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

2.2.7 · D2 · Physics › Fluid Mechanics › Buoyancy — Archimedes' principle, derivation from pressure d

Step 1 — "Pressure" ka matlab kya hota hai (push ki ek picture)

KYA. Kisi bhi formula se pehle, humein ek idea chahiye: ek fluid har us surface ko push karta hai jise woh touch karta hai, aur woh push us surface ke perpendicular hota hai. Us push ki strength ko hum pressure ke roop mein measure karte hain = force spread over area.

YEH IDEA PEHLE KYUN. Buoyancy ek net push hai. Pushes add karne ke liye humein pehle pressure ko force mein wapas convert karna aana chahiye: rearrange karne se milta hai $F_{⊥} = p A$ . Woh ek move — pressure times area equals force — poore derivation ka engine hai.

PICTURE. Chhoti surface patches dekho. Har arrow ek fluid push hai, hamesha surface ke straight andar point karta hai, kabhi bhi sideways nahi.

Step 2 — Pressure depth ke saath kyun badhta hai (upar ke paani ka weight)

KYA. Socho ek lamba, patla vertical column of fluid, jaise pool mein khada paani se bhara hua straw. Us column ke bottom pe fluid ko upar stacked saari fluid uthani padti hai. Upar zyada fluid ⟶ zyada weight neeche dabaata hai ⟶ zyada hard push. Toh pressure badhta hai jab tum deeper jaate ho.

KYUN. Hum ek difference dhundh rahe hain — ek deep face aur ek shallow face ke beech push mein. Woh difference possible hi nahi jab tak pressure depth pe depend nahi karta. Step 2 wahi dependence paida karta hai.

Chalte hain weight count karte hain. Ek column lo jiska cross-section area $A$ hai aur jo surface se depth $h$ tak jaata hai.

Yeh formula poori tarah Pressure in fluids — hydrostatic pressure mein derive kiya gaya hai; yahan humein sirf iska shape chahiye: pressure = surface pressure + (density × gravity × depth).

PICTURE. Fluid ka column; gauge jitna deep, utna zyada fluid weight carry karta hai, utna lamba pressure arrow.

Step 3 — Setup: ek block jiske top face aur bottom face hain

KYA. Ab ek rectangular block fluid mein poora duba do. Use do:

horizontal cross-section area $A$ (flat top aur bottom dono ka yahi area hai),
height $H$ (block kitna tall hai, top se bottom tak),
top face depth $h_{1}$ par, bottom face depth $h_{2} = h_{1} + H$ par.

BLOCK KYUN? Kyunki block ke flat horizontal faces hote hain, aur Step 2 kehta hai ki ek single depth pe ek flat face ek uniform pressure feel karta hai. Flat faces se bookkeeping clean rehti hai. (Hum "block" assumption Step 8 mein hataate hain — answer care nahi karega.)

PICTURE. Block, dono depths dashed lines se marked, height $H$ unke beech bracketed.

Step 4 — Top vs bottom face pe pressure

KYA. Har depth ko Step 2 ke formula mein daalo.

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

$p_{top}$ — upper face pe push per area; chhoti depth $h_{1}$ use karta hai, isliye yeh weaker push hai.
$p_{bottom}$ — lower face pe push per area; badi depth $h_{2}$ use karta hai, isliye yeh stronger push hai.

KYUN. Yahi buoyancy ka dil hai ek nazar mein: bottom pressure top pressure se zyada hota hai kyunki $h_{2} > h_{1}$ . Iske baad sab kuch sirf us inequality ko ek number mein badalna hai.

PICTURE. Wahi block; bottom pe upar ka arrow top pe neeche ke arrow se lamba draw kiya gaya hai — kyunki $p_{bottom} > p_{top}$ .

Step 5 — Har pressure ko force mein badlo (aur sides cancel karo)

KYA. Top aur bottom faces pe Step 1 ka rule $F = p A$ use karo.

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

$F_{down}$ — upar wala fluid block ko neeche press kar raha hai.
$F_{up}$ — neeche wala fluid block ko upar press kar raha hai.
$A$ — shared face area; dono ko same $A$ multiply karta hai, isliye sirf pressure difference decide karega ki kaun jeetta hai.

SIDES KYUN DROP HOTE HAIN. Block ke chaar vertical side faces opposite-facing pairs mein aate hain. Ek left-facing patch aur directly saamne wala right-facing patch same depth pe hote hain, toh woh equal pressure feel karte hain aur equal, opposite horizontal forces se push karte hain. Woh exactly cancel ho jaate hain. Sirf top aur bottom vertical net ke roop mein bachte hain. (Isliye humein kabhi sides ke liye formula ki zarurat nahi padi.)

PICTURE. Side-force pairs equal opposing arrows ke roop mein dikhaye gaye jo annihilate ho jaate hain; sirf vertical top/bottom arrows baachte hain.

Step 6 — Subtract karo: net upward force

KYA. Buoyancy woh bacha hua hai jab up aur down ladte hain.

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

Ab Step 4 ke pressures substitute karo:

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

$p_{0}$ terms ko milte dekho:

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

$p_{0} - p_{0} = 0$ — atmosphere equally top aur bottom pe press karta hai, isliye woh kabhi net vertical push produce nahi kar sakta. Woh disappear ho jaata hai.
$ρ_{f} g (h_{2} - h_{1})$ — depth-difference pressure, akela survivor.
$A$ — face area abhi bhi saath chal raha hai.

