bon Posted July 16, 2013 Share Posted July 16, 2013 (edited) hi, can somebody help me My task: Determine the force with which you have to pull the two suspend sorts at least on paper straight down, so that the paper unrolls. Allows you to choose which suspension is therefore easier. My approach: sin(∂)=Fn/T cos(∂)=Fg/T Fn = Fg/cos(∂) * sin(∂) = Fg * tan(∂) Fr= μ * Fn = μ * Fg * tan(∂) Fr=Fz Fz=μ * Fg * tan(∂) I think the difference between the two systems is the force exerted by the pulling force on the wall. Why is one easier than the other. I have no idea, how I can calculate the force which is produced by the pulling force. thanks already in advance Edited July 16, 2013 by bon Link to comment Share on other sites More sharing options...
studiot Posted July 16, 2013 Share Posted July 16, 2013 Maybe it's the weather, but I'm sorry, I have looked at you diagram three times and I still don't know what is what. Please try to explain in no more than 5 lines what exactly makes up your system or systems. For instance what is a suspended sort What are the circles on your diagram? What is mg the weight of? You have already done a lot of calculation so don't do any more until we understand your setup. Link to comment Share on other sites More sharing options...
bon Posted July 16, 2013 Author Share Posted July 16, 2013 Hanging of toilet rolls: The one they depend on so that the loose end hanging down the front, but also many other way around. It raises the question, in which the arrangements less force must be applied to roll. For this purpose, a simplified model be considered, in which the clones paper roll of mass m, as in the figure outline is considered to be a solid cylinder having a radius R, which is a rod of length L fastened to 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. Link to comment Share on other sites More sharing options...
studiot Posted July 16, 2013 Share Posted July 16, 2013 I don't see that the tension in the string or weight of the roll has any real relevance to the forces of unrolling. And you haven't shown any pull force on the paper. Link to comment Share on other sites More sharing options...
bon Posted July 16, 2013 Author Share Posted July 16, 2013 (edited) The pulling force is directed vertically downwards. Once on the one side (at the wall), and on the other side. Can no one help me? Edited July 16, 2013 by bon Link to comment Share on other sites More sharing options...
studiot Posted July 16, 2013 Share Posted July 16, 2013 The pulling force is directed vertically downwards. So why not show it on your diagram? Can no one help me? I am helping you, but you are totally off track with your analysis. If you are to pull sheets off the roll what must happen? That what must the roll do? Link to comment Share on other sites More sharing options...
bon Posted July 16, 2013 Author Share Posted July 16, 2013 The tensile force must be greater than the static frictional force, so that the roller rotates. The question is interrested me where exactly is the difference whether I prefer the other vertical wall or down on the open side. ??? Link to comment Share on other sites More sharing options...
studiot Posted July 16, 2013 Share Posted July 16, 2013 (edited) The tensile force must be greater than the static frictional force, so that the roller rotates. What tensile force? And, like I already asked you, why did you not show it on your diagram? Edited July 16, 2013 by studiot Link to comment Share on other sites More sharing options...
md65536 Posted July 16, 2013 Share Posted July 16, 2013 I'll try to explain what seems to be the problem: The paper roll is on a rod hanging freely from a string attached to the wall, letting it press against the wall, with some amount of force due to its weight. Depending on how the roll is oriented, pulling on the paper to unroll it will affect the force with which the roll is pressing against the wall. It is hypothesized that this difference in force is the essence of what makes one way easier than the other. So the question is how to determine the force. Link to comment Share on other sites More sharing options...
studiot Posted July 16, 2013 Share Posted July 16, 2013 I'll try to explain what seems to be the problem: The paper roll is on a rod hanging freely from a string attached to the wall, letting it press against the wall, with some amount of force due to its weight. Depending on how the roll is oriented, pulling on the paper to unroll it will affect the force with which the roll is pressing against the wall. It is hypothesized that this difference in force is the essence of what makes one way easier than the other. So the question is how to determine the force. Thank you, that is as I read it. However bon has not included the force pulling on the roll of paper in his diagram, so cannot hope to achieve a solution. This is homework help, and I believe this is set work, so I cannot do the problem for him, only prompt for direction. Link to comment Share on other sites More sharing options...
md65536 Posted July 16, 2013 Share Posted July 16, 2013 Thank you, that is as I read it. However bon has not included the force pulling on the roll of paper in his diagram, so cannot hope to achieve a solution. This is homework help, and I believe this is set work, so I cannot do the problem for him, only prompt for direction. Ah, I see. After I wrote the previous response, I figured I may have misread the thread. So the problem is not that bon needs to *explain* more what the details are or mean, but must *specify* them (in the diagram or maths)? Link to comment Share on other sites More sharing options...
