..

>asses: eachormalrefore;enter

urved

gas ISforceslrface

jected

-rtiallyA net:orcesan bees onlersedraw aNote)f theghtof~rtingmcel,

:2.21)

dEq.

2.22)V2.6

:ctionhown

'eight

I

2.11 Buoyancy, Flotation, and Stability 75

11 "'-'.: :--: ::::::-::_-:::::~ -:: :-:--:-:'=:-:::::::::~~~::::::::::::::

I ~1

h2l

[~ - --- B

;L,I --I II

I Y~ I xJ

--

Centroidof displaced

volume

(d)

D c

(a)Centroid

A

. FIGURE 2.24Buoyant force onsubmerged and float-ing bodies.

F3-+-

D

(e)

c

(b)

of the fluid displaced by the body and is directed vertically upward. This result is commonlyreferred to as Archimedes' principle in honor of Archimedes (287-212 B.C.), a Greek mech-anician and mathematician who first enunciated the basic ideas associated with hydrostatics.

The location of the line of action of the buoyant force can be determined by summingmoments of the forces shown on the free-body diagram in Fig. 2.24b with respect to someconvenient axis. For example, summing moments about an axis perpendicular to the paper

through point D we have .

FBYe = F2YI - FIYI - 'WY2

and on substitution for the various forces

VYe = VTYI - (VT - V)Y2 (2.23)

where VT is the total volume (h2 - hI)A. The right-hand side of Eq. 2.23 is the first momentof the displaced volume V with respect to the x-z plane so that Ycis equal to the Y coordinateof the centroid of the volume V. In a similar fashion it can be shown that the x coordinateof the buoyant force coincides with the x coordinate of the centroid. Thus, we conclude thatthe buoyantforce passes through the centroid of the displaced volume as shown in Fig. 2.24c.The point through which the buoyant force acts is called the center of buoyancy.

These same results apply to floating bodies which are only partially submerged, asillustrated in Fig. 2.24d, if the specific weight of the fluid above the liquid surface is verysmall compared with the liquid in which the body floats. Since the fluid above the surface isusually air, for practical purposes this condition is satisfied.

In the derivations presented above, the fluid is assumed to have a constant specificweight, y. If a body is immersed in a fluid in which 'Yvaries with depth, such as in a layeredfluid, the magnitude of the buoyant force remains equal to the weight of the displaced fluid.However, the buoyant force does not pass through the centroid of the displaced volume, butrather, it passes through the center of gravity of the displaced volume.

76 Chapter 2 / Fluid Statics

EXAMPLE2.10

A spherical buoy has a diameter of 1.5 m, weighs 8.50 kN, and is anchored to the sea floorwith a cable as is shown in Fig. E2.l0a. Although the buoy normally floats on the surface,at certain times the water depth increases so that the buoy is completely immersed as illus-trated. For this condition what is the tension of the cable?

Pressureenvelope

(a) (b) (c) .. FIGURE E2.10

SOLUTION

We first draw a free-body diagram of the buoy as is shown in Fig. E2.1Ob,where FEis thebuoyant force acting on the buoy, OWis the weight of the buoy, and T is the tension in thecable. For equilibrium it follows that

T = FE - OW

From Eq. 2.22

FE = ')IV

and for seawater with ')I = 10.1 kN/m3 and V = 1Td3/6then

FE = (10.1 X 103 N/m3)((1T/6)(lSm)3] = 1.785 X 104 N

The tension in the cable can now be calculated as

T = 1.785 X 104N - 0.850 X 104 N = 9.35 kN (Ans)

Note that we replacedthe effectof the hydrostaticpressureforce on the bodyby thebuoyant force, FE, Another correct free-body diagram of the buoy is shown in Fig. E2.1Oc.The net effect of.the pressure forces on the surface of the buoy is equal to the upward forceof magnitude, FE(the buoyant force). Do not include both the buoyarit force and the hydro-static pressure effects in your calculations-use one or the other.

