Occasionally an historian will publish a list of choices for the greatest scientific geniuses of all time. Invariably on these lists appears the name of Archimedes, a Greek of the third century B.C. Among other things, Archimedes developed integral calculus, discovered the principles of the lever, inclined plane and screw, and laid down the foundations of fluid statics.

This last achievement occurred on the island of Sicily, then the home of the Greek colony of Syracuse. Archimedes had been given the task by the local monarch of determining if his crown was of pure gold or of some less valuable alloy. Although Archimedes recognized that this boiled down to finding the density of a very irregular object, the actual solution eluded him for a period of time, as an accurate volume measurement seemed impossible. Suddenly, while floating in his bath, the answer came to him. Archimedes thrust aside his rubber duckies and ran naked through the streets of Syracuse shouting ";Eureka!"; (";I have found it!";). Such eccentricities were accepted in those days, so Archimedes returned to his workshop to test the theory.

As it turned out, the crown had something less than the proper amount of gold, and the unhappy goldsmith was brought to justice.

Here is Archimedes' approach to the problem:

1) When an object is placed in water, it appears to be lighter.
2) As the object is lowered into the water, the level rises in the tank. That is, the object displaces a volume of water equal to its own volume below the surface.
3) Finally the Great Synthesis: The weight of water displaced is equal to the weight loss of the object! This is crucial and not obvious. It means that if Archimedes weighed 175 pounds and carefully stepped into a tub full to the brim, he would spill over, say, 165 pounds of water onto the floor. His apparent weight in the tub, then would be 175 - 165 = 10 pounds. That is, his weight loss would be 165 pounds.

Archimedes's Principle applies to any fluid. The buoyant force on the object is equal to the weight of the displaced fluid.

You are probably still unsure about what all this means, so try some problems.

1. A 12 pound rock is lowered into water and is found to weigh 8 pounds submerged. What weight of water is displaced?

Buoyant force = weight of the displaced fluid. The rock lost four pounds, so the water it displaced weighed: 4 lbs

2. An old vase, found in a shipwreck, weighs 43 newtons submerged. It displaces 21 newtons of water, so what will its weight be when taken out into the air?

By displacing 21 newtons, it loses that much weight under water. Therefore, above water it must weight 43 nt + 21 nt: weight in air = 64 nt

3. A 480 pound boat is lowered into a lake. What weight of water does it displace?

Assuming the boat floats, it must displace its full weight in water. weight of water displaced = 480 lbs

I hope you've noticed something about the floatability of objects: If the object is bulky enough to displace its entire weight in water, it will float if not it will sink. 10 pounds of wood, for example, has enough volume to displace more than 10 pounds of water. It will therefore settle into the water until it displaces exactly 10 pounds, balancing forces, and float. 10 pounds of lead, however, has a much smaller volume and will displace only 0.88 pounds of water. Thus it sinks to the bottom.

Before we pursue this marvelous subject, take a few minutes to review the concept of density. Here are a few densities you will need in this chapter:

substance	density
water	1.00 gm/cm3
mercury	13.6 gm/cm3
alcohol	0.8 gm/cm3
oil	0.9 gm/cm3
aluminum	2.7 gm/cm3
copper	8.9 gm/cm3
cork	0.4 gm/cm3
gold	19.3 gm/cm3
ice	0.9 gm/cm3
iron	7.6 gm/cm3
lead	11.3 gm/cm3
wood (pine)	0.42 gm/cm3
air	1.3 x 10-3 gm/cm3

4. What is the mass of 17.4 cm³ of lead?

m = pV m = (11.3 gm/cm3)(17.4 cm3) m = 197 gm

5. A slab of rock, 5 cm long, 4 cm wide and 2 cm thick masses 277 gm. What is its density?

p = m/v p = (277 gm)/[(5 cm)(4 cm)(2 cm)] p = 6.93 gm/cm3

6. What is the weight of a roomful of air? The room measure 7.3 m by 5.8 m by 2.6 m.

pair = 1.3 * 10-3 gm/cm3 pair = 1.3 kg/m3 W = pVg W = (1.3 kg/m3)(7.3 * 5.8 * 2.6 m3)(9.8 m/sec2) W = 1.40 * 103 nt note: this is around 315 lb!

