Pascal's law — pressure transmits equally: Foundations

2.2.6 · D1 · Physics › Fluid Mechanics › Pascal's law — pressure transmits equally

Iss idea par believe karne se pehle, tumhe kuch symbols aur pictures mein fluent hona padega. Parent note mein $p$ , $A$ , $F$ , $ρ$ , $g$ , $h$ , $Δ p$ , $θ$ , $d A$ , wedge geometry, aur "isotropic" ka use hota hai. Yeh page har ek ko zero se build karta hai, uss order mein jis order mein woh ek doosre par depend karte hain, taaki koi bhi cheez use hone se pehle samjhi ja sake.

1. Force — ek push, measure ki gayi

Yeh topic ko kyun chahiye? Kyunki Pascal's law ka pura payoff — "ek choti push ek car ko lift karti hai" — forces ke baare mein ek statement hai. Hum precisely keh sakein "yeh force 100 N neeche push kar rahi hai."

2. Piston — woh sliding lid jo fluid ko press karti hai

Ise abhi kyun introduce karein? Kyunki agla idea (area) ko kuch concrete chahiye jiska area matter karta hai. Ek piston ek circle hai, toh iska area $π r^{2}$ hai — aur do pistons ke alag-alag areas hi lift ki puri trick hain.

3. Area — push kitni surface par land karti hai

Circular piston of radius $r$ ke face ka area: $A = π r^{2} .$

Figure — Pascal's law — pressure transmits equally

Figure s01 — Scale par drawn do circles of radius 1 aur 2; bada wala chaar times area cover karta hai jabki uska radius sirf double hua hai. Isliye $A = π r^{2}$ bade pistons ko enormously stronger banata hai.

Yeh topic ko kyun chahiye? Kyunki pressure force ka ek area par spread hona hai, aur lift ka magic yeh hai ki bade piston ka bahut bada area hota hai. Radius-squared wahi hai jahan worked example mein 100 number aata hai.

Ab hum upar ke do arrows ko combine karte hain.

"Force per area" aur sirf "force" kyun nahi? Kyunki fluid tumhare total push ki parwah nahi karta — yeh respond karta hai ki woh push kitni concentrated hai. Do alag-size pistons jo same pressure feel karte hain woh alag forces produce karte hain. Woh ek fact hi hydraulic press hai. Pressure — force per unit area dekho.

5. Hydraulic press formula $F_{2} = F_{1} (A_{2} / A_{1})$

Humhare paas ab exactly woh sab hai jisse hum dekh saken kyun ek tiny push heavy load uthati hai — poore topic ki destination. Do pistons lo jo confined fluid se jude hain: ek chota (area $A_{1}$ ) jise tum force $F_{1}$ se push karte ho, aur ek bada (area $A_{2}$ ) jo load $F_{2}$ carry karta hai.

Yahan $F_{1}, A_{1}$ small (input) piston label karte hain aur $F_{2}, A_{2}$ big (output) piston label karte hain. Is formula ka har symbol upar define kiya gaya hai; equal-pressure step woh ek cheez hai jo abhi baki hai, aur §10 woh provide karta hai.

6. Density — matter kitna tightly packed hai

Yeh topic ko kyun chahiye? Kyunki fluid ka taller column zyada weighs karta hai, aur kitna zyada yeh depend karta hai ki fluid per volume kitna heavy hai — woh $ρ$ hai. Yeh aage aata hai, depth term mein.

7. Gravity $g$ aur depth $h$ — upar ke fluid ka weight

Figure s02 — Surface $h = 0$ marked kiye ek water tank; ek yellow arrow depth $h$ ko neeche measure karta hai, aur bottom par pink arrows upar ke column ka weight neeche press hote dikhate hain. Aur neeche matlab taller, heavier column, isliye zyada pressure.

$g$ aur $h$ topic ko kyun chahiye? $ρ$ ke saath milke yeh fluid ke apne weight se add hone wale pressure ko measure karte hain — woh pressure ka part jo Pascal's law untouched rehne deta hai jab tum top par extra press karte ho.

8. Hydrostatic pressure — $p = p_{0} + ρ g h$ derive karna

Hum sirf yeh formula quote nahi karte — hum ise fluid ka weight slab by slab stack karke build karte hain.

Figure s03 — Depth $y$ par fluid ki ek single thin slab: blue arrows dikhate hain pressure $p (y)$ top par neeche push karta aur $p (y + d y)$ bottom par upar push karta, ek pink arrow slab ka weight $ρ A g d y$ neeche kheenchta dikhata hai. Inhe teeno balance karne se $d p / d y = ρ g$ milta hai.

Derive kyun karein? Kyunki transmission argument yeh notice karne se kaam karta hai ki jab tum $p_{0}$ badhate ho, toh stacked-up $ρ g h$ piece nahi badalta. Hydrostatic pressure — p = p0 + ρgh dekho.

9. Change symbol $Δ p$ — "kitna badh gaya"

10. Isotropic pressure aur tiny wedge — $θ$ , $d A$ , $p_{x} = p_{y} = p_{n}$

Upar (§5) humne assume kiya ki small piston ka pressure big piston tak undiminished aur equal pohonchta hai. Iska reason yeh hai ki ek point par pressure har direction mein same hota hai. Iske liye word hai isotropic (Greek: "same in all turns"). Yahi proof hai.

