# Soap bubbles

**1. The problem statement, all variables and given/known data**

We have got two bubbles with r_{1} and r_{2}at p_{0}=1bar and T_{0}=20°C, made of two different soapy waters. We pierce them with a straw with length l=5cm and r_{0}=2mm. There are 2g of air inside the system. (Surface tensions are γ_{1}= 0,025 N/m^{2}, γ_{2}= 0,01 N/m^{2})

a) What are the stationary states and determine the shape of the membranes on both sides.

b) What are the radiuses before we pierce the system with a straw if the bubbles fall with the same velocity? (Use linear law for resistance)

a) So I assume that the stationary states are when the bubbles have the same radius, otherwise the other stationary state is when one is full and the other is empty (just a membrane on the straw). This is because the system wants lower energies (E=γ_{1}*S_{1}+γ_{2}*S_{1})

Is there a formula to determine the right shape or is enough if I say that on one side is a bubble on the other is half a sphere? I tried to find shape with formulas F=2γ*(2∏r)*cosθ=p_{inside}*S but I have problems, because I dont know the angle and the pressure inside.

b) Fg-F_{buoyancy}-F_{resistance}=m*a=m*dv/dt

=mg-ρgV-6∏rηv=m*dv/dt from here on I don’t have a clue how to cope with it.

Can anyone help me please, thank you in advance. 🙂

**1. The problem statement, all variables and given/known data**

**2. Relevant equations**

**3. The attempt at a solution**

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