2.11.2 Stability

Another interesting and important problem associated with submerged or floating bodies isconcerned with the stability of the bodies. A body is said to be in a stable equilibrium positionif, when displaced, it returns to its equilibrium position. Conversely, it is in an unstableequilibrium position if, when displaced (even slightly), it moves to a new equilibrium position.Stability considerations are particularly important for submerged or floating bodies since thecenters of buoyancy and gravity do not necessarily coincide. A small rotation can result ineither a restoring or overturning couple. For example, for the completely submerged body

)r

v,

s-

hehe

1S)

theDc.rcero-

;isionbleon.thetin)dy

I

~ -~~_u_--~-_u_~--------

V2.7

:=-=-=-=-= =-~:~=:=====:-~-:-=-=-=-=-~cY=-:-=-:-=-=-:==:=:::.:=:~:=:

Stable

Restoringcouple

. FIGURE 2.25

Stability of a completely im-mersed body-center of gravitybelow centroid.

2.11 Buoyancy, Flotation, and Stability 77

-:-:~~-= =-:-=-:-:-~c-:-=:.:_=-:--_'!:-=::====:=====:=:::==:=c

Unstable

Overturningcouple

. FIGURE 2.26

Stability of a completely im-mersed body-center of gravityabove centroid.

shown in Fig. 2.25, which has a center of gravity below the center of buoyancy, a rotationfrom its equilibrium position will create a restoring couple formed by the weight, OW,and thebuoyant force, FE,which causes the body to rotate back to its original position. Thus, for thisconfiguration the body is stable. It is to be noted that as long as the center of gravity fallsbelow the center of buoyancy, this will always be true; that is, the body is in a stable equi-librium position with respect to small rotations.-However, as is illustrated in Fig. 2.26, if thecenter of gravity is above the center of buoyancy, the resulting couple formed by the weightand the buoyant force will cause the body to overturn and move to a new equilibrium position.Thus, a completely submerged body with its center of gravity above its center of buoyancyis in an unstable equilibrium position.

For floating bodies the stability problem is more complicated, since as the body rotatesthe location of the center of buoyancy (which passes through the centroid of the displacedvolume) may change. As is shown in Fig. 2.27, a floating body such as a barge that rides lowin the water can be stable even though the center of gravity lies above the center of buoyancy.This is true since as the body rotates the buoyant force, FE,shifts to pass through the centroidof the newly formed displaced volume and, as illustrated, combines with the weight, OW,toform a couple which will cause the body to return to its original equilibrium position. How-ever, for the relatively tall, slender body shown in Fig. 2.28, a small rotational displacementcan cause the buoyant force and the weight to form an overturning couple as illustrated.

It is clear from these simple examples that the determination of the stability of sub-merged or floating bodies can be difficult since the analysis depends in a complicated fashionon the particular geometry and weight distribution of the body. The problem can be furthercomplicated by the necessary inclusion of other types of external forces such as those inducedby wind gusts or currents. Stability considerations are obviously of great importance in thedesign of ships, submarines, bathyscaphes, and so forth, and such considerations playa sig-nificantrole in the workof navalarchitects(see, for example,Ref.6).

c = centroid of originaldisplacedvolume

n . FIGURE 2.27Stability of a float-ing body-stableconfiguration.

c' = centroid of newdisplacedvolume

Restoringcouple

Stable

I

"

78 Chapter 2 / Fluid Statics

()c = centroid of original c' = centroid of new Overturning

displacedvolume displacedvolume coupleUnstable

. FIG U R E 2.28 Stability of a float-ing body-unstable configuration.

2.12 Pressure Variation in a Fluid with Rigid-Body Motion

Although in this chapter we have been primarily concerned with fluids at rest, the generalequation of motion (Eq. 2.2)

- Vp - -yk = pa

was developed for both fluids at rest and fluids in motion, with the only stipulation being thatthere were no shearing stresses present. Equation 2.2 in component form, based on rectangularcoordinates with the positive z axis being verticaUy upward, can be expressed as

- apax = pax

ap-ay=pay

- ap = l' + pazaz(2.24)

Even though afluid.may be in.motion, if..' ..j. .,. , ,

it moves' as a rigidbody there will be

,noshearlng stressespresent.

A general class of problems involving fluid motion in which there are no shearingstresses occurs when a mass of fluid undergoes rigid-body motion. For example, if a containerof fluid accelerates along a straight path, the fluid will move as a rigid mass (after the initialsloshing motion has died out) with each particle having the same acceleration. Since there isno deformation, there will be no shearing stresses and, therefore, Eq. 2.2 applies. Similarly,if a fluid is contained in a tank that rotates about a fixed axis, the fluid will simply rotate withthe tank as a rigid body, and again Eq. 2.2 can be applied to obtain the pressure distributionthroughout the moving fluid. Specific results for these two cases (rigid-body uniform motionand rigid-body rotation) are developed in the following two sections. Although problemsrelating to fluidshaving rigid-body motion are not, strictly speaking, "fluid statics" problems,they are included in this chapter because, as we will see, the analysis and resulting pressurerelationships are similar to those for fluids at rest.

