2.2.6 · Physics › Fluid Mechanics
Paani se bhari sealed bag ko kahin bhi dabo, aur andar ka har point us dabav ko feel karta hai — sirf woh jagah nahi jahan tumne dabaya. Ek confined fluid extra pressure ko locally "store" nahi karta; woh use fluid ke har hisse aur walls tak bina kisi kami ke pass kar deta hai . Isliye ek patli tube par thodi si push se bhari car lift ho sakti hai: wahi pressure ek bade area par act karta hai, jisse bada force milta hai.
Ek enclosed (confined) incompressible fluid par apply kiya gaya pressure change, fluid ke har hisse aur container ki walls tak bina kisi kami ke transmit hota hai.
Is ek sentence ke andar do ideas chhupe hain:
Ek level par har jagah same pressure — ek height par static fluid mein, pressure har direction mein aur us level ke har point par equal hota hai.
Changes poori tarah transmit hote hain — agar ek point par pressure Δ p badhao, toh har point par wahi Δ p badhta hai.
Intuition "Har direction mein equal" kyun?
Fluid bahta hai agar koi bhi unbalanced sideways force ho. Static equilibrium mein kuch nahi bahta, isliye kisi bhi chote se chunk par net force zero hona chahiye. Hum prove karenge ki yeh cheez pressure ko majboor karti hai: pressure tab hi balanced ho sakta hai jab pressure har direction mein same ho ek point par.
Setup: Ek chote se wedge-shaped fluid element (triangular prism) lo thickness ke saath, ek static fluid ke andar. Teen faces par pressures ho: p x (vertical face par), p y (horizontal face par), p n (slanted face par). Element itna chota hai ki hum gravity ignore kar sakte hain (volume → 0 , area se zyada tezi se).
Maan lo slanted face angle θ banata hai, areas ke saath:
vertical face area = d A x
horizontal face area = d A y
slant face area = d A n
Wedge ki geometry se milta hai:
d A x = d A n sin θ , d A y = d A n cos θ
Horizontal balance (x): vertical face par pressure force = slant face par force ka horizontal component
p x d A x = p n d A n sin θ
Yeh step kyun? Pressure ek surface ke andar perpendicular push karta hai; sirf slant-face force ka horizontal component, vertical-face force ko oppose karta hai.
d A x = d A n sin θ substitute karo:
p x d A n sin θ = p n d A n sin θ ⇒ p x = p n
Vertical balance (y): isi tarah
p y d A y = p n d A n cos θ ⇒ p y = p n
Isliye:
p x = p y = p n
Ab transmission wala part. Hydrostatic relation yaad karo (height h ke fluid column se):
p = p 0 + ρ g h
Agar hum external/applied pressure p 0 → p 0 + Δ p badhate hain, toh har depth h par:
p new = ( p 0 + Δ p ) + ρ g h = p old + Δ p
ρ g h term unchanged rehta hai (same fluid, same heights), isliye increase Δ p har point par identical hota hai . Yahi Pascal's law hai.
Worked example 1. Hydraulic lift force multiplication
Chota piston: radius r 1 = 2 cm . Bada piston: radius r 2 = 20 cm . Tum F 1 = 100 N se push karte ho. Kitna weight lift ho sakta hai?
A 1 = π r 1 2 , A 2 = π r 2 2 ⇒ A 2 / A 1 = ( r 2 / r 1 ) 2 = ( 20/2 ) 2 = 100 .
Squared kyun? Area, radius² ke saath scale karta hai — yahan se hi bada boost milta hai.
F 2 = F 1 A 1 A 2 = 100 × 100 = 10 , 000 N ≈ 1020 kg .
Worked example 2. Distance / work check
Ex.1 mein bada piston d 2 = 1 cm utha chahiye. Chote piston ko kitna push karna hoga?
Volume in = volume out: A 1 d 1 = A 2 d 2 ⇒ d 1 = d 2 ( A 2 / A 1 ) = 1 × 100 = 100 cm .
Kyun? Incompressible fluid — jo ek cylinder se nikalta hai woh doosre mein jaata hai.
Work check: F 1 d 1 = 100 × 1 m = 100 J ; F 2 d 2 = 10000 × 0.01 m = 100 J . ✓ Energy conserved — koi free lunch nahi.
Worked example 3. Top par pressure add karna
Paani ke ek closed tank par upar piston hai. Shuruaat mein pressure neeche (depth 5 m par) p 0 + ρ g h hai. Tum piston dabate ho, surface pressure Δ p = 4000 Pa badhata hai. Neeche ka pressure kitna badhega?
Δ p bottom = 4000 Pa — bilkul utna hi .
