Pascal's law — pressure transmits equally: Visual walkthrough

2.2.6 · D2 · Physics › Fluid Mechanics › Pascal's law — pressure transmits equally

Humein sirf do prerequisites chahiye, aur dono ko hum scratch se dobara explain karenge:

Pressure — force per unit area — "pressure" ka matlab kya hota hai.
Hydrostatic pressure — p = p0 + ρgh — pressure depth ke saath kyun badhta hai.

Step 1 — "Pressure" ka matlab kya hai (woh cheez jo hum spread karne wale hain)

KYA HAI. Pressure transmit karne se pehle, humein pata hona chahiye ki uski ek unit kya hoti hai. Ek flat patch lo jiska area $A$ hai. Usse touch karta hua fluid us patch par kuch total force $F$ lagate hai, jo seedha patch ke andar (perpendicular) aim karti hai. Hum define karte hain:

Per area kyun? Kyunki ek jaisi total push ek pin-tip par aur ek dinner plate par bahut alag feel hoti hai. $A$ se divide karne par "patch kitna bada hai" woh factor nikal jaata hai aur sirf "fluid kitna hard press kar raha hai" bachta hai — yeh point ki property hai, patch ki nahi.

PICTURE. Figure mein, same fluid ek chhote patch aur ek bade patch par push karta hai. Arrows (force) hamesha surface ke andar perpendicular aim karte hain — yeh "hamesha perpendicular" wali baat poori derivation ki neenv hai.

Figure — Pascal's law — pressure transmits equally

Step 2 — Fluid force perpendicular kyun honi chahiye

KYA HAI. Claim: ek fluid at rest sirf seedha andar push kar sakta hai kisi surface mein, kabhi sideways nahi.

KYUN. Maan lo fluid ek tilt par push kare. Us tilted push mein surface ke saath ek component hoga (ek tangential, sliding push). Fluid mein koi rigidity nahi hoti — zameen si bhi sliding push use flow kara deti hai. Lekin humne kaha ki fluid at rest hai (static). To sliding component zero hona chahiye. Jo bachta hai woh seedha andar point karta hai.

PICTURE. Lal tilted arrow ko ek green perpendicular part aur ek yellow tangential part mein split kiya gaya hai. Yellow part fluid ko slide karata — at rest fluid mein yeh forbidden hai — to sirf green bachta hai.

Recall Why does this matter so much?

Static fluid force perpendicular to any surface ::: Kyunki koi bhi tangential (sideways) push fluid ko flow karati, jo "at rest" ko contradict karta.

Step 3 — Tiny wedge banao (hamara test element)

KYA HAI. Yeh samajhne ke liye ki ek point par pressure alag-alag directions mein kaisa behave karta hai, hum fluid ka ek microscopic wedge isolate karte hain: ek triangular prism. Iske ek vertical face hai, ek horizontal face hai, aur ek slanted face hai. Hum ise ek single point ki taraf shrink karenge.

Teen faces par teen pressures ke naam rakho:

$p_{x}$ vertical face par (iska push horizontally, $x$ direction mein point karta hai),
$p_{y}$ horizontal face par (iska push vertically, $y$ direction mein point karta hai),
$p_{n}$ slanted face par (iska push slant ke perpendicular point karta hai).

The slant ek angle $θ$ (theta = "slanted face ka tilt") banata hai. Face areas ko $d A_{x}$ , $d A_{y}$ , $d A_{n}$ kaho (chhota " $d$ " sirf tiny matlab hai).

Wedge kyun? Kyunki wedge ke teen faces teen alag directions mein point karte hain. Agar pressure secretly direction ke saath badlta, to yeh shape use ek leftover unbalanced force ke roop mein reveal kar deta. Ek cube (saare faces axis-aligned) ek slanted direction test nahi kar sakta; wedge kar sakta hai.

PICTURE. Wedge apne teen labelled faces ke saath, base par angle $θ$ , aur ek perpendicular pressure-arrow har face ke andar push karta hua.