2.12.1 Linear Motion

We first consider an open container of a liquid that is translating along a straight path with aconstant acceleration a as illustrated in Fig. 2.29. Since ax = 0 it follows from the first ofEqs. 2.24 that the pressure gradient in the x direction is zero (ap/ax = 0). In the y and zdirections

ap -ay - - pay

apaz = -peg + az)

(2.25)

(2.26)

,a float-

-meral

g that19u1ar

[2.24)

~aringtainerinitialtere is

Harly,~with}utionlotionblems

Jlems,~ssure

withairst ofand z

(2.25)

(2.26)

I

--

2.12 Pressure Variation in aFluid with Rigid-Body Motion 79

Freesurfaceslope=dzldy a'L~:

ayPI Constantpz pressureP3 lines . FIGURE 2.29

Linear acceleration ofa liquid with a freesurface.

The change in pressure between two closely spaced points located at y, z, and y + dy, z + dzcan be expressed as

ap apdp = - dy + - dz

ay az

or in terms of the results from Eqs. 2.25 and 2.26

dp = - pay dy - p(g + az) dz (2.27)

Along a line of constant pressure, dp = 0, and therefore from Eq. 2.27 it follows that theslope of this line is given by the relationship

dz = -~dy g + az

(2.28)

'ressure distri-

in afluidat is accel-alonga

ht path is not'static.

Along a free surface the pressure is constant, so that for the accelerating mass shown in Fig.2.29 the free surface will be inclined if ay 0;6O.In addition, all lines of constant pressure willbe parallel to the free surface as illustrated.

For the specialcircumstancein whichay = 0, az 0;60, which corresponds to the massof fluid accelerating in the vertical direction, Eq. 2.28 indicates that the fluid surface will behorizontal. However, from Eq. 2.26 we see that the pressure distribution is not hydrostatic,but is given by the equation

dpdz = - p(g + az)

For fluids of constant density this equation shows that the pressure will vary linearly withdepth, but the variation is due to the combined effects of gravity and the externally inducedacceleration, p(g + az), rather than simply the specific weight pg. Thus, for example, thepressure along the bottom of a liquid-filled tank which is resting on the floor of an elevatorthat is accelerating upward will be increased over that which exists when the tank is at rest(or moving with a constant velocity). It is to be noted that for afreely falling fluid mass (az =- g), the pressure gradients in all three coordinate directions are zero, which means that ifthe pressure surrounding the mass is zero, the pressure throughout will be zero. The pressurethroughout a "blob" of orange juice floating in an orbiting space shuttle (a form of free fall)is zero. The only force holding the liquid together is surface tension (see Section 1.9).

80 Chapter 2 / Fluid Statics

EXAMPLE2.11

The cross section for the fuel tank of an experimental vehicle is shown in Fig. E2.ll. Therectangular tank is vented to the atmosphere, and a pressure transducer is located in its sideas illustrated. During testing of the vehicle, the tank is subjected to a constant linear accel-eration, ay. (a) Determine an expression that relates ay and the pressure (in lb/ft2) at thetransducer for a fuel with a SG = 0.65. (b) What is the maximum acceleration that can occurbefore the fuel level drops below the transducer?

ay .ZL

Y

I--0,75ft ---+--- 0,75ft--! . FIGURE E2.11

SOLUTION

(a) For a constant horizontal acceleration the fuel will move as a rigid body, and from Eq.2.28 the slope of the fuel surface can be expressed as

dz ay

dy g

since az = O.Thus, for some arbitrary ay, the change in depth, z\, of liquid on the rightside of the tank can be found from the equation

z ay1-- = --0.75 ft g

or

z\ = (0.75 ft) (~)Since there is no acceleration in the vertical, z, direction, the pressure along the wallvaries hydrostatically as shown by Eq. 2.26. Thus, the pressure at the transducer is givenby the relationship

p = yh

where h is the depth of fuel above the transducer, and therefote

p = (0.65)(62.4 lb/ft3 )[0.5 ft - (0.75 ft)(ay/ g)]

ay= 20.3 - 30.4-g

for z\ :S 0.5 ft. As written, p would be given in lb/ft2.

