Applied Fluid Mechanics - 05 Bouyancy and Stability

5 Buoyancy and Stability

5.1 OBJECTIVES

Buoyancy is the tendency of a fluid to exert a supporting force on a body placed in the fluid. StabUity refers to the ability of a body to return to its original position after being tilted about a horizontal axis. Figure 5.1 shows several devices for which an analysis of buoyancy and stability should be completed. The buoy {a) and the ship (e) must be designed to float stably. 1he instrument package (b) would tend to float if not restrained by the anchoring cable. The diving bell (G) must be supported by the crane qn the ship to keep it from sinking. However, the submarine (d) has the ability to hover atany depth, dive deeper into the water, or rise to the smface and float. Mter completing tbs chapter, you should be able to:

1. Write the equation for the buoyant force. 2. Analyze the case of todies floating on a fluid. 3. Use the principle of static equilibriuIlll to solve for the forces involved in buoyancy problems. 4. Define the ·conditions that must b-e met for a body to be stable when completely submergej in a fluid. 5. Define the conditions that must be met for a body to be stable when floating on a fluid. 6. Define the terin metacenter and compute its location.

5.2 A body in a fluid, whether floating or subme-rged, is' buoyed up by a force BUOYANCY

c:>

equal to the weight of tt.e fluid displaced. The buoyant force acts vertically upward through the centroid of the displaced volume and ~an be defined mathem~tically by Archimedes' principle as stated below. (5-1)

BUOYANTFORCE

where

Fb :::; buoyant force 'Yf = specific weight of the fluid Vd = displaced volume -of the fluid When a body is floating freely, it displaces a sufficient volume of fluid to just balance its own weight. The analysis of problems dealing with buoyancy requires the application of the equation ·o f static equilibrium in .the vertical direction, L Fu = 0, 115

Chapter 5 Buoyancy and Stability

116 FIGURE 5.1 Examples of buoyancy problem types.

assuming the object is at rest in the fluid. The following protedure is recommended for all problems, whether they involve floating or bwerged bodies: PROCEDURE FOR SOLVING BUOYANCY PROBLEMS

1. Determine the objective of the problem solution. Ar.e - y. to find a force, weight, volume; or specific weight? 2. Draw a free body diagram of the object in the fluid. S -ow all forces that act on the free body in the vertical direction, includin: the weight of the body, the buoyant force, and all external forces. If the direction of some force is not known, "assume the most probable direction and show it on "the free body. 3. Write the equation of static equilibrium in the verticallirection, L Fv = d, assuming the positive direction to be upward. 4. Solve for the desired force, weight, volume, or specBc weight, remembering the following concepts: a. The buoyant for~ ·s ~aJcu.lated from F'b = Yt Vd. b. The weight f ~ solid objeot is the product of its btat volume and its specificweigllt; t at is, W = yV. c. An object with ~n av€-rage pecific weight less than that of the fluid will tend to float be ause w < Eb with the object suboerged. d. An object with an average specific weigiht greater 1 han that of the fluid " wiU tend to sink because w > Fb with the object :mbmerged. ~ Neutral buoyancy occurs when a body stays in a ~iven position where er i! is submerged in the fluid. An object whose average specific weigfit is equal to that of the fluid would be neutl""ally buoyant.

P -

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EXAMPLE PROBLEM 5.1

= AMMED EXAMPLE PROBLEMS

A cube, 0.50 m on a side, is made of bronze having a specific weight of 86.9 kN/m 3 • Determine the magnitUde and direction of the force required to hold the cube in

5.2 Buoyancy

117

equilibrium completely sUJmerged (a) in water and (b) in mercury. The specific gravity bf mercury is I3 .Sf. Consider part (a) fir~t. Imagine the cube of bronze submerged in water. Now do Step I of the procedure. Assuming that the bronze cube will not stay in equilibrium by itself, some external force is required. The objective is to find the magnitude of this forc e and the direction in whioch it wouU act-that is, up or down. Now do Step 2 of t"e procedure before looking at the next panel. The free body is sillply the cube itself. There are three forces acting on the cube in the vertical direction, as shown in Fig. 5.2: the weight of the cube w, acting downward through its ceder of gravity; the buoyant force F b , acting upward through the centroid of the displaced volume; and the externally applied supporting force F~ .

