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Question Regarding Vectors

  1. Sep 27, 2009 #1
    This is a net force problem that I have solved, I would just like to clarify some aspects of my work. I will not post the details of the question, but it involves calculating the net force of two charges on another charge. I have defined up and right as positive.

    I did this solution using both the component method and also the trigonometric method, and the solution for the component method involves something I really haven't come across before in physics. My experience with vectors is fairly limited, so I turn here. These are the values for the composite x and y magnitudes, obtained from vector addition and subtraction (where the direction was negative):

    Fx=-2.2x10-23 N
    So, the horizontal force acts left or west.
    Fy= 1.0x10-22 N
    So, the vertical force acts up or north.

    Fnet=1.0x10-22 N as determined by the Pythagorean Theorem and acts northwest.

    So I proceeded to calculate the angle from the horizontal of the net force.

    My First Attempt:
    tanx= Fy/Fx
    tanx= |1.0x10-22 N| / |-2.2x10-23 N| (used absolute values)
    x = 78 degrees (approx.)

    Initially, I thought I got the question wrong because the textbook (as well as my trigonometric solution, using the cosine law) gave a different answer (the angle should be between 12-13degrees).

    At first I thought maybe I shouldn't use absolute values for trig ratios, but dropping the absolute value brackets ended up giving me an angle of -78*. As can be seen, though, 78* and 12* are complementary angles, so I thought that since the direction of the net force is northwest, the vector would be located in the second quadrant of a Cartesian grid (where y=+, x=-) so I thought I should adjust my calculated angle to this fact. By adding 90* to the angle 78*, the value of tan would remain the same and the angle with the x-axis would simply be 180 - (90+78)=12*.

    It has been my experience with questions involving vectors that you get the angle with the x-axis by calculating the trig ratios with absolute values, without ever having to adjust for what quadrant the vector is located in. Is there something unusual with my question, then, or in my solution, or is this something typical that must just be kept in mind? Could I have avoided this by doing something differently, or does my work look sound?

    Thanks.
     
  2. jcsd
  3. Sep 27, 2009 #2

    Pengwuino

    User Avatar
    Gold Member

    My advice to students is to just KNOW where your vector should point and approximately what your angle should be. In your question, the y-component was much greater than the x-component so you should expect a resultant vector that points almost parallel to the y-axis. When you got the result of 78 degrees, you should have figured you calculated 78 degrees off the -x axis.
     
  4. Sep 27, 2009 #3
    That is sound advice. I am usually not mindful of what my angle should approximately be and accordingly entrust that task to my calculator. If the solution had eluded me for longer than it had, I would eventually make use of that to make sense of it - but thanks for the advice; I think it is worth considering in vector problems, should I run into similar issues.

    Nevertheless, I would like to know whether there is something unique to this problem or whether I will find myself having to compensate for calculated angles depending on their direction (in other words, their position in a cartesian grid).
     
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