7. What volume of gold will weigh one pound? (1 lb = 4.45 nt)

V = m/p V = (w/g)/p V = [(4.45 nt)/(9.8 m/sec2)]/(19.3 * 103 kg/m3) V = 2.35 * 10-5 m3 note: less than a 3 cm cube = 23.5 cm3

To help you convert densities to other unit systems, here are the densities of water: 1.00 gm/cm³ = 1000 kg/m³ = 1.94 slug/ft³

8. What is the weight of 7 cubic feet of cork?

W = mg W = pVg W = [(.4 gm/cm3)[(1.94 slug/ft3)/(1.00 gm/cm3)]](7 ft3) (32 ft/sec2) W = 174 lb

9. A block of metal measure 3 inches by 7 inches by 1/2 inch. If it weighs 2.3 lb, what is its density in gm/cm³?

V = l * w * h V = (3/12 ft)(7/12 ft)(.5/12 ft) V = 6.076 * 10-3 ft3 m = W/g m = (2.3 lb)/(32 ft/sec2) m = .0719 slug p = m/V p = (.0719)/(6.076 * 10-3 ft3) p = (11.8 slug/ft3)[(1.00 gm/cm3)/(1.94 slug/ft3)] p = 6.04 gm/cm3

Now we return to Archimedes's Principle. If you ever run into trouble in this section, just say the words, "Archimedes, HELP me!". If you listen carefully, you'll faintly hear "The buoyant force equals the weight of the displaced fluid." These words of wisdom will help you through the most trying problems.

10. An object, lowered into a cup of water, is found to displace 18.3 cm³. How much weight did it lose when submerged?

Buoyant force = weight of the displaced fluid W = Vpg W = (18.3 cm3)(1.0 gm/cm3)(980 cm/sec2) W = 1.79 * 104 dyne

11. A rock weighs 87,000 dyne in air and 62,400 dyne when submerged in water. What is its volume?

Buoyant force = 87,000 - 62,400 Buoyant force = 24,000 dyne water m = W/g m = (24,600)/(980) m = 25.1 gm water Vrock = Vwater Vrock = mwater/pwater Vrock = 25.1/1.00 Vrock = 25.1 cm3

12. A 13.4 lb block of plastic weighs 2.6 lb submerged in water. What is its volume?

Weight loss = 13.4 - 2.6 Weight loss = 10.8 lb water displaced We need only find the volume of this much water: W = mg W = pVg V = W/pg V = (10.8 lb)/[(1.94 slug/ft3)(32 ft/sec)] V = 0.174 ft3

13. A 47 pound block of aluminum has a volume of 0.23 ft³. How much will it weigh at the bottom of a lake?

It will displace 0.23 ft3 of water, or a weight of: W = pVg W = (1.94)(.23)(32) W = 14.3 lb Thus the new weight will be: weight at bottom of lake = 47 - 14.3 weight at bottom of lake = 32.7 lb

14. The engine from a Honda bike weighs 850 pounds in air and 670 pounds when submerged in water. What is its volume?

850 - 670 = 180 lb of water displaced V = m/p V = (180/32 slug)/(1.94 slug/ft3) V = 2.90 ft3

15. Mason Jarr weighs 207 pounds out of water, and when he allows himself to sink To the bottom, he finds he weighs only 4 pounds. What is his volume?

Mason displaces: 207 lb - 4 lb = 203 lb of water V = m/p V = (203/32 slug)/(1.94 slug/ft3) V = 3.27 ft3

16. What is the volume of 4.9 kg of iron? What would it weigh out of water? What would it weigh submerged?

V = m/p V = (4.9 kg)/(7.6 * 103 kg/m3) V = 6.45 * 10-4 m3 Wdisplaced = pVg Wdisplaced = (1 * 103)(6.45 * 10-9)(9.8) Wdisplaced = 6.32 nt Wout = mg Wout = (4.9 kg)(9.8 m/sec2) Wout = 48.02 nt Win = Wout - Wdisplaced Win = 48.02 nt - 6.32 nt Win = 41.7 nt

17. What is the weight of 3 cm³ of gold? What would it weigh submerged in water?

Wgold = mg Wgold = pVg Wgold = (19.3)(3)(980) Wgold = 5.67 * 104 dyne Wwater displaced = pVg Wwater displaced = (1)(3)(980) Wwater displaced = .29 * 104 dyne Wsubmerged = 5.67 * 104 - .29 * 104 Wsubmerged = 5.38 * 104 dyne

