How do I proceed with two different trig functions containing x on the left?

[DerbisEternal]

Homework Statement

The problem is this: two 10 g balls hang from 1 m strings attached to the same point on the ceiling. They are both charged with 100 nC. What angle are they separated to?

Relevant Equations

Fe=(k*q1*q2)/(r^2)

Given the total angles in the x direction, I set up this:
(mg/cos(x))*sin(x)-Fe=0
then isolated for x:

mgtan(x)=(kq^2)/(2*sin^2x)

sin^2(x)*tan(x)=(kq^2)/(2mg)

From here I am stuck. How do I go forward when x is contained in two different trig functions on the left?

 

[jedishrfu]

you should start by drawing a force diagram and using a variable not x for the angle.

wrt to the trig functions sin^2 is the same as (1-cos^2) and tan is sin/cos so you could change it to a sin cos term and from there find other Trig identities To reduce it further.

 

[haruspex]

Homework Statement:: The problem is this: two 10 g balls hang from 1 m strings attached to the same point on the ceiling. They are both charged with 100 nC. What angle are they separated to?
Relevant Equations:: Fe=(k*q1*q2)/(r^2)

Given the total angles in the x direction, I set up this:
(mg/cos(x))*sin(x)-Fe=0
then isolated for x:

mgtan(x)=(kq^2)/(2*sin^2x)

sin^2(x)*tan(x)=(kq^2)/(2mg)

From here I am stuck. How do I go forward when x is contained in two different trig functions on the left?

You can express all the trig in terms of tan, but it will give you a nontrivial cubic. So you could try a small angle approximation for both sin and cos, giving a trivial cubic instead. Having solved it, you can figure out whether the angle is small enough to justify the approximation.

Btw - magnetic??

 

[Delta2]

Homework Statement:: The problem is this: two 10 g balls hang from 1 m strings attached to the same point on the ceiling. They are both charged with 100 nC. What angle are they separated to?
Relevant Equations:: Fe=(k*q1*q2)/(r^2)

mgtan(x)=(kq^2)/(2*sin^2x)

I believe you have a small mistake here, it should be $$mg\tan x=k\frac{q^2}{4\sin^2 x}$$ as I think the distance between the two balls is $2\sin x$ if I have done the figure correctly myself. $x$ here is half the angle between the two strings.

 

[Delta2]

That trigonometric equation might be hard to solve exactly (unless you use the small angle approximation as @haruspex suggested). I plugged this trigonometry equation at wolfram and I get very long and complex solutions. A workaround for this is the follow:

Using similarity of triangles (sorry i should have posted a figure) I seem to get the purely algebraic equation
$$k\frac{q^2}{4mgy^2}=\frac{y}{\sqrt{1-y^2}}$$
where $y$ is half the distance between the two balls. (it is $y=\sin x$ to connect it with the previous post).

Letting $a=k\frac{q^2}{4mg}$ and after some algebra we seem to get $$y^6+a^2y^2-a^2=0$$ which can be reduced to a regular cubic equation using the substitution $z=y^2$ which can then be solved exactly using typical formulas. After solving for z , you get $y=\pm\sqrt{z}$ and $x=\arcsin y$.

 

[haruspex]

That trigonometric equation might be hard to solve exactly (unless you use the small angle approximation as @haruspex suggested). I plugged this trigonometry equation at wolfram and I get very long and complex solutions. A workaround for this is the follow:

Using similarity of triangles (sorry i should have posted a figure) I seem to get the purely algebraic equation
$$k\frac{q^2}{4mgy^2}=\frac{y}{\sqrt{1-y^2}}$$
where $y$ is half the distance between the two balls. (it is $y=\sin x$ to connect it with the previous post).

Letting $a=k\frac{q^2}{4mg}$ and after some algebra we seem to get $$y^6+a^2y^2-a^2=0$$ which can be reduced to a regular cubic equation using the substitution $z=y^2$ which can then be solved exactly using typical formulas. After solving for z , you get $y=\pm\sqrt{z}$ and $x=\arcsin y$.

I got the cubic from the equation in post #1 by writing t for tan(x) and using $\sin^2(x)=\frac {\sin^2(x)}{\cos^2(x)+\sin^2(x)}=\frac{t^2}{1+t^2}$.

 

[Delta2]

I got the cubic from the equation in post #1 by writing t for tan(x) and using $\sin^2(x)=\frac {\sin^2(x)}{\cos^2(x)+\sin^2(x)}=\frac{t^2}{1+t^2}$.

I see, interesting, probably the two equations are equivalent. Do you agree on my post#4?

 

[haruspex]

probably the two equations are equivalent

Almost certainly.

 

[Delta2]

Almost certainly.

Well, they turn out to be equivalent if we consider positive values of tangent and sine. Because in my equations it is $y=\sin x$ while in yours, $y^2=\sin^2x=\frac{t^2}{1+t^2}$ which if we solve for $t$ we get for the positive value of t $t=\frac{y}{\sqrt{1-y^2}}$ and then your equation becomes $$t\sin^2x=\alpha\Rightarrow \frac{y}{\sqrt{1-y^2}}y^2=\alpha$$ which is my equation :D