studiot Posted July 17, 2013 Share Posted July 17, 2013 I don't see that the tension in the string or weight of the roll has any real relevance to the forces of unrolling. I am trying to get bon to think about the mechanics of the situation since his approach is fundamentally wrong (and also far too complicated, making life difficult for himself). Link to comment Share on other sites More sharing options...
bon Posted July 17, 2013 Author Share Posted July 17, 2013 Here are only the two different situations presented. I think I have to use the lever rule. Only the two points of attack are different. It is helpful to split Fz in two different forces? Can anyone give a specific tip, how I should go on? Toiletpaper.pdf Link to comment Share on other sites More sharing options...
studiot Posted July 17, 2013 Share Posted July 17, 2013 Bon, I will try one last time. If you want to make progress, please simply reply to the questions or comments I (or perhaps others) make. That is the point of a forum. I have told you twice that your approach is fundamentally wrong, but you haven't asked me about it. So one last time. What do you know about the moment of a force about a point the force passes through? You may call it the lever rule. Link to comment Share on other sites More sharing options...
bon Posted July 17, 2013 Author Share Posted July 17, 2013 Moment of a force: M1=Fz*R M2=Fr*T M1=M2 Fz*R=Fr*R Fz=Fr right? What / Why is my approach wrong? Link to comment Share on other sites More sharing options...
studiot Posted July 17, 2013 Share Posted July 17, 2013 (edited) Oh dear oh dear, You still haven't responded to my question. I have no idea what Fz and Fr or T might be. I am guessing that R is the radius of the roll. Here is a block resting on a table. The weight of the block acts at the C of G of the block and passes through point C as shown. What is the moment of W about C? This is absolutely fundamental to your understanding not only of this problem but mechanics more generally. Edited July 17, 2013 by studiot Link to comment Share on other sites More sharing options...
bon Posted July 17, 2013 Author Share Posted July 17, 2013 Fz - traction force Fri - friction force T - force on the rod (all forces are also shown in the first illustration) What is the moment of W about C? Newton's first law: Balance of forces Fg (graviton) = Fn (normal) Link to comment Share on other sites More sharing options...
studiot Posted July 17, 2013 Share Posted July 17, 2013 bon, do you understand what a turning moment is? Link to comment Share on other sites More sharing options...
bon Posted July 17, 2013 Author Share Posted July 17, 2013 turning movement is the action of a force about a fixed point at a distance from that point i.e move it in circular motionthis movement is called as torqueT=Fxrwhere F is forced applied , r is distance of application of force from that fixed point Link to comment Share on other sites More sharing options...
studiot Posted July 17, 2013 Share Posted July 17, 2013 So in my sketch of the block on the table, if C is the fixed point and W is the force what is the moment of W about C? Link to comment Share on other sites More sharing options...
bon Posted July 17, 2013 Author Share Posted July 17, 2013 T=W*r W - force r - distance from C Link to comment Share on other sites More sharing options...
studiot Posted July 17, 2013 Share Posted July 17, 2013 (edited) So what is the distance from C since since the line of action of W passes through C? In other words, what is the value of r? Edited July 17, 2013 by studiot Link to comment Share on other sites More sharing options...
bon Posted July 17, 2013 Author Share Posted July 17, 2013 I don`t know what you mean, maybe: Gravitational force from the focus on the table Normal force of the table upward, both are gleoch great, here we also have the difference r of the points. In my task, r, is the radius from the roll Link to comment Share on other sites More sharing options...
studiot Posted July 17, 2013 Share Posted July 17, 2013 You have stated that the moment of a force is the force times the distance from the force to some point. This is not true, because it is imcomplete and the missing bits are very important. The moment is the force times the perpendicular distance from the point of application to the line of action of the force. Do you understand what this means, because your responses suggest you do not. Do you have any course notes? I suggest you revise moments or torque or turning effect before proceeding. Link to comment Share on other sites More sharing options...
bon Posted July 17, 2013 Author Share Posted July 17, 2013 (edited) Just as a force is a push or a pull, a torque can be thought of as a twist to an object. Mathematically, torque is defined as the cross product of the lever-arm distance and force, which tends to produce rotation. τ is the torque vector and τ is the magnitude of the torque, r is the displacement vector (a vector from the point from which torque is measured to the point where force is applied), and r is the length (or magnitude) of the lever arm vector, F is the force vector, and F is the magnitude of the force Edited July 17, 2013 by bon Link to comment Share on other sites More sharing options...
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