A cone with height h = 1 m and a base circle of radius r = 1 m is formed from a sector-shaped sheet of paper. The sheet is of such a size and shape that its two straight edges almost touch on the slop

Question

A cone with height h = 1 m and a base circle of radius r = 1 m is formed from a sector-shaped sheet of paper. The sheet is of such a size and shape that its two straight edges almost touch on the sloping surface of the cone. In this state the cone is stress-free. The cone is placed on a horizontal, slippery table-top, and loaded at its apex with a vertical force of magnitude w = 2 π , without collapsing. The splaying of the cone is opposed by a pair of forces of magnitude F acting tangentially at the join in the base circle (see figure). Ignoring any frictional or bending effects in the paper, find the value of F .

TestHub · Accepted Answer

Imagine the apex of the cone being depressed by $Δ h$ , and the radius of its base consequently increasing by $Δ r$ , i.e. the perimeter of the base circle increases by $2 π Δ r$ . Then the total work done by the external forces would be:

$w Δ h - F (2 π Δ r) = 0 \Rightarrow F = \frac{w}{2 π} \frac{Δ h}{Δ r}$

However, the changes in height and base circle radius are not independent, the connection between them being determined by the fixed length of one of the straight edges of the sheet. Using Pythagoras's theorem:

$(r + Δ r)^{2} + (h - Δ h)^{2} = ℓ^{2} = r^{2} + h^{2}$

from which we have

$\frac{Δ h}{Δ r} = \frac{2 r + Δ r}{2 h - Δ h} \approx \frac{r}{h}$

So the force in question has magnitude

$F = \frac{w}{2 π} \frac{r}{h} = \frac{2 π}{2 π} \times \frac{1}{1} = 1$

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