KYUN YEH PUNCHLINE HAI. Buoyancy absolute depth ( $h_{1}$ ya $h_{2}$ akele) ki parwah nahi karta aur atmosphere ki bhi nahi. Woh sirf gap $h_{2} - h_{1}$ ki parwah karta hai. Block ko aur deep daalo: dono depths badhti hain, lekin unka difference fixed rehta hai — toh $F_{B}$ fixed rehta hai.

PICTURE. Do lambe pressure bars ( $p_{top}$ , $p_{bottom}$ ) side by side draw kiye gaye; shared $p_{0}$ chunk dono pe identically shaded aur struck out; sirf bottom bar pe extra slab — depth-difference — glowing bachi hai.

Step 7 — Formula mein chhupa volume pehchano

KYA. Step 3 se, $h_{2} - h_{1} = H$ . Substitute karo:

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

$H A$ — height times cross-section = block ka volume $V$ .
Woh $V$ exactly us fluid ka volume hai jise block ne raaste se hataya = displaced volume $V_{disp}$ .

KYUN. Notice karo final formula mein koi $A$ nahi, koi $H$ nahi, koi $h_{1}$ nahi, koi $p_{0}$ nahi — "box" aur "depth" ka har trace gone hai. Sirf fluid density, displaced volume, aur gravity bachi hai. Yeh signal hai ki box sirf scaffolding tha.

PICTURE. Block ek transparent "displaced fluid" ke blob mein dissolve hota hai same volume ke saath, $F_{B} = ρ_{f} V g$ upar point karte hue tagged.

Step 8 — Edge case: koi bhi shape, sirf box nahi

KYA. Real object ko, apne dimaag mein, surrounding fluid se bane ek identical blob se replace karo.

KYUN KAAM KARTA HAI. Woh fluid blob wahan equilibrium mein baitha rehta hai — woh sink ya rise nahi karta, kyunki woh fluid hi hai. Newton's first law se, surrounding fluid ko us blob ko upar ek force se push karna chahiye jo exactly blob ke weight ke equal ho, $ρ_{f} V g$ . Lekin surrounding fluid nahi bata sakta ki us boundary ke andar kya hai — woh sirf boundary ki shape feel karta hai. Fluid blob ki jagah same shape ka real object daalo: surrounding fluid same force se push karta hai. Toh $F_{B} = ρ_{f} V g$ kisi bhi shape ke liye hold karta hai — ek sphere, ek machhli, ek crumpled can.

PICTURE. Left: paani ke andar ek odd wavy shape. Right: same outline fluid se bhara hua, balance mein floating, neeche weight aur upar buoyancy equal aur opposite.

Step 9 — Degenerate cases: extremes pe kya hota hai?

Har scenario cover karo taaki baad mein kuch surprise na kare.

Ek-picture summary

Upar sab kuch, ek single canvas pe: pressure depth ke saath badhta hai (bars), bottom push top push se beat karta hai, atmosphere cancel ho jaata hai, aur jo bachta hai woh displaced fluid ka weight hai seedha upar point karta hua.

Recall Poore walkthrough ki Feynman retelling

Ek fluid har us cheez ko push karta hai jise woh touch karta hai, aur push wahan harder hota hai jahan fluid deeper hota hai — kyunki deeper fluid apne kandhe par zyada fluid utha raha hota hai. Pool mein ek block daalo. Uska bottom uske top se neeche hota hai, isliye fluid bottom ko upar harder dhakelta hai jitna woh top ko neeche dhakelta hai. Block ki walls pe sideways pushes sab pairs mein cancel ho jaate hain, toh hum sirf top vs bottom ki parwah karte hain. Dono subtract karo: atmosphere ka push dono pe same hai, toh woh disappear ho jaata hai, aur jo bachta hai woh sirf is baat pe depend karta hai ki block kitna tall hai — uske faces ke beech depth gap. Use face area se multiply karo aur tumhara block ka volume milta hai. Toh bachi hui upward force equals (fluid density) × (volume) × (gravity): exactly us fluid ka weight jo block ne side mein hataya. Aur kyunki bahar wala fluid nahi dekh sakta ki boundary ke andar kya hai, wahi ek machhli, ek ship, ya ek rubber duck ke liye bhi sach hai.

Active Recall

Recall Pictures test karo

Kaun sa face zyada strong push feel karta hai, top ya bottom, aur kyun? ::: Bottom — woh deeper hai, isliye uske upar zyada fluid weight baith raha hai, jisse higher pressure milta hai. $p_{0}$ $F_{B}$ se kyun vanish ho jaata hai? ::: Woh top aur bottom pe equally press karta hai, toh subtraction $p_{bottom} - p_{top}$ mein woh cancel ho jaata hai. Side forces kyun appear nahi karte? ::: Equal depth pe opposite side patches equal, opposite horizontal forces se push karte hain jo pairs mein cancel ho jaate hain. Final formula mein $h_{2} - h_{1}$ ki jagah kya aata hai? ::: Block ki height $H$ , aur $H A$ uska volume $V$ hai. Box shape kyun matter nahi karta? ::: Object ko ek equal fluid blob se replace karo jo balance mein float karta hai; surrounding fluid same shape ki boundary pe same push karta hai. $F_{B}$ kya hota hai jab object fluid se bahar uthaya jaata hai? ::: Zero — koi displaced volume nahi.

Connections

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