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Vectors and Charges

  1. Sep 19, 2008 #1
    1. The problem statement, all variables and given/known data

    Four electrical charges are located at the corners of a rectangle. Like charges, you will recall, repel each other while opposite charges attract. Charge B exerts a repulsive force (directly away from B) on charge A of 3.3 N. Charge C exerts an attractive force (directly toward C) on charge A of 6.6 N. Finally, charge D exerts an attractive force of 2.2 N on charge A.

    The Dimensions of the rectangle are 141 cm x 100 cm

    A B
    C D (That's how the rectangle is laid out)

    Assuming that forces are vectors, what is the magnitude of the net force \vec{F}_{\rm net} exerted on charge A?

    What is the direction of the net force \vec{F}_{\rm net} exerted on charge A?

    2. Relevant equations

    I would think it would be Ax= Acostheta Av=Asintheta

    3. The attempt at a solution

    I've been working on this one for about 4 hours ( I know =[).

    Now, I thought we were supposed to break up the components into an X and Y component, but I'm not sure where the negative signs go (if any) and which components don't need to be broken down.

    I tried calculating the angles and bisecting the rectangle in two then using the B charge and C charge as they were, and finding the components for the D charge. Im still not sure if this is the right way to go about it. Any help?
  2. jcsd
  3. Sep 19, 2008 #2
    Well, so far I think i should do this in vector notation form so ihat= -3.3N jhat= -6.6N and K hat is what needs to be broken up? Am i on the right track?
  4. Sep 19, 2008 #3
    You need to define the x and y axes. Let's call positive y upwards, positive x to the right. If B exerts a repulsive force on A, the force vector is pointing to the left, so the x component of the force is negative, and the y component is zero. Repeat this for the forces due to C and D, then add all the components up. The magnitude of F is

    [tex] \sqrt{F_x^2+F_y^2} [/tex]

    For the direction, you should draw a right angled triangle where the hypotenuse is of length |F|, and the other two lengths are the components of F. Use trigonometry to find the angles of the triangle, and write the answer as something like "the net force makes an angle of ?? with the x axis".
  5. Sep 19, 2008 #4
    My problem is that im unsure of how to break up the "D" component of the force. I have -3.3N going to A -6.6N going to C but i'm unsure of the sign of the D vector and how to break it's components up. Could you give some insight as to how to do this please?
  6. Sep 19, 2008 #5
    The force on A due to D is attractive, so the force vector lies on the line connecting A and D, and points towards D. Draw the rectangle, and draw the line between A and D. You'll see that there are two right angled triangles now. You can use trigonometry to find the ratios of side lengths in the triangle. 2.2N is the magnitude of the force, so it corresponds to the hypotenuse. The x and y components of the force correspond to, respectively, the horizontal and vertical sides of the triangle.
  7. Sep 21, 2009 #6
    I have the exact same problem, but I just can't seem to break up the components properly. I just get random numbers that have no relation to the problem. (like I got 9.8 for the forces pushing in the negative x-direction. Am I doing this wrong?
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