18. What is the volume of 80 gm of aluminum? What would it weigh submerged in water?

VAl = mAl/pAl VAl = (80 gm)/(2.7 gm/cm3) VAl = 29.6 cm3 Wsubmerged = mAlgAl - pwaterVwater gwater Wsubmerged = (80 * 980) - (1 * 296 * 980) Wsubmerged = 4.94 * 104 dyne

19. What will 23.6 gm of copper weigh when submerged in water?

First, the find the volume: Vwater = Vcopper Vcopper = mCu/pCu Vcopper = 23.6/8.9 Vcopper = 2.65 cm3 W = mg - pVg W = (23.5 * 980) - (1 * 2.65 * 980) W = 2.05 * 104 dyne

20. An object of mass m has a density p. What is its weight in cgs units when submerged in water?

Vobject= m/p Wsubmerged = mg - pVg Wsubmerged = mg - [1 * (m/p) * g) Wsubmerged = mg(1 - 1/p)

Thus far we have dealt only with submersion in water. In fact, Archimedes's Principle works for all fluids. Think about it: it's easier for you to float in salt water because its increased density your body weight without sinking so deeply. Think of how easy it would be to displace your weight in mercury, which has a density of 13.6 gm/cm3! On the other hand, if you were ever to fall into a pool of alcohol (density 0.8 gm/cm3), there is no way you'd be able to stay afloat.

21. A marble whose volume is 6.3 cm³ is dropped into a glass of alcohol. How much weight does it lose when submerged?

Buoyant force = weight of the displaced fluid weight submerged = pVg weight submerged = 0.8 * 6.3 * 980 weight submerged = 4.94 * 103 dyne

22. A 46,200 dyne block of aluminum is lowered into a glass of alcohol. What does it weigh submerged?

Valcohol = Valuminum Valuminum = mAl/pAl Valuminum = (46200/980)/(2.7) Valuminum = 17.5 cm3 Wsubmerged = W - pVg Wsubmerged = 46,200 - (.8 * 17.5 * 980) Wsubmerged = 3.25 * 104 dyne

23. Gold has a density of 19.3 gm/cm³. If 240 gm of it is lowered into a pool of mercury, what will be its submerged weight?

Vgold = m/p Vgold = 24.0/19.3 Vgold = 12.4 cm3 Wdisplaced mercury = pVg Wdisplaced mercury = (13.6)(12.4)(980) Wdisplaced mercury = 1.65 * 105 dyne W = Wgold - Wdisplaced W = mg - pVg W = (240 * 980) - (1.65 * 105) W = 7.02 * 104 dyne

24. An object weighs 5.7 x 10⁴ dyne out of water and 2.1 x 10⁴ dyne submerged. What is its volume? ...its density?

Weight loss = (5.7 * 104) - (2.1 * 104) Weight loss = 3.6 * 104 dyne V = mwater/pwater V = [(3.6 * 104)/(980)]/(1.00) V = 36.7 cm3 pobject = mobject/Vobject pobject = [(5.7 * 104)/(980)]/(36.7) pobject = 1.58 gm/cm3

25. A rock weighs 54 pounds out of water and 47 in water. What is its density?

Wwater = 54 - 47 Wwater = 7 lb Vwater = mwater/pwater Vwater = (7/2 slug)/(1.94 slug/ft3) Vwater = .113 ft3 pobject = mobject/Vobject pobject = (54/32)/(.113) pobject = 15.0 slug/ft3

26. A 740 nt statue of Arnold Swartzenegger is lowered into a large aquarium where it is found to weigh 611 nt. What is the density of the statue?

mwater = W/g mwater = (740 - 611)/(9.8) mwater = 13.2 kg Vobject = Vwater Vwater = mwater/pwater Vwater = (13.2 kg)/(1 * 103 kg/m3) Vwater = 13.2 * 10-3 m3 pobject = m/V pobject = (740/9.8)/(13.2 * 10-3) pobject = 5.74 * 103 kg/m3

27. A large olive weighs 5.6 nt when submerged in alcohol. If the density of the olive is 1.2 x 10³ kg/m³, what will its weight be in air?

weight - buoyant force = submerged weight poVog - pAVog = 5.6 nt Vo = (5.6)/[(po - pA)(g)] Vo = (5.6)/[(1.2 * 103 - .8 * 103)(9.8)] Vo = 1.43 * 10-3 m3 Wo = poVog Wo = (1.2 * 103)(1.43 * 10-3)(9.8) Wo = 16.8 nt

