Chapter 4
4.2 Buoyancy forces. Archimedes’s law states that the buoyant force exerted on a submerged body is equal to the weight of the displaced fluid and acts in direction opposite to the gravity force. Consider the buoyant force on a submerged circular cylinder of height h and crosssectional area A , shown in Figure 4.2-1.
P
A h
Solid
Liquid z
Figure 4.2-1 Buoyant force on a submerged cylinder.
Making a force balance in the vertical ( z ) direction, the net upward force due to the difference between the opposing pressures on the bottom and top surfaces is
F b
= ( P +
 gh

P ) A =
 ghA
This net upward force is the buoyant force ( F b
) that is the weight of the displaced liquid.
Thus the effective net weight of a submerged body is its actual weight less the weight of an equal volume of the fluid. This result also applies to a body submerged in a fluid that is subject to any acceleration. For example, a solid particle of volume
~
V s
submerged in a fluid within a centrifuge at a point r where the angular velocity is

is subjected to a net radial force equal to (
 s
  f
)

2 r
~
V s
.
Example 4.2-1 .
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Consider the situation shown in Figure 4.2-1, in which a person and a rock rest on a boat that floats in a lake. The rock is then pushed off the boat. If the rock sinks to the bottom, determine whether the water level in the lake will rise, fall, or remain constant, relative to the initial level in (a).
V i
V b
Lake Lake
(a) Initial (b) Final V r
Figure 4.2-1 A person with (a) and without (b) a rock in a boat floats in a lake.
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Solution ------------------------------------------------------------------------------------------
Let the mass of the boat and the person be M b
and the mass of the rock be M r
. Let V i be the initial volume of water displaced then
( M b
+ M r
) g = V i
 g (E-1)
When the boat does not contain the rock, the volume of water displaced by the boat is given by
M b g = V b
 g (E-2)
Since the rock sinks, the weight of the water displaced by the rock no longer suffices to support the weight of the rock. Therefore
M r g > V r
 g (E-3)
Adding (E-2) and (E-3) we have
( M b
+ M r
) g > ( V b
+ V r
)
 g
Comparison (E-1) with (E-4) shows that
(E-4)
V i
> V b
+ V r
Since the volume of water in the lake is constant, and the total displaced volume is reduced, the level of the surface falls.
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4.3 Pressure in Response to External Forces
The only stress that can exist in a fluid at rest is pressure. This also applies to fluids in motion provided there is no relative motion within the fluid since the shear stresses are determined by the velocity gradients or relative motion within the fluid. However the fluid will experience additional normal force if the motion involves acceleration.
Vertical Acceleration
Consider the cylinder of fluid illustrated in Figure 4.3-1. The fluid is accelerating upward with an acceleration of a z
. Applying a momentum or force balance in the zdirection on the control volume A
 z we obtain

F z
= ma z
AP| z

AP| z+
 z
  gA
 z =
 gAa z
Dividing the equation by the control volume A
 z and letting
 z approach zero we obtain
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lim
 z

0
P | z
  z
 z

P | z = dP dz
=
 
( g + a z
) a z
P |
A
 z
(4.3-1) z
P | z
Figure 4.3-1 Vertical accelerating fluid.
The effect of a superimposed vertical acceleration ( a z
) is equivalent to an increase in the gravitational acceleration by a z
. In general, an acceleration ( a i
) in the i direction will result in a pressure gradient within the fluid in the
 i direction, of magnitude
  a i

P
 x i
=
  a i
(4.3-2)
Horizontally Accelerating Free Surface y a x
 x
Figure 4.3-2 Horizontally accelerating car.
Suppose we have a liquid in the bed of a pickup truck that accelerates in the xdirection with constant acceleration a x
. The pressure within the fluid is now a function of both x and y since it experiences both the acceleration of gravity in the ydirection and the car acceleration in the xdirection.
P = P ( x , y )
 dP =

P
 x dx +

P
 y dy
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The slope of the water surface can be determined since dP | s
= 0 along the surface.
  a x dx | s
  gdy | s
= 0
Therefore dP =
  a x dx
  gdy dy dx s
=
 a x g
= tan

The slope of the surface is independent of fluid properties.
Surface of a Rotating Liquid
Consider a liquid in a cylindrical container, as shown in Figure 4.3-3. The system is at steady state with an angular velocity

(radian/s), and the axis is vertical. The pressure within the fluid is a function of z and r , therefore
P = P ( r , z )
 dP =

P
 r dr +

P
 z
dz z o z r

Figure 4.3-3 A container of liquid rotates about its axis.
The pressure gradient in the rdirection is due to the centripetal acceleration (
 r
 2
) experienced by an element of liquid at radial position r . Therefore dP =

r
 2 dr
  gdz
This equation can be integrated along the free surface where dP | s
= 0
 2

0 r rdr

g
 z o z dz = 0
 z = z o
+ r
2

2
2 g
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