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I have to say thank you for your great help, I have already solve the problem

true: Fz=Fr (because both have got the same distance from the graviton point) Fr=Fn*μ=Fz ???

no it isn`t. I think only the same direction because of the held levers: F1*a1 = F2*a2

He shall produce the same amount and the same direction. M1=M2 Fr*r=Fz*r r - radius mmh, I don´t understand, what I should do???

two forces which can exert a moment about the centre line of the roll: Fz and Fr=Fn*μ The tensile force can be split into two components of force in the tangential force and the pressure force is directed to the center of gravity exerts an additional force on the wall. The frictional force is equal to the pulling force.

M=m*g*sin(∂)*r ???

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

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

T=W*r W - force r - distance from C

turning movement is the action of a force about a fixed point at a distance from that point i.e move it in circular motion this movement is called as torque T=Fxr where F is forced applied , r is distance of application of force from that fixed point

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)

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?

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

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. ???

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?

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.

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