Modularized Instruction In Hydrostatics

MODULARIZED INSTRUCTION IN HYDROSTATICS

Prepared by: TRYON R. GABRIEL October, 2004

Lesson 1: MASS DENSITY Objective: To be able to understand the concept of mass density and to use its definition into computation. The mass density of a liquid or gas is one factor that determines its behavior as a fluid. Definition of Mass Density The mass density ρ is the mass m of a substance divided by its volume V: ρ=m∕V SI Unit of Mass Density: kg m-3 Equal volumes of different substances generally have different masses, so the density depends on the nature of the material. A convenient way to compare densities is to use the concept of specific gravity. The specific gravity (or relative density) of a substance is its density divided by the density of standard reference material, usually chosen to be water at 4˚C. sp. grav. =

density of substance

= density of substance density of water at 4˚C 1.000 x 103 kg∙m-3

Being the ratio of two densities, specific gravity has no unit. Examples: 1. Numerous jewelry items of solid silver are melted down and cast into solid circular disk that is 0.02 m thick. The total mass of the jewelry is 10.0 kg. Find the radius of the disk. (ρ = 10500 kg∙m-3) Solution: Since ρ = m/V, the volume of the disk therefore is V = m ∕ ρ = 10.0 kg ∕ 10500 kg∙m-3 = 9.524 x 10-4 m3 Also, the volume of the disk is given by V = πr2t; hence, the radius of the disk is r = √ (V ∕ πt) = √ [9.524 x 10-4 m 3 ∕ π(0.02 m)] = 0.123 m. 2. A 2 x 10-4 m3 flask is filled with water at 4˚C; when the flask is heated to 80˚C, 6 x 10 kg of water spill out. What is the density of water at 80˚C? (Assume that the expansion of the flask is negligible.) -3

Solution: The original mass of water in the flask at 4˚C is m = ρV = (103 kg∙m-3)(2 x 10-4 m3) = 0.200 kg and the mass of water remaining after 6 x 10-3 kg spill out is m’ = m - 6 x 10-3 kg = 0.200 kg – 6 x 10-3 kg = 0.194 kg. Hence, the density of water at 80˚C will be ρ’ = m’ = 0.194 kg V 2 x 10-4 m3

-3

= 970 kg∙m .

3. The body of a man whose weight is about 690 N typically contains about 5.2 x 10-3 m3 of blood. (a) Calculate the weight of the blood and (b) express it as a percentage of the body weight. Solution: (a) The density of blood is 1060 kg∙m3, so the mass and weight of the blood are m = ρV = (1060 kg∙m3) (5.2 x 10-3 m3) = 5.5 kg W = mg = (5.5 kg) (9.80 m∙s-2) (b) The percentage of body weight contributed by the blood is Percentage = 54 N X 100 = 7.8% 690 N

Exercises: 1. A water bed has dimension of 1.83 m x 2.13 m x 0.229 m. The floor of the bedroom will tolerate an additional load of no more than 6660 N. Find the weight of the bed and determine whether it should be purchased. (ρw = 103 kg∙m-3) 2. A 60-ml flask is filled with mercury at 0˚C. When the temperature rises to 80˚C, 1.47 gm of mercury spills out of the flask. Assuming that the volume of the flask is constant, find the density of mercury at 80˚C if its density at 0˚C is 13,645 kg∙m-3.

3. Find the mass and weight of the air in a living room with a 4.0 m x 5.0 m floor and a ceiling 3.0 m high. What is the mass and weight of an equal volume of water?

(Note: The answers to the exercises above are given at the end of the last lesson. If the student did not perform satisfactorily on the exercises, he/she is advised to go over the lesson again and carefully study the examples.)

