Buoyancy — Archimedes' principle, derivation from pressure difference: Visual walkthrough

Step 1 — What "pressure" even means (a picture of pushing)

WHAT. Before any formula, we need one idea: a fluid pushes on every surface it touches, and it pushes perpendicular to that surface. We measure the strength of that push as pressure = force spread over area.

WHY this idea first. Buoyancy is a net push. To add up pushes we must first be able to turn a pressure back into a force: rearranging gives $F_{⊥} = p A$ . That single move — pressure times area equals force — is the engine of the whole derivation.

PICTURE. Look at the little surface patches. Every arrow is a fluid push, always pointing straight into the surface, never sideways along it.

Step 2 — Why pressure grows with depth (weight of the water above)

WHAT. Imagine a tall, thin vertical column of fluid, like a straw full of water standing in the pool. The fluid at the bottom of that column has to hold up all the fluid stacked above it. More fluid above ⟶ more weight bearing down ⟶ harder push. So pressure climbs as you go deeper.

WHY. We are chasing a difference in push between a deep face and a shallow face. That difference is impossible unless pressure depends on depth. Step 2 is where that dependence is born.

Let us count the weight. Take a column of cross-section area $A$ reaching from the surface down to depth $h$ .

This formula is derived in full in Pressure in fluids — hydrostatic pressure; here we just need its shape: pressure = surface pressure + (density × gravity × depth).

PICTURE. The column of fluid; the deeper the gauge, the more fluid weight it carries, the longer the pressure arrow.

Step 3 — The setup: a block with a top face and a bottom face

WHAT. Now drop a rectangular block fully into the fluid. Give it:

horizontal cross-section area $A$ (the flat top and bottom both have this area),
height $H$ (how tall the block is, top to bottom),
top face at depth $h_{1}$ , bottom face at depth $h_{2} = h_{1} + H$ .

WHY a block? Because a block has flat horizontal faces, and Step 2 says a flat face at a single depth feels one uniform pressure. Flat faces make the bookkeeping clean. (We remove the "block" assumption in Step 8 — the answer will not care.)

PICTURE. The block, the two depths marked with dashed lines, the height $H$ bracketed between them.

Step 4 — Pressure on the top vs the bottom face

WHAT. Feed each depth into the Step 2 formula.

$p_{top} = p_{0} + ρ_{f} g h_{1} p_{bottom} = p_{0} + ρ_{f} g h_{2}$

$p_{top}$ — push per area on the upper face; uses the smaller depth $h_{1}$ , so it is the weaker push.
$p_{bottom}$ — push per area on the lower face; uses the larger depth $h_{2}$ , so it is the stronger push.

WHY. This is the heart of buoyancy in one glance: bottom pressure beats top pressure because $h_{2} > h_{1}$ . Everything after this is just turning that inequality into a number.

PICTURE. Same block; the arrow up on the bottom is drawn longer than the arrow down on the top — because $p_{bottom} > p_{top}$ .

Step 5 — Turn each pressure into a force (and cancel the sides)

WHAT. Use the Step 1 rule $F = p A$ on the top and bottom faces.

$F_{down} = p_{top} A F_{up} = p_{bottom} A$

$F_{down}$ — the fluid on top pressing the block downward.
$F_{up}$ — the fluid below pressing the block upward.
$A$ — the shared face area; the same $A$ multiplies both, which is why the pressure difference alone will decide who wins.

WHY the sides drop out. The block's four vertical side faces come in opposite-facing pairs. A left-facing patch and the right-facing patch directly across from it sit at the same depth, so they feel equal pressure and push with equal, opposite horizontal forces. They cancel exactly. Only the top and bottom survive as a vertical net. (This is why we never needed a formula for the sides.)

PICTURE. The side-force pairs shown as equal opposing arrows that annihilate; only the vertical top/bottom arrows remain.

Step 6 — Subtract: the net upward force

WHAT. Buoyancy is the leftover after up fights down.

$F_{B} = F_{up} - F_{down} = (p_{bottom} - p_{top}) A$

Now substitute the Step 4 pressures:

$F_{B} = [(p_{0} + ρ_{f} g h_{2}) - (p_{0} + ρ_{f} g h_{1})] A$

Watch the $p_{0}$ terms meet:

$F_{B} = [= 0 p_{0} - p_{0} + ρ_{f} g h_{2} - ρ_{f} g h_{1}] A = ρ_{f} g (h_{2} - h_{1}) A$

$p_{0} - p_{0} = 0$ — the atmosphere presses equally on top and bottom, so it can never produce a net vertical push. It vanishes.
$ρ_{f} g (h_{2} - h_{1})$ — the depth-difference pressure, the only survivor.
$A$ — the face area still riding along.

