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Hydrostatic Forces and Buoyancy in Fluid Mechanics

Understand the principles of hydrostatic forces, pressure variation, and buoyancy in fluid mechanics with detailed examples and calculations provided by Professor Lin Zai Xing. Learn how to calculate resultant forces and buoyant forces on curved and submerged surfaces.

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Hydrostatic Forces and Buoyancy in Fluid Mechanics

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  1. §2.9 Pressure Prism Area of the plane is rectangular Since p0=0 at surface p = p0+ρgh at bottom FR= pCGA = (p0+ρghCG)A =ρg (h/2) A where p0=0 =ρg (h/2) b·h = 1/2 (ρgh)·b·h = (Volume of pressure prism) The resultant force must pass through the centroid of the pressure prism (1/3 h above the base) 林再興教授編

  2. FR= (Volume of trapezoidal pressure prism ) = (volume of pressure prism ABDE) +(volume of pressure prism BCD) = F1+F2= (ρgh1A) + [(1/2)ρg(h2-h1)A] FRyR = F1y1+F2y2 (F1+F2) yR = F1[(h2-h1)/2]+F2 [(h2-h1)/3] Only if the area of plane is rectangular 林再興教授編

  3. Example 2.8 Given：as figure SG=0.9(oil) square 0.6m * 0.6m Pgage = 50Kpa = PS Find: the magnitude and location of the resultant force on the attached plate. Solution： Trapezoidal pressure prism, ABCDO FR = (vol. of trapezoidal pressure prism) = (vol. of ABDO)+(Vol. of BCD) = F1+F2 B A D C 林再興教授編

  4. F1=(PS+ρgh1)A = {50*103+[0.9*(9.81*103N/m3)*2m]}* 0.6 * 0.6 = 24.4 * 103 N F2=(Vol. of BCD)=[ ] A = *(0.6m*0.6m) =0.954*103N FR=F1+F2=25.4*103N By summing moments around an axis through point O,so that FRy0=F1*0.3m+F2*0.2m (25.4*103N)y0=24.4*103*0.3+0.954*103N*0.2m  y0 = 0.296m 林再興教授編

  5. §2.10 Hydrostatic Force on a Curved Surface FH = F2 = ρghCACAAC = PCG(AV)proj FV = F1 + w = ρg (aboveAB) +ρg (ABC) FR = 林再興教授編

  6. Example2.6 Given : Length(b) =1m Determine : The magnitude and line of FR Solution: FH = F2 = ρghCACAAC = 62.4 FV=W=ρg(ABC)= 林再興教授編

  7. §2.11 Buoyancy, Flotation, and Stability §2.11.1 Archimedes’ Principle   ─Buoyant force : when a body is completely submerged in a fluid, or floating so that it is only partially submerged, the resultant fluid force acting on the body is called the buoyant force. ─Buoyant force direction: upward vertical force because pressure increases with depth and pressure forces acting from below are larger than the pressure force acting from above. 林再興教授編

  8. ─The first law (immersed body) In equilibrium  dFB= dFup –dFdown =ρgy2dA–ρgy1dA =ρg(y2-y1)dA = ρghdA = γhdA = weight of fluid equilibrium to body volume  For unform density of body immersed  point B = Center of mass For variable density of body immersed  point B≠ Center of mass 林再興教授編

  9. ─The second law (floating body) dFB=dFup-dFdown =[γ1h+ γ2(y2-h)]dA- γ1y1dA = γ1(h-y1)dA+ γ2(y2-h)dA = γ1dV1+ γ2dV2 For a layered fluid 林再興教授編

  10. Example 2.10 Given: As the figures on right (ρg)seawater =10.1*103N/m3 Note:(ρg)fresh water = 9.80*103N/m3 Weight of buoy = 8.50*103N Dbuoy = 1.5m Determine: What is the tension of the cable? 林再興教授編

  11. Solution: For equilibrium ΣF=0  T+W – FB = 0 T = FB – W Where FB = (ρg)waterBuoy = (10.1*103N/m3)(4/3)π*(1.5/2)3(m3) = 1.785*104N W = 8.50*103N  T = 1.785*104 – 8.50*103 = 9.35*103N 林再興教授編

  12. §2.11.2 Stability “Distrub” the floating body slightly (a)develops a restoring moment which will return it to its origin positionstable (b)otherwise  unstable The basic principle of the static-stability calculations (1)The basic floating position (2)The body is titled a small angle Δθ 林再興教授編

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  14. M MG>0(M above CG) =>Stable MG<0(M below CG) =>Unstable M 林再興教授編

  15. §2.12 pressure variation in a Fluid with Rigid-Body Motion The general equation of motion, No shearing stress 林再興教授編

  16. §2.12.1 Linear Motion Free surface slope = dz/dy Fluid at rest P1 constant P3P2 pressure lines Fluid with acceleration az ay 林再興教授編

  17. No shearing stress 林再興教授編

  18. Along a line of constant pressure , dp=0 All lines of constant pressure will be parallel to the free surface If from Eq(2.26) 林再興教授編

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  23. Example 2.11 Given:Right figure ay only Questions: (1)P = (ay) at point 1 (2)ay = ? if fuel level on point 1 Solution: 林再興教授編

  24. since 林再興教授編

  25. §2.12.2 Rigid-Body Rotation since ……(2.30 ) 林再興教授編

  26. at (r,z) = (0,0), P=P0 林再興教授編

  27. For a constant pressure surface at air liquid surface, P=P1 a e b d c volume(ace) Volume(abcde) 林再興教授編

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