FIGURE 5.2 Free body diagram of cube.

Fe=Extemal supporting force

w

Centroid of volume

Fb=Buoyant force (a) Forces acting on the cube

(b) Two dimensional free . body diagram

Part (a) of Fig. 5.2 shows the cube as a three-dimensional object with the three forces. acting alO)1g a vertical line through the centroid of the volume. This is the preferred visualization of the free body diagram. However, for most problems it is suitable to use a simplified two-dimensional sketch as shown in part (b). How do we know to draw the force F~ in the upward direction? We really do not know for certain. However, experience should indicate that without an external fOlce the solid bronze cube would tend to sink in water. Therefore, an upward force seems to be required to hold the cube in equilibrium. If our choice is wrong. the final result will indicate that to us. Now, assuming that the forces are as shown in Fig. 5.2, go on to Step 3.

118

Chapter 5

Buoyancy and Stability

The equation should look like this. (Assume that positive forces act upward .)

2: Fa = 0 (5-2) As a part of Step 4, solve this equation algebraically for the desired term.

You should now have F~

=

(5-3)

Fb

w -

since the objective is to find the external force. How do we calculate the weight of the cube w? Item b under Step 4 of the procedure indicates that w = yBV, where YB is the specific weight of the bronze cube and V is its total volume. For the cube, since each side is 0.50 rn, we have V = (0.50 m)3

= 0.125

m1

and w = YB V

= (86.9 kN/m1)(0. 125 m3) = 10.86 kN

There is another unknown on the right side of Eq. (5-3). How do we calculate

Check item a under Step 4 of the procedure if you have forgotten.

Fb = yfVd In this case Yf is the specific weight of the water (9.81 kN/m 3), and the displaced volume Vd is equal to the total volume of the cube that we already know to be 0.125 ml. Then, we have

F b-= yfVd

= (9.81

kN/m1)(0.125 m3)

Now we can complete our solution for

= 1.23 kN

F~.

The solution is F~

=w

- Fb

= 10.86 kN

- 1.23 kN

= 9.~3

kN

Notice that the result is positive. This means that our assumecl dir~tion for p~ was 'Correct. Then the solution to the problem is that an upward foree Of 9.63 kN is required to hold the block of bronze in equilibrium under water. rcury? What about part (b) of the problem, where the cube is submerge.d in Our objective. is the same as before-to determine the magnilude ang dire tion of the force required to hold the cube in equilibrium. Now do Step 2 of the procedure.

5.2

Buoyancy

119

FIGURE 5.3 Two possible free body diagrams.

L

L

Fb

Fb

{a) Assuming cube would sink

(b) Assuming cube would float

Either of two free bociy diagrams is correct as shown in Fig. 5.3, depending on the assumed direction for :he external force F~ . The solution for the two diagrams will be carried out simulta,eously so you can check your work regardless of which diagram looks like yours :and to demonstrate that either approach will yield the correct answer. Now do Step 3 of the procedure. These are the corroct equations of equilibrium. Notice the differences and relate them to the figures.

Now solv~ fQ! Ft .

YQ!:! shQuld now ha'we Ft" == w - F/J Since the magnitude of wand Fb are the same for each equation, they can now be calculated. As in part (a) of the prob:,..em, the weight of the cube is w

= 1'8;/ = (86.9 kN/m3)(0.125 m3) =

10.86 kN

For the buoyant force F b • you should have

where the subscript m re::ers to mercury. We then have

Fb

= (0.54)(9.81

kN/m 3)(0.125 m 3)

Now go on with th:! solution for Ft.

= 16.60 kN

no

Chapter 5 Buoyancy and Stability The correct answers are F,

= w - Fb = 10.86 kN - 16.60 kN = - 5.74 kN

Ft!

