Two charges hanging on strings

[catch22]

Homework Statement

Homework Equations

The Attempt at a Solution

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?

 

[TSny]

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

d^3 = 2k q1 q2 = 0.0121 m.

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.

 

[catch22]

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.

whoops, sorry, the length of d is 0.0121 m after solving for it.

 

[TSny]

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?

 

[catch22]

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.

How did you express the tangent of the angle in terms of d and L?

my math was off. sorry, I will look at it over.

 

[catch22]

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?

wow, the math is much longer
I'm getting :

2.25d³ - (d⁵/4) = 1.14 x 10^-11

 

[TSny]

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?

 

[catch22]

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?

tanθ = (d/2) / sqrt (L² - (d/2)²)

 

[TSny]

tanθ = (d/2) / sqrt (L² - (d/2)²)

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.

 

[catch22]

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.

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.

 

[TSny]

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.