(Ans)

Theside

ccel-t theIccur

Eg.

ght

allen

s)

2.12 Pressure Variation in a Fluid with Rigid-Body Mot/on 81

(b) The limiting value for ay (when the fuel level reaches the transducer) can be found fromthe equation

0.5 ft = (0.75 ft) [(aY~max]or

2g(~)max = 3

and for standard acceleration of gravity

(ay)max = 1(32.2 ft/s2) = 21.5 ft/s2 (Ans)

Note that the pressure in horizontal layers is not constant in this example since ap/ay =- pay ¥' O.Thus, for example, PI ¥' P2'

2.12.2 Rigid-Body Rotation

After an initial "start-up" transient, a fluid contained in a tank that rotates with a constantangular velocity w about an axis as is shown in Fig. 2.30 will rotate with the tank as a rigidbody. It is known from elementary particle dynamics that the acceleration of a fluid particlelocated at a distance r from the axis of rotation is equal in magnitude to rw2, and the directionof the acceleration is toward the axis of rotation as is illustrated in the figure. Since the pathsof the fluid particles are circular, it is convenient to use cylindrical polar coordinates r, e, andz, defined in the insert in Fig. 2.30. It will be shown in Chapter 6 that in terms of cylindricalcoordinates the pressure gradient V'pcan be expressed as

t'7 ap A 1 ap A ap A

vp = - e + - - Ce+ - Car r r ae az Z

Thus, in terms of this coordinate system

(2.29)

ar = - rw2 er ae = 0 az = 0and from Eq. 2.2

apar = prw2 ap = 0ae

ap -az - - 'Y (2.30)

I Axis of, rotation

I,I,

~a,= r(JJ2

y

e'"

r i?/e

'\'"e,

x . FIGURE 2.30Rigid-body rotationof a liquid in a tank.

ap apdp = - dr + - dzar az

82 Chapter 2 / Fluid Statics

These results show that for this type of rigid-body rotation, the pressure is a function of twovariables rand z, and therefore the differential pressure is

or

dp = prul dr - 'Ydz(2.31) ~

Along a surface of constant pressure, such as the free surface, dp = 0, so that from Eq. 2.31(using 'Y = pg)

w2r2

z = 2g + constant(2.32)

dz rw2---dr g

urface inliquidis'herthan

and, therefore, the equation for surfaces of constant pressure is

This equation reveals that these surfaces of constant pressure are parabolic as illustrated inFig. 2.31.

Integration of Eq. 2.31 yields

I dp = pW2 I r dr - 'YI dz

or

pw2r2p = ~ - 'YZ+ constant

(2.33)

where the constant of integration can be expressed in terms of a specified pressure at somearbitrary point ro, zo° This result shows that the pressure varies with the distance from theaxis of rotation, but at a fixed radius, the pressure varies hydrostatically in the vertical directionas shown in Fig. 2.31.

PI

Constant P2'

pressurelines P3

P4

r~

P3 - I~

(j)2r2

2g

ry . FIG U R E 2. 3 1 Pressure

distribution in a rotating liquid.x

I

2.12 Pressure Variation in a Fluid with Rigid-Body Motion 83

}ftwo XAMPLE2.12

It has been suggested that the angular velocity, w, of a rotating body or shaft can be measuredby attaching an open cylinder of liquid, as shown in Fig. E2.l2a, and measuring with sometype of depth gage the change in the fluid level, H - ho' caused by the rotation of the fluid.Determine the relationship between this change in fluid level and the angular velocity.

~R---1

(2.31)

1111

dr

1.2.31

(2.32)

(a)

SOLUTION

(b) . FIGURE E2.12

ated in The height, h, of the free surface above the tank bottom can be determined from Eq. 2.32,and it follows that

w2r2h = - + ho

2g

The initial volume of fluid in the tank, Vi, is equal to

Vi = l7R2H

(2.33)

The volume of the fluid with the rotating tank can be found with the aid of the differentia]element shown in Fig. E2.l2b. This cylindrical shell is taken at some arbitrary radius, r, aneits volume is

dV = 21T1"hdrIt some'om theirection

The total volume is, therefore

IR

(w2r2 )

l7W2R4

V = 217 r - + ho dr = - + l7R2ho° 2g 4g

Since the volume of the fluid in the tank must remain constant (assuming that none spillover the top), it follows that

l7W2R4l7R2H = - + l7R2h(l

4g .

orw2R2

H-ho=- 4g

This is the relationship we were looking for. It shows that the change in depth could inde(be used to determine the rotational speed, although the relationship between the change

depth and speed is not a linear one.