Yeh step kyun? Pascal: ρ g h wala part unchanged rehta hai; applied increase bina kami ke transmit hota hai.
Common mistake "Pressure jahaan maine dabaya us se door ya neeche jaane par kam ho jaata hai."
Kyun sahi lagta hai: Ek solid ya gas mein, force spread out hota hai aur weak ho jaata hai (springs, foam locally squish hote hain). Aur total pressure depth ke saath badhta hai (ρ g h ), toh lagta hai "neeche zyada ho raha hai."
Fix: Pascal, tumhare apply kiye gaye change Δ p ke baare mein hai, absolute value ke baare mein nahi. Applied change bina kami ke transmit hota hai; ρ g h wala part ek alag, fixed contribution hai jo decay bhi nahi karta.
Common mistake "Lift free energy deta hai kyunki
F 2 ≫ F 1 ."
Kyun sahi lagta hai: Chhoti si input force se bahut bada output force milta hai.
Fix: Tum distance ko force ke saath trade karte ho. Bada piston A 1 / A 2 times kam move karta hai. F 1 d 1 = F 2 d 2 — work conserved hai (friction ignore karte hue).
Common mistake "Pascal's law kisi bhi fluid ke liye kaam karta hai."
Kyun sahi lagta hai: Paani aur oil dono push back karte hain.
Fix: Iske liye strictly enclosed, incompressible fluid chahiye. Ek compressible gas change ka kuch hissa compress hokar absorb kar leta hai, isliye transmission perfectly undiminished/instant nahi hoti.
Pascal's law ek sentence mein Ek enclosed incompressible fluid par apply kiya gaya pressure change, fluid ke har hisse aur container walls tak bina kisi kami ke transmit hota hai.
Ek static fluid mein kisi point par pressure har direction mein equal kyun hota hai? Kyunki equilibrium mein ek tiny wedge element par net force zero hota hai; wedge geometry force karta hai ki p x = p y = p n .
Hydraulic press force relation F 2 = F 1 A 2 / A 1 (kyunki p 1 = p 2 ).
Hydraulic lift mein chhoti si force bada load kyun lift karti hai? Wahi pressure ek bade area par act karta hai; force = pressure × area, aur A 2 / A 1 bada hota hai.
Kya hydraulic press free energy deta hai? Nahi — F 1 d 1 = F 2 d 2 ; tum distance ko force ke saath trade karte ho.
Agar tum surface pressure Δ p badhate ho, toh depth h par pressure kitna badhega? Exactly Δ p se (ρ g h term unchanged rehta hai).
Pascal's law ke liye do conditions Fluid enclosed (confined) aur incompressible hona chahiye.
1 cm aur 10 cm radii wale pistons ke liye mechanical advantage? ( 10/1 ) 2 = 100 .
Recall Feynman: 12-saal ke bacche ko samjhao
Socho ek balloon poori tarah paani se bhara hai, bina kisi haawa ke. Agar tum ek taraf apni ungli se poke karo, toh poora balloon har jagah se bahar aata hai — sirf wahan se nahi jahan tumne dabaya. Paani squash nahi ho sakta, isliye tumhara push kahi na kahi jaana chahiye, aur woh paani ke rubber ke har bit ko equally push karta hai. Car lift mein yahi trick use hoti hai: tum paani ki ek patli straw ko dabate ho, aur wahi push ek mote platform par phail jaata hai. Ek choti jagah par chhoti push ek badi jagah par giant push ban jaati hai — lekin tumhein apni patli straw ko bahut loooong door push karna padta hai car ko thoda sa lift karne ke liye. Kuch bhi free nahi; tum sirf "kitna zyada" aur "kitna door" ko swap karte ho.
"PEEW" — P ressure, E nclosed, E qual, W alls: Pressure ek Enclosed fluid mein har point aur Walls tak Equally jaata hai. Aur lift ke liye: "Zyada area, zyada force; zyada force, kam distance."
Hydrostatic pressure — p = p0 + ρgh — woh ρ g h term deta hai jise humne constant rakha.
Pressure — force per unit area — basic definition p = F / A jo press mein use hoti hai.
Hydraulic systems — brakes, lifts, jacks — direct engineering application.
Archimedes' principle — buoyancy — isotropic fluid pressure par bhi rely karta hai.
Conservation of energy in machines — explain karta hai kyun press free-energy device nahi hai.
Incompressibility & continuity equation — kyun A 1 d 1 = A 2 d 2 .
same pressure bigger area
Static equilibrium no flow
Pressure isotropic px=py=pn
Delta p same at every depth
Small push lifts big load
Perpendicular into surface