28. A rock weighs 57 pounds out of water and 46 submerged. What is its density?

Vwater = mw/pw Vwater = [(57 - 46)/(32)]/(1.94) Vwater = 0.177 ft3 prock = mrock/Vrock prock = (57/32)/(.177) prock = 10.1 slug/ft3

29. A block of walnut weighs 90 newtons in air and 12 newtons when submerged in oil What is its density?

Voil = moil/poil Voil = [(90 - 12)/(9.8)]/(.9 * 103) Voil = 8.84 * 10-3 m3 p = m/V p = (90/9.8)/(8.84 * 10-3) p = 1.04 * 103 kg/m3

30. A piece of plastic whose weight is 15,000 dyne and volume is 5.0 cm³ is lowered into a fluid of unknown density. If its submerged weight of 8,000 dyne, what is the density of the fluid?

mfluid = Wfluid/g mfluid = (15000 - 8000)/(9800) mfluid = 7.14 gm pfluid = mfluid/Vfluid pfluid = (7.14 gm)/(5 cm3) pfluid = 1.43 gm/cm3

31*. An object weighs 3,000 dyne in water and 3250 dyne in alcohol. What is the volume of the object?

Wobject - pwaterVobjectg Wobject = 3000 dyne Wobject - palcoholVobjectg Wobject = 3250 dyne subtracting: (pwater - palcohol)Vobjectg = 250 dyne Vobject = (250)/[(1.0 - .8)(980)] Vobject = 1.28 cm3

32. A block of wood measures 20 cm by 15 cm by 5 cm and it floats on water. If it has a mass of 1000 gm, how deep will it press into the water?

Since it floats, the wood displaces its weight (and therefore its mass) in water. pwoodVwood = pwaterVwater 1000 gm = (1.0)(2.0 *15 * x) x = 3.33 cm

33. A rod with a 3 cm² cross-sectional area is lowered vertically into a pool of water until it floats. If 15 cm of it are then below the surface, what does the thing weigh?

Vsubmerged = (3 cm2)(15 cm) Vsubmerged = 45 cm3 Wfluid = pVg Wfluid = (1.0)(45)(980) Wfluid = 4.41 * 104 dyne

34. A cylinder of cross-sectional area A, height H, and density p_c, is lowered into a fluid of density p_f. How deep will it sink?

pcVcg = pfVfg But: Vc = AH Vf = Ax So: pcAHg = pfAxg x = (pc/pf)H

35. A barge measuring 8 ft wide by 13 ft long sinks 5 inches deeper into a lake when a car is rolled onto it. What is the weight of the car?

Weight of additional displaced water: W = pVg W = (1.94 slug/ft3)[(8 ft)(13 ft)(5/12 ft)](32 ft/sec2) W = 2690 lb

We have studied Archimedes's Principle without delving into why it works. What is it that causes objects to buoy up when submerged? You can get a feeling for it by trying to press an empty glass into a bucket of water. As the glass descends deeper into the water, an increasing force is felt pressing up against the glass. Since the water pressure is evidently getting greater at greater depths, a submerged object will feel a greater force pushing up on its bottom than it will pressing down on its top. Hence it feels lighter in water.

We are now ready to investigate the nature of pressures in fluids. First, a few principles:

1) Pressure in a fluid at any given point is exerted equally in all directions up, down, and sideways. (It must do this, or the molecules of the fluid itself would not be in equilibrium.)
2) The pressure exerted by a fluid is a result only of the depth below the surface the shape of the vessel is not important. (It takes a while for this to make physical sense, but the idea will grow on you.)
3) Pressure is defined as force per unit area. Thus it is measured in units like pounds/in² or dyne/cm² or nt/m².

36. A fish tank holds 40.3 pounds of water and its bottom measures 18 inches by 9 inches. What is the pressure on the bottom?

P = F/A P = (40.3 lb)/(18 in * 9 in) P = .249 lb/in2

37. The atmosphere exerts a pressure of 14.6 pounds/in². What is the total force exerted by air on the top of this page? (8 1/2"; x 11";)

F = P * A F = (14.6 lb/in2)(8.5 in * 11 in) F = 1.37 * 103 lb

38. In the late '50s, spike heels were popular. What would be the pressure on a heel whose tip measures 0.15 inches in radius supporting a force of 73 pounds?

P = F/A P = (73 lb)/[(3.14)(.152 in2)] P = 1033 lb/in2

39. Even greater pressures can easily be generated: A thumb tack has a square point measuring 0.02 cm on a side. What pressure is exerted by a thumb pushing the tack with a 23 nt force?

