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Homework Help: Require Help.

  1. Aug 28, 2007 #1
    1. The problem statement, all variables and given/known data
    Vector A has a magnitude of 11 and points in the positive x-direction. Vector B has a magnitude of 22 and makes an angle of 32 degrees with the positive x-axis.
    What is the magnitude of Vector A minus Vector B?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 28, 2007 #2

    learningphysics

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    Can you describe where you're getting stuck and what you tried?
     
  4. Aug 28, 2007 #3
    Another question.

    I also have another question whereas it states: Use the method of components to find the magnitude and direction of the vector sum R1 where R1= A + B. The Vector A= 15.2 m at an angle alpha= 180 degrees from the positive horizontal axis, and Vector B = 17.2 m at an angle Beta= 41.3 degrees from the positive horizontal axis. Answer in meters. Answer in units of m.

    What is the angle, theta 1, from the positive horizontal axis of the vector sum, R 1?

    What is the magnitude of the vector difference R 2 where R 2= Vector A - Vector B?

    What is the angle, theta 2, of the resulting vector?

    What is the magnitude of the vector difference R 3 where R 3= B - A?

    What is the angle, theta 3, of the resulting vector?
     
    Last edited: Aug 28, 2007
  5. Aug 28, 2007 #4
    I've only sketched it to get a graphical representation of the vectors, but other than that I have only gotten the resultant magnitude of A + B, but don't know how to get the resultant magnitude for A - B.
     
  6. Aug 28, 2007 #5
    draw a llgm with two vectors A and B,
    you have found one diagonal, just find other one, and it would be A - B

    or,
    reverse the direction of vector B
     
  7. Aug 28, 2007 #6

    learningphysics

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    A - B is the resultant of A and (-B). Use the same approach as before except use -B instead of B...

    another idea... draw both vectors originating from the same point... what is the vector joining the arrow ends of B and A?
     
  8. Aug 28, 2007 #7

    Yes, I know what you are referring to about the -B pertaining to its opposite direction, but the problem is I only know how to find the resultant magnitude for A + B and don't know how to find it for A - B because it's different since it's only two vectors and not more.
     
  9. Aug 28, 2007 #8
    Ok, so but isn't finding the magnitude of the resultant vector using basically the pythagorean theorem? If I use -B it will come out to the same answer. The thing that I'm confused about is that when I draw it graphically the A + B resultant vector looks shorter than the A - B resultant vector. So the problem is that I don't know how to find that A - B resultant vector, I only know the A + B resultant vector.
     
  10. Aug 28, 2007 #9

    learningphysics

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    So you have drawn A - B, but are just having trouble calculating the magnitude?
     
  11. Aug 28, 2007 #10

    learningphysics

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    No, the pythagorean theorem is only for right triangles... this isn't a right triangle... pythagorean theorem won't work for A-B or A + B.
     
  12. Aug 28, 2007 #11
    Yes, the way I depicted it was with Vector A going towards the right with a magnitude of 11 along the x-axis, while the -B is at 212 degrees going in the southwest direction with a magnitude of 22.
     
  13. Aug 28, 2007 #12
    ? But I've used it before to determine the magnitude value of three vectors with their x and y coordinates provided.
     
  14. Aug 28, 2007 #13

    learningphysics

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    Yes... from the x and y components, you can calculate the magnitude using the pythagorean theorem... That is because the x component is a vector in the x direction... the y component is a vector in the y direction... they are perpendicular and represent a right triangle...

    if you calculate the x component (call it x) and y component (call it y) of A+B... then you can use the pythagorean theorem to calculate the magnitude... so maginitude = [tex]\sqrt{x^2 + y^2}[/tex] but this is not equal to [tex]\sqrt{A^2 + B^2}[/tex]

    ie: A and B are not the x and y components of A+B...
     
  15. Aug 28, 2007 #14
    Oh, I see, so then how do I find the resultant magnitude of A - B?
     
  16. Aug 28, 2007 #15

    learningphysics

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    Have you studied the law of cosines and the law of sines for triangles (non-right triangles)?

    EDIT: You can also calculate the x-component of A-B, and the y-component of A-B... then use pythogrean theorem.

    Both approaches are good... it's up to you...
     
    Last edited: Aug 28, 2007
  17. Aug 28, 2007 #16
    Yes, but I'm not very familiar with using it with the vectors.
     
  18. Aug 28, 2007 #17
    So the x/y component of Vector A would be <11,0>? and the vector component to find -B would be cos(212)= x/22? By which I would then use tan to find the y-component value?
     
    Last edited: Aug 28, 2007
  19. Aug 28, 2007 #18

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    Once you see the triangle you need... just approach it like triangles now... at this point don't think about them being vectors...

    you've got the triangle made up of |A|, |-B|, and |A-B|... You already know |A|, |-B| and the angle between these two...

    So you've got a triangle where you know two sides, and the angle between the two sides... you want the third side... do you see how to use the cosine rule here to calculate the third side?
     
  20. Aug 28, 2007 #19

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    No you'd use sin to calculate the y component... so once you get the x and y components of -B... just add A and -B... then get the magnitude of that vector using pythagorean theorem...

    I highly recommend using the other approach also for the purpose of learning... you should know how to do it both ways.
     
    Last edited: Aug 28, 2007
  21. Aug 28, 2007 #20
    Ok, so far pertaining to my -B vector I received as the x-component -18.6571 and the y-component as -11.6582. Is this correct? and to my A vector, x-component is just 11 and y-component is zero?
     
  22. Aug 28, 2007 #21

    learningphysics

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    Yes, that looks correct. add the two vectors...
     
  23. Aug 28, 2007 #22
    Final Answer for magnitude resultant of A - B is 13.94795?
     
  24. Aug 28, 2007 #23

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    Yes, that's correct. Now with the other approach:

    You've got the triangle composed of A, -B, and A-B...

    This is a triangle with sides, 11, 22 and an angle of 32degrees between them... using the cosine rule I can directly calculate the third side which is the magnitude of A-B:

    r^2 = 11^2 + 22^2 - 2(11)(22)cos(32)
    r^2 = 194.54
    r=13.9479

    I personally prefer this method...
     
  25. Aug 28, 2007 #24
    Thank you so much for your help. I really appreciate it and I was wondering if you could help me out on one more problem?

    If so, the problem is Four vectors, each of magnitude 62 m, lie along the sides of a parallelogram as shown in the figure. The angle between vector A and B is 49 degrees.
    What is the magnitude of the vector sum of the four vectors? Answer in units of m.

    So far, I have drawn the picture and made out with a parallelogram with the left side as vector A, the top vector C, the right vector D, and the bottom vector B with all of the sides congruent to each other respective to 62 m, and also the angle with between vector A and B is 49 degrees which also means that between C and D is the same while between both A and C and B and D is 131 degrees in between.

    What I've been thinking was since I have to find the magnitude of the vector sum of the four vectors, wouldn't it be the same as finding the magnitude diagonal of the parallelogram? or am I off the tee here?
     
  26. Aug 29, 2007 #25

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    Hmmm... so the way I understand the problem, the vector A = the vector D... and the vector B = the vector C ? Remember that a vector is determined by its magnitude and direction... so if two vectors are parallel (ie they have the same direction), and they have the same magnitude, then they are equal vectors,,,

    So A+B+C+D = 2A + 2B = 2(A+B) ?

    So just calculate the resultant A+B, and then 2(A+B)... is in the same direction but with double the magnitude...

    Hope I'm understanding the problem correctly.
     
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