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Coulomb's law and String Tension

  1. Jan 28, 2008 #1
    1. The problem statement, all variables and given/known data

    Two small pith balls, each of mass m = 12 g, are suspended by 1.2 m fine (so that we can neglect their mass in this problem) strings and are not moving. If the angle that each string makes with the vertical is q = 42.6, and the charges on the two balls are equal, what is that charge (in C)?

    2. Relevant equations/attempt at solution

    I have tried to solve this by using string tension and setting it equal to mg, but I am missing something and can't figure it out. I tried originally to just use Coulomb's law, but I just hit a dead end there. I did find that that length between the two balls is 12.8 meters, but I'm not sure now if that is relevant. I have worked on this problem for over a half hour and have essentially gotten no where. I am very stuck, please help me.

    Thanks,
    Aphrael
     
  2. jcsd
  3. Jan 28, 2008 #2
    Draw a free body diagram for one of the balls. What forces are acting on the ball? Also, what can you say about the acceleration of the ball in the vertical and horizontal directions, and how can you use Newton's 2nd law in this situation?
     
  4. Jan 28, 2008 #3
    I know that the forces acting on the balls are gravity and string tension, but I do not think they are moving. There is a diagram with the problem that shows the two balls hanging, each with a 42.6 degree angle from vertical. Doesn't this mean that they are suspended on some sort of wire and not moving?
     
  5. Jan 28, 2008 #4
    Well lets just focus one one ball initially. You've noted the force on the ball due to gravity, and the tension force (acting at an angle), but what about the electric force? Choose one ball, and determine the direction that the electric force is acting. Now you have three forces acting on the ball, and as you've noted, the ball does not move, thus its acceleration is zero.

    Once you've determined each direction of the three forces, use Newton's 2nd law to sum the forces in the horizontal and vertical directions. You know that these forces must sum to zero in the x-direction, and they must sum to zero in the y-direction. Can you show me how you would set up the equations?
     
  6. Jan 28, 2008 #5
    Since the balls are equal in charge the force is directed away from each other.

    F(y) =mg =(.012)(9.81) and F(x) =mgcos(q) =(.012)(9.81)cos(42.6)

    Is this even close?
     
  7. Jan 28, 2008 #6
    It's somewhat close. Let's hold off on plugging in values for the moment and deal with variables only.

    Both the forces in the x-direction, and the forces in the y-direction need to add to zero, as there is zero acceleration for the ball.

    In the x-direction, you have the electric force, and a component of the tension force.

    In the y-direction, you have the other component of tension force, and the weight force.

    You want to define a coordinate system if you havent already done so, and see which forces act in the positive direction, and which act in the negetive direction, and sum them according to Newton's 2nd law.
     
  8. Jan 28, 2008 #7
    Alright I tried to do what you told me.

    x-direction: F(tension)cos(q)-F(electric)= 0
    y-direction: F(tension)sin(q)-mg= 0

    Solving for F(tension) I can then use F(electric)=kq^2/r^2 to solve for q. Q=(F(electric)r^2/k)^1/2. However something I am doing is wrong because when I plug in numbers I get 1.617x10^-10 which I know is wrong because I know the problem does not require scientific notation, besides the fact that that is just a ridiculously small amount of charge.

    I am just really bewildered by what you mean by the third force. Where is that coming from and how does it apply?
     
  9. Jan 28, 2008 #8
    Hold on a second, I'm going to make a quick diagram so you can see what I'm talking about. I don;t think my diagram should differ much from your problem, but it'll help us get on the same page. The reason I'm doing this is because I believe you sin(q) and cos(q) and the direction of your electric force may be mixed up. I'll post it in a sec and you can let me know if I'm picturing your problem correctly.
     
  10. Jan 28, 2008 #9
    Alright thanks.
     
  11. Jan 28, 2008 #10
    Ok here we go:

    http://s62.photobucket.com/albums/h116/pepsi_in_a_can/?action=view&current=untitled11.jpg

    If your problem is similar to this diagram, the ball is being repeled to the right by the electric force.

    Your net force in the horizontal direction would be:

    [tex] F_x = F_E - Tsin \theta = 0 [/tex]

    Your net force in the vertical direction would be:

    [tex] F_y = Tcos \theta - mg = 0 [/tex]

    This is what I picture from your problem. Does the above diagram differ from your perspective?
     
  12. Jan 28, 2008 #11
    Mine differs just slightly. There are two balls that form a triangle and the angle taken is from the horizontal splitting the angle formed by the two strings, but the angles equal. I noticed your first equation uses the opposite sign of what I used, but is that significant enough to change my answer?
     
  13. Jan 28, 2008 #12
    Could be. It sounds like you have the right idea for the most part. If we just use my example above, you see that we solve for the tension T using the second equation. Then substitute T into the first equation and solve for the electric force. Once we know the magnitude of the electric force, we can use Coulomb's Law to solve for the charge q.

    If this doesn't help you, post your calculations, and maybe we can spot an error. It can be a bit confusing if I don't know what your free body diagram looks like, but if you can post it or explain it better in addition to posting your calculations, that'd be a great help.
     
  14. Jan 28, 2008 #13
    Alright it sounds like we're on the same tract so I'll post my equations.

    F(x)=F(e)-Tsin(q)=0
    F(e)=Tsin(q)=mgsin(q)=(.012)(9.81)sin(42.6)=.0797 N (I just realized my calculator was in Radians, not degrees from my Calc class, but let's please continue because I want to make sure I understand this fully)

    How does this look so far?
     
  15. Jan 28, 2008 #14
    I'm a bit lost with your equations in the above post. Are you giving me the Fx and Fy equations? If not, go ahead and give me the Fx and Fy equations first, I'll fix us up a diagram, and you can tell me if we're on track. Then we'll be able to hammer out the rest of the problem quite easily.
     
  16. Jan 28, 2008 #15
    I was then using q=(F(e)r^2/k)^1/2. I knew that the length of my string was 1.2 meters, so I used tan(42.6)(1.2)=1.10 for half the radius between the two balls. However when I plug all these numbers in I get 1/95x10^-11 which is too small.
     
  17. Jan 28, 2008 #16
    I was using the Fx equation. That was my first one. However I wasn't using Fy. I'm not sure how to use it in this senario. F(e) I was using as electric force. I don't have a theta key so I just used 'q' as a dummy variable.
     
  18. Jan 28, 2008 #17
    I don't believe it's too small, as 1 Coulomb of charge consists of billions and billions of elementary charges. But the thing is, I won't know if you're on the right track unless you give me the Fx and Fy equations, and compare a free body diagram. If you're up for that, we can do that. If not that's ok too. It seems like you have the right idea though, so if you're comfortable with your answer, let me know.
     
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