WHY this is the punchline. Buoyancy does not care about absolute depth ( $h_{1}$ or $h_{2}$ alone) and not about the atmosphere. It cares only about the gap $h_{2} - h_{1}$ . Sink the block deeper: both depths grow, but their difference stays fixed — so $F_{B}$ stays fixed.

PICTURE. The two long pressure bars ( $p_{top}$ , $p_{bottom}$ ) drawn side by side; the shared $p_{0}$ chunk shaded identically on both and struck out; only the extra slab on the bottom bar — the depth-difference — is left glowing.

Step 7 — Recognise the volume hiding in the formula

WHAT. From Step 3, $h_{2} - h_{1} = H$ . Substitute:

$F_{B} = ρ_{f} g (h_{2} - h_{1}) A = ρ_{f} g (H A) = ρ_{f} g V$

$H A$ — height times cross-section = the volume $V$ of the block.
That $V$ is exactly the volume of fluid the block shoved out of the way = the displaced volume $V_{disp}$ .

WHY. Notice the final formula has no $A$ , no $H$ , no $h_{1}$ , no $p_{0}$ — every trace of "box" and "depth" is gone. Only fluid density, displaced volume, and gravity remain. That is the signal that the box was just scaffolding.

PICTURE. The block dissolves into a transparent blob of "displaced fluid" of the same volume, tagged $F_{B} = ρ_{f} V g$ pointing up.

Step 8 — Edge case: any shape, not just a box

WHAT. Replace the real object, in your mind, with an identical blob made of the surrounding fluid itself.

WHY it works. That fluid blob just sits there in equilibrium — it does not sink or rise, because it is the fluid. By Newton's first law, the surrounding fluid must be pushing up on it with a force exactly equal to the blob's weight, $ρ_{f} V g$ . But the surrounding fluid cannot tell what is inside that boundary — it only feels the boundary's shape. Swap the fluid blob for a real object of the same shape: the surrounding fluid pushes with the same force. So $F_{B} = ρ_{f} V g$ holds for any shape — a sphere, a fish, a crumpled can.

PICTURE. Left: an odd wavy shape underwater. Right: the same outline filled with fluid, floating in balance, weight down and buoyancy up equal and opposite.

Step 9 — Degenerate cases: what happens at the extremes?

Cover every scenario so nothing surprises you later.

The one-picture summary

Everything above, on a single canvas: pressure grows with depth (bars), the bottom push beats the top push, the atmosphere cancels, and what survives is the weight of the displaced fluid pointing straight up.

Recall Feynman retelling of the whole walkthrough

A fluid pushes on everything it touches, and the push is harder where the fluid is deeper — because deeper fluid is carrying more fluid on its shoulders. Slip a block into the pool. Its bottom sits lower than its top, so the fluid shoves the bottom up harder than it shoves the top down. The sideways pushes on the block's walls all cancel in pairs, so we only care about top vs bottom. Subtract the two: the atmosphere's push is the same on both, so it disappears, and what is left depends only on how tall the block is — the depth gap between its faces. Multiply that by the face area and you get the block's volume. So the leftover upward force equals (fluid density) × (volume) × (gravity): exactly the weight of the fluid the block pushed aside. And since the fluid outside can't see what's inside the boundary, the same is true for a fish, a ship, or a rubber duck.

Active Recall

Recall Test the pictures

Which face feels the stronger push, top or bottom, and why? ::: The bottom — it is deeper, so more fluid weight sits above it, giving higher pressure. Why does $p_{0}$ vanish from $F_{B}$ ? ::: It presses equally on top and bottom, so in the subtraction $p_{bottom} - p_{top}$ it cancels. Why do the side forces not appear? ::: Opposite side patches at equal depth push with equal, opposite horizontal forces that cancel in pairs. What replaces $h_{2} - h_{1}$ in the final formula? ::: The block's height $H$ , and $H A$ is its volume $V$ . Why does the box shape not matter? ::: Replace the object by an equal fluid blob that floats in balance; the surrounding fluid pushes the same on any shape with the same boundary. What is $F_{B}$ when the object is lifted out of the fluid? ::: Zero — no displaced volume.

Connections

2.2.07 Buoyancy — Archimedes' principle, derivation from pressure difference (Hinglish) — parent topic
Pressure in fluids — hydrostatic pressure — source of $p = p_{0} + ρ_{f} g h$ used in Step 2
Density and relative density — meaning of $ρ_{f}$
Newton's laws — equilibrium of forces — the floating-blob argument in Step 8
Floating bodies and stability — metacentre — where the partly-submerged case leads
Apparent weight and weighing methods — buoyancy as "lost" weight
Pascal's principle — companion pressure law