Walker4 Ism Ch15

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15 – 1

Chapter 15: Fluids

Answers to Even-Numbered Conceptual Questions 2. No, because the Moon has no atmosphere to press down on the surface of the liquid.

4. A suction cup is held in place by atmospheric pressure. When the cup is applied, you push it flat against the surface for which you want it to stick. This expels most of the air in the cup, and leads to a larger pressure on the outside of the cup. Thus, atmospheric pressure pushes the outside of the cup against the surface.

6. Mercury is more practical in a barometer than water because of its greater density. With such a large density, the height of the mercury column is only about 0.760 m. The density of water is less than that of mercury by roughly a factor of 14. Therefore, the height of a water column in a barometer would be about 10 m—the height of a three-story building.

8. In a hot-air balloon, vertical motion is controlled by adding heat to the air in the balloon, or by letting it cool off. As the temperature of the air in the balloon changes, so too does its density. By controlling the overall density of the balloon, one can control whether it rises, falls, or is neutrally buoyant.

10. The physics in this case is pretty ugly. Ice floats in water, whether it is a house-sized iceberg, a car-sized chunk, or a thimble-sized ice cube. If the Earth is warming and icebergs are breaking up into smaller pieces, each of the smaller pieces will be just as buoyant as the original berg.

12. A metal boat can float if it displaces a volume of water whose weight is equal to the weight of the boat. This can be accomplished by giving the boat a bowl-like shape, as illustrated in Figure 15-11.

14. As wind blows across the top of the chimney, a pressure difference is established between the top and bottom of the chimney, with the top having the lower pressure. This will cause smoke to rise more rapidly.

16. If a ball is placed in the stream of air such that the speed of air over its upper surface is greater than the speed across its lower surface, the result will be a lower pressure at the top of the ball. This results in an upward force that can equal the weight of the ball.

Solutions to Problems and Conceptual Exercises

1. Picture the Problem: Air, although light in comparison to water or solid objects, has mass. We wish to calculate the

mass of air in a typical classroom. Strategy: The volume of air in a typical classroom is on the order of 2 310 m . Multiply the mass by the acceleration of

gravity to calculate the weight. Solve equation 15-1 for the mass in terms of density and volume. The density of air is given in Table 15-1.

Solution: Use equation 15-1 to write the weight in terms of density and volume: ( )( )( )3 2 2 31.29 kg/m 10 m 9.81 m/s 10 NW mg Vgρ 3= = = =

Insight: The air in a physics classroom weighs about the same as two students. 2. Picture the Problem: A 25-gallon aquarium is filled with water. We wish to calculate the weight of the water.

Strategy: Multiply the mass times the acceleration of gravity to calculate the weight. Solve equation 15-1 for the mass in terms of density and volume. The density of water is given in Table 15-1.

Solution: Use equation 15-1 to write the weight in terms of density and volume: ( )( )( )( )3 3 3 21000 kg/m 25 gal 3.79 10 m /gal 9.81 m/s 0.93 kN

W mg Vgρ−

= =

= × =

Insight: The 25-gallon tank weighs about 200 lbs when filled.

Chapter 15: Fluids James S. Walker, Physics, 4th Edition


15 – 2

3. Picture the Problem: A gold ring has a density equal to its mass divided by its volume.

Strategy: If the ring is pure gold, its density will be equal to the density of gold. Since the mass and volume of the ring are known, use equation 15-1 to calculate the density. Compare the result with the density of gold given in Table 15-1.

Solution: 1. Divide the volume by the mass: 3

3

0.014 g 6.4 g cm0.0022 cm

mV

ρ = = =

2. Compare with the density of gold from Table 15-1: 3gold 19.3 g cmρ = . No, the ring is not solid gold.

Insight: If the ring were pure gold of the same volume given in the problem, its mass would be 42.5 g.

4. Picture the Problem: A treasure chest is filled with gold doubloons. A typical chest might measure 8 inches by 8 inches by 12 inches and have a volume on the order of 10-1 m3

Strategy: Solve equation 15-1 for the mass of the chest in terms of the density of gold and the volume of the chest. Calculate the weight of the chest by multiplying its mass times the acceleration of gravity. The density of gold is given in Table 15-1.

Solution: Use equation 15-1 to write the weight in terms of density and volume: ( )( )( )4 3 –1 3 2 41.93 10 kg/m 10 m 9.81 m/s 10 NW mg Vgρ= = = × =

Insight: Even though this chest seems fairly small, it weighs a ton!

5. Picture the Problem: A cube has a mass of 0.347 kg and sides of 3.21 cm each.

Strategy: Use equation 15-1 to calculate the density of the cube. Compare the resulting density with the densities given in Table 15-1 to determine the likely composition.

Solution: 1. Calculate the density of the cube:

( )4 3

31 m100 cm

0.347 kg 1.05 10 kg m3.21 cm

mV

ρ = = = ×⎡ ⎤⎣ ⎦

2. Compare with the densities in Table 15-1: The cube has the density of silver.

Insight: Cubes made of different materials could have considerably different masses. For example, a cube of gold (with the same volume as the silver cube) would have a mass 0.638 kg, while a cube of aluminum would have a mass of 0.089 kg.

6. Picture the Problem: The football field has an area given by its length multiplied by width. Atmospheric pressure produces a downward force on the field.

Strategy: Solve equation 15-2 for the force exerted over the area of the football field. Set the pressure equal to atmospheric pressure, 1.01×105 N/m2.

Solution: Solve equation 15-2 for F: ( )( )( )

25 2 8m1.01 10 N/m 360 ft 160 ft 5.40 10 N

3.281 ftF PA ⎛ ⎞= = × = ×⎜ ⎟

⎝ ⎠

Insight: This weight is over 50,000 tons!

7. Picture the Problem: A pressure of 1 dyne per square centimeter needs to be converted to the units of pascals and atmospheres.

Strategy: The pressure is given as a force divided by area. Convert the units of force and area to the standard SI units of Newton and square meter to write the pressure in pascals. Then use equation 15-3 to convert to atmospheres.

Solution: 1. (a) Convert the pressure to pascals:

25–1

2

1 dyne 10 N 100 cm 10 Padyne mcm

−⎛ ⎞⎛ ⎞ ⎛ ⎞ =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠



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2. (b) Use equation 15-3 to convert to atmospheres: –1 –6

3

1 atm10 Pa 10 atm101.3 10 Pa⎛ ⎞ =⎜ ⎟×⎝ ⎠

Insight: A dyne/cm2 is the standard unit of pressure in the cgs system of units (see section 1-2 of the text), which is commonly used in chemistry. The pascal is the standard unit of pressure in the mks system.

8. Picture the Problem: When a person sits in a four-legged chair the weight of the person and chair is distributed over

each leg of the chair, increasing the pressure each leg exerts on the ground.