KYA HAI. Humein areas linked chahiye. Maan lo slant face ka area $d A_{n}$ hai. Basic right-triangle trigonometry (yahan $sin θ$ = opposite-over-hypotenuse, $cos θ$ = adjacent-over-hypotenuse, wedge ke triangular cross-section par measure kiya gaya) deta hai:

$d A_{x} = d A_{n} sin θ, d A_{y} = d A_{n} cos θ$

Term by term: $d A_{x}$ (vertical face) slant face ka vertical par "shadow" hai — yeh $sin θ$ factor se shrink hota hai. $d A_{y}$ (horizontal face) slant ka horizontal par "shadow" hai — factor $cos θ$ .

Yeh kyun chahiye? Teen faces ki alag-alag sizes hain, isliye unke pressure forces ( $p \times A$ ) compare nahi ho sakte jab tak hum size ratios na jaanein. Yeh do relations wahi ratios hain.

PICTURE. Triangular cross-section jisme hypotenuse $d A_{n}$ hai, aur uske do shadows $d A_{x}$ (vertical) aur $d A_{y}$ (horizontal), har ek apne $sin θ$ / $cos θ$ factor ke saath labelled.

Step 5 — Horizontal forces balance karo → $p_{x} = p_{n}$

KYA HAI. Wedge at rest hai, isliye har direction mein net force zero hai. Horizontal ( $x$ ) direction dekho. Do faces horizontally push karti hain:

vertical face $p_{x} d A_{x}$ force se push karti hai (poori tarah horizontal),
slanted face ki push ka horizontal component $p_{n} d A_{n} sin θ$ hai.

Dono ko equal set karo (woh ek doosre ka virodh karte hain): $p_{x} d A_{x} = p_{n} d A_{n} sin θ$

Ab Step 4 se $d A_{x} = d A_{n} sin θ$ substitute karo: $p_{x} (d A_{n} sin θ) = p_{n} d A_{n} sin θ$

$d A_{n} sin θ$ dono sides se cancel ho jaata hai (kisi bhi real wedge ke liye yeh nonzero hai): $p_{x} = p_{n}$

Slant force ka sirf horizontal component kyun count karta hai? Kyunki hum sirf $x$ -direction balance kar rahe hain. Slant push slant ke perpendicular point karti hai; sirf uska $x$ -axis par shadow, jo $sin θ$ fraction hai, vertical face ki horizontal push se lad sakta hai.

Gravity kyun nahi aayi? Gravity wedge ke volume $\sim (length)^{3}$ par pull karti hai, lekin pressure forces areas $\sim (length)^{2}$ par kaam karte hain. Jab hum wedge ko ek point tak shrink karte hain, volume area se zyada tezi se khatam hota hai, isliye weight term pressure forces ke samne negligible ho jaata hai. Exactly isliye result ek point ke baare mein ek statement hai.

PICTURE. Horizontal force arrows: green $p_{x} d A_{x}$ right push karta hai, red slant force apne dashed green horizontal shadow ke saath left push karta hai, dono equal drawn hain.

Step 6 — Vertical forces balance karo → $p_{y} = p_{n}$ , aur punchline

KYA HAI. Vertical ( $y$ ) direction mein repeat karo: $p_{y} d A_{y} = p_{n} d A_{n} cos θ$ $d A_{y} = d A_{n} cos θ$ substitute karo; $d A_{n} cos θ$ cancel ho jaata hai: $p_{y} = p_{n}$

Steps 5 aur 6 combine karo:

Yeh poori foundation kyun hai. Humne $θ$ arbitrarily choose kiya. Kyunki $p_{n}$ (koi bhi slanted direction) hamesha $p_{x}$ aur $p_{y}$ ke barabar hota hai, us point par pressure ek single number hai, direction-blind. Isliye jab tum fluid ko push karte ho, koi direction nahi hai jahan woh extra pressure "chupa" sake — use har jagah har direction mein equally dikhna hi hoga.

