Visual walkthrough — Hydrostatics — pressure = ρgh, derivation

2.2.5 · D2 · Physics › Fluid Mechanics › Hydrostatics — pressure = ρgh, derivation

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Step 0 — In words ka matlab kya hai?

Koi bhi physics se pehle, do ideas ko pin down karte hain jo poori story carry karti hain: pressure aur depth. Ek smart 12-year-old jisne ye kabhi nahi dekha, uske liye bhi theek rahega.

Picture dekho: wahi haath ek wide board par aur ek thin nail par push kar raha hai. Same force $F$, lekin nail ka tiny area bahut bada pressure banata hai.

YE KYUN SHURU KAREIN? Baad ki har equation aslaan yahi sentence hai: "force per area." Agar "pressure" fuzzy rahti hai, toh algebra meaningless hai. Ab nahi rahegi.

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Step 1 — Paani ka ek imaginary column draw karo

Poore ocean par ek saath force-balance nahi kar sakte. Toh hum ek aisi shape cut karte hain jis par reason kar sakein: fluid ka ek seedha vertical cylinder.

KYA: Socho fluid ka ek tube-shaped chunk — cross-section area $A$, free surface (top, khula hawa mein) se seedha neeche depth $h$ tak khada hai.

YE SHAPE KYUN: Ek vertical cylinder mein sirf do flat faces hain jo upar/neeche point karti hain (top aur bottom). Jo bhi vertical hota hai woh sirf unhi do faces par hota hai. Iski side wall curved aur vertical hai, toh uس par pushes sideways hain — woh cancel ho jaayenge aur humein kabhi pareshan nahi karenge. Simple geometry = simple physics.

PICTURE: Lal cylinder hamaara chosen "body" hai. Yeh ordinary paani se bana hai; humne sirf uske aas-paas ek imaginary boundary draw ki hai.

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Step 2 — Column ko weigh karo

Pressure weight se aati hai, aur weight ke liye mass chahiye. Apne lal cylinder ka mass nikaalte hain.

KYA: Iski volume height $\times$ area hai, aur mass density $\times$ volume hai:

$$\underset{\text{volume }({\text{m}}^3)}{\underbrace{V}}=\underset{\text{area of face}}{\underbrace{A}}\times \underset{\text{height of column}}{\underbrace{h}}$$ $$\underset{\text{mass }(\text{kg})}{\underbrace{m}}=\underset{\begin{array}{c}\text{density}\\ {\text{kg/m}}^3\end{array}}{\underbrace{\rho }}\times \underset{Ah}{\underbrace{V}}=\rho Ah$$

KYUN: Density $\rho$ ka jawab hai "kitne kilograms per cubic metre?" Humare paas kitne cubic metres hain ($Ah$) se multiply karo aur kilograms nikalta hai. (Density Density and Specific Gravity se aati hai.)

PICTURE: Column chhote unit cubes mein split hai — cubes count karo ($V$), har cube ka weight $\rho$ per unit volume hai, total mass $\rho Ah$ ki tarah stack hoti hai.

Ab mass ko weight mein convert karo (neeche ki taraf gravity force):

$$\underset{\text{weight }(\text{N})}{\underbrace{W}}=\underset{\rho Ah}{\underbrace{m}}\underset{\begin{array}{c}\text{gravity}\\ 9.8{\text{m/s}}^2\end{array}}{\underbrace{g}}=\rho Ahg$$

Symbol $g$ ek kaam kar raha hai: kilograms ko newtons of downward pull mein convert karna.

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Step 3 — Column par har vertical force ka naam rakho

Ab saare up/down forces list karo. Exactly teen hain, aur har ek pressure $\times$ area hai.

