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thank you so much for your help!

Yes, you've understood the task. How can I compare the two different equations? I'm finished or have I to do something?

It is easier to pull when the paper hang near to the wall. because by the first equation: Fz+Fg=Fr+T*cos(θ) we can cut out Fr and Fz: Fg=T*cos(θ)Is this the solution for my task? I asked myself where ever the difference is between the two processes and it is just the point of action, is...

Fz+Fg+Fr = T*cos(θ) ?

Fr acts the same way like Fz. Both acts down.

It applies to when the paper hang next to the wall. hang away from wall: ?

I would like to choose option 3. I don´t know which case we discuss already, but I thought we speak about where the paper is hanging over. But I need both cases, because I should say which is from the physical side better.

M1=Fr*R M2=Fz*R M1=M2 Fz=Fr Is this right?

horizontal components: Fn - T*sin(θ)=0 vertical components: Fz+Fr+Fg-T*cos(θ)=Fz+Fr?

I have to say thank you to you, I have already the solution and it is the same as in the other method. Thank you, Thank you for helping me :) :)

T ---------------------------- T*cos(θ) --------------------- T*sin(θ) And then?

Vc=12V*(C1-C2)/(C1+C2+Cc) Vc*(C1+C2+Cc)=12V*(C1-C2) how can I get C1 and C2 to the other site?

Can you say me at which step, I have done a mistake. ?

Vc = 12V * ((C1-C2)/(C1+C2+Cc)) Vc*Cc=12V * ((C1-C2)/(C1+C2) Vc*(Qc/Vc)=12V * ((C1-C2)/(C1+C2) Qc=12V * ((C1-C2)/(C1+C2)=12V*((2,46*10^-11As/V - 2,01*10^-11As/V) / (2,46*10^-11As/V+2,01*10^11As/V) = 1,21As Cc=(Qc/Vc)=1,21As/0,15V=8,05As/V Is this right? Why is it not the same capacity as in...

12V * C1/(C1+C2+Cc) - 12V * C2/(C1+C2+Cc)) = Vc 12V * (C1/(C1+C2+Cc) - (C2/(C1+C2+Cc)) = Vc Vc = 12V * ((C1-C2)/(C1+C2+Cc)) ?