Lesson 2: PRESSURE

Objective: To be able to understand that pressure is developed at a surface by the perpendicular force acting on it; vis-à-vis, pressure generates perpendicular force at a surface. Definition of Pressure The pressure P is the magnitude F of the force acting perpendicular to a surface divided by the area A over which the force acts: P=F∕A SI Unit of Pressure: N∙m-2 = Pascal (Pa) A pressure of 1Pa is a small amount. Many common situations involve pressure of approximately 105Pa, an amount referred to as 1bar of pressure. Alternatively, pressure can also be measured in pounds per square inch (psi). While fluid pressure can generate a force, pressure itself is not a vector quantity, as it the force. Because in the definition, P = F ∕ A, the symbol F refers only to the magnitude of force, and hence, pressure has no directional characteristics. The force generated by the pressure of a static fluid is always perpendicular to the surface on which the fluid acts. In general, a static fluid cannot produce a force parallel to a surface, for if it did, the surface would apply a reaction force to the fluid, consistent with Newton’s 3rd law. In response, the fluid would flow and would not then be static. The air above the surface of the earth also creates pressure at sea level known as atmospheric pressure that has a value: 1.013 x 105 Pa. This amount of pressure corresponds to 14.7 lb∙in-2 (psi) and is referred to as one atmosphere (atm.). Examples: 1. In a room with 4.0 m x 5.0 m floor and a ceiling 3.0 m high, what is the total downward force on the floor surface due to air pressure of 1.00 atm? Solution: The floor area is A = (4.0 m) (5.0 m) = 20 m 2. The total downward force, therefore, is F = pA = (1.013 x 105 Pa.) (20 m2) = 2.0 x 106 N 2. A brick weighs 17.8 N and is resting on the ground. The dimensions of the brick are 0.203 m x 0.0890 m x 0.0570 m. A number of bricks are then stacked on top of this one. What is the smallest number of bricks (including the one on the ground) that could be used, so that their weight creates a pressure of at least one atmosphere (1.013 x 105 Pa) on the ground beneath the first brick? Solution: It is obvious that the face of the brick with smallest area that is in contact with the ground will give the smallest number of bricks to be used and the area is

A = width x thickness = (0.0890 m) (0.0570 m) = 0.005 m2. The total number of bricks N is equal to the total weight Wt of the stack of bricks divided by the weight W of each brick; that is N = Wt ∕ W. But Wt equals the product PA; that is, 5 2 N = PA = (1.013 x 10 Pa) (0.005 m ) = 28.456 W 17.8 N

Hence, the smallest number of bricks to be used is 29. 3. A cylinder (with circular ends) and a hemisphere are solid throughout and made from the same material. They are resting on the ground, the cylinder on one of its ends and the hemisphere on its flat end. The weight of each causes the same pressure to act on the ground. The cylinder is 0.500 m high. What is the radius of the hemisphere? Solution: The pressure cause by the cylinder and hemisphere are Pc = Wc ∕ Ac and PH = WH ∕ AH, respectively. Since Pc = PH, then Wc ∕ Ac = WH ∕ AH. But W = mg and A = πr 2 (both areas are circle as shown below); therefore, mcg = mHg πrc2 πrH2 or

mc = mH rc2 rH2

rc

h

rH

Note also that m = ρV, again the above becomes

ρcVc = ρHVH rc2 rH2 But Vc = πrc2h, VH = ⅔πrH3, and ρc = ρH = ρ; hence, ρπrc2h = 2ρπrH3 rc2

3rH2

then solving for rH, we have rH = 3h = 3(0.50 m) = 0.75 m 2 2 Exercises: 1. The top of a card table is 80 cm x 80 cm. What is the force exerted on it by the atmosphere? Why doesn’t the table collapse? 2. When a woman in high heels takes a step, she momentarily places her entire weight on one heel of her shoe, which has a radius of 0.40 cm. If her mass is 56 kg, what is the pressure exerted on the floor by the heel? 3. A swimming pool measures 5.0 m long x 4.0 m wide x 3.0 m deep. Compute the force exerted by the water against the bottom.

(Note: The answers to the exercises above are given at the end of the last lesson. If the student did not perform satisfactorily on the exercises, he/she is advised to go over the lesson again and carefully study the examples.)

