Homework Help: Electrostatics balancing gravitational, coulomb forces and tension

1. Oct 6, 2012

FatPhysicsBoy

1. The problem statement, all variables and given/known data
Hi, I'm on the first part of this question and I've got to the end but I have an extra factor of two which I can't seem to explain! Any insight would be much appreciated, thank you!

2. Relevant equations

3. The attempt at a solution

I resolved the forces around one ball and came up with the following:

Resolving vertically: mg = Tcosθ, Resolving horizontally: Tsinθ = F2,1

Then using pythagoras I found sinθ = r/2l and cosθ = √(4l2 - r2)

Putting it all together I get: 2F2,1 = mgr/√(4l2 - r2)

I just don't understand the extra factor of 2. I thought it might be because it's at half the separation but this isn't correct. Any help would be much appreciated. Thank you!

2. Oct 6, 2012

BruceW

cosθ = √(4l2 - r2)
This bit isn't quite right. It looks like you are trying to skip ahead to the next step before you have written things down properly. Maybe if you go through it again, writing down each step then you will get the right answer.

3. Oct 6, 2012

FatPhysicsBoy

I don't understand?

I split the "triangle" up with hypotenuse l and adjacent r/2 and then used pythagoras to get:

a2 = l2 - (r2/4)

a = √(l2 - (r2/4)

Then I just multiplied both terms by 4 under the square root sign to get a = √(4l2 - (r2)

4. Oct 6, 2012

vela

Staff Emeritus
This like saying you have $x=\sqrt{1/4}$, and you multiplied by 4 under the square root sign to get $x=\sqrt{1}=1$.

Your expression for the cosine can't be right because it has units of length. The result of taking the cosine of an angle should be unitless.

5. Oct 6, 2012

FatPhysicsBoy

Thank you for your help guys, I've figured it out now!

so cosθ= [√(l2-(r2/4))]/l

Then fast forward to 2F= mgr/√(l2-(r2/4)) take the 2 across and rewrite as √4 then push into the √(l2-(r2/4)) term to be left with mgr/√(4l2-(r2))

Thank You!!