PICTURE. Wedge ke shrinking tip par point, jisme equal-length arrows har direction mein radiate kar rahe hain — pressure same chahe tum kisi bhi direction mein dekho.

Step 7 — "Har direction mein equal" se "undiminished transmit hota hai" tak

KYA HAI. Isotropy ek point par directions ke baare mein hai. Pascal ka poora law change ko har point tak spread karne ke baare mein hai. Hum dono ko hydrostatic relation se bridge karte hain. $p_{0}$ pressure par rakhi gai surface ke neeche depth $h$ ke fluid column ke liye: $p = p_{0} + ρ g h$ Term by term: $p_{0}$ = top surface par pressure. $ρ$ (rho) = fluid ki density (mass per volume). $g$ = gravity ki strength. $h$ = surface ke neeche depth. To $ρ g h$ = depth $h$ par upar stacked fluid ke weight se extra squeeze.

Ab surface pressure ko $Δ p$ (delta-p = "pressure mein jo change humne apply kiya") se badhao. Same depth $h$ par naya pressure hai: $p_{new} = (p_{0} + Δ p) + ρ g h = p_{old} (p_{0} + ρ g h) + Δ p$

Subtract karo: $p_{new} - p_{old} = Δ p har depth h par.$

Yeh har jagah equal kyun nikalta hai? Dekho kya badla aur kya nahi. Term $ρ g h$ fluid ke apne weight aur depth par depend karta hai — humne dono nahi badla, isliye yeh frozen hai. Sirf $p_{0} \to p_{0} + Δ p$ move kiya, aur woh same $Δ p$ se move kiya chahe $h$ kuch bhi ho. To har point exactly $Δ p$ se utha. Yahi Pascal's law hai.

PICTURE. Do identical water columns side by side; right wale ka top $+ Δ p$ dabaya gaya hai. Teen depths par pressure bars dono same yellow chunk $Δ p$ se badhte hain, jabki blue $ρ g h$ part unchanged hai.

Step 8 — Edge aur degenerate cases (koi gap mat chhodo)

KYA / KYUN, case by case:

$θ \to 0$ (slant horizontal ban jaata hai). Tab $sin θ \to 0$ , aur $d A_{x} \to 0$ : vertical face khatam ho jaata hai. Horizontal balance $p_{x} = p_{n}$ ban jaata hai $0 = 0$ — vacuously true, koi information nahi gayi; wedge ek flat slab ban jaata hai aur $y$ -balance phir bhi $p_{y} = p_{n}$ deta hai. ✅
$θ \to 9 0^{\circ}$ (slant vertical ban jaata hai). Ab $cos θ \to 0$ , $d A_{y} \to 0$ : horizontal face khatam ho jaata hai. Symmetry se $x$ -balance bachta hai aur $p_{x} = p_{n}$ phir bhi hold karta hai. ✅ Milakar dono limits dikhate hain ki result ek tilt ka accident nahi hai.
Compressible fluid (gas). Agar fluid ko squash kiya ja sake, to tumhara applied $Δ p$ ka kuch hissa fluid ko compress karne mein jaata hai bajaaye far wall par push karne ke — transmission ab perfectly undiminished ya instant nahi rahta. Pascal's law ko incompressible fluid chahiye. Dekho Incompressibility & continuity equation.
Enclosed nahi (open surface, escape path). Agar fluid bahar nikal sake, to push karne se sirf fluid move hota hai pressure har jagah raise hone ki bajaaye. Law ko ek enclosed (confined) fluid chahiye.
Zero applied change ( $Δ p = 0$ ). Tab $p_{new} = p_{old}$ har jagah — law trivially hold karta hai aur koi change predict nahi karta, exactly jaisa expect tha.

PICTURE. Teen mini-panels: flat-wedge limit, ek gas jo push ko partly absorb karta hai (spring-jaisi), aur ek open container jo push ko leak kar deta hai.

