Archimedes force and floating

Pressure in a fluid grows with depth according to $p=\rho_f g h$, so the fluid pushes harder on the lower surface of a submerged body than on the upper one, and the net pressure force points vertically upwards. This upward force is called the Archimedes force:

$$F_A=\rho_f g V_{\text{disp}},$$

where $\rho_f$ is the fluid density, $g$ is the gravitational acceleration, and $V_{\text{disp}}$ is the displaced volume, equal to the volume of the submerged part of the body.

A body floats in equilibrium when the Archimedes force balances its weight: $F_A=mg$. Substituting $m=\rho_b V$ and $F_A=\rho_f g V_{\text{disp}}$ gives the floatation condition $\dfrac{V_{\text{disp}}}{V}=\dfrac{\rho_b}{\rho_f}$. If $\rho_b>\rho_f$, the body sinks, because its weight exceeds the largest possible Archimedes force; if $\rho_b=\rho_f$, it floats fully submerged without touching the bottom.

A cube of volume $V=2.0\times10^{-3}\,\text{m}^3$ is fully submerged in water of density $\rho_f=1000\,\text{kg/m}^3$. Find the Archimedes force acting on the cube. Take $g=9.8\,\text{m/s}^2$.

A wooden block of density $\rho_b=600\,\text{kg/m}^3$ floats in water ($\rho_f=1000\,\text{kg/m}^3$). What fraction of the block's volume is submerged? What does this fraction become in kerosene ($\rho_f=800\,\text{kg/m}^3$)?

A hollow sealed sphere has total mass $m=2.0\,\text{kg}$ and total volume $V=4.0\times10^{-3}\,\text{m}^3$. It is placed in water ($\rho_f=1000\,\text{kg/m}^3$, $g=9.8\,\text{m/s}^2$). (a) Will it sink? (b) What fraction of its volume will be under water? (c) What minimum mass must be added inside the sphere so that it becomes fully submerged (neglect the volume of the added mass)?

Self-check in a simulation (safe format). Open the PhET simulation "Buoyancy" and work only inside it: do not run experiments with open flame, acids, or household liquids of unknown composition.

1. Set the fluid density to $\rho_f=1000\,\text{kg/m}^3$. Take 5 materials with densities from 200 to 2000 $\text{kg/m}^3$ and record for each in a table: $\rho_b$, whether it floats, and the observed submerged volume fraction.
2. For each floating body compute the theoretical fraction $\dfrac{V_{\text{disp}}}{V}=\dfrac{\rho_b}{\rho_f}$ and compare it with the observation.
3. Repeat with fluids of $\rho_f=800\,\text{kg/m}^3$ and $\rho_f=1300\,\text{kg/m}^3$ and state how fluid density affects the depth of immersion.
4. Test the limiting cases: $\rho_b=\rho_f$ and $\rho_b>\rho_f$.

Show that the Archimedes force arises from the pressure difference between the lower and upper surfaces of a body, and that the submerged volume fraction depends linearly on the density ratio $\rho_b/\rho_f$.

Diagram and graph. On the left, a sketch of a cube in a fluid: a downward force $p_{\text{top}}A$ acts on the top face and an upward force $p_{\text{bottom}}A$ acts on the bottom face, with the height $\Delta h$ marked between the faces; the resultant is shown as an upward arrow $F_A$. On the right, a graph with $V_{\text{disp}}/V$ on the vertical axis and $\rho_b/\rho_f$ on the horizontal axis; the line runs from $(0,0)$ to $(1,1)$, and the region $\rho_b/\rho_f>1$ is marked with a dashed line as the sinking zone.

A person of mass $70$ kg floats on water. Their weight is $mg = 70 \cdot 9.81 = 687$ N. In fresh pool water ($\rho = 1000$ kg/m³) balancing this weight requires displacing a volume

$$V = \frac{mg}{\rho g} = \frac{687}{1000 \cdot 9.81} = 0.070 \text{ m}^3.$$

In Dead Sea water ($\rho = 1240$ kg/m³) the same weight is balanced by a smaller volume:

$$V = \frac{687}{1240 \cdot 9.81} = 0.0565 \text{ m}^3.$$

That is why in salt water the body is submerged less — the water pushes it up more strongly.

This uses the floating condition $F_A = mg$ together with the Archimedes force formula $F_A = \rho g V$: at equilibrium the displaced volume is set by the ratio of the body's mass to the liquid's density.

A ship of mass $200,000$ t $= 2.0 \cdot 10^8$ kg has a weight of $1.96 \cdot 10^9$ N. In a river ($\rho = 1000$ kg/m³) it displaces

$$V = \frac{1.96 \cdot 10^9}{1000 \cdot 9.81} = 2.00 \cdot 10^5 \text{ m}^3,$$

while in sea water ($\rho = 1025$ kg/m³) it displaces

$$V = \frac{1.96 \cdot 10^9}{1025 \cdot 9.81} = 1.95 \cdot 10^5 \text{ m}^3.$$

The difference is about $5.1 \cdot 10^3$ m³; with a waterplane area of $5000$ m² the draft decreases by roughly $1.0$ m.

The situation tests the ship's floating condition: the Archimedes force $F_A = \rho g V$ balances a constant weight, so moving into denser water requires a smaller displaced volume and a smaller draft.

An ice cube of mass $20$ g $= 0.020$ kg and density $917$ kg/m³ occupies a volume of $21.8$ cm³ and weighs $mg = 0.020 \cdot 9.81 = 0.196$ N. To stay in equilibrium it displaces

$$V = \frac{0.196}{1000 \cdot 9.81} = 2.00 \cdot 10^{-5} \text{ m}^3 = 20 \text{ cm}^3,$$

so $20/21.8 = 0.917$ of its volume lies below the surface, while about $1.8$ cm³ (roughly $8\%$) sticks out above.

This is a direct application of the floating condition $F_A = mg$: exactly the volume of water whose mass equals the ice's mass is displaced, so the submerged fraction equals the density ratio $\rho_{\text{ice}}/\rho_{\text{water}}$.