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Please help me with this easy problem

  1. Sep 5, 2004 #1
    I have no idea why I can't do it.. I must have done it a million times before:

    Two cork balls of mass .2g hang from the same support point by massless insulating threads of length 20cm. A total positive charge of 3.0x10^-8 C is added to the system. Half of this charge is taken up by each balls, and the balls spread apart to a new equilibrium position. (There is a value of theta in the diagram, which looks roughly like: /\ (each line holds a ball with a charge). Theta is between one of those lines and a line down the middle, so between /| or |\

    Find the tension in the threads.

    Now I get
    Fx = Tsin(theta) - Fq = 0
    Fy = Tcos(theta) - mg = 0

    I don't know how to solve it from there - to find Fq I need r, which is hard to find since it involves theta (I think). And I can't solve out for T because there is a sin and cosine function.

    There must be an easier way to solve this - what am I missing?
     
  2. jcsd
  3. Sep 5, 2004 #2

    Tide

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    Your equations aren't quite right - if [itex]\theta[/itex] is the full angle between the strings you should have [itex]\frac{\theta}{2}[/tex] instead of [itex]\theta[/itex].

    What you want to do is rewrite each equation with the trig function on one side then divide one by the other. From that you can determint [itex]\theta[/itex] and the rest should be simple.
     
  4. Sep 5, 2004 #3
    No, theta isn't the full angle, it's the angle halfway between each of the strings (I don't think I explained that well)

    Hrm... I'll give that a shot, thanks.

    edit: How do I find r though? (The distance between the two balls)
     
    Last edited: Sep 5, 2004
  5. Sep 5, 2004 #4

    Tide

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    [tex]r = 2L \sin \theta[/tex]
     
  6. Sep 6, 2004 #5
    OK, that's what I get.. but then i'm left to solve (using the method you told me):
    sin^3(theta)/cos(theta)= a bunch of values I have known

    Isn't there a way to do it without using theta, since i'm asked to find that in the next part of the problem?
     
  7. Sep 6, 2004 #6

    Tide

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    That [itex]\sin^3 \theta[/itex] does complicate matters!

    I don't see a way of getting rid of it but you might consider that "bunch of values." If it's small enough you might use a small angle approximation to the trig functions or try to solve the equation numerically.
     
  8. Sep 6, 2004 #7
    I don't think i'm supposed to get that complicated, this is a fairly easy problem (it's one of the first) and it's under the section that hasn't even gotten to Coulomb's law. I must be missing something
     
  9. Sep 6, 2004 #8

    Alkatran

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    Look at it this way: Gravity is pushing down, so the "down" part of your triangle is going to be 9.8 * whateverthemasswas

    Now, you just need to figure out at what point the force of tension in the cord towards the center equals the force of repulsion of the charges. Just remember that the upwards force of tension must always be 9.8 * mass, and that the force towards center will be proportional to it via TAN.
     
  10. Sep 6, 2004 #9
    I understand what you're saying until the last sentence. I can't solve the 2-equation-2 unknowns system since I have 3 unknowns.
     
  11. Sep 6, 2004 #10

    arildno

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    [tex]\frac{\sin^{3}\theta}{\cos\theta}=tan\theta(1-\cos^{2}\theta)=[/tex]
    [tex]tan\theta(1-\frac{1}{\frac{1}{\cos^{2}\theta}})=tan\theta(1-\frac{1}{tan^{2}\theta+1})=[/tex]
    [tex]\frac{tan^{3}\theta}{tan^{2}\theta+1}[/tex]
     
  12. Sep 6, 2004 #11

    Alkatran

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    let's assume the mass is 1, for simplicity. The angle, theta, is the angle between the center line and the rope. (x is opposite, y is adjacent)
    Rope (tension):
    Fy = 9.8
    Fx = Fy * tan(theta) = 9.8 * tan(theta)

    Charges:
    Distance = x (the opposite of the triangle) = .2 * sin(theta)
    Fx = charge^2/distance^2 = (1.5*10^-8)^2/(.2*sin(theta))^2
    tan(theta) = 2.25*10^-14 / 4*sin(theta)^2

    sin(theta)^3/cos(theta) = 2.25*10^-14/4

    Looks like we need some trig identites... you were right. My mistake
     
  13. Sep 6, 2004 #12
    Anyone? I'm really stumped.
     
  14. Sep 6, 2004 #13
    Pleeeeeease, it's due tomorrow and I can't figure it out for the life of me.
     
  15. Sep 6, 2004 #14

    Alkatran

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    I punched the equation into my TI-89 and it solved it. But that was using the 1 kg mass, etc...

    I won't post the solution since I don't know the work.
     
  16. Sep 6, 2004 #15

    Tide

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    Is it possible the text is simply asking for T in terms of [itex]\theta[/itex] without actually having to specify the value of [itex]\theta[/itex]?
     
  17. Sep 7, 2004 #16

    ehild

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    No, this is not an easy problem and you need to know Coulomb's Law.
    [tex]F_q=kq^2/(2Lsin\theta )^2 [/tex]
    [tex]k = 9 \cdot 10^9 Nm^2/C^2[/tex]
    [tex]tan(\theta) = \frac{kq^2}{4L^2mgsin^2(\theta)}[/tex]
    [tex] \rightarrow \frac{sin^3(\theta)}{cos(\theta)}=\frac{kq^2}{4L^2mg}=0.02583[/tex]

    You can solve it numerically. [tex]\theta\sim 17^o[/tex]

    ehild
     
  18. Sep 21, 2004 #17
    Try taking moment around the tied end. It should be easier as u know the length of the string.
     
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