Strategy: Use equation 15-2 to calculate the pressure each leg exerts on the floor. Set the force equal to the sum of the weights of the person and chair and the area equal to four times the cross-sectional area of each leg.

Solution: 1. Set the pressure equal to the weight divided by area:

22

42

F mg mgPA dd π

π

= = =⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟

⎝ ⎠⎢ ⎥⎣ ⎦

2. Insert given data: ( ) ( )

( )

26

2

79 kg 3.7 kg 9.81 m/s1.5 10 Pa 1.5 MPa

0.013 mP

π

+= = × =

Insight: Leaning back in the chair, so that is rests on only two legs, doubles the pressure those legs exert on the floor. 9. Picture the Problem: When walking with crutches, a person supports a large portion of her weight on the crutch. If the

end of the crutch did not have a rubber tip, the entire weight would be supported over the small area of the crutch. The rubber tip increases the area over which the weight is distributed, thus decreasing the pressure.

Strategy: Let Pwo represent the pressure without the rubber tip and Pw represent the pressure with the rubber tip. Calculate the ratio of Pw to Pwo using equation 15-2 for the pressure. The force on the crutches is the same and the cross-sectional area is the area of a circle.

Solution: 1. Find the ratio of the pressure with the tip to the pressure without:

( )( )

w

wo

22w wo wo

2 2wo w w

1.2 cm0.23

2.5 cm

FA

FA

P A rP A r

ππ

= = = = =

2. Invert the ratio to find the factor by which the pressure decreases:

1 4.30.23

=

Insight: Because area is proportional to the square of the radius, the pressure is decreased by the square of the fractional increase in radius.

10. Picture the Problem: The gauge pressure on a basketball is taken by measuring the difference between the pressure

inside the ball and the pressure outside the ball. Strategy: Calculate the absolute pressure inside the ball by adding the gauge pressure to the atmospheric pressure.

Atmospheric pressure in British units is 14.7 lb/in2. Solution: Add atmospheric pressure to the gauge pressure: 2 2 2

atm g 14.7 lb/in 9.9 lb/in 24.6 lb/inP P P= + = + =

Insight: Moving the basketball to a location of high external pressure (such as the bottom of a swimming pool) or low external pressure (such as at the top of a mountain) will change the measured gauge pressure, but will not change the absolute pressure inside the ball.



15 – 4

11. Picture the Problem: When you ride a bicycle, your weight and the weight of the bicycle are supported by the air

pressure in both tires spread out over the area of contact between the tires and the road. Strategy: To calculate your weight, first solve equation 15-2 for supporting force of the air pressure on the tires. Set

this force equal to the sum of your weight and the weight of the bicycle. Subtract the weight of the bicycle to determine your weight.

Solution: 1. Multiply the tire pressure by the contact area to calculate the supporting force on the bicycle:

( ) ( )25

2 22

1.01 10 Pa 1 m70.5 lb/in 2 7.13 cm100 cm14.7 lb/in

690.7 N

F PA⎛ ⎞× ⎛ ⎞= = ×⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠=

2. Set the supporting force equal to the sum of your weight and the weight of the bicycle: you bicycle you bicycleF W W W m g= + = +

3. Solve for your weight: ( )2you bicycle 690.7 7.7kg 9.8m/s 615 NW F m g N= − = − =

Insight: When “popping a wheelie” on the bicycle, such that only one wheel is touching the ground, that wheel must support the entire weight of the bicycle and rider. Therefore, because the tire pressure has not changed, the area of contact for the single tire would double. In this problem the area would increase to 14.26 cm2.

12. Picture the Problem: The weight of the car is supported by the air pressure in all four tires spread out over the area of contact between the tires and the road.

Strategy: Use equation 15-2 to find the necessary contact area to support the weight of the car for the given tire pressure. Divide the area by four to calculate the area of contact for each tire. Because the weight of the car does not change significantly as the tire pressure is increased, equation 15-2 shows that the tire pressure and contact area are inversely proportional to each other. Note that it is the gauge pressure that supports the weight of the car; atmospheric pressure inside the tire simply balances the air pressure outside the tire. We assume the rubber tire itself is very flexible and supports essentially no weight by itself.

Solution: 1. (a) Solve equation 15-2 for the total contact area:

F mgAP P

= =

2. Solve for the contact area on one tire: tire tire4

4mg mgA A AP P

= = ⇒ =

3. Insert given values:

( )( )( )( )5

2

22 2

tire 1.01 10 Pa214.7 lb/in

1420 kg 9.81 m/s0.0145 m 145 cm

4 35.0 lb/inA

×= = =

4. (b) Because the area and pressure are inversely proportional, as the pressure increases the area of contact decreases.

5. (c) Solve for the pressure and insert the given values:

( )( )( )

( )

2

22 1 mtire100 cm

25 2

5

1420 kg 9.81 m/s

4 4 116 cm

14.7 lb/in3.00 10 Pa 43.7 lb/in1.01 10 Pa

mgPA

= =⎡ ⎤× ⎣ ⎦

⎛ ⎞= × =⎜ ⎟×⎝ ⎠

Insight: The pressure in part (c) was greater than the pressure given in part (a). As predicted, the increase in pressure resulted in a decrease in contact area.

13. Picture the Problem: Two drinking glasses, 1 and 2, are filled with water to the same depth. Glass 1 has twice the diameter of glass 2.

Strategy: Use the principles of density and pressure in a static fluid to answer the conceptual questions.

Solution: 1. (a) Glass 1 has a larger diameter but is filled with water to the same depth, so it contains a larger volume and a larger mass of fluid. We conclude that the weight of the water in glass 1 is greater than the weight of the water in glass 2.



15 – 5

2. (b) Each glass is filled with water to the same depth h, so P g hρΔ = (equation 15-7) predicts the pressure at the bottom of glass 1 is equal to the pressure at the bottom of glass 2.

Insight: Recall that pressure is force per area. Glass 1 exerts four times as much force on its base, because it holds four times as much water, but it also has a base that is four times greater in area. These two factors of four cancel in determining the pressure.

14. Picture the Problem: The figure at right shows four containers, each filled with water to the same level.

Strategy: Use equation 15-7 to determine the ranking of the pressures at depth h in each of the containers.

Solution: We are to predict the pressure in each container at the same depth h, so P g hρΔ = (equation 15-7) predicts the pressures are all the same. The ranking is A = B = C = D.

Insight: The pressure at a given depth is independent of the shape of the container. If the containers had been filled with fluids of different densities, the one with the highest density fluid would have the highest pressure at depth h.

15. Picture the Problem: The water pressure in Hoover dam increases with depth.

Strategy: We need to solve for the pressure at the base of the dam, 221 meters below the surface. Because the problem asks for water pressure only, we combine equations 15-5 and 15-6 to find the gauge pressure at the base of the dam.