= Fb - W = 16.60 kN - 10.86 kN = + 5.74 kN

Notice that both solutions yield the same numerical "alue , but they have opposite signs . The negative sign for the solution on the left means that the assumed direction for F, in Fig. 5.3(a) was wrong. Therefore, both approaches give the same result. The required external force is a downward force of 5. 7~ kN. How could you have reasoned from the start that a dowLward force would be required? Items c and d of Step 4 of the procedure suggest that the specific weight of the cube and the fluid be compared. In this case: For the bronze cube, 'YB = 86.9 kN/m J For the fluid (mercury); 'Ynr

= (13.54)(9.81 = 132.8

H/m 3)

kN/m J

Since the specific weight of the cube is less than that of the meJICury , it would tend to float without an external force. Therefore, a downward forc:! , as pictured in Fig. 5. 3(b), would be required to hold it in equilibrium under the surface of the mercury. This example-problem is concluded.

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EXAMPLE PROBLEM 5.2

A certain solid metal object has such an irregular shape that it _s difficult to calc\JIlate its volume by geometry. We can use the principle of buo~ancy to calc~late iifs volume . First, the weight of the object is determined in the nomal manner to be 60 lb. Then, using a setup similar to that in Fig. 5.4, we find its apparent weight while submerged in water to be 46.5 lb. Using these data and the p~ocedure for analyziag buoyancy problems, we can find the volume of the object.

FIGURE 5.4 Metal object suspended in a fluid.

Total v.eight =46.5 Ib

5.2

121

Buoyancy

Now draw the free body diagram of the object while it is suspended in the water. The free body diagram of the object while it is suspended in the water should look like Fig . 5.5. In this figure, what are the two forces F .. and w?

1. FIGURE 5.5 gram.

From the problem statement we should know that w = 60 Ib, the weight of the object in air, and F~ = 46.5 Ib , the supporting force exerted by the balance shown in Fig. 5.4. Now do Step 3 of :he procedure.

Free body dia-

Using 2: Fv = 0, we get Fb

+ F~

- w = 0

Our objective is to find the total volume V of the object. How can we get V from this equation? w~

use this eqIJatbn: Fb

where 'Yf is

= '}'fV

·ght of t~ water, 62.4 tb/ft]. this into the prece.ding equation and solve for V.

the speCific \1

Sul.? ~ titute

You should now Cave

=0 - w =0 'YfV = w

Fb + F~ - w '}'fV

+

F~

F~

-

V = w '- F~ 'Yf

Now we can put in the known values and calculate V. The result is V = 0.216 ft3. This is how it is done: V .

=w'}'f

F,

= (60

_ 46.5)lb (

ft3

62.4 Ib

)

= 13.5 ft] = 0.216 ft3 62.4

Now that the voh:me of the object is known, the specific weight of the material can be found. - ~ 60 lb _ 278 .. If 3 '}' - V - 0.216 ft3 ) t

This is approximately 1he specific weight of a titanium alloy.

Chapter 5

il2

Buoyancy and Stability

The next two problems are worked out in detail aid should serve to check your ability to solve buoyancy problems . After reading the problem statement, you should complete the solution yourself ::Jefore reading the panel on which a correct solution is given. Be sure te read the problem carefully and use the proper units in your calculations . Although there is more than one way to solve some problems, it is possibl:e to get the correct answer by the wrong method. If your method is different from that given, be sure yours is based on sound principles before assumin~ it is correct.

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EXAMPLE PROBLEM 5.3

Solution

A cube, 80 mm on a side, is made of a rigid foam material and oats in water with 60 mm below the surface. Calculate the magnitude and direction o:::tlire fOlice required to hold it completely submerged in glycerine, which has a specilac glrav ·~ y .e f 1.26. Complete the solution before looking at the next panel. First calculate the weight of the cube, then the force required to hold the cube ' submerged in glycerine. Use the free body diagrams in Fig. 5.116: (a) onbe 600 ing