What is the Optimal Traction Power for Unrolling Toilet Paper?

[2013]

Homework Statement

Hanging rolls of toilet paper
The one they depend on so that the loose end hanging down the front, but many of the other way around. It raises the question, in which the orders are less force must be applied to roll.
For this purpose, a simplified model are considered, in which the clones
mass m paper roll regarded as a full cylinder with a radius R, which is attached to a rod of length L on the wall. The roller is free to rotate on its axis cylinder. The maximum coefficient of static friction between the paper and the wall is denoted by μ.

Determine the force with which you have to pull the two Aufhängearten at least on paper straight down, so that the paper unrolls. Allows you to choose which suspension is therefore better.


2. The attempt at a solution

- better is that the loose end hang down the front

I need:
- Weight force
- Friction force
- Normal force (pressure force)

How can I calculate the traction power? Which formula is it?

thanks in advance

 

[barryj]

The problem is confusing. an you show a diagram of what you are saying.

 

[2013]

I should say which possibility is better and calculate the tracing power for each.

Could you please help me?

 

[barryj]

First, I don't think you are asking for power as there is no indication of how fast you are pulling the paper. So, I think what you are really asking is the force required to pull the paper.

Think about this. When you pull the paper, where is the retarding force being generated? You will have to use the radius, length of rod and mass of the paper to find the required forces. Give it a try.

 

[2013]

Yes, you are right, I search the force.
Is this the formula?
F=m*g*sin(a)-(μ*Fn)

Can you please help me?
How can I calculate the force for the one possibility and the other possibility?

 

[barryj]

I think you have the right idea. Do diagram B, the one on the right, first and let's see what you get.

 

[2013]

How should I do that?
I have no idea.

 

[barryj]

Well, The force to pull the paper, F, must only overcome friction between the paper and the wall, right? So how do you calculate the friction. It's uFn. So calculate Fn based on the mass of the roll and the angles that you will have to calculate or find a relation for.

 

[2013]

Fn=m*g*cos(a)
F = F(friction)

and then?

 

[barryj]

You might be better using moments. What you are doing is OK for diagram B but I think using moments will be better for diagram A. The CW moment about the point where the rod connects to the wall due to the mass is mgR. What is the CCW moment due to Fn where the roll touches the wall.

 

[haruspex]

For each case, list all the forces, assign symbols to them, and identify the direction and line of action. (Or just post a diagram showing them.) There should be five.
Assuming no acceleration, resolving vertically and horizontally gives you two equations, taking moments gives a third. At limiting friction, you know the ratio of two of them. That should give you enough to express the strength of pull in terms of the mass of the roll and the dimensions.

 

[2013]

I have drawn the forces, but how do I get to the equations?

 

[haruspex]

I have drawn the forces, but how do I get to the equations?

OK. You've labelled two forces Fz, so let's disambiguate that by labelling the force from the rod Fa instead.
In terms of the given radius and rod length, can you resolve Fa into its vertical and horizontal components?
When you've done that, you can write down the two statics equations representing ƩFx = 0 and ƩFy = 0.
Next, you need to choose a point to take moments about. In principle, it doesn't matter which point you choose. Pick one that several forces pass through (so have no moment).

 

[2013]

m*g*sin(a)-(μ*Fn)=0
m*g*cos(a)=0

Is this right?
I haven`t got any other idea.
Sorry, but I haven`t understand what you mean and what I should do.
Can you show me what you have done?

 

[haruspex]

m*g*sin(a)-(μ*Fn)=0
m*g*cos(a)=0

Is this right?

Not even close, I'm afraid.
Let's start with what I'm calling Fa, the force along the rod.
The rod has length L and the cylinder has radius R. There is a right angled triangle, the rod being the hypotenuse and R the length of the horizontal side.
If the angle between those is θ, what is the horizontal component of Fa?
How can you write that using Fa, R and L instead of Fa and θ?

 

[2013]

Sorry that I am so bad.

Fa=R/L

Do you mean this? Is this right?
How must I get on?
How can I calculate the different forces for the two different systems?