Solution: Combine equations 15-5 and 15-6: ( )( )( )

g atm atm atm

3 2 61000 kg/m 9.81 m/s (221 m) 2.17 10 Pa

P P P P gh P ghρ ρ= − = + − =

= = ×

Insight: The water pressure at the base of the dam is 22.5 times atmospheric pressure.

16. Picture the Problem: A soda can has zero pressure inside and atmospheric pressure pushing inward from the outside.

Strategy: Calculate the net force using equation 15-2, with the pressure being atmospheric pressure and the area the area of a cylinder, .A D hπ=

Solution: 1. Calculate vertical area: ( ) ( ) 20.065 m 0.12 m 0.0245 mA D hπ π= = =

2. Solve for inward force: ( )5 2 2at 1.01 10 N/m 0.0245 m 2.5 kNF P A= = × =

Insight: This force is equal to over 500 lbs, which will easily crush the can.

17. Picture the Problem: Atmospheric pressure will cause a column of mercury to rise 760 mm into a vacuum. Changes in air pressure are measured by the changes in height of the barometer. Atmospheric pressure will cause a column of water to rise much higher because of is lower density.

Strategy: Convert the height of the mercury column to pascals by using equation 15-7, taking the density of mercury from Table 15-1. Then use equation 15-7 again to calculate the height h of the equivalent water column. Assume P1 = 0 at the top of the barometer.

Solution: 1. (a) Convert the height of the mercury column to pascals: ( )( )( )3 2

2 1 13,600 kg/m 9.81 m/s 0.736 m 98.2 kPaP P ghρ− = = =

2. (b) Solve equation 15-7 for the h:

( )( )

2 14

2 13 2

9.82 10 Pa 0 10.0 m1000 kg/m 9.81 m/s

P P ghP P

hg

ρ

ρ

= +

− × −= = =

Insight: The water column rises over 13 times farther than the mercury column because its density is over 13 times smaller than the density of mercury. The height of the water column makes it impractical to use as a barometer.



15 – 6

18. Picture the Problem: Two pistons are supported by a fluid, as shown

in the figure. The pressure in the fluid at the bottom of the left piston is equal to the pressure in the right piston at the same vertical level, which is a distance h below the right piston.

Strategy: Set the pressures in the two columns equal at the depth h. Use equation 15-1 to calculate the pressure due to the pistons and equation 15-7 to calculate the increase in pressure due to the fluid in the right-hand column.

Solution: 1. Set the pressures equal: L R

L R

L R

P Pm g m g

ghA A

ρ

=

= +

2. Solve for the height h: L R L R L R

2 2 2 2L R L R L R4 4

1 1 4m m m m m mh

A A D D D Dπ πρ ρ πρ⎛ ⎞ ⎛ ⎞⎛ ⎞

= − = − = −⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

3. Insert the given values:

( ) ( ) ( )2 23

4 1.8 kg 3.2 kg 1.2 m750kg/m 0.044 m 0.12 m

hπ

⎡ ⎤= − =⎢ ⎥

⎢ ⎥⎣ ⎦

Insight: The distance h does not depend on the overall height of the pistons. 19. Picture the Problem: As water is poured into the tube shown in the figure, the

pressure inside the barrel increases. When the upward force on the barrel lid exceeds 643 N, the barrel will burst.

Strategy: The force on the barrel top is the pressure at the surface times the area of the top. Use equation 15-6 to calculate the height of the water column when the barrel will burst. Calculate the weight of the water column from the height, cross-sectional area, and density of water using equation 15-1.

Solution: 1. Calculate the bursting pressure of the lid: ( )2

643 N 1460 Pa0.75 m 2

FPA π

Δ = = =

2. Solve equation 15-6 for the height of the water column:

( )3 2

1455 Pa 0.148 m1000 kg/m 9.81 m/s

P ghPhg

ρ

ρ

Δ =Δ

= = =

3. Use equation 15-1 to solve for the weight:

( )( )

2

23 2

2

0.010 m1000 kg/m 0.148 m 9.81 m/s 0.11 N2

dW mg hgρ π

π

⎡ ⎤⎛ ⎞= = ⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦⎡ ⎤⎛ ⎞= =⎢ ⎥⎜ ⎟

⎝ ⎠⎢ ⎥⎣ ⎦

Insight: A very short column of water is able to increase the pressure sufficiently to burst the barrel. This is one reason why rain barrels always have a hole at the top to allow excess water to flow out.



15 – 7

20. Picture the Problem: A cylinder is filled with a fluid, as shown in the diagram. The

pressure at the bottom of the fluid is greater than the atmospheric pressure at the top. We wish to find the depth of the fluid that will result in a pressure at the bottom of 116 kPa. Adding additional fluid to the container will increase the pressure at the bottom. We wish to calculate the increase in pressure when 2.05 × 10-3 m3 are added.

Strategy: Solve equation 15-6 to calculate the depth of the fluid. To calculate the pressure when additional fluid has been added, divide the volume of the fluid by the cross-sectional area to find the additional height of the fluid. Then insert the total height into equation 15-6 for the total pressure.

Solution: 1. (a) Solve equation 15-6 for h: at

at P P

P P gh hg

ρρ−

= + ⇒ =

2. Insert given values:

( )( )3 5

3 2

116 10 Pa 1.01 10 Pa 1.90 m806 kg/m 9.81 m/s

h × − ×= =

3. (b) Divide the volume by A:

3 3

2 4 2

2.05 10 m 0.314 m65.2 10 m

h−

−

×= =

×

4. Add the heights: total 2 1.897 m 0.314 m 2.211 mh h h= + = + =

5. Insert data into equation 15-6:

( )( )at

5 3 21.01 10 Pa 806 kg/m 9.81 m/s (2.211 m) 118 kPa

P P ghρ= +

= × + =

Insight: Part (b) could also have been solved by adding the additional depth to the bottom of the cylinder, such as: ( )( )( )5 3 21.16 10 Pa 806 kg/m 9.81 m/s 0.314 m 118 kPa .P = × + =

21. Picture the Problem: As a submarine dives, the pressure difference between the interior and exterior increases. To be safe, this pressure difference cannot exceed 10.0 N/mm2. We need to solve for the maximum depth that the submarine can dive.

Strategy: Solve equation 15-7 for the depth to which the submarine can descend. Use the density of sea water from Table 15-1. Examine the resulting equation to determine how the density affects the maximum depth.

Solution: 1. (a) Solve equation 15-7 for h: atm

atm ww

P P

P P gh hg

ρρ−

= + ⇒ =


( )( )( )

3 22 10 mmm

3 2

10.0 N/mm995 m

1025 kg/m 9.81 m/sh = =

3. (b) Fresh water is less dense than sea water, so the maximum safe depth in fresh water is greater than in salt water.

Insight: The maximum depth in fresh water is 1020 m.

22. Picture the Problem: A water tower is filled with water. The pressure in the tank increases as the water depth increases. We wish to calculate the pressure at specific depths.