KYA: teen forces —

$$\begin{array}{ll}\text{Weight, down:}& W=\rho Ahg\\ \text{Air on the top, down:}& F_{\text{top}}=\underset{\begin{array}{c}\text{pressure at}\\ \text{surface}\end{array}}{\underbrace{P_0}}\underset{\text{top face}}{\underbrace{A}}\\ \text{Fluid below, up:}& F_{\text{bot}}=\underset{\begin{array}{c}\text{pressure at}\\ \text{depth }h\end{array}}{\underbrace{P}}\underset{\text{bottom face}}{\underbrace{A}}\end{array}$$

YE TEEN KYUN, AUR NAHI: Top face atmosphere ko feel karta hai jo neeche press kar raha hai ($P_0$ surface par pressure hai). Bottom face neeche ke fluid ko feel karta hai jo upar press kar raha hai ($P$ — yahi mystery hai jise hum solve kar rahe hain). Gravity poori mass ko neeche kheenchti hai. Aur kuch vertically point nahi karta.

PICTURE: Vertical forces ke liye teen kaale arrows, aur wall par chhote grey sideways arrows — sirf dikhane ke liye draw kiye hain ki woh cancel ho jaate hain (har left push usi depth par ek equal right push se match hoti hai). Isliye humein kabhi unki zaroorat nahi padti.

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Step 4 — Forces balance karo (yahan physics hoti hai)

Column rest mein hai (yahi "hydrostatic" ka matlab hai). Newton's first law: rest mein ⇒ net force $=0$ ⇒ up-forces equal down-forces.

KYA:

$$\underset{\text{only up-force}}{\underbrace{F_{\text{bot}}}}=\underset{\text{air down}}{\underbrace{F_{\text{top}}}}+\underset{\text{weight down}}{\underbrace{W}}$$

Step 3 se har ek substitute karo:

$$\underset{\text{fluid pushes up}}{\underbrace{PA}}=\underset{\text{air pushes down}}{\underbrace{P_{0}A}}+\underset{\text{weight of column}}{\underbrace{\rho Ahg}}$$

KYUN: Agar up-push bada hota, toh column upar shoot kar jaata; agar chhota, toh neeche sink ho jaata. Dono mein se kuch nahi hota — yeh bas baithta hai — toh dono sides exactly equal hain. Yeh ek equality poora answer carry karti hai.

PICTURE: Ek see-saw / balance beam. Left pan par $PA$ (up) baithta hai; right pan par $P_{0}A$ aur $\rho Ahg$ (down) baithte hain. Beam level hai — perfectly balanced.

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Step 5 — Area cancel karo, law padho

Har term mein ek $A$ hai. Column ki thickness ko depth par pressure decide nahi karni chahiye, toh use hatate hain.

KYA: Poori equation ko $A$ se divide karo:

$$\frac{PA}{A}=\frac{P_{0}A}{A}+\frac{\rho Ahg}{A}⟹\boxed{P=P_0+\rho gh}$$

$A$ se divide karna kyun allowed aur meaningful hai: $A$ sirf hamaari imaginary tube ka size tha — humne ise invent kiya tha, toh yeh physical answer se gayab ho jaana chahiye. Iska gayab hona ek promise hai: pressure is par depend nahi karta ki tumne column kitna mota draw kiya. (Yeh hydrostatic paradox hai, Step 7 mein aata hai.)

Term by term, final law padha jaata hai:

$$\underset{\begin{array}{c}\text{total pressure}\\ \text{at depth }h\end{array}}{\underbrace{P}}=\underset{\begin{array}{c}\text{pressure at}\\ \text{the surface}\end{array}}{\underbrace{P_0}}+\underset{\text{how heavy}}{\underbrace{\rho }}\underset{\text{gravity}}{\underbrace{g}}\underset{\begin{array}{c}\text{how deep}\end{array}}{\underbrace{h}}$$

PICTURE: Ek vertical depth-axis ruler ke saath. $h=0$ par pressure bar ki height $P_0$ hai. Jaise jaise neeche jaate ho, ek lal wedge $\rho gh$ upar add hoti hai — bar depth ke saath perfectly straight line mein badhti hai.

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Step 6 — Edge cases (koi bhi scenario bina draw kiye mat chhodna)

Ek law jis par tum trust karte ho usse apni extremes survive karni chahiye. Sabko ek figure par check karte hain.

