GR9277 #83



Alternate Solutions 
cjohnson415 20130819 01:40:11  Here's an alternative solution that skips the trig altogether:
Due to small , we can consider each as a simple pendulum, a special case of the simple harmonic oscillator:
Thus , which is the restoring force constant (), so:
The rest is the same algebra, just a different/potentially quicker/potentially prettier way to arrive at the force balance.   neo55378008 20120818 12:27:15  Dimensional analysis helps!
Since is in N, will have units of
Mg has units of N
so will have units of
This leaves only A and B. Knowing that solving for will give half of d, take the solution with an extra 2 in it! (A)   redmomatt 20111106 08:25:21  Heres another take without the trig functions...
Since is small, we can remove all of the trig functions. Thus, we just need to equate the outward Coulomb force to the force of gravity along the horizontal .
, since .
Hence,
, which is (B) .   Andresito 20060329 14:56:46  It is a bit faster to equate F(electric) = T*Sin(theta), m*g = T*Cos(theta) and divide them. Tan(theta) ~ theta  

Comments 
astropolo 20140729 11:55:16  I have a quick question about the small angle approximations. Why can you approximate that Cos~1 while you use the geometry for Sin rather than saying sin~theta?
JoshWasHere 20140820 08:20:22 
When you expand cosine and sine, you see that the first two terms in the cosine and sine expansion are cos = 1  and sin =  . Because of the small angle, we select only the first order terms, and drop all higher order terms (with the condition that < 5 degrees).

  cjohnson415 20130819 01:40:11  Here's an alternative solution that skips the trig altogether:
Due to small , we can consider each as a simple pendulum, a special case of the simple harmonic oscillator:
Thus , which is the restoring force constant (), so:
The rest is the same algebra, just a different/potentially quicker/potentially prettier way to arrive at the force balance.   greatspirits 20121102 15:49:28  Energy conservation anyone? The change in gravitational potential energy should equate to potential between the charges. So... should equate to... hmmm well I usually prefer the energy method but the force way seems to be more effecient here.   neo55378008 20120818 12:27:15  Dimensional analysis helps!
Since is in N, will have units of
Mg has units of N
so will have units of
This leaves only A and B. Knowing that solving for will give half of d, take the solution with an extra 2 in it! (A)
henryb 20160912 18:14:40 
thx a lot. this is the simplest solution

  redmomatt 20111106 08:25:21  Heres another take without the trig functions...
Since is small, we can remove all of the trig functions. Thus, we just need to equate the outward Coulomb force to the force of gravity along the horizontal .
, since .
Hence,
, which is (B) .   Dodobird 20101112 11:57:31  tan(theta) = d/2L
tan(theta) = kq*q/d*d*MG
d/2L = kq*q/d*d*MG
gives answer A.   FortranMan 20081003 14:02:49  Also note because theta is so small here, we can approximate the vector component of the gravity that's NOT inducing tension to be d/2.   JAS 20071101 09:20:45  It seems easier to equate torques around the pivot point, where the torque on the right mass due to the repulsive electric force isrnrn counterclockwisernrnand the torque on the right mass due to its weight is:rnrn clockwisernrn and rnrnFrom this, choice A follows.   relain 20071023 19:21:51  typo in the question, option D should be a square not a cube root. It's currently identical to option B.   Andresito 20060329 14:56:46  It is a bit faster to equate F(electric) = T*Sin(theta), m*g = T*Cos(theta) and divide them. Tan(theta) ~ theta  

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