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Coulomb's Law and Tension

  1. Jan 23, 2012 #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 θ = 42.6°, and the charges on the two balls are equal, what is that charge (in μC)?

    2. Relevant equations

    F(T)+F(g)+F(E)=0
    F(Ty)=F(g)=mg
    F(Tx)=k(q^2/r^2)

    3. The attempt at a solution

    I really don't know how to manipulate these formulas to find an answer. I know you find F(Ty) first, but then I don't know how to get F(Tx). I think I am missing a formula somewhere. Please Help Thank you
     
  2. jcsd
  3. Jan 23, 2012 #2
    You're pretty close, but the problem has an additional piece of information that you haven't used yet. Hint: it's the angle!
     
  4. Jan 23, 2012 #3

    Simon Bridge

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    Probably best to draw free-body diagrams first perhaps?
    Hi Dancer, welcome to PF.
     
  5. Jan 24, 2012 #4
    The force diagram is attached


    F(Ty)=cos(θ)*F(T) - - - F(T)=F(Ty)/cos(θ)
    F(Tx)=sin(θ)*F(T) - - - F(T)=F(Tx)/sin(θ)

    So, F(Ty)/cos(θ)=F(Tx)/sin(θ) - - - F(Tx)=[F(Ty)*sin(θ)]/cos(θ) - - - F(Tx)=F(Ty)*tan(θ)

    Did I do all of that correctly?
     
  6. Jan 24, 2012 #5
    Sorry here is the force diagram actually
     

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  7. Jan 24, 2012 #6

    Simon Bridge

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    The next step after drawing the diagram is to add up the forces in each direction. Start with just the left-hand ball. eg. [tex]F_T\cos\theta=F_g[/tex]... do the horizontal one, then write out what each of the forces actually are. eg. [tex]F_e=\frac{kq^2}{r^2}[/tex]... and r is related to the angle... see?
     
  8. Jan 24, 2012 #7
    I understand that
     
  9. Jan 24, 2012 #8
    Yeah, that's right. tanØ=Tx/Ty

    So you're set, work out the algebra from here. Don't forget, as Simon mentioned, to get the EM force in terms of "x distance."
     
  10. Jan 25, 2012 #9
    F(Tx)=mg*tan(θ)

    q=√((F(Tx)*r^2)/k)

    would r be mg/tan(θ)??
     
  11. Jan 25, 2012 #10

    Simon Bridge

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    Complete:
    if the pith balls were at positions (±x,-y) ...
    then there is a triangle l:y:x with angle θ between l and y (l=length of the string)
    From this picture:

    the distance between the balls is ____________.

    and tanθ=_x/y_, sinθ=_____, and cosθ=_____.

    which of the three is likely to help you?
     
    Last edited: Jan 25, 2012
  12. Jan 25, 2012 #11
    Nah, that would be the horizontal force, which is the coulomb force you're in part trying to solve for. You've a string length you haven't used yet. ;)
     
  13. Jan 26, 2012 #12
    Ok so sin(θ)*string length=r

    SO..
    F(Ty)=F(g)=mg
    F(Tx)=k(q^2/r^2)
    F(Tx)=F(Ty)*tan(θ)
    q=√((F(Tx)*r^2)/k)

    F(Tx)=mg*tan(θ) - - - - - - - - - F(Tx)=.012kg*9.81m/s^2=.11772N→A
    q=√((F(Tx)*(sin(θ)*L)^2)/k)----q=√(A*(sin(42.6)*1.2)^2/9E9)=2.938E-6C or 2.9μC

    Is 2.9 μC the correct answer or did I make a mistake?
     
  14. Jan 26, 2012 #13
    Careful, you want the total distance between the pith balls. You just found one half, one leg.
     
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