How did they divide the equations ?

[aleksbooker]

How did they "divide the equations"?

Homework Statement

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.

Homework Equations

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

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}

 

[Doc Al]

Divide this equation:

T\sin{\theta}=\frac{Kq^2}{4sin^2{\theta} }

By this one:

T\cos{\theta}=mg

Divide the left sides by each other and the right sides by each other.

 

[tiny-tim]

hi aleksbooker!

T_x=T\sin{\theta}=F_{2on1}=\frac{Kq^2}{4sin^2{\theta} }
T_y=T\cos{\theta}=mg
…
What I don't know is how they "divided the two equations"

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

(hmm … there seems to be an L² missing )

 

[Doc Al]

(hmm … there seems to be an L² missing )

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

 

[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).

 

[aleksbooker]

Yea, L was the length of the string.

 

[Doc Al]

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).

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