Lesson 3: THE RELATION BETWEEN PRESSURE AND DEPTH IN STATIC FLUID

Objective: To be able to understand that depth contributes to the development of pressure at a point. To determine the relation between pressure and depth, the figure below shows a container of fluid and considers one particular column of the fluid. Pressure = P1 Area = A

F1 mg h F2 Free-body diagram of the column

Pressure = P2 Area = A

The free-body diagram in the figure above shows all the vertical forces acting on the column. On the top surface (area = A), the fluid pressure P 1 generates a downward force whose magnitude is P1A. Similarly, on the bottom face, the pressure P2 generates an upward force of magnitude P2A. The pressure P2 is greater than the pressure P1, because the bottom face supports an excess weight that is exactly equal to the weight mg of the fluid within the column, where m is the mass of the fluid and g is the magnitude of the acceleration due to gravity. The fluid being at rest, is in equilibrium, so the net force acting on the column must be zero; that is, ∑F = P2A − P1A − mg = 0 or P2A = P1A + mg. The mass m is related to the density ρ and volume V of the column by m = ρV. Since the volume is the cross-sectional area A times the vertical dimension h, we have m = ρAh. With this substitution, the equation above becomes P2A = P1 + ρAhg. The area A can be eliminated algebraically from this expression, with the result that P2 = P1 + ρgh

In determining the pressure increment ρgh, we assumed that the density ρ is the same at any vertical distance h or say that the fluid is incompressible. When applied to gases, the relation P2 = P1 + ρgh can be used only when h is small enough that any variation in ρ can be neglected because, for gases, densities varies with the vertical distance. If we let P1 = Patm then the difference P2 −Patm is called the gauge pressure; that is, P2 −Patm = ρgh = Pgauge. The gauge pressure is the amount by which the pressure at a point exceeds atmospheric pressure. Also, the actual value for P2 is called absolute pressure and is given by Pabs = Patm + Pgauge. Examples: 1. Find the pressure at a depth of 10 m below the surface of a lake if the pressure at the surface is 1 atmosphere (1.013 x 105 Pa). Solution: In the equation P2 = P1 + ρgh, let P2 = Pbottom and P1 = Psurface; that is, Pbottom = Psurface + ρgh = 1.013 x 105 Pa + (103 kg∙m-3) (9.8 m∙s-2) (10 m) = 1.993 x 105 Pa. 2. The Mariana Trench is located in the Pacific Ocean and has a depth of approximately 11,000 m. The density of the seawater is 1025 kg∙m-3 and the pressure at the surface of the ocean is 101 kPa. If a diving chamber were to explore such depths, what force would the water exert on the chamber’s observation window (radius = 0.10 m)? Solution: The force exerted by the water on the window is F = PA. But P2 = Pbottom and P1 = Psurface, then the relation P2 = P1 + ρgh will become Pbottom = Psurface + ρgh. Therefore, F = PbottomA = (Psurface + ρgh) (πr2) = [1.013 x 105 Pa + (1025 kg∙m-3) (9.8 m∙s-2) (11,000 m)] [π (0.10 m) 2] = 3.47 x 106 N. 3. What gauge pressure must a pump produce to pump water from the bottom of the Grand Canyon (elevation 730 m) to Indian Garden (1370 m)? Solution: The height h is just the difference in elevations; that is, h ==1370 m − 730 m = 640 m. Using now the relation P2 − Patm = ρgh we have Pgauge = ρgh = (103 kg∙m-3) (9.8 m∙s-2) (640 m)

= 6.27 x 106 Pa 4. A solar water-heating system uses solar panels on the roof, 12.0 m above the storage tank. The water pressure at the level of the panels is one atmosphere (1.013 x 105 Pa). (a) What is the absolute pressure in the tank? (b) The gauge pressure? Solution: (a) The absolute pressure is Pabs = Patm + Pgauge = Patm + ρgh = 1.013 x 105 Pa + (103 kg∙m-3) (9.8 m∙s-2) (12.0 m) = 2.19 x 105 Pa (b) The gauge pressure is Pgauge = Pabs − Patm = 2.19 x 105 Pa −1.013 x 105 Pa = 1.18 x 105 Pa.

