# How did they divide the equations ?

1. Mar 1, 2014

### aleksbooker

How did they "divide the equations"?

1. The problem statement, all variables and given/known data

Two 5.0g point charges on 1.0m-long threads repel each other after being charged to +100nC.
What is angle theta? You can assume theta is a small angle.

2. Relevant equations
$K=9.0*10^9 Nm^2/C^2$
$g=9.8m/s^2$
$q=\frac{K|q_1||q_2|}{d^2}$

3. The attempt at a solution

I determined my unknowns and translated everything into standard units, and drew out the free-body diagram.

Translated everything into standard units.
5.0g = 5.0*10^(-3) kg
100nc = 100*10^(-9) C

$\frac{1}{2}d=\sin{\theta}$
$d=2\sin{\theta}$
$T_x=T\sin{\theta}=F_{2on1}=\frac{Kq^2}{d^2}$
$T_x=T\sin{\theta}=F_{2on1}=\frac{Kq^2}{(2\sin{\theta})^2}$
$T_x=T\sin{\theta}=F_{2on1}=\frac{Kq^2}{4sin^2{\theta} }$
$T_y=T\cos{\theta}=mg$

The "missing" step:
The solutions manual says that the next step from here is to "divide the two equations and solve for q". What I don't know is how they "divided the two equations" to get this:

$\sin^2{\theta}\tan{\theta}=\frac{Kq^2}{4L^2mg}=4.59*10^{-4}$

2. Mar 1, 2014

### Staff: Mentor

Divide this equation:
By this one:
Divide the left sides by each other and the right sides by each other.

3. Mar 1, 2014

### tiny-tim

hi aleksbooker!
you divide Tsinθ by Tcosθ, to give you Ttanθ = … ?

(hmm … there seems to be an L2 missing )

4. Mar 1, 2014

### Staff: Mentor

Apparently, L is the length of the string, which is given as 1 m.

5. Mar 1, 2014

### aleksbooker

Mind blown. Thanks @Doc Al. Okay, that works - I'm assuming it only makes to do something like this when you can use it to eliminate a variable (like T in this problem).

6. Mar 1, 2014

### aleksbooker

Yea, L was the length of the string.

7. Mar 1, 2014

### Staff: Mentor

Think of it as just another tool in your bag of tricks. A useful one.