§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) 2017/4/25 林再興教授編

= F1+F2= (ρgh1A) + [(1/2)ρg(h2-h1)A] 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 2017/4/25 林再興教授編

B A D C 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 2017/4/25 林再興教授編

F1=(PS+ρgh1)A = {50. 103+[0. 9. (9. 81. 103N/m3). 2m]}. 6. 6 = 24. 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 2017/4/25 林再興教授編

§2.10 Hydrostatic Force on a Curved Surface FH = F2 = ρghCACAAC = PCG(AV)proj FV = F1 + w = ρg (aboveAB) +ρg ( ABC) FR = 2017/4/25 林再興教授編

Example2.6 Given : Determine : The magnitude and line of FR Solution: Length(b) =1m Determine : The magnitude and line of FR Solution: FH = F2 = ρghCACAAC = 62.4 FV=W=ρg(ABC)= 2017/4/25 林再興教授編

§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. 2017/4/25 林再興教授編

For unform density of body immersed  point B = Center of mass ─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 2017/4/25 林再興教授編

─The second law (floating body) dFB=dFup-dFdown =[γ1h+ γ2(y2-h)]dA- γ 1y1dA = γ1(h-y1)dA+ γ2(y2-h)dA = γ1dV1+ γ 2dV2 For a layered fluid 2017/4/25 林再興教授編

Example 2.10 As the figures on right (ρg)seawater =10.1*103N/m3  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? 2017/4/25 林再興教授編

Where FB = (ρg)water Buoy = (10.1*103N/m3)(4/3)π*(1.5/2)3(m3) 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 2017/4/25 林再興教授編

§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 2017/4/25 林再興教授編

2017/4/25 林再興教授編

M MG>0(M above CG) =>Stable MG<0(M below CG) =>Unstable M 2017/4/25 林再興教授編

§2.12 pressure variation in a Fluid with Rigid-Body Motion  The general equation of motion, No shearing stress 2017/4/25 林再興教授編

§2.12.1 Linear Motion Free surface slope = dz/dy acceleration az ay Fluid at rest P1 constant P3 P2 pressure lines Fluid with acceleration az ay 2017/4/25 林再興教授編

No shearing stress 2017/4/25 林再興教授編

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) 2017/4/25 林再興教授編

2017/4/25 林再興教授編

2017/4/25 林再興教授編

2017/4/25 林再興教授編

2017/4/25 林再興教授編

Example 2.11 Given:Right figure ay only Questions: (1) P = (ay) at point 1 (2) ay = ? if fuel level on point 1 Solution: 2017/4/25 林再興教授編

since 2017/4/25 林再興教授編

§2.12.2 Rigid-Body Rotation   since ……(2.30 ) 2017/4/25 林再興教授編

at (r,z) = (0,0), P=P0 2017/4/25 林再興教授編

For a constant pressure surface at air liquid surface, P=P1 b d c volume(ace) Volume(abcde) 2017/4/25 林再興教授編

2017/4/25 林再興教授編

2017/4/25 林再興教授編