Exercises: 1. The pressure on the surface of a lake is atmospheric pressure Patm = 1.013 x 105 Pa. At what depth is the pressure twice atmospheric pressure? 2. You are designing a diving bell to withstand the pressure of seawater at a depth of 250 m, (a) what is the gauge pressure at this depth? (b) At this depth, what force due to the gauge pressure of the water outside is exerted on the circular window 30.0 cm in diameter? 3. Find (a) the absolute pressure and (b) the gauge pressure at the bottom of the pool of depth 5.00 m.

(Note: The answers to the exercises above are given at the end of the last lesson. If the student did not perform satisfactorily on the exercises, he/she is advised to go over the lesson again and carefully study the examples.) Lesson 4: PRESSURE GAUGES Objective: (i) To be able to get familiar with different pressure gauges. (ii) To be able to use the operating equations for pressure gauges.

One of the simplest pressure gauges is the mercury barometer used for measuring atmospheric pressure. As the figure below shows, this device is a tube sealed at one end, filled completely with mercury, and then inverted, so that the open end is under the surface of a pool of mercury. The space above the mercury in the tube is empty and the pressure P1 is nearly zero there. The pressure P2 at point A at the bottom of the mercury column is the same as that at point B, namely, atmospheric pressure, for these two points are at the same level.

Empty (P1=0) Mercury: ρ = 13.6 x 3

10 kg∙m h

A B ▪ ▪

-3

Point B (P2 = atmospheric pressure)

With P1 = 0 and P2 = Patm, it follows from the equation above that Patm = 0 + ρgh. Thus, the atmospheric pressure can be determined from the height h of mercury in the tube, the density ρ of mercury, and the acceleration due to gravity. Usually weather forecasters report in terms of the height h (=Patm ∕ ρg), expressing it in millimeters (or inches) of mercury; thus, Patm = 1.013 x 105 Pa can be expressed as 760 mm (or 29.9 in) of mercury. Another kind of pressure gauge is the open-tube manometer where one side of the U-tube is open to atmospheric pressure. The tube contains a liquid, often mercury, and its other side is connected to the container whose pressure P2 is to be measured, as shown below.

P1 = Patm

P1 = Patm

h P2 = Patm

P2 > Patm B

(a)

A

(b)

The figure above indicates that when the pressure in the container is equal to the atmospheric pressure, the liquid’s levels in both sides of the U-tube are the same. When the pressure in the container is greater than the atmospheric pressure, the liquid in the tube is pushed downward on the left side and upward on the right side. The relation P2 = P1 + ρgh can be used to determine the container pressure. Atmospheric pressure exists at the top of the right column, so that P1 = Patm. The pressure P2 is the same at points A and B, so find that P2 = Patm + ρgh Examples: 1. How high does a mercury barometer stand on a day when atmospheric pressure is 98.6 kPa? Solution: Using the equation Patm = ρgh, the height, therefore, is 98.6 x 103 N∙m2 h = Patm = = 0.74 m 3 3 2 ρg (13.6 x 10 kg∙m ) (9.8 m∙s ) 2. A U-shaped tube open to the air at both ends contains some mercury. A quantity of water is carefully poured into the left arm of the U-shaped tube until the vertical height

of the water column is 15.0 cm. (a) What is the gauge pressure at the water-mercury interface? (b) Calculate the vertical distance h from the top of the mercury in the righthand arm of the tube to the top of the water in the left-hand arm.