Strategy: Use equation 15-6 to solve for the pressure at the given depths. Use the density of water given in Table 15-1.

Solution: 1. (a) Use equation 15-6:

( )( )at

5 3 2 51.01 10 Pa 1000 kg/m 9.81 m/s 4.5 m 1.45 10 Pa

P P ghρ= +

= × + = ×

2. (b) Repeat for a depth of 5.5 m: ( )( )5 3 2 51.01 10 Pa 1000 kg/m 9.81 m/s 5.5 m 1.54 10 PaP = × + = ×

3. (c) The bands are closer together near to bottom because pressure increases with depth. A greater confining force is needed near the bottom than near the surface of the water.

Insight: The pressure at the bottom of the tank reaches a maximum of 1.64×105 Pa.



15 – 8

23. Picture the Problem: Water is held in a glass on an elevator that is accelerating upward. A free body

diagram for the water is shown at right. We want to calculate the additional pressure at the bottom of the glass due to the acceleration of the elevator.

Strategy: Calculate the acceleration of the water from its change in speed and time. Use Newton’s Second Law to write the additional force necessary to accelerate the water. Divide this force by the cross-sectional area of the glass to find the added pressure in the water.

Solution: 1. (a) The force that the glass exerts upward on the water is greater than the weight of the water in order to provide upward acceleration. By Newton’s Third Law, the water exerts an equal force downward on the glass bottom, so the pressure is greater than it was before the elevator began to move.

2. (b) Calculate the acceleration of the elevator: 22.4 m/s 0 0.75 m/s

3.2 svat

Δ −= = =Δ

3. Write the change in pressure as the added force divided by the area of the glass:

F maPA A

Δ = =

4. Write the mass as density times volume: ( )

( )( )( )3 21000 kg/m 0.069 m 0.75 m/s 52 Pa

AhP a ha

Aρ

ρΔ = =

= =

Insight: If the elevator were accelerating downward at 0.75 m/s2 the pressure in the glass would decrease by 52 Pa.

24. Picture the Problem: A 12-cm-tall column of water lies under a 7.2-cm-tall column of olive oil, as shown in the figure. We wish to calculate the pressure at the bottom of the water.

Strategy: Equation 15-7 gives relationship between pressure and depth within a fluid of known density. The pressure under the oil, P1, can be calculated using equation 15-7 with the density of the oil and height of the oil. To find the pressure at the bottom of the water, insert the pressure P1 into equation 15-7 and add the pressure change from the density of water and height of the water.

Solution: 1. Calculate the pressure at the bottom of the oil: ( )( ) ( )

1 atm

5 3 2

5

1.013 10 Pa 920kg/m 9.81m/s 0.072m

1.0195 10 Pa

P P ghρ= +

= × +

= ×

2. Calculate the pressure at the bottom of the water: ( )( )( )

1 water water

5 3 2 51.0195 10 Pa 1000 kg/m 9.81 m/s 0.12 m 1.03 10 Pa

P P ghρ= +

= × + = ×

Insight: Because the density of oil is less than the density of water, the pressure at the bottom of the water is slightly less than the pressure would be if the entire column were water.

25. Picture the Problem: Oil is placed on top of water on both sides of a U-tube, as shown in the figure. On the right side the depth of the oil is 5.00 cm and on the left side it is 3.00 cm.

Strategy: We want to find h4, the difference in heights of the fluids. The pressure in the tube at points A and B is the same. Use equation 15-7 to calculate the pressure at points A and B. Set the equations equal and solve for h3, the height of the water above point A. Subtract the height h3 and the 3.00 cm of oil from h2 to calculate the difference in heights of the two columns.

Solution: 1. Set the pressure at point A equal to the pressure at point B.

A B

at w 3 oil 1 at oil 2

P PP gh gh P ghρ ρ ρ

=+ + = +

2. Solve for the height of the water above point A:

( )3oil 2 1

3 3w

920 kg/m (0.0500 0.0300 m)( )0.0184m

1000 kg/mh h

hρ

ρ

−−= = =



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3. Calculate the difference in heights of the two columns: ( ) ( )4 2 1 3 0.0500 m 0.0300 0.0184 m 0.16 cmh h h h= − + = − + =

Insight: Because the oil is less dense than the water, the column of oil must be higher than the column of water and oil.

26. Picture the Problem: A straw sits in a glass of water. When you suck on the straw, the water rises in the water. We want to know, theoretically, what is the highest the water can rise in the straw.

Strategy: The minimum pressure that you could cause inside the straw would be a pure vacuum. The pressure outside the straw is atmospheric pressure. Solve equation 15-7 for the height of the water in the straw with no pressure above the water and atmospheric pressure at the bottom.

Solution: 1. (a) The atmospheric pressure that is exerted on the surface of the water creates an upward force on the water column in the straw that overcomes the force of gravity.

2. (b) Set the pressure above the fluid equal to zero in equation 15-7: atP P gh ghρ ρ= + =

3. Solve for the height of the water column: ( )( )

5at

3 2

1.01 10 Pa 10.3 m1000 kg/m 9.81 m/s

Ph

gρ×

= = =

Insight: No amount of suction can cause the water to rise higher than 10.3 meters in the straw.

27. Picture the Problem: An IV solution is elevated above the injection point. The pressure in the bag is atmospheric pressure, while the pressure at the injection point is 109 kPa.

Strategy: Solve equation 15-6 for the height of the solution above the injection point.

Solution: 1. (a) Solve equation 15-6 for h: at

at g

P PP P gh hρ

ρ−

= + ⇒ =

2. Insert the given data: ( )( )3 2

109 kPa 101.3 kPa 0.770 m1020 kg/m 9.81 m/s

h −= =

3. (b) From the equation in step 1, we see that the height is inversely proportional to the density of the fluid. Therefore if a less dense fluid is used, the height must be increased.

Insight: If the density of the fluid were reduced to 920 kg/m3, the bag would need to be suspended at a height of 0.853 m, which is, as predicted, higher than the 0.770 meters.

28. Picture the Problem: A cylinder is filled with mercury up to a depth d, and then filled the rest of the way with water, as shown in the figure. The pressure at the bottom of the cylinder is two atmospheres.

Strategy: Set the pressure at the bottom of the cylinder equal to the pressure at the top (atmospheric) plus the pressure increases due to the water and the mercury. Use equation 15-7 to calculate the pressure increases. The height of the water is one meter minus the height of the mercury, 1.0 m .wh d= −

Solution: 1. Use equation 15-7 to write the pressure at the bottom of the cylinder: ( )

at w w Hg Hg

at at w Hg2 1.0 m

P P g h g h

P P g d g d

ρ ρ

ρ ρ

= + +

= + − +

2. Solve for d: ( )at w

Hg w

1.0 mP gd

ρρ ρ−

=−

3. Insert the given values: ( )

53

2

4 3 3

1.01 10 Pa 1000 kg/m 1.0 m9.81 m/s 0.74 m1.36 10 kg/m 1000 kg/m

d

×−

= =× −

Insight: Atmospheric pressure is 760 mmHg. Because the density of mercury is much greater than the density of water, the height of the mercury is almost the same as if it were a vacuum above the mercury column.



