What is the Optimal Traction Power for Unrolling Toilet Paper?

[haruspex]

Maybe it is the best way, that you also draw your solution.

You already have two perfectly good diagrams you drew earlier. We just have to obtain the equations that come from them.
Please try to answer my questions in Post #27.

 

[2013]

T=Fn*(L^2-R^2)
Fn=mgr/(L^2-R^2) ?

What is Fz?

 

[barryj]

Fn = mgR/sqrt(L^2-R^2)

Now that you have Fn, how is the force to pull the paper related to Fn? Something to do with u perhaps?

 

[2013]

What does "sqrt" mean?

 

[barryj]

square root.

 

[2013]

thank you

Is this the complete solution?

 

[haruspex]

T=Fn*(L^2-R^2)
Fn=mgr/(L^2-R^2) ?

I asked

- what equation can you write relating θ, L and R?
- what is the horizontal component of T?

Neither of what you've written above appear to be attempted answers to those questions.

 

[2013]

I am so sorry,but I don´t know what I should do.
This was my task:
Determine the force with which you have to pull the two suspensions at least on paper straight down, so that the paper unrolls. Allows you to choose which suspension is therefore better.

How do I get to the solution?

 

[haruspex]

The centre of the roll, the point where the roll touches the wall, and the point where the rod is attached to the wall form a triangle, yes? The angle at the top is theta. What variable gives the length of the base? What is the length of the the longest side? What is the angle opposite the longest side?

 

[2013]

Is this right?
How should I go on?

 

[barryj]

You did not show the force due to gravity on the roll. There should be a force, mg, that points down from the center of the roll. Now you have three forces, T acting up and to the left, Fn acting to the right, and mg acting down. Now, find Fn based on mg and T. Then once you have Fn, you can fine the force that is required to pull the paper because friction is uFn and it acts on the paper, not the roll.

 

[2013]

Is it right?

 

[haruspex]

Is this right?

[In response to post #40, which I'm assuming is a reply to post #39]
No, my question was just about the geometry. I want you to forget about forces for the moment and get the geometric equation right. The diagram itself is fine, but put in distances, not forces. I'm looking for the equation that relates θ, L and R.

 

[2013]

And yet?

 

[haruspex]

And yet?

If those (T, Fn, Fg) were the only forces then your first equation would be right, but the second is wrong. But please, please, stop trying to jump ahead. Take one step at a time, as I keep asking you. It'll be quicker in the long run.
Forget the forces for now: what is the relationship between θ, L and R? I don't want to see any mention of T, Fn etc. in the answer.

 

[barryj]

Maybe it would help to really go back to the basics. I have attached a simple diagram. The question is, what is the value of T and Fn in this diagram?

 

[2013]

When drawn from the outside in addition to the weight force acts, the tensile force.
Same size and in the same direction, the frictional force down (due to the deflection).
So this results in an effective force to bottom:
F = F (g) + 2 F (r), this causes the pushing force F on the wall (E) = F tan (alpha),
from which the friction force F (r) = mu * F (D) results.
so:
mu * (F (g) + 2F (r)) (alpha) = F (r) * tan, switch to F (r), done.Is it right?
Excuse me but I have only two days time to solve the problem, otherwise I fall through the semesters and must repeat the year.
Can you please help me?

 

[haruspex]

When drawn from the outside in addition to the weight force acts, the tensile force.
Same size and in the same direction, the frictional force down (due to the deflection).
So this results in an effective force to bottom:
F = F (g) + 2 F (r), this causes the pushing force F on the wall (E) = F tan (alpha),
from which the friction force F (r) = mu * F (D) results.
so:
mu * (F (g) + 2F (r)) (alpha) = F (r) * tan, switch to F (r), done.


Is it right?
Excuse me but I have only two days time to solve the problem, otherwise I fall through the semesters and must repeat the year.
Can you please help me?

I'm unable to comprehend what you've posted above. Suddenly adopting a new set of symbols for the forces doesn't help. Please stick to:
θ = angle of rod to vertical,
L = rod length,
R = roll radius,
T = tension in the rod,
Fn = normal force from wall,
Fg = gravitational force,
Fz = vertical pull from hand
Fr = frictional force,
μ = coefficient of static friction.
One reason this thread is taking so long is that instead of following guidance provided you repeatedly posted wild guesses at equations. I'll give it one last go. Please post an answer to each of the following questions as far as you can:
1. Forgetting the forces for the moment, what is the relationship between θ, L and R?
2. In the horizontal direction, what are the components of the five forces listed above?
3. What equation does that give you?
4. In the vertical direction, what are the components of the five forces listed above?
5. What equation does that give you?

 

[2013]

My ideas, I´m not sure if they are right:

1. sin(θ) = R/L
L = R/sin(θ)
R = L* sin(θ)

2.+3.
Fn = sin(θ) * L

4.+5.
Fg = L * cos (θ)
Fn = Fg/cos(θ) * sin(θ)

I don`t know what I also should do.
How should I go on?

I have to say, thank you so much for your help and for your patience.

 

[haruspex]

1. sin(θ) = R/L

Hooray!

2.+3.
Fn = sin(θ) * L

I asked for the horizontal components of 5 forces, so there should be 5 items before the equation that combines them. Pls don't skip questions I ask, I ask them for a reason.
What is the horizontal component of Fn? This one is easy: it's Fn, right? Now do the other four. In each case, the answer should be that force multiplied by something; the something could be 0 or 1 or sin(θ) or cos(θ) or tan(θ).
Fg = gravitational force,
Fz = vertical pull from hand
Fr = frictional force.

The equation you posted makes no sense anyway. The LHS is a force, the RHS is a length.

 

[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?