A rock feels lighter under water, and a steel ship floats. Both come from pressure growing with depth: the water pushes harder on the bottom of an object than on its top.
Why it works
Take a box of height $L$ and top area $A$, held under water with its top at depth $h$. The water pushes down on the top with $(P_0 + \rho g h)A$ and up on the bottom with $(P_0 + \rho g (h + L))A$. The side pushes cancel. The net push is up: $$F_b = \rho g L A = \rho_{\text{fluid}} V g,$$ the weight of the fluid that would fill the box's space. Another way to see it: replace the box with water. That water would float in place, so the surrounding water must push up on it with exactly its weight. The surrounding water cannot tell what fills the space.
Key idea
Archimedes' principle: the buoyant force equals the weight of the fluid pushed aside, $F_b = \rho_{\text{fluid}} V_{\text{sub}} g$. An object denser than the fluid sinks; a less dense one floats with the fraction $\rho_{\text{object}}/\rho_{\text{fluid}}$ of its volume under the surface.
Worked examples
Worked example: A block on a scale
A $1.0\,\text{L}$ ($0.0010\,\text{m}^3$) aluminum block ($2700\,\text{kg/m}^3$) hangs from a spring scale, fully under fresh water. What does the scale read?
- Weight: $W = \rho_{\text{Al}} V g = (2700)(0.0010)(9.8) = 26.5\,\text{N}$.
- Buoyant force: $F_b = \rho_{\text{water}} V g = (1000)(0.0010)(9.8) = 9.8\,\text{N}$, the weight of one liter of water.
- Forces balance: scale + buoyant force = weight, so the scale reads $26.5 - 9.8 = 16.7\,\text{N}$.
Answer: $16.7\,\text{N}$
Worked example: The tip of the iceberg
Ice ($917\,\text{kg/m}^3$) floats in seawater ($1025\,\text{kg/m}^3$). What fraction of an iceberg is under the surface?
- Floating, the buoyant force equals the weight: $\rho_{\text{sea}} V_{\text{sub}} g = \rho_{\text{ice}} V g$.
- $\dfrac{V_{\text{sub}}}{V} = \dfrac{917}{1025} = 0.895$.
- About $89\%$ is under water; only about a tenth shows.
Answer: $0.895$, about $89\%$
Worked example: Finding a density
A metal sample weighs $30.0\,\text{N}$ in air and $27.0\,\text{N}$ fully under fresh water. What is its density?
- The buoyant force is the lost weight: $30.0 - 27.0 = 3.0\,\text{N}$.
- Volume: $V = \dfrac{F_b}{\rho_{\text{water}} g} = \dfrac{3.0}{(1000)(9.8)} = 3.06 \times 10^{-4}\,\text{m}^3$. Mass: $m = 30.0/9.8 = 3.06\,\text{kg}$.
- $\rho = m/V = 10{,}000\,\text{kg/m}^3$, ten times water, since it lost a tenth of its weight.
Answer: $1.0 \times 10^{4}\,\text{kg/m}^3$
Common mistake
Thinking the buoyant force depends on the object's own weight or density. It depends only on the volume under the surface and the fluid's density. An aluminum block and a lead block of the same size feel the same buoyant force; the lead sinks because its weight is bigger, not because the water pushes less. Check: $F_b = \rho_{\text{fluid}} V_{\text{sub}} g$ has nothing about the object in it except its submerged volume.