Problem — Pressure on a paper cone

Problem

A paper cone resting on a table has base $A = 100$ cm$^2$. A pressure of $10^5$ Pa would give $F = pA = 10^3$ N on one side of the base: why isn’t the cone crushed? Show the vector balance of forces.

Solution

The mistake is treating pressure as if it acted in a single direction. Pressure is isotropic: it presses perpendicularly on every surface element, in all directions.

Force is a vector. The value $F=pA=10^5\cdot10^{-2}=10^3$ N is only the magnitude of the force on a flat area of 100 cm². But on a closed surface, like the cone, the force vectors point in different directions and must be added vectorially, not arithmetically.

Balance on the cone. Each slanted face receives from the outside air a push $\vec F_\text{est}$ directed inward, perpendicular to the surface. But the air inside the cone (and below, through the paper and the contact with the table) exerts an equal and opposite push $\vec F_\text{int}$. The horizontal components of the pushes on the two faces are symmetric and cancel; the vertical components from above are balanced by those from below. $\sum \vec F = \oint_{\text{closed surface}} (-p\,\hat n)\,\dd A = \vec 0$ For a fluid at uniform pressure, the integral of pressure over any closed surface is zero: the forces cancel pair by pair. $\vec F_\text{est}+\vec F_\text{int}=\vec 0\quad\text{on every element}\ \Longrightarrow\ \vec F_\text{tot}=\vec 0$ The cone isn’t crushed because the air presses on it evenly from every side: the resultant is zero. (Only if we created a vacuum inside the cone would the outside push no longer be balanced, and only then would it get crushed.) $\ev{\text{isotropic pressure}\Rightarrow \vec F_\text{tot}=\vec 0\ \text{on a closed surface: the cone stays intact}}$

Connections

Topics: Teoria cinetica dei gas Concepts: Pressione Skills: Analisi di casi limite e fantafisica · Interpretazione micro-macro