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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 1 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Engineering 36 Chp09: FluidStatics

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 2 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Fluid Statics The Fish Feels More “Compressed” The Deeper it goes – An example of HydroStatic Pressure Definition of Pressure Pressure is defined as the amount of force exerted on a unit area of a surface:

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 3 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Direction of fluid pressure on boundaries Furnace ductPipe or tube Heat exchanger Dam Pressure is a Normal Force i.e., it acts perpendicular to surfaces It is also called a Surface Force

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 4 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Absolute and Gage Pressure Absolute Pressure, p: The pressure of a fluid is expressed relative to that of a vacuum (absence of any substance) Gage Pressure, p g : Pressure expressed as the difference between the pressure of the fluid and that of the surrounding atmosphere (the BaseLine Pressure) which has an Absolute pressure, p 0. p 0 ≡ BaseLine Pressure:

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 5 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Pressure Units

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 6 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Static Fluid Pressure Distribution Consider a Vertical Column of NonMoving Fluid with density, ρ, and Geometry at Right Taking a slice of fluid at vertical position-z note that by Equilibrium: The sum of the z-directed forces acting on the fluid- slice must equal zero

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 7 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics The z-Direction Fluid Forces Consider the Fluid Slice at z under the influence of gravity S is the Top & Bot Slice-Surface Area x y z Let p z and p z+Δz denote the pressures at the base and top of the slice, where the elevations are z and z+Δz respectively

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 8 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics The z-Direction Fluid Forces A Force Balance in the z-Dir x y z Solving for Δp/Δz and Taking the limit Δz→0

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 9 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Constant Density Case For Constant ρ Next Consider the typical Case of a “Free Surface” at Atmospheric Pressure, p 0, and total Fluid Depth, H If h is the DEPTH, then the z↔h reln:

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 10 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Constant Density Case Then Sub into Δp Eqn Now if at h = 0, p = p 0, and Δh = h 2 – 0 = h In this Case

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 11 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Constant Density Case The Gage Pressure, p g, is that pressure which is Above the p 0 Baseline Thus the typical Formulation for Liquids at atmospheric pressure The Gage Pressure is used for Liquid-Tanks or Dams

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 12 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Specific Weight Since the acceleration of gravity, g, is almost always regarded as constant, the ρ liq g product is often called the Specific Weight Then the pressure eqns

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 13 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Bouyancy A body immersed in a fluid experiences a vertical buoyant force equal to the weight of the fluid it displaces A floating body displaces its own weight in the fluid in which it floats Free liq surf F1F1 F2F2 h1h1 h2h2 ΔhΔh

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 14 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Bouyancy The upper surface of the body is subjected to a smaller force than the lower surface Thus the net bouyant force acts UPwards Free liq surf F1F1 F2F2 h1h1 h2h2 ΔhΔh The net force due to pressure in the vertical direction F B = F 2 - F 1 = (p bottom - p top )(ΔxΔy) = Δp(ΔxΔy)

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 15 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Bouyancy The pressure difference Subbing for Δp in the F B Eqn F B = (ρgΔh)(ΔxΔy) But ΔhΔxΔy is the Volume, V, of the fluid element So Finally the Bouyant Force Eqn Free liq surf F1F1 F2F2 h1h1 h2h2 ΔhΔh p bottom – p top = ρg(h 2 -h 1 ) = ρgΔh Δp = ρgΔh

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 16 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 17 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Equivalent Point Load Consider Fluid Pressure acting on a submerged FLAT Surface with any azimuthal angle θ In this situation the Net Force acting on the Flat Surface is the pressure at the Flat Surface’s GEOMETRIC Centroid times the Surface area i.e., F liq = the AVERAGE Pressure times the Area

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 18 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Equivalent Point Load F liq is NOT Positioned at the Geometric Centroid; Instead it is located at the CENTER of PRESSURE Calculation of the Center of Pressure Requires Moment Analysis as was described last lecture

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 19 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Force on a Submerged Surface Pressure acts as a function of depth Then the Magnitude of the Resultant Force is Equal to the Area under the curve y 0 dd Resultant, R d ΔyΔy Face Width of Structure INTO the Screen = b

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 20 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Force on a Submerged Surface The location of the Resultant of the pressure dist. is found the by same technique as used on the horizontal beam Thus the Location of the Resultant Hydrostatic force Passes Thru the Pressure-Area Centroid y 0 dd Resultant, R d S The Distance OS is Also Called the Center of Pressure

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 21 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Angled Submerged Surface How does one find the forces on a SUBMERGED surface (red) at an angle? Construct the FBD Detach The Triangular Fluid Volume from the bulk fluid to Reveal Forces: Weight of the Volume Pressure at NORMAL Surfaces Can also ATTACH the Fluid to the Body if desired

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 22 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Angled SubMerged Surface The resulting force distribution without the weight of the water would show Equal to the Trapezoidal Area Under the Pressure Curve Times the Width b

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 23 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics NonLinear Submerged Surface Consider The P-distribution on a non-linear surface. Liquid Generates Resultant, R on The Surface Use Detached Fluid Volume as F.B.D. The Force Exerted by the Surface is Simply −R