15.0 cm

Water

h

Mercury

Solution: (a) The pressure at the interface is Pinterface = ρwaterghwater = (1.00 x 103 kg∙m-3) (9.80 m∙s-2) (15.0 x 10-2 m) = 1.47 x 103 Pa. (b) The gauge pressure at a depth of 15.0 cm – h below the top of the mercury column must be that found in part (a); that is, ρHgg (15.0 cm – h) = ρwaterg (15.0 cm) Then, solving for h, we have h = 13.9 cm. 3. The liquid in the open-tube manometer in the figure below is mercury (ρ = 13.6 x 103 kg∙m-3), y1 = 3.00 cm, and y2 = 7.00 cm. Atmospheric pressure is 1.013 x 105 Pa. (a) What is the absolute pressure at the bottom of the U-tube? (b) What is the absolute pressure of the gas in the tank?

h y2

Ptank

y1

Solution: (a) Using the relation P2 = P1 + ρgh, where P2 = Pbottom, P1 = Psurface = Patm, and h = y2, we have Pbottom = Patm + ρgy2 = 1.013 x 105 Pa + (13.6 x 103 kg∙m-3) (9.8 m∙s-2) (0.07 m) = 1.11 x 105 Pa. Note that Pbottom in the above computation is already absolute. (b) Using the relation P2 = Patm + ρgh where P2 = Ptank and h = y2 − y1, we have Ptank = Patm + ρg(y2 − y1) = 1.013 x 105 Pa + (13.6 x 103 kg∙m-3) (9.8 m∙s-2) (0.07 m − 0.03 m) = 1.07 x 105 Pa. Note also that Ptank is already absolute.

Exercises: 1. Compute the atmospheric pressure Patm on a day when the height of mercury in a barometer is 760 mm. 2. Referring to the example above, compute the gauge pressure (a) at the bottom of the U-tube;(b) of the gas in tank.

3. (a) A mercury barometer is used at a location where the atmospheric pressure is 1.013 x 105 Pa and the acceleration due to gravity is 9.78 m∙s-2. How high (in mm) does the mercury rise in the barometer? (b) At another location the atmospheric pressure is also 1.013 x 105 Pa, but the acceleration due to gravity is 9.83 m∙s-2. How high (in mm) does the mercury rise here?

(Note: The answers to the exercises above are given at the end of the last lesson. If the student did not perform satisfactorily on the exercises, he/she is advised to go over the lesson again and carefully study the examples.)

Lesson 5: PASCAL’S PRINCIPLE Objective: To be able to understand its application to hydraulic lifts and its operating equation.

Pascal’s Principle Any change in the pressure applied to a completely enclosed fluid is transmitted undiminished to all parts of the fluid and the enclosing walls. The drawing below shows two interconnected cylinder chambers. The chambers have different diameters and together with the connecting tube, are completely filled with a liquid. The larger chamber is sealed at the top with a cap, while the smaller one on the left is filled with a movable piston.

F1

A1 F2

A2

What determines the pressure P1 at a point immediately beneath the piston is the magnitude F1 of the external force divided by the area A1 of the piston: P1 = F1 ∕ A1. If it is necessary to know the pressure P2 at any other place in the liquid, we just add to the value of P1 the increment ρgh, which takes into account the depth below the piston: P2 = P1 + ρgh. Notice that pressure P1 adds to the pressure ρgh due to the depth of the liquid at any point, whether that point is in the smaller chamber the connecting tube, or larger chamber. This behavior is described by Pascal’s principle. Now, when the tops of the left and right chambers are at the same level, the pressure increment ρgh is zero, so that the relationship P2 = P1 + ρgh becomes P2 = P1. Consequently, F2 ∕ A2 = F1 ∕ A1, and

F1 = F2 (A1 ∕ A2) where F2 is the force applied by the liquid to the cap, A 2 is the area of the cap, and P 2 is the pressure there. Example: 1. In a hydraulic press used in a trash compactor, the radii of the input piston and the output plunger are 6.4 x 10-3 m and 5.1 x 10-2 m, respectively. If the height difference between the input piston and the output plunger can be neglected, what force is applied to the trash when the input force is 330 N? Solution: The force applied to the trash equals the force applied to the output plunger (F2). Using F1/A1 = F2/A2, we find that F2 = F1 (A2/A1) = 330 N (5.1 x 10-2 m / 6.4 x 10-3) = 2.63 x 103 N 2. In the process of changing a flat tire, a motorist uses a hydraulic jack. She begins by applying a force F1 = 5.0 N to the input piston. As a result, the output plunger applies a force of F2 = 340 N to the car. The height difference between the input piston and the output plunger can be neglected. What is the ratio A2/A1 of the plunger and piston area? Solution: Using F1/A1 = F2/A2, the needed ratio will be A2/A1 = F2/F1 = 340N/5.0 N = 68 3. In a hydraulic car lift, the input piston has a radius of r 1 = 0.0120 m and a negligible weight. The output plunger has a radius of r2 = 0.150 m. The combined weight of the car and the output plunger is F2 = 20500 N. The lift uses hydraulic oil that has a density of 8.00 x 102 kg∙m-3. What input force F1 is needed to support the car and the output plunger when the bottom surfaces of the piston and the plunger are at (a) the same level and (b) different levels as shown in the figure below with h = 1.10 m?

