Problem about the lever on a swing-top beer bottle. (Classical Mechanics)

Question

I'm faced with the following problem as stated in the image:

Once I find the geometry part, where I find what are the angles and the lengths and so on, to solve the forces I use 2 equations. The first ones is the equilibrium of forces in the y axis, and the second one is the equilibrium of momenta from the top point on the cork. Now the important thing I use which is the one I doubt is that I impose that the force on the bottom left point is $textit{perpendicular}$ to the $L_1$ arm, and therefore parallel to the $L_2$ one. I had to do this since the equilibrium of momenta and forces in the 2 axis gives me 3 equations, and If I had left general horizontal and vertical forces in the two left points I'd have 4 unknowns, so in order to relate two of them I used that. I put down here what I did and what I obtained

Lastly, sorry about uploading it like this but this I wrote initially in a table, and thanks in advance for any insight about what I may have missed or do wrong.

Eli · Accepted Answer

Take the sum of the torques about point A , you obtain:
$$vec tau=vec r_Ftimes vec F+vec r_Htimes vec F_H=vec 0$$
with :
$$vec r_F=begin{bmatrix}
  sqrt{L_1^2+L_2^2} 
   0
   0
end{bmatrix}$$
$$vec F=Fbegin{bmatrix}
   0
   1
   0
end{bmatrix}$$
$$vec r_H=L_1begin{bmatrix}
   cos(varphi)
   -sin(varphi)
   0
end{bmatrix}$$
$$vec F_H=F_Hbegin{bmatrix}
   0
   1
   0
end{bmatrix}$$
and
$$varphi=arctanleft(frac {L_2}{L_1}right)$$
$Rightarrow$
$$vec tau=left[ begin {array}{c} 0 0 
sqrt {{L_{{1}}}^{2}+{L_{{2}}}^{2}}F+L_{{1}}cos left( varphi 
 right) F_{{H}}end {array} right] 
=vec 0$$
$$|F_H|=F,frac{sqrt{L_1^2+L_2^2}}{L_1,cos(varphi)}$$
thus :
$$F_1=|F_H|,cos(psi)$$
with
$$tan(psi)=frac {x}{y}=frac{L_1,cos(varphi)}{L_3-L_1,sin(varphi)}$$

Problem about the lever on a swing-top beer bottle. (Classical Mechanics)

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