(An:

Pressureliquid.

I

84 Chapter 2 / Fluid Statics

References

1. The U.S. Standard Atmosphere, 1962, U.S. Government Printing Office, Washington,D.C., 1962.

2. The U.S. Standard Atmosphere, 1976, U.S. Government Printing Office, Washington,D.C., 1976.

3. Benedict, R. P., Fundamentals of Temperature, Pressure, and Flow Measurements, 3rdEd., Wiley, New York, 1984.

4. Dally, J. W., Riley, W. F., and McConnell, K. G., Instrumentation for EngineeringMeasurements, 2nd Ed., Wiley, New York, 1993.

5. Holman, J. P., Experimental Methodsfor Engineers, 4th Ed., McGraw-Hill, New York,1983.

6. Comstock, J. P., ed., Principles of Naval Architecture, Society of Naval Architects and,Marine Engineers, New York, 1967.

7. Hasler, A. F., Pierce, H., Morris, K. R., and Dodge, J., "Meteorological Data Fields'In Perspective' ", Bulletin of the American Meteorological Society, Vol. 66, No.7,July 1985.

Review Problems

Note: Problems designated with (R) are review problems. Thephrases within parentheses refer to the main topics to be usedin solving the problems. Complete, detailed solutions to thesereview problems can be found in the supplement titled StudentSolution Manual for Fundamentals of Fluid Mechanics byMunson, Young, and Okiishi (John Wiley and Sons, New York,1997).

2.1R (Pressure head) Compare the column heights ofwater, carbon tetrachloride, and mercury corresponding to apressure of 50 kPa. Express your answer in meters.(ANS: 5.10 m; 3.21 m; 0.376 m)

2.2R (Pressure-depth relationship) A closed tank is par-tially filled with glycerin. If the air pressure in the tank is 6Ib/in.2 and the depth of glycerin is 10 ft, what is the pressurein Ib/ff at the bottom of the tank?

(ANS: 1650 Ib/ft2)

2.3R (Gage-absolute pressure) On the inlet side of a pumpa Bourdon pressure gage reads 600 Ib/ft2 vacuum. What is thecorresponding absolute pressure if the local atmospheric pres-sure is 14.7 psia? .

(ANS: 10.5 psia)

2.4R (Manometer) A tank is constructed of a series of cyl-inders having diameters of 0.30,0.25, and 0.15 m as shown inFig. P2.4R. The tank contains oil, water, and glycerin and a

mercury manometer is attached to the bottom as illustrated. Cal-culate the manometer reading, h.

(ANS: 0.0327 m)

t0.1 m

+-0.1 m

t-0.1m

t-0.1m.t

Ih

~Mercury

. FIGURE P2.4R

2.5R (Manometer) A mercury manometer is used to mea-sure the pressure difference in the two pipelines of Fig. P2.5R.Fuel oil (specific weight = 53.0 Ib/ft3) is flowing in A andSAE 30 lube oil (specific weight = 57.0 Ib/ft3) is flowing inB. An air pocket has become entrapped in the lube oil as indi-cated. Determine the pressure in pipe B if the pressure in A is15.3 psi.

(ANS: 18.2 psi)

i!j

Ij

ij,i

IIjj!

J

'

j

I

I

I' J

.,;~1

I ~

I

11i ~Ii

I j

Air bubble-m,

m, SAE36 oil

rd

ng

ok,P2.5R

lld 2.6R (Manometer) Determine the angle 8 of the inclinedtube shown in Fig. P2.6R if the pressure at A is 1 psi greaterthan that at B. .

ds

7,

B.Air

. FIGURE P2.6R

- 2.7R (Force on plane surface) A swimmingpoolis 18mlong and 7 m wide. Determine the magnitude and location ofthe resultant force of the water on the vertical end of the poolwhere the depth is 2.5 m.(ANS: 214 kN on centerline, 1.67 m below surface)

11-

2.8R (Force on plane surface) The vertical cross sectionof a 7-m-long closed storage tank is shown in Fig. P2.8R. Thetank contains ethyl alcohol and the air pressure is 40 kPa. De-termine the magnitude of the resultant fluid force acting on oneend of the tank.