F1

h

A1

F2

A2

Solution: (a) using A = πr2 for the circular areas of the piston and plunger, we find that F1 = F2 (A1 ∕ A2) = F2 (πr12 ∕ πr22) = (20500) [(0.0120)2 ∕ (0.150)2] = 131 N (b) Considering the figure above, we can apply P2 = P1 + ρgh, with P2 = 2 F2 ∕ πr2 and P1 = F1 ∕ πr12; therefore, F2

= 2

πr2

F1

+ ρgh

2

πr1

Solving for F1 gives F1 = F2 (r12 ∕ r22) − ρgh (πr12) = 131 N − (8.00 x 10-2 kg∙m-3) (9.8 m∙s-2) (1.10 m) π (0.0120 m2) = 127 N

Exercises: 1. The large piston in a hydraulic lift has a radius of 20 cm. What force must be applied to the small piston of radius 2.0 cm to raise the car of mass 1,500 kg? Assume that the lower surfaces of the pistons are at the same level. 2. The drawing shows two blocks (masses m1 and m2) resting on hydraulic cylinders. The bottom surfaces of the cylinders are at the same level. The radius of the cylinder on the right is three times that of the cylinder on the left. Find the ratio m 1/m2 of the masses.

m1

m2

3. For the system shown in the figure below, the cylinder on the left, at L, has a mass of 600 kg and cross – sectional area of 800 cm2. The piston on the right, at R, has cross – sectional area 25 cm2 and negligible weight. If the apparatus is filled with oil (ρ = 780 kg∙m-3), what is the force required to hold the system in equilibrium as shown?

F

R

8m L

(Note: The answers to the exercises above are given at the end of the last lesson. If the student did not perform satisfactorily on the exercises, he/she is advised to go over the lesson again and carefully study the examples.)

Lesson 6: ARCHIMEDES’ PRINCIPLE Objective: To be able to apply the principle in understanding the mechanics of buoyancy.

Archimedes’ Principle Any fluid applies a buoyant force to an object that is partially or completely immersed in it; the magnitude of the buoyant force equals the weight of the fluid that the object displaces: FB magnitude of the buoyant force

=

Wdisp. fluid weight of the displaced fluid

Buoyant force is the upward force exerted by all fluids to objects that are immersed in them. It exists because fluid pressure is larger at greater depths. In the figure below, a cylinder of height h is being held under the surface of a liquid. The pressure P1 on the top face generates the downward force P1A, where A is the area of the face. Similarly, the pressure P2 on the bottom face generates the upward force P2A.

A

P1A h P2A

Since the pressure is greater at greater depths, the upward force exceeds the downward force. Consequently, the liquid applies to the cylinder a net upward force, buoyant force, whose magnitude FB is