15 – 10

29. Picture the Problem: A Bathysphere descends deep below the surface of the ocean. Strategy: Use Archimedes’ principle to answer the conceptual question. Solution: 1. (a) The buoyant force equals the weight of the displaced water. Assuming the steel Bathysphere

compresses very little, it displaces the same volume of water at each depth. Furthermore, we assume the mass of the displaced water remains constant because water is a nearly incompressible fluid. We conclude that the buoyant force exerted on it at a depth of 10 ft was equal to the buoyant force exerted on it at a depth of 50 ft.

2. (b) The best explanation is I. The buoyant force depends on the density of the water, which is essentially the same at 10 ft and 50 ft. Statement II is false because the buoyant force depends upon differences in pressure, not the absolute pressure. Statement III is also false.

Insight: Water does compress a little bit, becoming about 1.6% denser at a depth 2000 m = 6560 ft.

30. Picture the Problem: A solid lead sphere and a solid aluminum sphere are each immersed in water. Strategy: Use Archimedes’ principle to answer the conceptual questions. Solution: 1. (a) The two spheres displace the same volume and therefore the same weight of water. Archimides’

principle states that the buoyant force equals the weight of the displaced water. We conclude that the buoyant force on the solid lead sphere is equal to the buoyant force on a solid aluminum sphere of the same diameter.

2. (b) Although the magnitude of the buoyant force varies from fluid to fluid (because of their different densities) the buoyant force is the same on any two objects that displace the same volume. Therefore, no, the answer to part (a) does not depend on the fluid that is causing the buoyant force.

Insight: Although each sphere experiences the same buoyant force, the lead sphere is more massive than the aluminum sphere and therefore the lead sphere still weighs more while submerged than does the aluminum sphere.

31. Picture the Problem: A fish carrying a pebble in its mouth swims with a small, constant velocity in a small bowl. The fish then drops the pebble to the bottom of the bowl.

Strategy: Use Archimedes’ principle to answer the conceptual questions. Solution: The fish is neutrally buoyant as it holds the pebble; therefore, it displaces a volume of water equal to its

weight plus the weight of the pebble. When the fish drops the pebble, it readjusts its swim bladder to be neutrally buoyant again. In this case, the fish displaces water equal to its weight, but the pebble displaces a volume of water equal only to its own volume. This is a smaller volume of water than was displaced before the pebble was dropped, and we conclude that the water level will fall when the fish drops the pebble.

Insight: Suppose the density of the rock were 3 g/cm3 and its volume is 1 cm3. When the fish is holding the pebble it must displace 3 g = 3 cm3 of water to supply an upward force on the pebble to keep it from falling. Once the pebble is released it will only displace 1 cm3, so the water level will go down by an amount equivalent to 3 – 1 cm3 = 2 cm3.

32. Picture the Problem: The weight of a horse is supported by the additional buoyant force provided by the raft sinking 2.7 cm into the water.

Strategy: Set the weight of the horse equal to the increase in the buoyant force from the additional submerged volume of the raft.

Solution: Set the weight equal to the buoyant force using equation 15-9: ( )( )( )( )( )

b

3 21000 kg/m 4.2 m 6.5 m 0.027 m 9.81 m/s 7.2 kN

F W VgρΔ = =

= =

Insight: The raft’s weight is supported by the buoyant force from the portion of the raft that was submerged before the horse entered the raft. The additional volume that is submerged supports the weight of the horse.

33. Picture the Problem: Two water-walking boots are needed to support the weight of a 75-kg person. The buoyant force from each boot will need to support one-half of the person’s weight. We are given the width and height of each boot and need to determine the necessary length.

Strategy: Set one-half of the person’s weight equal to the buoyant force of the boot, given by equation 15-9. Solve the resulting equation for the length of the boot.

Solution: 1. Set half the weight equal to the buoyant force: ( )B2 2

W mg F Vg l wh gρ ρ= = = =



15 – 11

2. Solve for the length:

( )( )( )3

75 kg 0.41 m 41 cm2 2 1000 kg/m 0.34 m 0.27 m

mlwhρ

= = = =

Insight: The boots displace a volume of water that weighs the same as the person. 34. Picture the Problem: A helium balloon displaces heavier air, causing it to rise. We want to calculate the maximum

additional weight that the balloon can lift. Strategy: Calculate the buoyant force on the balloon using equation 15-9 with the density of air and the volume of a

sphere. Subtract the weight of the balloon and the weight of the helium ( HeW Vgρ= ) to calculate the additional weight the balloon can lift.

Solution: 1. Calculate the buoyant force on the balloon:

( ) ( ) ( )33 3 2b air air

4 41.29 kg/m 4.9 m 9.81 m/s 6236 N3 3

F Vg r gρ ρ π π⎡ ⎤ ⎡ ⎤= = = =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

2. Subtract the weight of the balloon and of the helium:

( ) ( )

( ) ( ) ( )

2b balloon He He

33 2

6236 N 3.2 kg 9.81 m/s

40.179 kg/m 4.9 m 9.81 m/s 5.3 kN3

W F m g V gρ

π

= − − = −

⎡ ⎤− =⎢ ⎥⎣ ⎦

Insight: In this problem the weight of the balloon (31 N) is negligible. The additional mass that the balloon can lift is equal to the difference in densities of the air and helium times the volume of the balloon.

35. Picture the Problem: A hot air balloon remains at constant height because the net

force on the balloon is zero.

Strategy: Set the buoyant force (equation 15-9) on the balloon equal to the weight of the balloon and cargo plus the weight of the hot air in the balloon. Set the weight of the hot air equal to its density times the volume of the balloon and solve for the density of the hot air.

Solution: 1. Set the buoyant force equal to the weight of the balloon and the weight of the hot air:

b

air hot air

F Mg mgVg Mg Vgρ ρ

= +

= +

2. Solve for the density of the hot air: 3 3

hot air air 3

1890 kg1.29 kg/m 1.12 kg m11,430 m

MV

ρ ρ= − = − =

Insight: The height of the balloon does not enter into the solution of this problem. 36. Picture the Problem: A beaker, half-full of water rests on a scale. A metal

cube is lowered into the water and suspended by a string. We wish to calculate the increase in scale reading when the cube is submerged in the water.

Strategy: The forces acting on the water sum to zero because the water is in equilibrium. These forces include the weight of the water, the upward normal force from the scale, and the reaction force from the buoyant force on the cube. The change in reading on the scale is the change in normal force when the mass is added. Calculate the normal force without the cube and subtract that from the normal force with the cube to find the change in scale reading.