• Surface par, $h=0$: $P=P_0+\rho g(0)=P_0$. Sahi — tumhare upar koi fluid nahi, sirf hawa. Lal wedge ki height zero hai.
• Zero gravity, $g=0$: $P=P_0$ har jagah. Free-fall ya deep space mein, paani ka koi weight nahi, toh depth matter karna band ho jaati hai — pressure uniform hai. Wedge flat ho jaati hai.
• Bahut halka fluid, $\rho \to 0$ (thin gas ki tarah): $\rho gh$ tiny ho jaata hai; pressure depth ke saath barely change hoti hai. Isliye hum aam taur par ek room mein hawa ki "depth" ignore kar dete hain.
• Neeche ki jagah upar jaana ($h$ negative): surface ke upar $\rho gh$ negative hai — pressure girta hai. Wahi straight line, upar ki taraf extend hoti hai.

PICTURE: Pressure vs depth ka ek graph jismein char lines hain: normal fluid (steep red line), halka gas (almost flat), $g=0$ case (vertical — koi change nahi), aur surface point jahan sab $P_0$ par milte hain.

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Step 7 — Container ki shape kyun invisible hai

Area $A$ Step 5 mein cancel ho gaya. Chalte hain physically dekhte hain ki woh promise kya matlab rakhta hai.

KYA: Teen containers — ek thin straw, ek fat barrel, ek slanted funnel — sab same depth $h$ tak bhare hain. Teeno mein bottom pressure identical hai, chahe barrel hazaar guna zyada paani rakhta ho.

KYUN: Sirf ek point ke seedha upar ka fluid usi par press karta hai. Fat barrel mein, sides mein baaki chauda paani us point ke upar stacked nahi hai — uska weight bottom aur walls carry karte hain, beech ke neeche wale point se nahi. Toh volume irrelevant hai; sirf vertical depth count hoti hai.

PICTURE: Teen alag-alag shape ke vessels side by side, paani ka level sab mein bilkul flat, har bottom par ek lal dot jo same pressure $P_0+\rho gh$ ke saath label hai.

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Ek picture mein summary

Yahan poori derivation ek single frame mein compressed hai: lal column, uske teen vertical forces, balance, aur resulting straight pressure line — sab ek saath.

Recall Feynman retelling — poora walkthrough plain words mein

Ek pool socho. Main imagine karta hoon ki main paani ka ek patla invisible tube draw kar raha hoon, top surface se seedha neeche depth $h$ par mere paon tak. Woh tube paani nahi hil raha, toh perfectly balanced hona chahiye — jo ise upar rakhta hai woh exactly wahi hai jo ise neeche kheenchta hai. Neeche kheenchna: hawa iske top par press kar rahi hai ($P_{0}A$) aur iska apna weight ($\rho Ahg$ — density times kitna tall times gravity). Upar push karna: neeche ka paani, iske bottom face par press kar raha hai ($PA$). Up equal down set karo: $PA=P_{0}A+\rho Ahg$. Area $A$ sirf meri imaginary tube ki motaai thi — yeh real answer ka part nahi ho sakta, toh main ise divide karke hatata hoon aur paata hoon $P=P_0+\rho gh$. Bas itna hi. $\rho gh$ piece paani ke column ki extra squeeze hai; do guna gehre jaao aur yeh double ho jaata hai. Aur kyunki $A$ cancel ho gaya, kabhi matter nahi kiya tha ki container kitna chowda ya weirdly-shaped tha — sirf kitna neeche jaate ho. Surface par ($h=0$) tumhare upar koi paani nahi hai, toh sirf hawa $P_0$ feel hoti hai. Zero gravity mein paani weightless hai aur depth bilkul matter karna band ho jaati hai. Sab check out hota hai.

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Connections

• ← Parent: full derivation & examples
• Pascal's Law — cancelling-$A$ idea, aage push ki gayi
• Atmospheric Pressure & Barometer — $P_0$ kahan se aata hai
• Buoyancy & Archimedes' Principle — top aur bottom faces ke beech pressure difference
• Manometers — $\rho gh$ ko height ki tarah padhna
• Bernoulli's Equation — motion add karo; static limit yahi law wapas deta hai
• Density and Specific Gravity — $\rho$ supply karta hai