x-axis to signify a downward direction (-y-direction)? I'm trying to grasp this concept, it was discussed in class by using the Cartesian coordinate system.
    Perhaps I'm not visualizing these correctly. So for instance, part a would be a line lowered below the x-axis 20deg and to the right side of the y-axis, and you can find the x-component by solving for the (h)ypotenuse; IE: 100/cos(45)=h, which is approximately 141.42, then you could find the x-component by solving 141.42cos45=x-component which is 100.
    Is this the proper format for solving the first two problems? I've gotten answers already for part a and part b, but part c is confusing. I mean if acceleration is 5m/s/s straight down then the x-component would be zero. What would be the best way to visualize part c?
     
  11. Sep 13, 2013 #10

    SteamKing

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    No, a vector specified as d = (100m, 45 deg. below + x-axis) means that the length of the vector is 100m, not 100/cos(45). The components of this vector are x = 100 cos (-45) and y = sin (-45). Remember, the components of the vector must add up to the original vector.
     
  12. Sep 14, 2013 #11
    Can you explain why the angle would be -45 degrees?

    Some things in trig I'm really good at, other things, not so much, like understanding why my degree should've been -45. Is it simply because it is below the x-axis? What if it here 45 degrees in the -y-direction? Is that possible? Would the angle just be bounded between the line created and the y-axis and still be negative, which would basically give the same results?

    Also for this,
    d = (100m, 45 deg. below + x-axis)
    it's not the y-component or x-component that is 100m but the line created by the 45 degree angle below the +x-axis, correct?
     
    Last edited: Sep 14, 2013
  13. Sep 14, 2013 #12

    SteamKing

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    Angles are measured positive counterclockwise. An angle of zero degrees coincides with the positive x-axis. The positive y-axis is 90 degrees and the negative x-axis is +180 degrees or -180 degrees, your choice. The negative y-axis is -90 degrees. This convention should have been covered in elementary trigonometry.

    Vectors are specified either as pairs of component values or as a magnitude and a direction angle. Thus, vector A = (2,3) would have x-component = 2 and y-component = 3.
    Vector B = (100, ∠ -45°) would be a vector of magnitude 100 making an angle measured 45 degrees clockwise from the positive x-axis.
     
  14. Sep 14, 2013 #13
    Ok, that makes sense. There were parts of trig that were totally lost on me, and parts of trig that I really got. I struggled through the class, but managed to pass.

    Part C is still giving me issues though. I'm not understanding how to visualize it.

    Draw each of the following vectors, then find its x and y components.
    a) d=(100m, 45° below +x-axis.)
    b) v=(300m/s, 20° above +x-axis.)
    c) a=(5m/s/s, -y-direction)

    So it's 5m/s/s in the negative y-direction (the line coincides with 270deg). Does this mean that is coincides with the y-axis and therefore its x-component would be 0, and it's y-component would be 5? I'm not understanding how this line is formed as there is no degree value given, I'm assuming the above.
     
  15. Sep 14, 2013 #14

    SteamKing

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    The term '-y-direction' is one way of stating that the vector is pointing straight down, angle 270 deg CCW (or 90 deg. CW, if you prefer). Your text is trying to give you hints by using this somewhat unconventional method of indicating direction.
     
  16. Sep 14, 2013 #15
    But not necessarily running along the y-axis? I mean for instance it's x value wouldn't necessarily be 0?

    Or do I just solve 5sin(-90) and 5cos(-90)?
     
    Last edited: Sep 14, 2013
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