Figure s04 — Angle $θ$ wale still fluid ka ek tiny triangular wedge; blue arrows har face mein normal push karte hain ( $p_{x}$ vertical face mein, $p_{y}$ horizontal face mein, $p_{n}$ slant mein). Slant push horizontal aur vertical shares mein split hoti hai jo exactly dono ko balance karti hain, $p_{x} = p_{y} = p_{n}$ force karte hue.

Formula Balance likhna (yahi WHY hai)

Wedge ki geometry (triangle ki shape se): $d A_{x} = d A_{n} sin θ, d A_{y} = d A_{n} cos θ .$ Horizontal ( $x$ ) balance. Vertical face par push $p_{x} d A_{x}$ hai. Sirf slant push ka horizontal component ise oppose karta hai; slant push $p_{n} d A_{n}$ hai aur uska horizontal share $p_{n} d A_{n} sin θ$ hai. Equal aur opposite: $p_{x} d A_{x} = p_{n} d A_{n} sin θ .$ $d A_{x} = d A_{n} sin θ$ substitute karo: $p_{x} (d A_{n} sin θ) = p_{n} d A_{n} sin θ .$ Poora factor $d A_{n} sin θ$ dono sides par appear karta hai, toh hum ise divide kar dete hain — woh cancellation hi kyun areas aur angle vanish ho jaate hain, yeh chhod ke: $p_{x} = p_{n} .$ Vertical ( $y$ ) balance. Isi tarah slant push ka vertical component $p_{n} d A_{n} cos θ$ hai: $p_{y} d A_{y} = p_{n} d A_{n} cos θ \Rightarrow p_{y} (d A_{n} cos θ) = p_{n} d A_{n} cos θ \Rightarrow p_{y} = p_{n} .$ Isliye $p_{x} = p_{y} = p_{n}$ : point par pressure har direction mein same hai.

Yeh §5 ko kaam karne kyun deta hai? Kyunki pressure koi "direction prefer" nahi karta, jo pressure tum small piston ko push karne se create karte ho woh equally fluid par aur far piston par press karta hai — jo exactly "transmitted undiminished" ko chahiye.

Piston kya hai aur uski shape kyun matter karti hai?

Ek sliding disc jo cylinder mein fluid seal karti hai; yeh ek circle hai, toh uska area

π r^{2}

hai.

Symbol

r

ka matlab aur uske units?

Circular piston face ka radius — centre-to-edge distance, metres mein.

Kya mujhe circle ka area pata hai aur kyun yeh itni fast badhti hai?

A = π r^{2}

; area radius squared ke saath scale karta hai, toh

r

double karne par

A

chaar guna ho jaata hai.

Kya main pressure teen ways mein likh sakta hoon?

p = F / A

F = p A

A = F / p

; unit pascal hai,

1 Pa = 1 N / m^{2}

Kya main press formula derive kar sakta hoon?

Equal pressure

p_{1} = p_{2}

deta hai

F_{1} / A_{1} = F_{2} / A_{2}

, toh

F_{2} = F_{1} A_{2} / A_{1}

Mass aur weight mein kya fark hai?

Mass kg mein (stuff ki miqdar); weight N mein (force

= m g

);

1000

kg paani weighs karta hai

\approx 9.8

kN.

Hydrostatic derivation mein

y

aur

p_{0}

ka matlab?

y

depth coordinate hai (surface se neeche,

y = 0

top par);

p_{0}

uss surface par pressure hai.

d p / d y = + ρ g

positive kyun hai?

Kyunki

y

neeche measure hota hai, aur aur neeche jaane par pressure badhta hai, toh slope positive hai.

Kya main ek slab se

p = p_{0} + ρ g h

derive kar sakta hoon?

d p / d y = ρ g

paane ke liye weight vs pressure difference balance karo, phir

0

h

tak integrate karo.

Δ p

(aur ek subscript) ka matlab kya hai, aur uski key property?

Labeled jagah par increase

p_{new} - p_{old}

;

Δ p

har depth par same hota hai.

p_{x}

p_{y}

p_{n}

kya denote karte hain?

Woh pressures jo wedge ke vertical face, horizontal face, aur slanted face mein normal push karte hain.

Fluid pressure ek point par isotropic kyun hota hai?

Still tiny wedge mein zero net force hoti hai;

d A_{n} sin θ

divide karne se

p_{x} = p_{y} = p_{n}

force hota hai.

"Incompressible" humein kya deta hai?

Volume squash nahi ho sakta, toh

A_{1} d_{1} = A_{2} d_{2}

aur push undiminished transmit hoti hai.

Press free energy kyun nahi hai?

F_{1} d_{1} = F_{2} d_{2}

— tum force ke liye distance trade karte ho; work conserved hai.

Foundations — Pascal's law — pressure transmits equally

1. Force — ek push, measure ki gayi

2. Piston — woh sliding lid jo fluid ko press karti hai

3. Area — push kitni surface par land karti hai

5. Hydraulic press formula $F_{2} = F_{1} (A_{2} / A_{1})$

6. Density — matter kitna tightly packed hai

7. Gravity $g$ aur depth $h$ — upar ke fluid ka weight

8. Hydrostatic pressure — $p = p_{0} + ρ g h$ derive karna

9. Change symbol $Δ p$ — "kitna badh gaya"

10. Isotropic pressure aur tiny wedge — $θ$ , $d A$ , $p_{x} = p_{y} = p_{n}$

11. Incompressible — volume squash nahi ho sakta

12. Volume conservation aur no-free-lunch trade

Prerequisite map

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