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What is the magnitude of the resultant between these two vectors?

  1. Sep 30, 2005 #1

    IB

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    Vector A has a magnitude of 10 units and makes an angle of 30° with the horizontal
    x-axis. Vector B has a magnitude of 25 units and makes an angle of 50° with the
    negative x-axis. What is the magnitude of the resultant between these two vectors?

    I'm not sure how to do this one. Are you supposed to add 10 and 25 to get the resultant vector or something?
     
  2. jcsd
  3. Sep 30, 2005 #2

    Diane_

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    There are at least three ways to do this one. Try this: Draw a picture. Put the tail of vector A at the origin and sketch it out, then put the tail of vector B at the head of vector A and sketch it. The resultant will be the vector from the tail of A to the head of B. They will make a triangle. A bit of geometry and a bit of trigonometry should get you the solution.
     
  4. Sep 30, 2005 #3

    IB

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    What are the other ways I can do for this kind of problem?
     
  5. Sep 30, 2005 #4

    IB

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    I got that A is a leg B is hypotenuse and Vector C, our resultant vector, is a leg. So is it [tex] {C^2 = B^2 - A^2} [/tex]? That way i'll get x^x = 30^2 - 10^2 so x = 900 - 100 = 800 => x = [tex] {\sqrt 800} [/tex] = 28.
     
  6. Sep 30, 2005 #5

    Diane_

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    I don't see any right angles in there, luv. There is an 80 degree angle between A and B - perhaps that's what fooled you. I'm afraid Pythagoras won't be much use here - it's back to precalc and the Law of Cosines.
     
  7. Sep 30, 2005 #6

    IB

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    so if it isn't a right triangle how do i know which angle is 30, which angle is 50 after having moved vector B?
     
  8. Sep 30, 2005 #7

    Tide

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    Why don't you just resolve the vectors into components and add the components to find the resultant?
     
  9. Sep 30, 2005 #8

    Diane_

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    There are so many times I miss having a blackboard on these forums.

    I presume there are no problems sketching A, starting at the origin? OK, once you have that, lightly sketch in a new set of coordinate axes with the origin at the head of A. Use those to sketch in B. You'll have both sets of x-axes parallel, so all of the stuff you learned in geometry about "alternate interior angles" and so on will apply. I think you'll be able to see what the angles are fairly quickly.

    Edit: Tide, I thought about having him do that, but I'm not sure he'd learn as much that way as through a purely geometric approach. It's the teacher in me, I guess.
     
  10. Sep 30, 2005 #9

    IB

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    Hmm interesting! You're really a good teacher. Now I understand this vector problem. :)

    You should teach Higher Level Maths, Further Maths, and Multivariable Calculus at secondary school instead of C.Sc. :wink:
     
    Last edited: Sep 30, 2005
  11. Sep 30, 2005 #10

    Diane_

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    I did. Math and physical science in American high schools for thirteen years. I've been thinking about going back - I miss the teenagers.

    Anyway, glad to be of help.
     
  12. Sep 30, 2005 #11
    vector

     
  13. Sep 30, 2005 #12

    Tide

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    Diane,

    Without additional information it's often difficult to gauge what background students have when attempting to answer their questions particularly in an international forum such as this. In any case, IB seems happy with the results! :)
     
  14. Sep 30, 2005 #13

    div(f grad g)=?
    WE kow grad g= i dg/dx+j dg/dy+k dg/dy
     
    Last edited: Sep 30, 2005
  15. Sep 30, 2005 #14
    curl(A+B)=?
     
  16. Sep 30, 2005 #15

    IB

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    Euh...you may need to explain more...
     
  17. Sep 30, 2005 #16

    Diane_

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    Oh, you're absolutely right, luv. If it seemed like I was criticizing, I apologize. Given the problem he was having, I did make an estimate as to his background, but the suggestion you offered was certainly reasonable.
     
  18. Sep 30, 2005 #17

    IB

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    By the way, Diane, i somehow got sin20 = x/25 so x = 8.6 but i don't think that's true because the given choices are:
    a. 20
    b. 35
    c. 15
    d. 45
    e. 25

    so further help is appreciated! :)
     
  19. Sep 30, 2005 #18

    Tide

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    Diane,

    Oh, heavens no! I didn't take anything you said as criticism. It was merely an "editorial comment" on how difficult it is to teach with inadequate input from the student.
     
  20. Sep 30, 2005 #19

    Diane_

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    IB - Can you tell me what you did to get to that point? I don't see any way there right off the bat.

    Tide - True that. I think we understand each other.
     
  21. Sep 30, 2005 #20

    IB

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    I have 20 degrees because I got 90 for the right angle at the tail of vector B at origin, then I got 50 - 30 = 20; the 30 is from the other origin, i used geometry to deduce that.

    I think I'm doing it all wrong though...
     
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