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 24 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics HydroStatic Free Body Diagram Examine Support Structure for Non-Rectilinear Surfaces Detach From Surrounding Liquid, and And from the Structure, a LIQUID VOLUME that Exposes Flat X&Y Surfaces exposed to Liquid Pressure Pressure is –Constant at X-Surfaces (Horizontal) –Triangular or Trapezoidal at Y-Surfaces (Vertical)

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 25 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Centroids of Parabolas SectorSpandrel

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 26 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics WhtBd Example: P9-122 Determine the resultant horizontal and vertical force components that the water exerts on the side of the dam. Find for R the PoA on the Dam-Face The dam is 25 ft long SW for Water = 62.4 lb/cu-ft

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 27 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 28 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Resultant Location

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 29 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics % Bruce Mayer, PE % ENGR36 * 03Dec12 % ENGR36_Dam_Face_P9_122_1212.m % W = 260; % kip P = 487;.5 % kip % Deqn = [P/4 W -(25*P/3+30*W/8)] x = roots(Deqn) y = (x(2))^2/4 xp = linspace(0,10, 500); yp1 = polyval(Deqn, xp); yp = xp.^2/4; xD = x(2) yD = y plot(xD,yD,'s', xp, yp, 'LineWidth', 3), grid, xlabel('x'),... ylabel('y'), title('P9-122 Dam-Face Force Location') Solve By MATLAB Deqn = 1.0e+03 * 0.1218 0.2600 -5.0333 x = -7.5856 5.4500 y = 7.4257 xD = 5.4500 yD = 7.4257

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 30 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Example Small Dam Given Fresh Water dam with Geometry as shown If the Dam is 24ft wide (into screen) Find the Resultant of the pressure forces on the Dam face BC Assume Detach the Parabolic Sector of Water

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 31 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Example: Small Dam Create FBD for Detached Water-Chunk The gage Pressure at the Bottom

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 32 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Example: Small Dam The Force dF 18 at the base of the Dam with Face Width, b (into paper), consider a vertical Distance dh located at 18ft Deep

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 33 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Example: Small Dam Now to Covert dF to w (lb/vertical-Foot) Simply Divide by the vertical distance that generated dF In this case b = 1ft

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 34 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Example: Small Dam Digression For a Submerged Flat surface, with Face Width, b, in a constant density Fluid the Load per Unit-Length profile value, w(h) can found as In Units of lb/ft or lb/in or N/m

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 35 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Example: Small Dam The Load Profile is a TriAngle → A = ½BH B = 1123 lb/ft H = 18 ft Then the Total Water Push, P, is the Area under the Load Profile Previous Slide

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 36 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Example: Small Dam Then the Volume of the Detached Water And the Weight, W 4, of the Attached Water

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 37 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Example: Small Dam And the Location of the of P is the Center of Pressure which is located at the Centroid of the LOAD PROFILE In this case the CP is 6ft above the bottom Also W4 is applied at the CG of the parabolic sector From “Parabola” Slide

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 38 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Example: Small Dam Now the Dam also pushes on the Detached Water For Equilibrium the Push by the Dam on the Water-Chunk must be the negative of Resultant of the Load ON the Dam The Applied Loads are P and W 4 Then the FBD for the Water-Chunk

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 39 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Example: Small Dam The Water Chunk FBD Notice that This is a 3-Force Body Thus the Forces are CONCURRENT and no Moments are Generated The Force Triangle Must CLOSE Use to Find Mag & Dir for R

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 40 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Example: Small Dam The Force Triangle: The the Load per Unit Width of the Dam 12580 lb/ft-Width Directed 36.5° below the Horizontal and to the Left relative to the drawing

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 41 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Resultant Location on Dam Determine the CoOrdinates, Horizontally & Vertically, for the Point of Application of the 12 580 lb Take the ΣM C =0 about the upper-left Corner where the water-surface touches the dam

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 42 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 43 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics MATLAB Results W = 7488 P = 10109 k = 0.0309 Deqn = 1.0e+05 * 0.0023 0.1011 -1.6624 x = -56.4769 12.7360 y = 5.0064 xD = 12.7360 yD = 5.0064 x canNOT be Negative in this Physical Circumstance

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 44 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics PoA for R on Dam Face

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 45 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics MATLAB Code % Bruce Mayer, PE % ENGR36 * 23Jul12 % ENGR36_Dam_Face_1207.m % W = 7488 P = 10109 k = 10/18^2 Deqn = [k*W P -(6*W+12*P)] x = roots(Deqn) y = k*(x(2))^2 xp = linspace(0,18, 500); yp = polyval(Deqn, xp); yp = k*xp.^2; xD = x(2) yD = y plot(xD,yD,'s', xp, yp, 'LineWidth', 3), grid, xlabel('x'),... ylabel('y'), title('Dam Face Force Location‘)

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 46 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics WhiteBoard Work A 55-gallon, 23-inch diameter DRUM is placed on its side to act as a DAM in a 30-in wide freshwater channel. The drum is anchored to the sides of the channel. Determine the resultant of the pressure forces acting on the drum and anchors. Barrel Dam Problem ATTACH some Water Segments to the Drum

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 47 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 48 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 49 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 50 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 51 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 52 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 53 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Bruce Mayer, PE Registered Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Engineering 36 Appendix

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BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 54 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics

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