Solution: 1. Solve Newton’s Second Law without the cube:

2

2

1 H 0

1 H 0

0F N W

N W

= − =

=∑

2. Solve Newton’s Second Law including the cube:

2

2

2 H 0 B

2 H 0 B

0F N W F

N W F

= − − =

= +∑

3. Subtract the normal forces to find the difference: ( )2 2 22 1 H 0 B H 0 B H O cubeN N W F W F V gρ− = + − = =

Insight: The increase in scale reading does not depend on the mass or density of the cube, only on the cube volume. A gold cube and an aluminum cube of the same volume would cause the same increase.



15 – 12

37. Picture the Problem: A block of wood has a steel ball glued to one surface. The block can be floated with the ball

“high and dry” on its top surface. The block is inverted, immersing the ball in the water. Strategy: Use Archimedes’ principle to answer the conceptual question. Solution: 1. (a) The wood-steel ball system must displace its own weight in fluid to remain floating. When the ball is

“high and dry” it does not displace any volume of water, but it does when it is inverted. In that case a smaller volume of wood needs to be submerged in order to displace a sufficient volume of liquid. We conclude that when the block is inverted, and the ball is immersed in water, the volume of wood that is submerged will decrease.

2. (b) The best explanation is III. When the block is inverted the ball experiences a buoyant force, which reduces the buoyant force that must be provided by the wood. Statement I is false, and statement II is partly true (the same mass is supported in either case) but ignores the displacement of fluid by the submerged ball.

Insight: The arrangement with the ball submerged is mechanically more stable than the “high and dry” ball because the center of mass is lower.

38. Picture the Problem: A block of wood has a steel ball glued to one surface. The block can be floated with the ball

“high and dry” on its top surface. The block is inverted, immersing the ball in the water. Strategy: Use Archimedes’ principle to answer the conceptual question. Solution: 1. (a) Whether the block-ball system is upright or inverted, its weight is the same. Therefore, the volume of

water that must be displaced to float the block and ball is the same in either orientation. We conclude that when the block is inverted, and the ball is immersed in water, the water level in the tank will stay the same.

2. (b) The best explanation is II. The same mass is supported by the water in either case, and therefore the amount of displaced water is the same. Statements I and III are each false.

Insight: The result would be different if the ball were to separate from the block and sink to the bottom. In that case it would displace water equal to its own volume, but in the floating case it displaces water equal to its own mass.

39. Picture the Problem: The hydrometer shown at right measures fluid density. If the

hydrometer samples fluid 1, the small float inside the tube is submerged to the level 1. When fluid 2 is sampled, the float is submerged to level 2.

Strategy: Use Archimedes’ principle to answer the conceptual question. Solution: In both cases, the weight of displaced fluid is equal to the weight of the small

float. We conclude that the density of fluid 1 is less than the density of fluid 2 because a smaller volume of fluid 2 is displaced but it has the same mass as fluid 1.

Insight: The higher the small float rises above the surface of the fluid, the denser the fluid. In a similar fashion you will float higher in salt water than you do in fresh water because salt water is denser.

40. Picture the Problem: A piece of wood is tied with a string to the bottom of a

water-filled flask. The wood is completely immersed as shown. The flask with the wood tied to the bottom is placed on a scale. At some point the string breaks and the wood rises to the surface where it floats.

Strategy: Use Archimedes’ principle to answer the conceptual question.

Solution: 1. (a) The scale reading indicates the total weight of the objects placed upon it. The total mass contained in the flask is the same before and after the string breaks. It follows that when the wood is floating, the reading on the scale is equal toits previous reading.

2. (b) The best explanation is I. The same mass is supported by the scale before and after the string breaks, and therefore the reading on the scale remains the same. Statement II is partly true (the water level will drop) but the equilibrium scale reading will be unaffected. Statement III is false because the string exerts an internal force in the flask-water-wood system, not a net upward force on the scale.

Insight: During the time that the wood is accelerating toward the surface the center of mass of the flask-water-wood system accelerates downward. As it does so the force required to support the system will decrease and the scale reading will temporarily be smaller than its equilibrium value.



15 – 13

41. Picture the Problem: You float in a pool of water on a planet in a different solar system where the acceleration of

gravity is greater than it is on Earth.

Strategy: Use Archimedes’ principle to answer the conceptual question.

Solution: As far as floating goes, this planet and the Earth are the same. The increased acceleration of gravity increases your weight and the weight of the displaced water by the same factor. We conclude that you will float at the same levelon this planet as when you float in water on Earth.

Insight: You’d have to travel to a different solar system to try this because Earth has the highest surface gravity of any of the terrestrial planets. Jupiter and Neptune have higher “surface” gravities but they are gas/liquid planets.

42. Picture the Problem: An air mattress floats in fresh water. We wish to calculate the maximum mass that the air

mattress will support.

Strategy: Use equation 15-9 to calculate the buoyant force of the air mattress. Set the buoyant force equal to the weight of the air mattress and the maximum additional weight. Solve for the additional weight and divide by gravity to calculate the maximum mass that can be supported on the air mattress.

Solution: 1. Set the buoyant force equal to the weight supported:

2

B max am

max B am

max H O am am

F W WW F WW V g m gρ

= +

= −

= −

2. Divide by gravity to calculate the mass:

( )( )( )( )

2

maxmax H O am am

31000 kg/m 2.3 m 0.66 m 0.14 m 0.22 kg 210 kg

Wm V m

gρ= = −

= − =

Insight: The mass of the air mattress (including the air) is insignificant in this problem and could be ignored. 43. Picture the Problem: When a block is suspended from a scale, its

weight is equal to the tension on the scale. When the block is suspended in water, its weight is equal to the sum of the tension and buoyant force, as shown in the figure.

Strategy: We wish to calculate the volume and density of the block from the scale readings. Solve Newton’s Second Law for the mass suspended in the water to determine the volume of the block. The weight of the block is given from the scale when the block is suspended in air. The density of water is given in Table 15-1. Calculate the density by dividing the block’s mass by the volume.

Solution: 1. (a) Solve Newton’s Second Law for the buoyant force:

w

0w B

B w

F T F WF Vg W Tρ

= + − =

= = −∑

2. Solve for the block’s volume: ( )( )

4 3w3 2

w

20.0 N 17.7 N 2.34 10 m1000 kg/m 9.81 m/s

W TV

gρ−− −

= = = ×

3. (b) Calculate the density using equation 15-1: ( )( )

3 32 4 3

20.0 N 8.70 10 kg m9.81 m/s 2.345 10 m

m WV gV

ρ−

= = = = ××

Insight: Comparing the submerged and non-submerged weights of an irregularly shaped object is an effective way of determining its volume and density.



15 – 14

44. Picture the Problem: When a block is suspended in a fluid, the

weight of the block is supported by the string tension and the buoyancy of the fluid.

