1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Simple Vector Question

  1. Sep 28, 2009 #1
    So I have constructed this vector for a physics problem:

    http://img23.imageshack.us/img23/1753/vectorx.jpg [Broken]

    Why is it when I take the tangent of the labelled angle, I get an angle of 77*, which is the complementary pair for the correct angle (which is 13*).

    I can rationalize this by adding 90* to the calculated angle of 77*, the sum for which is 167*. 180*-167* = 13*, so I get the correct answer by accounting for the quadrant the vector is located in. But in my (admittedly limited) experience with vectors, I have never had to adjust the calculated angles for the quadrant (on a cartesian plane) where the vector is located; the trigonometric ratio between the two sides has always given the correct angle.
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 28, 2009 #2


    User Avatar
    Homework Helper
    Gold Member

    Can you state the problem the way it is phrased and show your work? It seems that the angle that you have interpreted as θ is the complementary angle of what the problem thinks is the correct θ.
  4. Sep 28, 2009 #3

    Three charges are arranged as shown:

    http://img8.imageshack.us/img8/1841/82829031.jpg [Broken]

    q1 = 2.5x10E-17
    q2 = 3.0x10E-17
    q3 = 3.5x10E-17
    (all are positively charged)

    Calculate the magnitude and direction of the force acting on q1.

    Let up and right be positive.

    21F = 2.7x10E-23 N
    31F = 8.8x10E-23 N
    (Both of these were calculated using Coloumb's Law).

    So, I then proceeded to resolve these forces into the horizontal and vertical components:
    Fx = 21Fcos37* + 31Fcos90*
    = -2.2x10E-23 N + 0N
    = -2.2x10E-23 N (negative because left has been defined as negative)
    Fy = 21Fsin37* + 31Fsin90*
    = 1.0x10^-22 N

    So using the Pythagorean Theorem, I calculated the net force to be: 1.0x10E-22 N

    This is where I had to now calculate the angle. I also performed a solution in terms of cosine line and got an angle of approx. 12-13* as I should get.
    Last edited by a moderator: May 4, 2017
  5. Sep 28, 2009 #4


    User Avatar
    Homework Helper
    Gold Member

    Your drawing shows two charges labeled q2 and no q1. Which one of the two q2 charges is q1? You don't need to repost the drawing, just describe it.
  6. Sep 28, 2009 #5
    Oh, I apologize for that. The top point is q1.
  7. Sep 28, 2009 #6


    User Avatar
    Homework Helper
    Gold Member

    Your method is fine. The correct answer could be many things depending on what your reference for measuring the angle is.

    It is 13o from the positive y-axis.
    It is 103o from the positive x-axis.
    It is 77o from the negative x-axis.

    If this is a web-based problem graded by the machine using a stored formula, then the reference line should have been made clear in the statement of the problem. If this is a homework problem graded by a sentient human being, any of the above three answers would be correct. BTW, I didn't redo your numbers, I assume you did them correctly.
  8. Sep 28, 2009 #7
    Thank you, that is reassuring.

    Do you suppose I could rationalize it the following way:

    Instead of using the angle of 37* for my calculations, I would use 143* (considering the vector for q2 points northwest and would lie in the second quadrant on a cartesian plane, so 180-37=143). Therefore, when I calculate the tan ratio as I did, it makes sense to add 90* in order to validate the answer for the second quadrant .
  9. Sep 28, 2009 #8


    User Avatar
    Homework Helper
    Gold Member

    No, don't bother. You understand how to do this problem and that's what counts. The line of reference of the angle is irrelevant. Personally, I prefer to express vectors in terms of their x and y components and forget about angles and magnitudes.
  10. Sep 28, 2009 #9
    Does my rationalization not make sense (or is inapplicable) or do you just think explanations along these lines are useless, provided I know how to approach the problems?

    It isn't really so much for this problem that I made this inquiry, but I don't want to be caught with my pants down again in a similar way in the future. For example, if I'm writing an exam and have limited time to check my answers, I think it might be useful to find a working approach - which, in this case, involves using angles from 0-360 (as in a cartesian plane) and adjusting the calculated angle (obtained from trig ratios) depending on the quadrant it would be located in a cartesian plane. Otherwise, I fear answering with the wrong angle again, since my initial answer for this question was the complementary angle to the correct one.
  11. Sep 28, 2009 #10


    User Avatar
    Homework Helper
    Gold Member

    As long as you specify with respect to what line you measure your angles and your calculations are correct, nobody in his/her right mind will mark you wrong. If you want to put your mind at ease, before you take an exam, talk to your instructor and ask "Is there a preferred line with respect to which angles are measured or can I specify it myself?" and see what (s)he says. If you asked me that, I would say "As long as you specify the reference line, and your answer is correct, you get full credit."
  12. Sep 28, 2009 #11
    I believe I should clarify myself a bit more.

    As you can see from my diagram in my original post, the calculated angle clearly uses the x-axis as the reference. So one would expect that using the tan trig ratio to calculate the angle, dividing the y-component by the x-component, the correct angle would be obtained. However, in my original solution, for some reason, that gave the complementary angle (with the reference axis clearly defined).

    So, I got the question marked wrong, since when I made the calculation, I thought my method would provide the correct angle. I mean, I did establish the x-axis as the reference, I had all the magnitudes correctly defined; yet, the trig ratios did not yield the correct answer and needed to be adjusted. Presumably, I think, because I have to account for the vectors' position on a cartesian plane.
  13. Sep 28, 2009 #12


    User Avatar
    Homework Helper
    Gold Member

    I redid your calculations and the original diagram you posted is correct. You say 77 degrees with respect to the negative x-axis, I say 76.13 degrees with respect to the negative x-axis. Your answer is correct to within roundoff errors.

    Who marked your answer wrong? You need to take the matter to your instructor and have him/her explain why your answer was marked wrong. I don't see anything wrong with it.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Simple Vector Question
  1. Simple Vector Question (Replies: 1)

  2. Simple Vector Question (Replies: 4)

  3. Simple vector question (Replies: 7)