(ANS: 847 kN)

r-2 m1

IAir

a-{.IdIII]-is

2m

t4m

1~4m~. FIGURE P2.8R

I

Review Problems 85

2.9R (Center of pressure) A 3-ft-diameter circular plate islocated in the vertical side of an open tank containing gasoline.The resultant force that the gasoline exerts on the plate acts 3.1in. below the centroid of the plate. What is the depth of theliquid above the centroid?(ANS: 2.18 ft)

2.10R (Force on plane surface) A gate having the trian-gular shape shown in Fig. P2.lOR is located in the vertical sideof an open tank. The gate is hinged about the horizontal axisAB. The force of the water on the gate creates a moment withrespect to the axis AB. Determine the magnitude of this mo-ment.

(ANS: 3890 kN.m)

T8m

I

T6m

11--6 m-1-7 m l

P2.10R

2.11R (Force on plane surface) The rectangular gate CDof Fig P2.11R is 1.8 m wide and 2.0 m long. Assuming thematerial of the gate to be homogeneous and neglecting frictionat the hinge C, determine the weight of the gate necessary tokeep it shut until the water level rises to 2.0 m above the hinge.(ANS: 180 kN)

. FIGUREP2.11R

2.12R (Force on curved surface) A gate in the form of apartial cylindrical surface (called a Tainter gate) holds backwater on top of a dam as shown in Fig. P2.12R. The radius ofthe surface is 22 ft, and its length is 36 ft. The gate can pivotabout point A, and the pivot point is 10 ft above the seat, C.

86 Chapter 2 / Fluid Statics

Determine the magnitude of the resultant water force on thegate. Will the resultant pass through the pivot? Explain.

(ANS: 118,000 Ib)

Taintergate '>..

A

. FIGURE P2.12R

2.13R (Force on curved surface) A conical plug is locatedin the side of a tank as shown in Fig. 2.13R. (a) Show that thehorizontal component of the force of the water on the plug doesnot depend on h. (b) For the depth indicated, what is the mag-nitude of this component?

(ANS: 735 Ib)

. FIGURE P2.13R

2.14R (Force on curved surface) The 9-ft-Iong cylinderof Fig. P2.14R floats in oil and rests against a wall. Determinethe horizontal force the cylinder exerts on the wall at the pointof contact, A.

(ANS: 2300 Ib)

II FIGURE P2.14R

I

2.1SR (Buoyancy) A hot-air balloon weighs 500 Ib, in-cluding the weight of the balloon, the basket, and one person.The air outside the balloon has a temperature of 80 of, and theheated air inside the balloon has a temperature of ISOof. As-sume the inside and outside air to be at standard atmosphericpressure of 14.7 psia. Dete~mine the required volume of theballoon to support the weight. If the balloon had a sphericalshape, what would be the required diameter?(ANS: 59,200 ft3; 48.3 ft)

2.16R (Buoyancy) An irregularly shaped piece of a solidmaterial weighs 8.05 lb in air and 5.26 Ib when completelysubmerged in water. Determine the density of the material.

(ANS: 5.60 slugs/ft3)

2.17R (Buoyancy, force on plane surface) A cube, 4 ft ona side, weighs 3000 Ib and floats half-submerged in an opentank as shown in Fig. P2.17R. For a liquid depth of 10 ft, de-termine the force of the liquid on the inclined section AB of thetank wall. The width of the wall is 8 ft. Show the magnitude,direction, and location of the force on a sketch.

(ANS: 75,000 Ib on centerline, 13.33 ft along wall from freesurface)

1-4 It-I

A

II FIGURE P2.17R

2.18R (Rigid, body motion) A container that is partiallyfilled with water is pulled with a constant acceleration along aplane horizontal surface. With this acceleration the water sur-face slopes downward at an angle of 40° with respect to thehorizontal. Determine the acceleration. Express your answer inm/s2.

(ANS: 8.23 m/s2)

2.19R (Rigid-body motion) An open, 2-ft-diameter tankcontains water to a depth of 3 ft when at rest. If the tank isrotated about its vertical axis with an angular velocity of 160rev/min, what is the minimum height of the tank walls to pre-vent water from spilling over the sides?

(ANS: 5.18 ft)