Buoyancy — Archimedes' principle, derivation from pressure difference

WHY does buoyancy exist at all?

WHAT is being claimed (Archimedes' principle):

HOW to derive it from pressure (first principles)

Take a rectangular block of horizontal cross-section area $A$ and height $H$ . Its top face is at depth $h_1$ , its bottom face at depth $h_2 = h_1 + H$ . The fluid has density $\rho_f$ .

Figure — Buoyancy — Archimedes' principle, derivation from pressure difference

Step 1 — Pressure on each horizontal face. $p_{\text{top}} = p_0 + \rho_f g h_1, \qquad p_{\text{bottom}} = p_0 + \rho_f g h_2$ Why this step? Hydrostatic pressure depends only on depth, so each flat face sees one uniform pressure.

Step 2 — Force = pressure × area (forces on the vertical sides cancel). $F_{\text{down}} = p_{\text{top}}\,A, \qquad F_{\text{up}} = p_{\text{bottom}}\,A$ Why this step? Side faces come in opposite-facing pairs at equal depth, so their horizontal forces cancel. Only top and bottom give a net vertical force.

Step 3 — Net upward force. $F_B = F_{\text{up}} - F_{\text{down}} = (p_{\text{bottom}} - p_{\text{top}})\,A$ $= \big[(p_0 + \rho_f g h_2) - (p_0 + \rho_f g h_1)\big]A = \rho_f g (h_2 - h_1) A$ Why this step? The $p_0$ atmospheric terms cancel — buoyancy comes purely from the depth difference.

Step 4 — Recognize the volume. $h_2 - h_1 = H \quad\Rightarrow\quad F_B = \rho_f g \, (H A) = \rho_f g \, V$ Why this step? $H A$ is exactly the block's volume = the volume of fluid it displaced.

Floating vs sinking (apply the principle)

For an object of density $\rho_o$ , volume $V$ , weight $W = \rho_o V g$ :

Sinks if $F_B < W \Rightarrow \rho_f < \rho_o$
Floats (rises) if $F_B > W \Rightarrow \rho_f > \rho_o$
Floats in equilibrium partly submerged: only fraction $f$ is underwater, and $\rho_f (fV) g = \rho_o V g \;\Rightarrow\; f = \frac{\rho_o}{\rho_f}$

Worked Examples

Worked example 3 — Apparent weight (weighing in water)

A rock weighs $50\,\text{N}$ in air. Submerged, a scale reads $30\,\text{N}$ . Find its volume and density.

$F_B = W_{\text{air}} - W_{\text{apparent}} = 50 - 30 = 20\,\text{N}$ . Why? The water's upthrust "removes" apparent weight. $V = \dfrac{F_B}{\rho_f g} = \dfrac{20}{1000 \times 9.8} = 2.04\times10^{-3}\,\text{m}^3$ . Why? Invert $F_B = \rho_f V g$ . Mass $= 50/9.8 = 5.10\,\text{kg}$ , so $\rho_o = 5.10 / 2.04\times10^{-3} \approx 2500\,\text{kg/m}^3$ .

Common Mistakes (Steel-manned)

Recall Feynman: explain to a 12-year-old

Water near the bottom of a pool is squeezed by all the water above it, so it pushes harder than water near the top. When you put a ball under, the harder push from below beats the gentler push from above, so the ball gets shoved up. The amount of that shove is exactly equal to the weight of the water the ball pushed out of its spot. If the ball is lighter than that water, up it pops; if heavier, down it goes.

Active Recall

What is Archimedes' principle?

An immersed object feels an upward force equal to the weight of fluid it displaces,

F_B=\rho_f V_{disp} g

Where does buoyancy come from physically?

The pressure on the bottom face is greater than on the top (deeper fluid = higher pressure); the net upward pressure force is buoyancy.

Why does atmospheric pressure

p_0

cancel in the buoyancy derivation?

It acts equally on top and bottom faces, so it subtracts out; only the depth-difference term

\rho_f g H

survives.

Which density appears in

F_B

, object or fluid?

The fluid's density

\rho_f

(the fluid does the pushing).

Does buoyant force depend on the object's depth?

No (for incompressible fluid, fully submerged); it depends on displaced volume via

p_{bottom}-p_{top}=\rho_f g H

Condition for an object to float?

\rho_{object} < \rho_{fluid}

(then

F_B

at full submersion exceeds weight).

Fraction submerged of a floating body?

f = \rho_{object}/\rho_{fluid}

Apparent weight of a submerged body?

W_{apparent}=W_{air}-F_B

Why does a steel ship float?

Its hull encloses air so the average density is below water's, displacing enough water to balance its weight.

Where does the buoyant force act?

At the centre of buoyancy = centroid of the displaced fluid volume.

Connections

Pressure in fluids — hydrostatic pressure (source of $p=p_0+\rho g h$ )
Density and relative density
Floating bodies and stability — metacentre
Pascal's principle
Newton's laws — equilibrium of forces
Apparent weight and weighing methods

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, buoyancy ka pura raaz ek hi baat hai: paani me jitne neeche jaoge, pressure utna zyada hota hai ( $p = p_0 + \rho_f g h$ ). Ab agar tum koi block paani me dubao, to uska neeche wala face zyada gehraai pe hai aur upar wala face kam gehraai pe. Isliye neeche se paani ka push (upar ki taraf) upar ke push (neeche ki taraf) se bada hota hai. Yeh difference hi upar ki taraf net force deta hai — yahi buoyant force hai.

Derivation simple hai: top par force $= p_{top} A$ niche, bottom par $= p_{bottom} A$ upar. Inka difference lo to $p_0$ cancel ho jaata hai (kyunki dono faces pe atmosphere barabar dabaata hai), aur bachta hai $F_B = \rho_f g H A = \rho_f V g$ . Yahaan $V$ wo volume hai jo object ne paani hata diya (displaced). Isi ko Archimedes' principle kehte hain: "upthrust = displaced fluid ka weight".

Important point: $F_B$ me fluid ki density aati hai, object ki nahi. Object ki density sirf yeh decide karti hai ki cheez doobegi ya tairegi. Agar $\rho_{object} < \rho_{fluid}$ to tairega, aur submerged fraction $= \rho_{object}/\rho_{fluid}$ . Isiliye iceberg ka sirf 10% bahar dikhta hai. Aur steel ka jahaaz isliye tairta hai kyunki uske andar hawa hone se average density paani se kam ho jaati hai.

Yeh concept exams aur real life dono me kaam aata hai — submarines, ships, hydrometer, hot air balloons sab isi principle pe chalte hain. Bas yaad rakho: buoyancy depth-difference se aati hai, absolute depth se nahi (incompressible fluid me), aur force hamesha upar ki taraf hoti hai.