0
$\begingroup$

I tried to solve the following problem: There are 2 objects . The object m1 with charge q and the object m2 with charge q.(same charge).The object m2 is connected with a rope to the ceiling.

at the initial state , m1 and m2 were on the same height. When the object m2 is placed in distance d from the object m1, the following situation occurs: enter image description here

Please ignore the force of gravity between the masses and ignore any kind of friction. just consider the tension , electrical force and mg.

m2 : ΣFx,y=0

You have to find x.

You can use the parameters k(Coulomb's constant), q , L , d , mg.

I was able to find only 2 equations that would help to solve the problem but one is still missing.

I wasn't able to calculate the components of the electrical force. The equations that I found are :

1) r ^ 2 = (L-Lcos a) ^ 2 + (x+d) ^ 2 (r is the distance between the two objects)

2) sin a = x / L

3) as you can see ΣFx = 0 and ΣFy = 0 (The object m2), but without the components of the electrical force I wansen't able to calculate anything.

Improve this question
$\endgroup$
3
  • $\begingroup$ is $m_1$ fixed? $\endgroup$
    – ramseysdream111
    Commented Jul 14, 2014 at 12:24
  • $\begingroup$ $x=l\sin {\alpha}$ $\endgroup$
    – ramseysdream111
    Commented Jul 14, 2014 at 12:30
  • $\begingroup$ Yes user , m1 is fixed. $\endgroup$
    – Michael
    Commented Jul 14, 2014 at 12:53

2 Answers 2

Reset to default
1
$\begingroup$

I worked the problem out a ways and it involves quite a bit of tedious algebra. The technique I used was to include the electric force in the Fx and Fy equations by looking at the angle that $ \hat{r}$ (the vector between the two charges) makes between the charges. For example, the equation i came up for Fx is $T_x - \frac{Kq^2}{r^2} \cos{\theta} =0$ where $T_x$ is the tension in the x direction and $\theta$ is the angle between the charges relative to the x axis.I then wrote the equation for Fy, used the tension terms to relate the equations and began changing the trig functions in the equation to relations involving only the allowed terms. Good Luck!

Improve this answer
$\endgroup$
1
$\begingroup$

This may be cheating, but I think the problem is easier if you use conservation of energy. If you set the gravitational potential energy reference to the height of m1, then initially you have $$ U_i = k\frac{q^2}{d}. $$ The final energy will have a gravitational potential energy and an electrical potential energy: $$ U_f = mgh + k\frac{q^2}{r} $$ A bit of trig will find the distances (h and r) you need. By equating the initial and the final energies you should get your answer.

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.