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Suppose a paper cone is made with height equal to its radius, only the two straight sides just touch each other and are not glued together. It is kept on a frictionless table and a vertical force is applied at its apex.

What force do I need to apply to the base of the cone at the point where the paper meets on straight edge, to prevent it from spreading out. Ignore friction and bending effects.

I tried making free body diagram with the table applying normal force along the surface, but that leads me to conclude the cone should shrink which is obviously not happening. What am I doing wrong?enter image description here

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    $\begingroup$ This is actually a very complicated problem. It requires much more information/assumptions to calculate. If you want to only calculate the force needed to prevent it from spreading it out, we should assume the paper is infinitely strong and rigid. That would be fairly doable and mostly involve a bunch of geometry calculations. However if you really want to calculate it based on paper it gets extremely complex, as it will deform and start buckling, the thickness and stiffness will matter and probably a bunch of factors I haven't thought of right this moment. $\endgroup$ – Bob van de Voort Jul 27 at 12:42
  • $\begingroup$ Actually this problem was asked in one of my entrance exams. Although i am less interested in getting answer to that problem, i want to know if calculating the force was even possible with the assumptions given in the question. One of my friends solved it by assuming as little deformation such that height changes by dh and circumference by 2πdr. Then he calculated the work done and equated them to zero. And somehow answer popped out. I am not satisfied by his approach. Can anyone provide more insight into his approach or an alternative method? $\endgroup$ – Prakhar Pratap Mall Aug 2 at 2:53
  • $\begingroup$ I just added a image of that specific question if that helps. $\endgroup$ – Prakhar Pratap Mall Aug 2 at 2:54
  • $\begingroup$ I dont know if its correct but By considering the radius and hight of cone, the only way to construct such cone is by cutting out a exact 1/4th of circle. This makes force $W$ to be 45° inclination to a reference horizontal. The two forces are at a angles of 90°, so by using resultant force and equating the two we get $F=\sqrt{2}\pi$. So is it a correct answer? $\endgroup$ – Proxy Aug 2 at 5:00
  • $\begingroup$ No. Answer is F=1 $\endgroup$ – Prakhar Pratap Mall Aug 6 at 1:27
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I think you need to create a slice of the cone and balance the force traveling down along the paper with a hoop force around the base.

Look at the problem from the side, and realize that the force through the paper is at an angle equal to $\tan \psi = \tfrac{r}{h}$ and it is split into a vertical component that is reacted by the floor, and a radial component that is reacted by the "hoop" stress on the base. The vertical component of a small slice ${\rm d}\theta$ is $$ {\rm d} V= \tfrac{w}{2 \pi} {\rm d}\theta$$ with the total load $V = w$

Looking from the top at this slice, the radial component of the force is ${\rm d}R = {\rm d}V \tan \psi$ or

$$ {\rm d}R = \left( \tfrac{r}{h} \tfrac{ w}{2 \pi} \right) {\rm d} \theta$$

The balance of forces on the horizontal plane is thus

fig1

$${\rm d}R = 2 F \left(\sin \tfrac{{\rm d}\theta}{2}\right) $$

$$ \left( \tfrac{r}{h} \tfrac{ w}{2 \pi} \right) {\rm d} \theta = F\, {\rm d} \theta $$

with the direct solution

$$ \boxed{ F = \frac{r}{2\pi h} w } $$


Update 1

To understand the internal forces along the part, look at a slice of the top part of the paper from two angles

fig2

On the right is an edge one view of the slice along with the vertical component ${\rm dV}$ that directly opposes $w$. On the left, you see the internal compressive forces ${\rm dT}$ act at an angle to $w$, and split into vertical and radial components ${\rm d}V$ and ${\rm dR}$.

It is the radial forces that need to be balanced by the "glue" on the rip of the paper.

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  • $\begingroup$ Though the answer gives the correct required value, I have a little doubt. How can you take vertical and radial component of a already vertical force( in this case w)? Even if we consider w to travel down the slope of cone how do we know it does it in the same amount?. Shouldn't a component of w travel along the slope of cone while there remains a force perpendicular to the slope? $\endgroup$ – Prakhar Pratap Mall Aug 6 at 1:26
  • $\begingroup$ See update 1. $\endgroup$ – John Alexiou Aug 6 at 3:38
  • $\begingroup$ Okay. I understand now. However this was when we considered a portion of the cone. If we consider the full cone, the internal forces are neglected and there is vertically upwards normal reaction from the floor balanced by w. How would then the forces tally? Please explain. $\endgroup$ – Prakhar Pratap Mall Aug 6 at 9:35
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    $\begingroup$ @madhubalasingh - consider only the bottom rim of the cone, and the force balance on it along the horizontal plane. There are radial forces coming from the internal loading, and the external tangential forces $F$ that need to balance those. That is my #3 equation above ${\rm d}R = 2 F \left(\sin \tfrac{{\rm d}\theta}{2}\right)$. $\endgroup$ – John Alexiou Aug 6 at 11:31
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Assumptions about the collapse of the cone (if it occurs):

  1. I assume the base of the cone remains a circular arc because the question states there is no bending of the paper. $r$ is the radius of this circle. This is the only form of collapse that doesn't change the geodesic distance between any two points on the cone.

This problem becomes quite easy if you try to find the KE of the cone. Let's say the apex of the cone collapses from the height of $h$ to $h+dh$ ($dh$ is negative). Then the radius changes form $r$ to $r+dr$. $$r^2+h^2=constant$$ Differentiating both sides, $$⇒2rdr+2hdh=0$$ $$⇒\boxed{-\frac{dh}{dr}=\frac{r}{h}}$$ Let $K$ be the kinetic energy of the cone. For non-spontaneity of collapse, $dK<0$. From the work-energy theorem: $$dK=w(-dh)+F(-2πdr)<0$$ $$⇒F(2πdr)>w(-dh)$$ Divide by $dr$ on both sides $(dr>0)$ $$2πF>w\left(-\frac{dh}{dr}\right)$$ $$⇒2πF>\frac{wr}{h}$$ $$⇒\boxed{F>\frac{wr}{2πh}}$$

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  • $\begingroup$ Please ignore any calculus atrocities $\endgroup$ – user220805 Aug 6 at 6:34

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