FB = P2A − P1A = (P2 − P1) A Recall that P2 = P1 + ρgh and hence, substituting P2 −P1 = ρgh to the above equation, we find that the buoyant force equals ρghA. In this result the hA is the volume of the liquid that the cylinder moves aside or displaces in being submerged, and ρ denotes the density of the liquid. Therefore, ρhA gives the mass m of the displaced fluid, so that the buoyant force equals mg, the weight of the displaced fluid. The cylindrical shape of the object in the figure is not important. No matter what shape, the buoyant force arises in a similar fashion. Note that if the weight of the object is less than or equal to the buoyant force, the object will float. And for any object that is solid throughout, if the density of the object is less that or equal to the density of the liquid, the object will float as well. Examples: 1. A block of material has a density ρ1 and floats three-fourths submerged in a liquid of unknown density. Show that the density ρ2 of the unknown liquid is given by ρ2 = 4/3 ρ1. Solution: The weight of the material equals the buoyant force exerted by the unknown liquid; that is W = FB But W=mg=ρ1Vg and FB=ρ2 (3V/4) g, therefore ρ1Vg = ρ2 (3V/4) g which gives ρ2 = 4/3 ρ1 2. A 15.0 kg solid gold statue is being raised from a sunken ship. What is the tension in the hoisting cable when the statue is at rest and completely immersed. (ρ = 1.03 x 103 kg∙m-3) Solution: The volume of the statue is V=

m ρgold

=

15.0 kg

-4

= 7.77 x 10 m

3

19.3 x 103 kg∙m-3

The buoyant force equals the weight of the displaced seawater; that is, FB = Wsw = mswg. Using m = ρV, we have FB = ρsw Vswg = ρswVg = (1.03 x 103 kg∙m-3) (7.77 x 10-4 m3) (9.8 m∙s-2)

= 7.84 N Since the statue is at rest, ∑Fy = FB + T − mg = 0 Solving for T, we have T = mg − FB = (15.0 kg) (9.8 m∙s-2) − 7.84 N = 147 N − 7.84 N = 139 N 3. A slab of ice floats on a freshwater lake. What minimum volume must the slab have for a 45.0-kg lady to be able to stand on it without getting her feet wet? (ρ = 920 kg∙m-3) Solution: Since we are looking for the minimum volume, this means that the top surface of the slab of ice is along the surface of the lake, making the volume of the slab equals the volume of the displaced water; that is Vi = Vdw. Hence, FB = Wdw = mdwg = ρwVwg = ρwVig Again, the whole system is in equilibrium; therefore, ∑Fy = FB − Wi − WL = 0 Or FB − Wi = WL ρwVig − mig = mLg ρwVi − ρiVi = mL Solving for Vi: Vi =

mL ρw − ρ i

=

45 kg________ 103 kg∙m-3 − 920 kg∙m-3

= 0.56 m3

Exercises: 1. A metal ball weighs 0.096 N. When suspended in water it has an apparent weight of 0.071 N. Find the density of the metal.

2. The density of ice is 917 kg∙m-3, and the approximate density of the seawater in which an iceberg floats is 1025 kg∙m-3. What fraction of the iceberg is beneath the water surface? 3. A piece of wood is 0.600 m long, 0.250 m wide, and 0.080 m thick. Its density is 600 kg∙m-3. What volume of lead must be fastened underneath to sink the wood in calm water so that its top is just even with the water level?

(Note: The answers to the exercises above are given at the end of the last lesson. If the student did not perform satisfactorily on the exercises, he/she is advised to go over the lesson again and carefully study the examples.)

Answers to the exercises: Lesson 1 1. 8751.4 N 2. 13,621 kg∙m-3 3. 72 kg, 700 N; 6.0 x 104 kg, 5.9 x 105 N Lesson 2 1. 6.48 x 104 N, The table doesn’t collapse because the atmosphere also exerts an upward force on the bottom surface of the table. 2. 1.09 x 107 Pa. 3. 5.88 x 105 N Lesson 3 1. 10.34 m 2. (a) 2.51 x 106 Pa; (b) 7.10 x 105 N 3. (a) 1.503 x 105 Pa; (b) 4.90 x 104 Pa Lesson 4 1. 1.01 x 105 Pa 2. (a) 9329.6 Pa; (b) 5331.2 Pa 3. (a) 762 mm; (b) 758 mm Lesson 5 1. 147 N 2. 1/9 3. 31 N Lesson 6 1. 3840 kg∙m-3 2. 0.89 3. 4.66 x 10-4 m3