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Calculate traction power

  1. Jun 28, 2013 #1
    1. The problem statement, all variables and given/known data

    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
  2. jcsd
  3. Jun 28, 2013 #2
    The problem is confusing. an you show a diagram of what you are saying.
  4. Jun 28, 2013 #3
    I should say which possibility is better and calculate the tracing power for each.

    Could you please help me?

    Attached Files:

  5. Jun 28, 2013 #4
    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.
  6. Jun 28, 2013 #5
    Yes, you are right, I search the force.
    Is this the formula?

    Can you please help me?
    How can I calculate the force for the one possibility and the other possibility?
  7. Jun 28, 2013 #6
    I think you have the right idea. Do diagram B, the one on the right, first and lets see what you get.
  8. Jun 28, 2013 #7
    How should I do that?
    I have no idea.
  9. Jun 28, 2013 #8
    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.
  10. Jun 28, 2013 #9
    F = F(friction)

    and then?
  11. Jun 28, 2013 #10
    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.
  12. Jun 28, 2013 #11


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    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.
  13. Jun 29, 2013 #12
    I have drawn the forces, but how do I get to the equations?

    Attached Files:

  14. Jun 29, 2013 #13


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    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).
  15. Jun 29, 2013 #14

    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?
  16. Jun 29, 2013 #15


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    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?
  17. Jun 29, 2013 #16
    Sorry that I am so bad.


    Do you mean this? Is this right?
    How must I get on?
    How can I calculate the different forces for the two different systems?
  18. Jun 29, 2013 #17


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    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.
  19. Jun 29, 2013 #18

    Sorry no other ideas.
    Can your draw your idea in a diagram?
  20. Jun 29, 2013 #19
    Maybe I can help. Lets 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.
  21. Jun 29, 2013 #20
    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.;

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