Strategy: We wish to determine the volume and density of a block by comparing the scale readings while suspended in water and in alcohol. Use Newton’s Second Law to write equations relating the forces when submerged in water and again when submerged in alcohol. This results in two equations with two unknown values (the weight and volume of the block). Solve the equations for the weight and volume. Use equation 15-1 to write the density.

Solution: 1. (a) Write Newton’s Second Law for the block submerged in water:

W BW

W BW W w

0F T F WW T F T Vgρ

= + − =

= + = +∑

2. Write Newton’s Second Law for the block submerged in alcohol:

A BA

A BA A A

0F T F W

W T F T Vgρ

= + − =

= + = +∑

3. Set the two equations equal and solve for the volume:

( )

W W A A

W A

A W

T Vg T VgT T

Vg

ρ ρ

ρ ρ

+ = +−

=−

4. Insert the given data:

( )( )4 3

3 3 2

25.0 N 25.7 N 3.68 10 m806 kg/m 1000 kg/m 9.81 m/s

V −−= = ×

−

5. (b) Use equation 15-1 for the density, and insert the weight from step 2:

block block W w wblock w

m m g T Vg TWV Vg gV gV gV

ρρ ρ

+= = = = = +


( )( )3 3 3

block 2 4 3

25.0 N 1000 kg/m 7.93 10 kg/m9.81 m/s 3.68 10 m

ρ−

= + = ××

Insight: By comparing the density with the densities in Table 15-1, we see that the block is most likely made of iron.

45. Picture the Problem: A person’s body-fat percentage is measured by comparing a person’s weight with their weight while submerged in water. In this problem we are given the body-fat percentage and want the find the person’s weight while submerged in water.

Strategy: Use Siri’s formula (page 512 of the text) to calculate the person’s overall density ( pρ ). Divide the person’s weight by the overall density and gravity to calculate volume. Finally, to calculate the submerged weight, subtract the buoyant force (equation 15-9) from the person’s weight.

Solution: 1. (a) Solve Siri’s formula for the overall density:

3

fp

3 33 3

pf

4950 kg/m – 4.50

4950 kg/m 4950 kg/m 1.04 10 kg m4.50 0.281 4.50

x

x

ρ

ρ

=

= = = ×+ +

2. (b) Solve equation 15-1 for volume:

( )( )3

p 3 2p p

1 756 N 0.0745 m1035kg/m 9.81 m/s

m WVgρ ρ

⎛ ⎞= = = =⎜ ⎟

⎝ ⎠

3. (c) Subtract the buoyant force from the weight:

( )( )( )a b w p

3 3 2756 N 1000 kg/m 0.0744 m 9.81 m/s 26 N

W W F W V gρ= − = −

= − =

Insight: As a person’s body-fat percentage increases, his weight while submerged decreases. The person is weightless in the water when his overall density is 1000 kg/m3, which occurs at a body-fat percentage of 45%.



15 – 15

46. Picture the Problem: A log floats in water with ¾ of the log submerged in the

water. Because the log is not accelerating up or down, the buoyant force must be equal to the log’s weight.

Strategy: Set the buoyant force equal to the weight. Use equation 15-9 to write the buoyant force in terms of the water’s density and equation 15-1 to write the weight of the log in terms of the log’s density. Solve the equation for the density of the log.

Solution: 1. (a) Set the forces equal and write in terms of the densities:

b

w submerged log

F WV g Vgρ ρ

==

2. Solve for the density of the log: submerged 3

log w 3

kg 31000 750 kg m4m

VV

ρ ρ⎛ ⎞ ⎛ ⎞⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠⎝ ⎠

3. (b) It increases because the buoyant force is proportional to the density of the displaced fluid and the density of salt water is greater than the density of fresh water.

Insight: You can verify that 26.8% of the log is above the surface if it is floating in salt water instead of fresh water. 47. Picture the Problem: A person is floating in water. Because the

person is not accelerating, the buoyant force must equal her weight. When an upward force is applied to the swimmer, her body rises out of the water such that the new buoyant force and the applied force are equal to her weight.

Strategy: Set the buoyant force (equation 15-9) equal to the person’s weight and solve for the submerged volume. Subtract the submerged volume from the total volume to calculate the volume above the surface. Set the sum of the buoyant force and the applied force equal to the weight and solve for the applied force.

Solution: 1. (a) Set the buoyant force equal to the weight and solve for volume submerged:

b

w sub subw

F WmV g mg Vρρ

=

= ⇒ =

2. Subtract submerged volume from total volume: above total sub total

w

3 33

81 kg0.089 m 0.008 m1000 kg/m

mV V V Vρ

= − = −

= − =

3. (b) Set the applied force and buoyant force equal to the weight and solve for the applied force:

applied b

applied b w sub2

F F W

F W F mg V gρ

+ =

= − = −

4. Set the submerged volume equal to the initially submerged volume minus the given change in volume:

( )

( )( )( )( )

3applied w sub

3w

3w

3 3 2applied

0.0018 m

0.0018 m

0.0018 m

1000 kg/m 0.0018 m 9.81 m/s 18 N

w

F mg V g

mmg g

g

F

ρ

ρρ

ρ

= − −

⎛ ⎞= − −⎜ ⎟

⎝ ⎠

=

= =

Insight: Note that the applied force equals the change in the buoyant force on the person, that is the density of the water times the change in volume submerged times gravity.



15 – 16

48. Picture the Problem: A block of wood floats at the interface between water and oil. We

want to calculate the fraction of the wood that is submerged in the water.

Strategy: Because the block is in equilibrium, the weight of the block of wood is equal to the buoyancy forces of the water and of the oil. Use the fraction of the wood submerged in the water (before the oil was added) to calculate the weight of the wood in terms of the density of water and the volume of the wood. Then set the weight equal to the buoyant forces when the water was added. Solve for the fraction submerged in the water.

Solution: 1. (a) The block is submerged in less water than before because the oil provides additional buoyant force.

2. (b) Set the weight equal to the buoyant force before the oil is added:

b w submerged w total w total(0.9 ) 0.9W V g V g V gρ ρ ρ= = =

3. Set the sum of the buoyant forces equal to the weight:

b,w b,oil

w in water oil in oil w total0.9bF F W

V g V g V gρ ρ ρ

+ =

+ =

4. Eliminate the volume in oil and solve for the fractional volume in water:

( )( ) ( )

w in water oil total in water w total

in water w oil total w oil

in water w oil

total w oil

0.9

0.90.9

V g V V g V g

V g V gVV

ρ ρ ρ

ρ ρ ρ ρρ ρ

ρ ρ

+ − =

− = −

−=

−

5. Insert the densities:

( )3 3in water

3 3total

0.9 1000 kg/m 875 kg/m0.2

1000 kg/m 875 kg/mVV

−= =

−

Insight: As predicted, less of the block is submerged in the water when the oil is added than was submerged before the oil was added.

