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Coulomb's Law - Vector Analysis

  1. Aug 1, 2009 #1
    1. The problem statement, all variables and given/known data

    I've taken the liberty to translate the problem, along with its corresponding diagram, into an image:

    http://img15.imageshack.us/img15/9695/questionm.jpg [Broken]

    angle q1 = 53*
    angle q2 = 37*
    angle q3 = 90*

    2. Relevant equations

    F21 = kq1q2/(r21)^2
    F31 = kq1q3/(r31)^2
    Sine Law

    3. The attempt at a solution

    I think I have a good idea on how to solve this problem. First, here is my vector FBD:

    http://img36.imageshack.us/img36/5599/vectorw.jpg [Broken]

    I'm not sure if I did it properly. The other issue is I'm not sure whether the angle x is 37 or 53 degrees. My personal take on it is that it is 53 degrees, because vector F21 is drawn at 37 degrees; because that 37 degrees isn't in the triangle, I believe the only way to get an angle is to consider the angle 53 (complimentary angle). But because of this, I am second-guessing whether my FBD is right in the first place or if I am wrong to infer the angle that way.

    Secondly, I have calculated:

    F21 = 2.7x10^-23 N
    F32 = 8.8x10^-23 N

    So, my approach to this problem is to apply the sine law (as soon as I can determine angle x) to get all the angles within the triangle and then to get a magnitude for the net force vector.

    1. Is my vector FBD correct?
    2. Is my inference of angle x correct?
    3. Is my approach to the problem sound?

    I'll appreciate any help I can get.
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 1, 2009 #2


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    Homework Helper

    1. Your vector diagram looks fine.

    2. Angle x is actually neither 53 nor 37 degrees. The angle between F21 and the horizontal direction is 53 degrees, and the angle between F21 and the vertical direction is 37 degrees, as you could easily tell from the triangle in the original problem. But your angle x is the angle between F21 and the resultant force, which is neither horizontal nor vertical.

    3. You could do the problem this way, but you'd have to start by figuring out angle y rather than x. You don't have enough information to figure out x until you've actually solved the problem. But that's not the way I'd do it. Instead, I would split each of the two individual force vectors into horizontal and vertical components, add up the components, and compute the magnitude and direction of the final vector from that.
  4. Aug 1, 2009 #3
    Yeah. I understand that to find Fnet, I need to figure out angle y, but given that I was so inclined toward a trigonometric solution, not being able to find x just made the whole problem a non-starter for me. I didn't know how to rationalize getting an angle for the diagram, so thank you for setting me on the right course.

    I'm kind of rusty with vector analysis, so let me see if I can put this together correctly:

    Fx=F21cos37 + F31cos90
    Fy=F21sin37 + F31sin90
    Fnet2= Fx2 + Fy2

    Does that look about right?
  5. Aug 1, 2009 #4


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    Homework Helper

    Yeah, that looks about right, if you have the positive x direction going to the left. If you wanted positive x going to the right, as is typical, you'd just need to use a negative sign, Fx = -F21 cos 37o - F31 cos 90o.

    Although technically, none of those F's you wrote (except perhaps the Fnet) should be bold, since you're referring to components, not full vectors.
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