Ek-picture summary

Upar sab kuch ek single flow mein: static fluid → forces perpendicular hain → tiny wedge → geometry face areas ko link karti hai → $x$ aur $y$ mein force balance $θ$ cancel karta hai → pressure isotropic ( $p_{x} = p_{y} = p_{n}$ ) → $p = p_{0} + ρ g h$ mein feed karo → applied $Δ p$ upar unchanged sawar rahta hai → har point tak undiminished transmit hota hai. Aur payoff: same $Δ p$ ek bade area $A$ par $\Rightarrow$ badi force — hydraulic press $F_{2} = F_{1} A_{2} / A_{1}$ (dekho Hydraulic systems — brakes, lifts, jacks).

Recall Feynman: poora walkthrough simple words mein

Paani ka ek tiny wedge imagine karo, itna chhota ki basically ek dot hai. Ruke hue paani mein sideways push nahi ho sakti (woh toh bas slide kar jaata), isliye woh sirf seedha andar push karta hai har face par. Is wedge ke liye "left push = right push" aur "upar push = neeche push" likhdo. Jab tum faces ki shapes plug in karte ho, tilt angle khud cancel ho jaata hai aur tum ek shocking, simple fact par pahunchte ho: paani us dot par har direction mein exactly utni hi hard push karta hai. Koi direction special nahi. Ab paani ka ek tall column stack karo — jitna neeche jaao, utna zyada weight upar baitha hai, $ρ g h$ add karta hua. Agar tum surface par ek piston ko kuch extra squeeze $Δ p$ se press karo, woh squeeze neeche jaate jaate fade nahi hoti, kyunki weight part nahi badla — sirf top pressure badla, aur woh har jagah $Δ p$ se move kiya. To same extra pressure har point par dikhta hai. Finally: ek patla piston press karo, aur wahi pressure ek bade area wale mote piston par utarta hai — pressure times bada area = bahut badi force. Tumne apni ungli se gaadi uthaayi. Pakda? Patla piston bahut door travel karta hai jabki mota barely hilta hai — tumne distance ko force ke liye trade kiya, aur energy honest rehti hai.

Recall Quick self-test

$p = p_{0} + ρ g h$ mein kaunsa term applied change $Δ p$ carry karta hai? ::: $p_{0}$ (surface term); $ρ g h$ unchanged rehta hai. Wedge result se $θ$ kyun gayab ho jaata hai? ::: Same factor $d A_{n} sin θ$ (ya $cos θ$ ) dono sides par hota hai aur cancel ho jaata hai. Pascal's law kya tod deta hai? ::: Ek compressible fluid (change absorb kar leta hai) ya un-enclosed fluid (use escape hone deta hai).

Connections

Parent: Pascal's law — woh summary jise is page ne derive kiya.
Pressure — force per unit area — Step 1 se definition $p = F / A$ .
Hydrostatic pressure — p = p0 + ρgh — Step 7 mein constant rakha gaya $ρ g h$ term provide karta hai.
Incompressibility & continuity equation — fluid compress kyun nahi hona chahiye.
Hydraulic systems — brakes, lifts, jacks — payoff $F_{2} = F_{1} A_{2} / A_{1}$ .
Conservation of energy in machines — distance-for-force energy ko honest kyun rakhta hai.
Archimedes' principle — buoyancy — isotropic fluid pressure par bhi tikaa hai.

Visual walkthrough — Pascal's law — pressure transmits equally

Step 1 — "Pressure" ka matlab kya hai (woh cheez jo hum spread karne wale hain)

Step 2 — Fluid force perpendicular kyun honi chahiye

Step 3 — Tiny wedge banao (hamara test element)

Step 5 — Horizontal forces balance karo → $p_{x} = p_{n}$

Step 6 — Vertical forces balance karo → $p_{y} = p_{n}$ , aur punchline

Step 7 — "Har direction mein equal" se "undiminished transmit hota hai" tak

Step 8 — Edge aur degenerate cases (koi gap mat chhodo)

Ek-picture summary

Connections

Concept Map