 

[haruspex]

Fa=R/L

No.
If there is a force F acting at angle θ above the horizontal, what are the horizontal and vertical components of that force? Hint: they are each F times something, and each something involves a trigonometric function of θ.

 

[2013]

Fa*r=Fr*r

Sorry no other ideas.
Can your draw your idea in a diagram?

 

[barryj]

Maybe I can help. Let's do the simple case first, B , where the paper is next to the wall , understand it, and then move to the more complicated case A where the paper is off the wall. For this case, B, there is not really any reason to do trig here. Just consider moments. So where should you take the moment? where the rod connects to the wall. There is a CW moment due to the mass, and a CCW moment due to Fn. See my post #10. Once you have Fn you can easily find the friction, uFn. Once you have a grasp on this problem, move on to the A case.

 

[barryj]

Here is a diagram for the simple case, B. Remember that for static condition, the clockwise moments must be equal to the counter clockwise moments. You have to select the point to calculate the moments as previously stated but think about the problem first, before you start plugging into formulas. In this diagram, ask yourself where does the system naturally pivot? Take moments about this point. Certainly, you can use angles, but you can also just do Moment = Force X moment arm.. If you want to use trig, it is OK but you will get the same answer.;

 

[2013]

thank you so much for your help,

But what I have to do now?
How can I calculate the force?

 

[barryj]

Look at the diagram. Let the pivot point be where the rod connects to the wall. What is the moment of the mass of the roll of paper. Generally, the moment is.. M = F X D . This assumes that the force and moment arm are perpendicular. This is why I extended the dotted line from the pivot point horizontally. This will be the cw moment. Now, ask yourself where is the force that opposes this moment. It has to be the Fn force. You will have to calculate the distance between the pivot point and where Fn is applied using the Pythagorean theorem. Now equate the two moments and solve for
Fn. Now, realize that the friction is between the paper and the wall. There is no friction to worry about between the paper and the roll. So, how do you calculate the friction force? It is u X Fn . Now you have the force required to pull the paper.

 

[2013]

Sorry I simply can not get to the solution.

I am simply overwhelmed.

How is the force?

 

[barryj]

Lets try a different approach. Remember, we are only discussing diagram B, the simple one. Look at my diagram. The force to pull the paper is uFn. So, you have to find Fn. I have shown the free body diagram. There are three forces acting on the roll of paper. There is mg acting downward. There is the tension, T, in the rod, and there is Fn that is pushing the roll from the wall. Knowing mg, you can find the necessary vertical force of T, and the T. Knowing T you can find Fn. This approach uses the angle theta shown on the diagram.

 

[2013]

Is Fz then:

Fz=T*cos( θ)?

 

[barryj]

2013, I am not sure what else I can do other than completely work the problem for you, which I have mostly done, and that is a no-no on this forum. I have tried to show two different approaches to solving this simple problem with the paper next to the wall. The other problem is more complicated and I feel if you don't understand this one, you will definitely not understand the other. I don't think you have a good grasp of forces and moments. If you are taking a class in physics, you might ask ypour teacher for some one on one tutoring to get a better understanding here. Good luck.

 

[haruspex]

There are three forces acting on the roll of paper.

No, as I stated earlier, there are five. You've left out friction and the user pulling on the paper.

Is Fz then: Fz=T*cos( θ)?

In barryj's three-force diagram, yes. You now correctly have that the vertical component of T (what I was calling Fa) is T cos(θ), where θ is the angle the rod makes to the vertical. Next steps:
- what equation can you write relating θ, L and R?
- what is the horizontal component of T?

 

[barryj]

No, in the right diagram, the friction is between the paper and the wall and doesn't apply to the roll.

 

[haruspex]

No, in the right diagram, the friction is between the paper and the wall and doesn't apply to the roll.

Whether you consider it as applying to the descending end of the paper or to the roll makes no difference. Either way it plays its part in the force with which the user will need to pull.

 

[2013]

Maybe it is the best way, that you also draw your solution.
That would be very nice haruspex, if you do it for me because I don´t understand anything where the problem is between your both argumentation's.
I think this would be the easiest way for me that I can continue my work.

 

[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.