The optimal traction power for unrolling toilet paper depends on the configuration of the roll and the forces acting on it. The discussion identifies two configurations: one where the loose end hangs down the front and another where it hangs off the wall. Key calculations involve determining the weight force (mg), friction force (uFn), and normal force (Fn) using the equations F = mg*sin(θ) - (μ*Fn) and Fn = mg*cos(θ). The analysis concludes that using moments to calculate forces provides a clearer understanding of which configuration requires less force to unroll the paper.
PREREQUISITESPhysics students, engineers, and anyone interested in understanding the mechanics of forces and moments in practical applications.
It could be, depending on which option you are adopting for the signs on the forces. Please choose one.2013 said:M1=Fr*R
M2=Fz*R
M1=M2
Fz=Fr
Is this right?
2013 said:I would like to choose option 3.
I don´t know which case we discuss already, but I thought we speak about where the paper is hanging over.
But I need both cases, because I should say which is from the physical side better.
Consider which way Fr acts when the paper hangs away from the wall.2013 said:It applies to when the paper hang next to the wall.
hang away from wall:
?
Yes, so how does that change the equation for vertical forces?2013 said:Fr acts the same way like Fz.
Both acts down.
2013 said:Fz+Fg+Fr = T*cos(θ)
?
Yes.2013 said:It is easier to pull when the paper hang near to the wall.
because by the first equation:
Fz+Fg=Fr+T*cos(θ)
True, but it's the two values of Fz that you need to compare.we can cut out Fr and Fz:
Fg=T*cos(θ)
In the OP you wroteIs this the solution for my task?
I asked myself where ever the difference is between the two processes and it is just the point of action, is different.
I must not calculate the force acting when the paper is not on the wall, on the wall by the pulling force?
which suggests you are actually supposed to determine the two values of Fz (as functions of the given variables). But since this is a translation it's hard for me to be sure.Determine the force with which you have to pull the two Aufhängearten
haruspex said:True, but it's the two values of Fz that you need to compare.
You have2013 said:How can I compare the two different equations?