Buoyancy, Flotation, and Stability Buoyant force: When a body is completely or partially submerged

& Upward vertical force exerting on a body due to pressure difference : Fbottom of body () > FTop ()

y Question: Determine the magnitude of Buoyant force. Consider a completely submerged body (Arbitrary shape and volume V ) Step 1. Enclose a body with a parallelepiped Step 2. Find all forces exerting on fluid

INSIDE a parallelepiped (INSTEAD of forces on the body).

1F & 2F : Forces on Horizontal planes

3F & 4F : Forces on Vertical planes

W : Weight ( = Volume of fluid in a parallelepiped NOT body !!) FB: Force of the body exerting on fluid

Archimedes Principle

Step 3. Equations of motion (Equilibrium) 1) Horizontal components

0 43 =+ FF 2) Vertical component

0 12 = BFWFF

& VVAhhAhAhFB == ])[() () ( 1212 where A: Top area (AB) of parallelepiped V: Volume of a body From the static (equilibrium) condition, BF : Body force exerting on the fluid

= Fluid force exerting on the body (Action - Reaction) VFB = : Buoyant force (Archimedes Principle) = Weight of fluid displaced by the body (Upward)

h1

h2

Area A

W

Volume of a parallelepiped

W

y Question: Determine the location of FB

Consider a moment equation about an axis passing through D (x axis) Moment = 0y c21112 = BFWyyFyF

0 ])[( 2121112 = cVyyVAhhAyhAyh

& 0 ])[( ) ( 212112 = cyVyVAhhyAhh

& 21 )( yVVyVyV TcT += (Equation of Volume center) cy : y coordinate of the centroid of displaced volume V By a similar manner, cx : x coordinate of the centroid of displaced volume V

Total volume of a parallelepiped, VT

Center of Total volume

Center of Displace volume

Center of Fluid volume in parallelepiped

What to be determined

Stability of submerged or floating bodies What does the stability of a body mean? - Stable: When slightly displaced, it returns to original position - Unstable: When displaced, it moves to new equilibrium position Question: Determine the stability of submerged or floating bodies

- Relation between the center of buoyancy and gravity

Type 1. Completely submerged body (1) If Center of gravity (CG) of body: Below center of buoyancy (CB) A small displacement FB W pair: Restoring force Return to the original position Stable equilibrium (2) Center of gravity (CG) of body: Above center of buoyancy (CB) A small displacement FB W pair: Overturning force Move to new position

Unstable equilibrium Applicable only to completely submerged bodies

Type 2. Floating body (Partially submerged body) - Hard to determine due to change of CB for a small rotation Stable situation although CG: above CB Unstable situation We must consider the geometry and weight distribution

- Important in the design of ships, submarines, barges, etc.

az

ay ar

Pressure Variation in a Fluid in motion - But still under a no shearing stress ( ) condition, only if all fluid particles are moving with same acceleration.

- Rigid-body motion & NOT interacting with the wall of container Not a static situation, but we can still use the equation from Sec 2.2,

akp r = : Derived only under the condition of no Valid not only for a static ( )0=ar situation but also a dynamic case ( )0ar )

Case 1. Linear Motion Consider a open container - Translating along a line - Constant acceleration ar Q. Determine the slope of free surface y Equations of motion (from akp r = )

xp = 0 (since ax = 0)

yayp =

)( zagzp +=

Related with change in p by changing the points of interest

Pressure difference between two points at (x,y,z) and (x, y+dy, z+dz)

dp = xp dx +

yp dy +

zp dz (because No p along x-axis)

dzagdya zy )( += If two points are on a special line, i.e. 0=dp (Constant pressure!) Lets find the line (or surface) of constant pressure. (e.g. free surface) Since dp = 0 & dzagdya zy )(0 +=

or =dydz

z

y

aga+ : Slope of a line of constant p

(= Slope of free surface) Special example. Vertical motion of a container (ay = 0, az 0)

dydz = 0 (Horizontal line)

)( zagdzdp += (p: vary linearly, but not hydrostatic)

(Compare to gdzdp = or ghp = : Hydrostatic pressure)

Case 2 Rigid-body Rotation Consider a situation shown - Rotating in a circular motion - Constant Q. Determine the slope of free surface (or the line of constant p) y Equation of motion in polar coordinates (r, , z) akp r =

& )(1 2 rzzrz ereezpep

re

rpep =

+

+=

or

2rrp = 0=

p =

zp

Pressure change for small changes in dr and dz

dp = dpdr

rp

+

+ dzdrrdzzp = 2

y Line (or surface) of constant pressure ( 0=dp )

drdz =

gr 2 (since g = ) : Slope of the free surface

ra = r 2

x

y

z

re

ze

e

By integrating the equation,

z = gr

2

22 + constant

: Line (or surface) of constant pressure (Parabolic surface ) Direct relation between pressure (p), distance from the axis of rotation (r), and vertical depth (z), By integrating the equation of motion, dzdrrdp = 2 = dzrdrdp 2

or p = zr 2

22

+ constant

i. p: Increases as the point moves away (r increases) ii. p: Hydrostatically increase along z at a fixed r