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Finding the magnitude and the angle

  1. Mar 5, 2008 #1
    1. The problem statement, all variables and given/known data
    A particle undergoes the following consecutive displacements 3.8m south, 8.4m northeast, and 14 m west. What is the resultant displacement and angle from the positive x-axis?

    2. Relevant equations

    R=square root (Rx^2+Ry^2+Rz^2)

    In this case the displacement is the same thing as the magnitude right? I know that the answer is 8.34m and 165°, but I'm not getting the right answer.

    This is what I did:


    square root (8.4^2+3.8^2+8.4^2)

    How would I find the angle? Would I just use tangent?

    Thank you very much
  2. jcsd
  3. Mar 5, 2008 #2
    Make sure you are using the right equation. The directions south, northeast, west, etc. can be displayed on a two-dimensional map. This problem is in two dimensions, not three. It would probably help to draw a picture of the motion of the particle.

    After you draw the motion, the displacement is the magnitude and you do find the angle using inverse tangent.
  4. Mar 5, 2008 #3

    Shooting Star

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    Homework Helper

    What are the three quantities here? There's no Rz.

    A displacement of R to the NE would mean a vector [itex]\frac{R}{\sqrt{2}}[/itex](i+j). Add all the i and j terms, and then apply the formula to find the magnitude of the net displacement.

    If you just draw a diagram, things will become clear.
    Last edited: Mar 6, 2008
  5. Mar 5, 2008 #4
    Thank you very much

    OK, I drew the picture. I'm not sure how you would find the resultant vector, though. Before, I was adding up the i's and the j's and plugging them into that formula. Could you tell me what formula I would use? Would I use the law of cosines?

    Thank you
  6. Mar 5, 2008 #5
    Remember that vector addition is head to tail. Put the tail of the next vector to be added to the head of the previous one. The resultant vector is the vector drawn from the tail of the first vector to the head of the last one. You can always find i and j of this vector, since you will find that it is just the i and j of each of the individual vectors added together (the components i and j are at right angles to each other, so you Pythagoras's Theorem is fine, no need for Law of Cosines).

    As Shooting Star says, northeast would involve a 45 degree angle.
  7. Mar 5, 2008 #6
    Thank you.I know what you have to do if there are 2 vectors. If you have three, do you add up the first 2 and then add on a third one? Is the resultant vector in the third quadrant? I took square root 14^2+3.8^2=14.5 Do you see what I did wrong?

    Thank you
  8. Mar 5, 2008 #7
    Yes. If you have three vectors, just add the third vector to the first two. You can add up the first two vectors first, then add the third vector, or you can just draw all the vectors head to tail, putting the tail of the next vector onto the head of the previous one. The total sum, also called the resultant vector, is the vector from the tail of the very first vector to the head of the very last vector added.
  9. Mar 5, 2008 #8
    Thank you

    Then wouldn't 14.5 be correct?
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