Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    Attached Files:

  7. Jan 24, 2012 #6

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper

    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


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

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper

    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


    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook