What is the Optimal Traction Power for Unrolling Toilet Paper?

[2013]

Fg=Fn/tan(θ)
Fz=Fr

I can not enter the vertical components of Fz and Fr.
What triangle3 should I draw to get to them?

 

[2013]

Maybe:
Fz=Fn*μ and Fr=Fn*μ How should I go on?

 

[haruspex]

Fg=Fn/tan(θ)
Fz=Fr

You're still not doing what I ask. Don't write any equations yet - you're not ready.
Here's an example of what I want you to do:

The force Fn acts horizontally. Therefore it has a vertical component 0 and a horizontal component Fn. The horizontal component acts away from the wall.

Try to write the corresponding statements for the other four forces: Fz, Fr, Fg, and T. The last one, T, will involve theta.

 

[2013]

------------------vertical component -------------------- horizontal component

Fn ---------------------------- 0 ---------------------------- Fn
Fz ---------------------------- Fz ---------------------------- 0
Fr ---------------------------- Fr ---------------------------- 0
Fg ---------------------------- Fg ---------------------------- 0
T ---------------------------- Fg/cos(θ) --------------------- Fn/sin(θ)
Is this right? Do you mean that? And yet?

 

[haruspex]

------------------vertical component -------------------- horizontal component

Fn ---------------------------- 0 ---------------------------- Fn
Fz ---------------------------- Fz ---------------------------- 0
Fr ---------------------------- Fr ---------------------------- 0
Fg ---------------------------- Fg ---------------------------- 0

All the above are correct

T ---------------------------- Fg/cos(θ) --------------------- Fn/sin(θ)

No. The answer should only involve T and θ.

 

[2013]

T ---------------------------- T*cos(θ) --------------------- T*sin(θ)

And then?

 

[haruspex]

T ---------------------------- T*cos(θ) --------------------- T*sin(θ)

And then?

Great, we're getting somewhere.
Now take all the horizontal components you've listed and add them up, paying attention to the signs. Since it's static, they'll add up to 0, right?
Then do the same for the vertical components.

 

[2013]

horizontal components:
Fn - T*sin(θ)=0

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

 

[haruspex]

horizontal components:
Fn - T*sin(θ)=0

Yes.

vertical components:
Fz+Fr+Fg-T*cos(θ)=Fz+Fr

No, it makes no sense to put the same force on both sides.

You have to decide how you're going to define the positive directions. There are several options, which makes it confusing.

Option 1: Up is always positive.
With this definition, you just write ƩF = 0 (since there's no acceleration). So here we have Fz+Fr+Fg+T*cos(θ)=0. It will, of course, then turn out that Fg and Fz are negative in value.

Option 2: Down is always positive. This is very little different from option 1.

Option 3: You know (or think you know) which direction each force will act, and define that to be the positive direction for that force. So here we would have Fz+Fg = Fr+T*cos(θ).

You choose. As long as you're consistent, each option will produce the right answer.

Next, we need another equation. Take moments about some point.

 

[2013]

M1=Fr*R
M2=Fz*R
M1=M2
Fz=Fr

Is this right?

 

[haruspex]

M1=Fr*R
M2=Fz*R
M1=M2
Fz=Fr

Is this right?

It could be, depending on which option you are adopting for the signs on the forces. Please choose one.
Also I'm not quite sure which of the two cases we're discussing here. In my previous post I was assuming the hanging paper is next to the wall.

 

[2013]

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.

 

[haruspex]

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.

OK.
Fz = Fr is correct with that option, and Fn = T*sin(θ).
But Fz+Fg = Fr+T*cos(θ) only applies to one of the two cases. Can you tell me which case it applies to (hang next to wall or hang away from wall), and what the equation is for the other case?

 

[2013]

It applies to when the paper hang next to the wall.

hang away from wall:
?

 

[haruspex]

It applies to when the paper hang next to the wall.

hang away from wall:
?

Consider which way Fr acts when the paper hangs away from the wall.

 

[2013]

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

 

[haruspex]

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

Yes, so how does that change the equation for vertical forces?

 

[2013]

Fz+Fg+Fr = T*cos(θ)

?

 

[haruspex]

Fz+Fg+Fr = T*cos(θ)

?

Yes! Can you answer the original problem now?

 

[2013]

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 different.
I must not calculate the force acting when the paper is not on the wall, on the wall by the pulling force?

 

[haruspex]

It is easier to pull when the paper hang near to the wall.

because by the first equation:
Fz+Fg=Fr+T*cos(θ)

Yes.

we can cut out Fr and Fz:
Fg=T*cos(θ)

True, but it's the two values of Fz that you need to compare.

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 different.
I must not calculate the force acting when the paper is not on the wall, on the wall by the pulling force?

In the OP you wrote

Determine the force with which you have to pull the two Aufhängearten

which suggests you are actually supposed to determine the two values of Fz (as functions of the given variables). But since this is a translation it's hard for me to be sure.

 

[2013]

Yes, you've understood the task.

True, but it's the two values of Fz that you need to compare.

How can I compare the two different equations?
I'm finished or have I to do something?

 

[haruspex]

How can I compare the two different equations?

You have
(1) Fn = T*sin(θ)
(2) Fr = Fn*μ
(3) Fz = Fr
(4) sin(θ) = R/L
When hanging paper next to wall:
(5A) Fz+Fg = Fr+T*cos(θ)
When hanging paper away from wall:
(5B) Fz+Fg+Fr = T*cos(θ)
The given data are Fg, L and R, so for each case you want an equation involving only those and Fz. Try using (1)-(4) and (5A) to get such an equation.

 

[2013]

thank you so much for your help!