I'm making an Arithmetic Error, Electrostatic force diagrams

[whitejac]

Homework Statement

Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 2.7 cm . Two of the particles have a negative charge: q1 = -6.0 nC and q2 = -12.0 nC. The remaining particle has a positive charge, q3= 8.0 nC . What is the net electric force acting on particle 3 due to particle 1 and particle 2?

Homework Equations

Fcos(θ) = Fx
Fsin(θ) = Fy
θ = Tanh(Fy/Fx)
F = (Fx^2 + Fy^2)^1/2
F(2on3) = K(q1q2)/r^2)

The Attempt at a Solution

Using the program's help function, I have found the components
(F1on3)x = 2.96x10^-4
(F1on3)y = 5.13x10^-4

So the only thing left was to find F(2on3) in the same way I found F(1on3).
Arranging the problem so that q3 was at the origin, q1 was at the top of the triangle and q2 was along the x-axis, I found this:

Fy = 0 (sin0=0)
Fx = (9x10^9)(12x10^-9)(8x10^-9) / (2.7x10^-2)^2
= 1.19x10^-3

This makes sense given that it should attract.

Here's where the problem falls off of the rails...

I need to find Theta

So Tanh (5.13x10^-4)/((2.96x10^-4)+(1.19x10^-3)) = 0.006

I really need this to be in the vicinity of 19degrees. I do not understand what mistake I am making... Does anybody else?

 

[SteamKing]

Homework Statement

Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 2.7 cm . Two of the particles have a negative charge: q1 = -6.0 nC and q2 = -12.0 nC. The remaining particle has a positive charge, q3= 8.0 nC . What is the net electric force acting on particle 3 due to particle 1 and particle 2?

Homework Equations

Fcos(θ) = Fx
Fsin(θ) = Fy
θ = Tanh(Fy/Fx)
F = (Fx^2 + Fy^2)^1/2
F(2on3) = K(q1q2)/r^2)

The Attempt at a Solution

Using the program's help function, I have found the components
(F1on3)x = 2.96x10^-4
(F1on3)y = 5.13x10^-4

So the only thing left was to find F(2on3) in the same way I found F(1on3).
Arranging the problem so that q3 was at the origin, q1 was at the top of the triangle and q2 was along the x-axis, I found this:

Fy = 0 (sin0=0)
Fx = (9x10^9)(12x10^-9)(8x10^-9) / (2.7x10^-2)^2
= 1.19x10^-3

This makes sense given that it should attract.

Here's where the problem falls off of the rails...

I need to find Theta

So Tanh (5.13x10^-4)/((2.96x10^-4)+(1.19x10^-3)) = 0.006

I really need this to be in the vicinity of 19degrees. I do not understand what mistake I am making... Does anybody else?

You should be careful here with your notation. Tan (θ) is not the same function as tanh (θ).

https://en.wikipedia.org/wiki/Hyperbolic_function

If θ = tan^-1 ((5.13x10^-4)/((2.96x10^-4)+(1.19x10^-3)) is what you are trying to calculate ...

 

[whitejac]

Really? I thought i'd looked it up and found it was the same, math-wise, just different in definition. My mistake I guess...
Scientific calculators only offer the tanh function though. Is that because they expect you to simply find Tan(Θ) and then find 1/x?

 

[SteamKing]

Really? I thought i'd looked it up and found it was the same, math-wise, just different in definition. My mistake I guess...
Scientific calculators only offer the tanh function though. Is that because they expect you to simply find Tan(Θ) and then find 1/x?

I'm confused here.

The standard trigonometric functions are abbreviated as sin, cos, tan, cot, sec, and csc. cot = 1/tan; sec = 1/cos; and csc = 1/sin.

The inverse trig functions are sin^-1, cos^-1, etc.

Note that sin^-1 ≠ 1 / sin. If y = sin (θ), then sin^-1(y) = θ; -1 ≤ y ≤ 1

The hyperbolic functions sinh, cosh, tanh, coth, sech, and csch are defined:

​

 

[whitejac]

Oh wow, yes I can see now how the hyperbolic functions are so much not the same thing...
Okay, I need to find tan^-1(Θ) in order to solve this problem and get me the angle between the force components? Sounds pretty straightforward now. I was making a mistake with my calculator because I didn't distinguish the notations.