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Two charges hanging on strings

  1. Oct 26, 2015 #1
    1. The problem statement, all variables and given/known data
    upload_2015-10-26_8-57-12.png

    2. Relevant equations


    3. The attempt at a solution
    upload_2015-10-26_8-58-6.png
    the answer for d comes out to be 0.017 m.

    BUT, if we kept with tan instead of switching to sin, we would have

    mg(d/2)/ mg =kq q / d2

    d^3 = 2k q q

    solving for d

    d = 0.0121 m

    clearly on a big scale, the two answers are about the same distance but the answer from using tan is more accurate, right?
     
    Last edited: Oct 26, 2015
  2. jcsd
  3. Oct 26, 2015 #2

    TSny

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    This equation doesn't make sense. On the left you have a length cubed. In the middle you have something with dimensions of Force times length squared. On the right you have just length.

    The answer using the tangent function should be very close to the answer using the sine function. You are right, that using the tangent function will give the more accurate answer. But it is harder to solve the equation with the tangent function and using the sine function approximation for this problem still gives a very accurate answer.
     
  4. Oct 26, 2015 #3
    whoops, sorry, the length of d is 0.0121 m after solving for it.
     
  5. Oct 26, 2015 #4

    TSny

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    Using the tangent, the answer for d will be .017 m to two significant figures. This is the same as using the sine function to two significant figures. In fact, using the tangent will give the same answer as using the sine to 5 significant figures (for this particular problem).

    How did you express the tangent of the angle in terms of d and L?
     
    Last edited: Oct 26, 2015
  6. Oct 26, 2015 #5
    my math was off. sorry, I will look at it over.
     
  7. Oct 26, 2015 #6
    wow, the math is much longer
    I'm getting :

    2.25d3 - (d5/4) = 1.14 x 10-11
     
  8. Oct 26, 2015 #7

    TSny

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    If I use the tangent function and make no approximations, I get an equation involving d6 and d2.

    How did you express the tanθ in terms of d and L?
     
  9. Oct 26, 2015 #8

    tanθ = (d/2) / sqrt (L2 - (d/2)2)
     
  10. Oct 26, 2015 #9

    TSny

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    OK. That looks good. Perhaps you then used an approximation for the square root part?

    But, if you don't want to make any approximations, then use your expression for tanθ in your equation

    tanθ = kq2/(mgd2)

    Square both sides to get rid of the square root.
     
  11. Oct 26, 2015 #10
    I try to avoid using approximations because I want the most accurate answer. But as you can see, the math does get a big long.
     
  12. Oct 26, 2015 #11

    TSny

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    Yes, it does get messy. Of course, that's why it is suggested to make the approximation of replacing the tangent by the sine.

    If you let x = d2, you can reduce the equation to the form x3 = bx +c for certain constants b and c. This has the form of a "depressed" cubic equation and the solution can be looked up.

    For example, see equations (2) and (4) here: https://sites.oxy.edu/ron/math/312/14/projects/Fernandez-Gosselin.pdf

    Or you can solve the cubic numerically to whatever accuracy you need.
     
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