Buoyancy and Archimedes' principle

Why the push exists. Pressure in a fluid increases with depth, because the fluid higher up presses down on the fluid below it. So the pressure pushing up on the bottom face of a submerged body is always greater than the pressure pushing down on its top face. The two forces do not cancel; the mismatch is a net force directed straight upward — the buoyant force or upthrust, $F_B$. This happens for any fluid, liquid or gas, and for any shape of body, though the size of the effect is easiest to see for liquids because their density is so much higher than air's.

What Archimedes' principle says. When a body is partly or wholly immersed in a fluid, it experiences an upthrust equal to the weight of the fluid it displaces:

where: $F_B$ is the upthrust in newtons, $V_{\text{displaced}}$ is the volume of fluid pushed aside (equal to the immersed volume of the body), $\rho_{\text{fluid}}$ is the density of the fluid in $\text{kg/m}^3$, and $g = 9.8\ \text{m/s}^2$. Notice what is absent from this formula: the body's own material, its own weight, its own density. Upthrust cares only about how much fluid got displaced — a solid gold block and a hollow plastic block of the same volume, fully immersed, feel exactly the same upthrust.

The three-case picture. Comparing $F_B$ (upthrust) with $W$ (the body's true weight) sorts every immersed body into one of three behaviours, and the syllabus expects all three to be explained, not just calculated.

Case I — body sinks. If the body's density exceeds the fluid's density ($d_{\text{body}} > d_{\text{fluid}}$), the upthrust when fully immersed is smaller than the true weight, $F_B < W$, and the body sinks. It still feels the upthrust, so on a spring balance it seems to lose weight: apparent weight = true weight minus $F_B$. This weight loss is exactly how the Relative Density of a solid is found experimentally.

Case II — body hovers, fully submerged. If the densities are equal ($d_{\text{body}} = d_{\text{fluid}}$), then $F_B = W$ everywhere inside the fluid, and the body stays suspended at whatever depth it is placed, with zero apparent weight.

Case III — body floats. If the body is less dense than the fluid, it cannot stay fully submerged; it rises until only part of its volume sits below the surface, just enough for the upthrust from that submerged part to balance the entire weight. (The exact relation between the densities and the submerged fraction is the topic of floatation, covered separately — here it is enough to know why partial immersion happens.)

A solid block of volume $2 \times 10^{-3}\ \text{m}^3$ and true weight $25\ \text{N}$ is lowered into water (density $1000\ \text{kg/m}^3$, $g = 9.8\ \text{m/s}^2$) until it is fully immersed. Find (a) the upthrust on the block, (b) its apparent weight while immersed, and (c) which case (sink, hover, or float) it belongs to.

Step 1. Find the volume of water displaced. Since the block is fully immersed, the displaced volume equals the block's own volume: $V_{\text{displaced}} = 2 \times 10^{-3}\ \text{m}^3$.

Step 2. Apply Archimedes' principle to get the upthrust. $F_B = V_{\text{displaced}} \times \rho_{\text{water}} \times g = 2\times10^{-3} \times 1000 \times 9.8 = 19.6\ \text{N}$.

Step 3. Find the apparent weight. Apparent weight equals true weight minus upthrust: apparent weight $= 25 - 19.6 = 5.4\ \text{N}$. Since the apparent weight is positive (not zero, not negative), the spring balance would indeed read a smaller but non-zero value while the block hangs immersed.

Step 4. Identify the case. Because the upthrust ($19.6\ \text{N}$) is less than the true weight ($25\ \text{N}$), $F_B < W$, so the block is denser than water and, if released, would sink — this is Case I.

The upthrust is 19.6 N, the apparent weight is 5.4 N, and the block belongs to Case I (it would sink).

Competitive stretch

The same block, still weighing $25\ \text{N}$ in air, is now made of a lighter material so that only $0.9\times10^{-3}\ \text{m}^3$ of it needs to be submerged for the upthrust to balance its weight exactly (Case III, floating). Check: $F_B = 0.9\times10^{-3} \times 1000 \times 9.8 = 8.82\ \text{N}$ — this does not equal $25\ \text{N}$, which is exactly the kind of quick sanity check worth running: if the computed upthrust for the stated submerged volume does not match the weight, the body has not yet reached its floating equilibrium, and more of it must go under (or the numbers given describe a different, non-floating scenario).

• Treating upthrust as depending on the body's material. It depends only on the fluid displaced — a lead ball and a wooden ball of the same volume, fully immersed, feel the same $F_B$.
• Using total volume instead of immersed volume. Only the submerged part of the body displaces fluid; for a partly immersed floating body, use the immersed volume, not the whole volume.
• Assuming "heavier always sinks." It is density, not absolute weight, that decides — a massive ship floats because its overall density (including the air inside its hull) is less than water's, while a small dense pebble sinks.
• Mistaking zero apparent weight for zero upthrust. In Case II the apparent weight is zero, but the upthrust is not zero — it exactly equals the true weight, which is precisely why the apparent weight vanishes.

Ask the tutor: Want to work through a trickier apparent-weight or floating-body numerical, or check which of the three cases a given problem falls under?