P = F/A P = (23 nt)/(.02 cm)2 P = 5.75 * 104 nt/cm2

40. A block of wood measures 5 cm by 3 cm by 11 cm, and has a mass of 108 gm.
a) What pressure does it exert when standing on the 5 x 3 end?
b) What pressure does it exert when on its 5 x 11 end?

a) P = F/A P = [(108)(980) dyne]/[(5)(3) cm2] P = 7.06 * 103 dyne/cm2

b) P = F/A P = [(108)(980) dyne]/[(5)(11) cm2] P = 1.92 * 103 dyne/cm2

41. A tank of fluid of density r has base dimensions l x w and height h.
a) What force does it exert on the base?
b) What pressure does it exert?

a) F = mg F = pVg F = p(lwh)g F = plwhg

b) P = F/A P = (plwhg)/(lw) P = pgh

42. What is the pressure at the bottom of a 17 cm glass of water?

P = pgh P = (1.0 gm/cm3)(980 cm/sec2)(17 cm) P = 1.67 * 104 dyne/cm2

43. What is the pressure 6.4 cm below the surface of a pool of mercury?

P = pgh P = (13.6 gm/cm3)(980 cm/sec2)(6.4 cm) P = 8.53 * 104 dyne/cm2

44. A submarine rests 17.3 m below the surface of a lake of alcohol. What total force is on its hatch which is 0.22 m in radius?

P = pgh P = (.8 * 103 kg/m3)(9.8 m/sec2)(17.3 m) P = 1.36 * 105 nt/m2 F = P * A F = (1.36 * 105 nt/m2)[(3.14)(.222 m2)] F = 2.06 * 104 nt

45. A marble, with a radius of 0.60 cm sits at the bottom of a swimming pool 4.8 m deep. What total force is exerted on its surface?

F = P * A F = (pgh)[(4)(3.14)(r2)] F = [(1 * 103)(9.8)(4.8)][(4)(3.14)(.6 * 10-2)2] F = 21.3 nt

46. Normal atmospheric pressure is 1.0 x 10⁶ dyne/cm². What depth of water would produce an equivalent pressure? (This will tell you how deep you have to dive to feel one additional atmosphere of pressure on your body.)

P = pgh h = P/pg h = (1.0 * 106 dyne/cm2)/[(1 gm/cm3)(980 cm/secnt/m2)] h = 1.02 * 103 cm

47. A tank of liquid is fitted with two pistons, one 2.0 cm in radius, the other 7.5 cm in radius (see picture). When the pressure on the tank builds up to 5.6 x 10³ dyne/cm², how much force is exerted on each piston?

small piston: F = P * A F = (5.6 * 103)[(3.14)(.22)] F = 7.04 * 104 dyne large piston: F = P * A F = (5.6 * 103)[(3.14)(7.52)] F = 9.90 * 105 dyne note: the larger piston feels a larger force

Problem 47 has an interesting practical application. By exerting a small force on a small area, a pressure may be established in a tank which will exert a far greater force on a large area. This is the principle of the hydraulic lift. A small piston pushes a fluid (usually oil) into a tank while a large piston is forced up by the fluid, exerting a great force.

48. A. Ward Winner decides it's time to rate the tires on his car. His hydraulic jack has a small piston of 0.60 cm radius which he uses to operate a large piston of 3.0 cm radius. With the lever on the pump, Ward can easily exert a force of 1500 nt on the small piston.
a) What pressure does this produce in the oil?
b) What force does this exert on the car?

a) P = F/A P = (1500 nt)/[(3.14)(.6 cm)2] P = 1326 nt/cm2

b) F = P * A F = (1326)[(3.14)(32)] F = 3.75 * 104 nt

49. An apple juice jug with a 1.2 inch diameter opening at the top is completely filled with juice. A cork fitted into the hole is struck sharply with a hammer, exerting a force of 27 pounds for the instant of impact. What force will be exerted on the bottom of the jug which is 8.3 inches in diameter?

We assume the pressure at the top and bottom are equal, which ignores the small pgh pressure. Ftop/Atop = Fbottom/Abottom (27 lb)/[(3.14)(.62 in2)] = Fbottom/ [(3.14)(4.152 in2)] Fbottom = 1292 lb This can pop the bottom out of the jug. Spectacular, but hazardous! note: radius = diameter/2