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Direction of cartesian vector

  1. Oct 12, 2015 #1
    1. The problem statement, all variables and given/known data
    if the alpha is larger than 90 degree ( which means the resultant F is lean towards -x axis , then the angle between the Ax and the line from Ax to A will be less than 90 , am i right ?

    2. Relevant equations


    3. The attempt at a solution
     

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  3. Oct 12, 2015 #2

    BvU

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    No. It will be exactly 90##^\circ##

    (the 90##^\circ## as indicated in figure 2-26 (a) )
     
  4. Oct 13, 2015 #3
    why , i cant visualize it??
     
  5. Oct 13, 2015 #4
    If the force is lean towards negative x axis , it means the angle between the force and positive x -axis is more than 90 degree, right ? then how is it possible for it to remain 90 degree?
     
  6. Oct 13, 2015 #5

    BvU

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    You were talking about the line from Ax to A, not about the line from the origin to A (i.e. ##\vec A## itself).
     
  7. Oct 13, 2015 #6
    ya , i mean the line from Ax to A, not about the line from the origin to A (i.e. A⃗ itself).
    if the force is lean towards the negative x-axis then the , we would have three line , right , namely the line joining origin to Ax , A(which is lean towards negative x axis ) , and a line which join A to negative x-axis , right ? if it is so , then the i an undersatnd why the angle is 90 degree. if it so , the triangle formed is at opposite of the current one .
     
  8. Oct 14, 2015 #7

    BvU

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    I think you've got it all right and correct, just that the wording in post #1 was rather unfortunate.
    With nowadays visualization possibilities it's almost a stone-age approach to put things in words, but let's do it anyway:

    let's look in the blue plane of figure 2-26 (a) for the case ##\alpha \in [\pi/2, \pi]## where all the angles in our story are
    [edit] I mean to say that all the angles are in that plane. Sorry for the ambiguity.​

    upload_2015-10-14_11-42-27.png

    so the vertical line Ayz is in the yz plane and that plane is perpendicular to the x axis.

    and as you can see the line from Ax to A (the green line) is parallel to that plane and therefore also perpendicular to the x axis. Hence my 90 ##^\circ##.

    You of course in your post #1 meant to refer to the angle between ##\vec A## and ##\vec A_x## which as you can see is ##\pi-\alpha ## and thereby in the range 0 to 90##^\circ##.

    Note that I forgot to draw the arrow above ##\vec A_x## and ##\vec A_{yz}##. Nobody is perfect, be we keep trying.
    To be specific: in ##\vec A_x = A_x\, \hat\imath\ ##, ##\ A_x## is a number (negative for the alpha in the picture: ##A_x = |\vec A | \cos\alpha##​
     
    Last edited: Oct 14, 2015
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