Problem — Iron cube in water

Problem

An iron cube ($\rho = 7{,}8\cdot 10^3\;\text{kg/m}^3$) of side $5\;\text{cm}$ is submerged in water ($\rho_\text{acqua} = 10^3\;\text{kg/m}^3$). Calculate the Archimedes’ thrust and determine whether the cube floats or sinks, justifying your answer. (Use $g = 9{,}8\;\text{m/s}^2$.)

Solution

Step 1 — Volume of the cube. The side is $\ell = 5\;\text{cm} = 0{,}05\;\text{m}$, so:

$V = \ell^3 = (0{,}05)^3 = 1{,}25\cdot 10^{-4}\;\text{m}^3$

Step 2 — Archimedes’ thrust. It is the weight of the displaced water, with the cube fully submerged:

$F_A = \rho_\text{acqua}\,V\,g = 10^3 \cdot 1{,}25\cdot 10^{-4} \cdot 9{,}8 \approx 1{,}2\;\text{N}$

Step 3 — Weight of the cube.

$P = \rho\,V\,g = 7{,}8\cdot 10^3 \cdot 1{,}25\cdot 10^{-4} \cdot 9{,}8 \approx 9{,}6\;\text{N}$

Step 4 — Comparison. Since $P \approx 9{,}6\;\text{N} > F_A \approx 1{,}2\;\text{N}$, the thrust is not enough to support the cube: the resultant points downward and the body sinks. Equivalently, since $\rho_\text{ferro} > \rho_\text{acqua}$, iron always sinks in water.

Result.

$\ev{F_A \approx 1{,}2\;\text{N} < P \approx 9{,}6\;\text{N}\ \Rightarrow\ \text{the cube sinks}}$

Note on the answer key $F_\text{Arch} \approx 0{,}98\;\text{N}$) does not match the data: with a $5\;\text{cm}$ side the volume is $V = 1{,}25\cdot 10^{-4}\;\text{m}^3$ and the thrust is $F_A = \rho_\text{acqua} V g \approx 1{,}2\;\text{N}$, not $0{,}98\;\text{N}$. The value $0{,}98\;\text{N}$ would correspond to $V = 10^{-4}\;\text{m}^3$ (side $\approx 4{,}6\;\text{cm}$): probably a rounding error on the volume in the key. The qualitative conclusion ("sinks") remains correct.

The answer printed in the margin (

Linked notes

Topics: Fluidostatica e fluidodinamica Concepts: Principio di Archimede · Forza peso