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Hydrostatic forces on curved surfaces by hcj

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									                        Hydrostatic forces on curved surfaces

(a) Liquid above the surface                                       Rule #1:
                                                                   The vertical component of pressure force on a curved surface is
Magnitude: We are interested in computing the hydrostatic force    equal to the weight of liquid vertically above the curved surface and
acting on the curved surface shown in the above figure. Consider   extending up to the free surface or its extension.
the differential element in the above figure. Let us analyze the
force acting on it.                                                The line of action of this force is a long the center of mass of the
                                                                   fluid volume and act downward.
Vertical Component:
                                        D                 C        Horizontal Component:

      dFy       
               dA 
                                                                                    dFx   h dAy
dAx               dF                             Fy
                  dAy                       h         
                                                                                    Fx    h dAy
                        dFx                          dF , P
                                                                                               h Ay  yAy
                                   A

                                                                   where
                                   C                               Ay is the horizontal projection of the curved surface onto a vertical
                                   CP
                                                Fx                 plane.
                                                              B
                                                                    h  y is the distance from the free surface to the centroid of
Consider then the forces acting on the differential                the projected area (Ay).
element shown in the figure.
                                                                 Rule #2:
            dF  PdA                                               The horizontal component of pressure force on a curved surface is
                                                                  equal to the pressure force exerted on the horizontal projection of
                hdA                                              the curved surface onto a vertical plane normal to the component.
            dFy  hdAx                                            The line of action is determined from the plane surface formula
            but hdAx  d
                                                                                                             I
            Fy    d                                                                    ycp  y 
                                                                                                            yA
             Fy     W 
                                                                   where
            (acting downward )                                     I is the second moment of the projected area along the Centroidal
                                                                   axis.
                                                                   A is the projected area, (Ay).


                                                                                                                        1
                                                                       above the curved surface and extending up to the free surface or its
(b) Liquid below the surface                                           extension.
                                                                       The line of action of this force is a long the center of mass of the
To analyze this problem, we vertically project the curved surface      imaginary fluid volume and acting upward.
onto the free surface extension. Then we imagine that the volume
ABCD is filled with an imaginary fluid similar to the real fluid.      Horizontal Component:
Consider then the differential element shown in the figure. The
               D                C                                      The magnitude and the line of action of the horizontal component is
                                                                       computed exactly as we did in Part A
Imaginary
  fluid
                                                               
                                                                 
                                                            P, dF     Final Note on the horizontal component
                                                             dA
                            h                P, dF



           A                                                                                         Closed Surface



                                  B                                             Curved surface                            Curved surface
                                                                              with equal elevation                    with different elevation
pressure at this element point is the same in all direction as was                   ends                                      ends

proved earlier. Hence,
                                                                        The horizontal component of pressure force on a closed surface or
                         P   P                                      a curved surface with equal elevation for both ends is always zero,
                                                                       since on opposite sides of the body the area-element projections
multiply both sides by dA                                              have opposite signs as shown on in the figures.

                                                                       However, if the ends of the curved surface do not have same
                    dF   P dA                                      elevations, then there will be a net horizontal force.
                           h dA
where h is the distance from the free surface (or its extension) to
the differential element.

Now we have a problem similar to part a. Hence,
               i.e. Fy   V  W 
                   (acting upward )
Rule #3:
The vertical component of pressure force due to liquid below a
curved surface is equal to the weight of imaginary liquid vertically



                                                                                                                                                 2
                          Buoyancy forces                                                   Concept of Hydrometry
        Having obtained the necessary tools to analyze curved                 A hydrometer uses the principle of buoyant force to
        surfaces, we are now able to compute the buoyancy                     determine the specific gravities of liquids.
        force acting on a totally (or partially) submerged body.

                               Q                          M



                                           B


                           A                                  C
                                                                                       stem
                                                                                                              h
                                               D
                                                                                                                             0
        We split the body into two surfaces, namely, ABC and
        ADC (due to liquid is above the former and below the                                             FB
        latter)

        (Fy)ABC = fluid  VABCMQ = VABCMQ                                               Whydrometr
        (Fy)ADC = fluid  VADCMQ = VADCMQ 
        (Fy)net = fluid  VABCD = VABCD 
                                                                              If we immerse a hydrometer in a fluid then it gives us
                     = weight of the displaced fluid.
                                                                              the reading S (specific gravity, = /water)

                                                                                                    stem  h  Astem
    Q                         M            Q                          M

           WABCMQ                                       WADCMQ
                 B                                                            A force balance in the y direction gives:

                                                                                     FB  Whydrometer
A                                  C   A                                  C

                                                              D

                                                   B                                (0  stem )  Whydrometer
                          A                            WABCD      C                 (0  stem ) S water  Whydrometer
                                                   D                                                Whydrometer
        The line of action is along the centroid of the displaced                    S
        volume of fluid.                                                                     (0  stem ) water


                                                                                                                                  3
Stability of Immersed Bodies
A body is said to be in a stable equilibrium
position if, when displaced, it returns to its                 Unstable Equilibrium Position
equilibrium position. Conversely, it is in an
unstable equilibrium position if, when displaced
(even slightly), it moves to a new equilibrium      If Center of gravity is above the centroid, the body
position.                                           is unstable. An overturning restoring moment will
                                                    bring the body to a new equilibrium position.
             Stable Equilibrium Position


If Center of gravity is below the centroid, the
body is stable. A restoring moment will bring the
body back to its original position.




                                                                                                      4
                                   Stability of Floating Bodies


                                                                                 M
                             G                                              G


                     C                                                                   Righting moment
                                                       Old position of               C
                         Fb                             the centroid

                                                          New position of       Fb
                                                           the centroid




New position of
Center of grav ity
                                                      M e t a c e n t e r ( M ) is d e fi ne d a s t he
                                                      intersection point of the line of action of the
                                 G                    buoyant force before and after the tilt.
                                      Ov er-turning
                                       Moment
                     M                                M etacenteric Hight (GM ) is defined as the
                                                      distance from G to M. If M is above G then the
                                                      body is stable otherwise it is not.

  Old position of        C
   the centroid

                              Fb




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