49. Picture the Problem: A lead hockey puck floats in a container of mercury. We

wish to calculate x, the depth that the puck sinks into the mercury. Strategy: Set the lead cylinder weight equal to the buoyant force provided by the

mercury and solve for the submerged height, x. Use the density of the lead and its volume to calculate its weight, and equation 15-9 to calculate the buoyant force.

Solution: 1. Set the buoyant force equal to the weight:

b Hg submerged

Hg submerged lead

F V g mg

V V

ρ

ρ ρ

= =

=

2. Write the volumes in terms of the diameter and heights:

2 2

Hg lead4 4d x d hπ πρ ρ

⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

3. Solve for the submerged height: ( )

3 3lead

3 3Hg

11.3 10 kg/m 2.5 cm = 2.1 cm13.6 10 kg/m

x hρρ

×= =

×

Insight: Because the mercury is denser than the lead, the lead puck will float on the mercury. The fraction submerged is equal to the ratio of the two densities.

50. Picture the Problem: A submerged lead weight is pulled upward by a fishing line.

Strategy: The figure shows the forces acting on the lead weight. Use Newton’s Second Law to write the acceleration in terms of the forces acting on the weight. Write the buoyant force using equation 15-9. Use equation 15-1 to calculate the mass. Solve the equation for the tension, using the known acceleration. For the last part of the problem, solve the Second Law equation for the acceleration using the given tension.

Solution: 1. (a) Solve Newton’s Second Law for the tension:

b

b b( )F T F mg maT mg ma F m g a F

Σ = + − == + − = + −



15 – 17

2. Use equation 15-1 to write the mass in terms of density and volume and equation 15-9 for the buoyant force:

( )( )( )( )( )( )

lead w

3 3 5 3 2 2

3 5 3 2

( )

11.3 10 kg/m 0.82 10 m 9.81 m/s 2.1 m/s

1000 kg/m 0.82 10 m 9.81 m/s

1.0 N

T V g a Vg

T

ρ ρ−

−

= + −

= × × +

− ×

=

3. (b) The tension stays the same because the only quantity that changes with depth is pressure. Neither the buoyant force nor the tension depend on pressure.

4. (c) Solve Newton’s Second Law for the acceleration:

( )( )( )( )

b

w w

lead

3 5 3 22

3 3 5 3

2

1.2 N 1000 kg/m 0.82 10 m 9.81 m/s9.81 m/s

11.3 10 kg/m 0.82 10 m

4.0 m/s , upward

ma T F mgT Vg T Vg

a g gm V

a

ρ ρρ

−

−

= + −+ +

= − = −

+ ×= −

× ×

=

Insight: Because the tension in part (c) is greater than the tension in part (a), the weight experiences a greater upward acceleration. The weight will experience a downward acceleration when the tension is less than 0.83 N.

51. Picture the Problem: Water flows through a hose at a constant volume flow rate. When the diameter of the hose is

decreased by partially blocking the end, the speed through the end increases so the same volume flows in the same time.

Strategy: We wish to find the speed through the partially blocked end. Write the continuity equation (15-12) in terms of the diameters of the hose and solve for the speed at the end.

Solution: 1. Write the continuity equation in terms of the diameters of the hose: ( ) ( )

1 1 2 2

2 21 1 2 2/ 4 / 4

A v A v

d v d vπ π

=

=

2. Solve for the end velocity: ( )

2 2

12 1

2

3.4 cm 1.1 m/s 39 m s0.57 cm

dv v

d⎛ ⎞ ⎛ ⎞

= = =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

Insight: The ratio of the speeds is inversely proportional to the square of the diameters. If the diameter were to be cut in half, the speed would increase by a factor of four.

52. Picture the Problem: Water flows through a pipe of given diameter at a given speed. We wish to calculate the mass

flow rate, or the mass of water that passes a given point in the pipe per second.

Strategy: Multiply the density of the water by the cross-sectional area of the pipe and the velocity to calculate the mass flow rate.

Solution: 1. Write the mass flow rate in terms of the density, diameter, and velocity:

2

4m dAv vt

πρ ρ⎛ ⎞Δ

= = ⎜ ⎟Δ ⎝ ⎠

2. Insert the given values: ( ) ( )

23 (0.038 m)1000 kg/m 2.1 m/s 2.4 kg s

4mt

π⎡ ⎤Δ= =⎢ ⎥Δ ⎣ ⎦

Insight: This flow rate is slightly over a ½ gallon per second.



15 – 18

53. Picture the Problem: A child’s pool is filled from a garden hose. The volume flow rate through the hose is equal to the

rate at which the pool fills. Strategy: In this problem we wish to calculate the time necessary to fill the pool. Use the volume continuity equation

pool pool hose hoseA v A v= (equation 15-11) to calculate the speed at which the water rises in the pool. Divide the depth of the pool by the speed of the rising water to determine the time necessary to fill the pool.

Solution: 1. Solve the continuity equation for the speed at which the pool fills:

( )

221hosehose hose4

pool hose hose hose21pool poolpool4

240.029 m 1.3 m/s 2.7 10 m/s

2.0 m

dA dv v v v

A ddππ

−

⎛ ⎞= = = ⎜ ⎟⎜ ⎟

⎝ ⎠

⎛ ⎞= = ×⎜ ⎟⎝ ⎠

2. Divide the height of the pool by the fill speed to find the required fill time: 4

pool

0.26 m 1 min960 s 16 min60 sec2.7 10 m/s

htv −

⎛ ⎞= = = × =⎜ ⎟× ⎝ ⎠

Insight: To decrease the time necessary to fill the pool, either the diameter of the hose or the speed of the water should be increased.

54. Picture the Problem: The volume rate of blood flow through the heart per minute is given and we are asked to

calculate the volume of blood that flows in a day. Strategy: Multiply the given volume flow rate by the number of minutes in a day to calculate the volume of blood that

passes though the heart each day. Multiply the volume of blood by the density to calculate the mass of blood that flows through the heart each day.

Solution: 1. Convert the flow rate from per minute to per day:

L 60 min 24 h5.00 7200 L/daymin h day

Vt

⎛ ⎞Δ⎛ ⎞ ⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟Δ⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠

2. Multiply the volume flow rate by the density of blood: ( )

33

3

m1060 kg/m (7200 L) 7630 kg/day10 L

m Vt t

ρ⎛ ⎞Δ Δ⎛ ⎞= = =⎜ ⎟⎜ ⎟Δ Δ⎝ ⎠ ⎝ ⎠

Insight: The average human heart pumps over 7½ tons of blood per day! 55. Picture the Problem: This solution contains an art image. Strategy: The volume flow rate is equal to the cross-sectional area of the blood vessel times the velocity of the blood

through the blood vessel. Divide the flow rate by the cross-sectional area to calculate the speed. In the capillaries the flow rate through each capillary will be equal to the total flow rate divided by the number of capillaries.

Solution: 1. (a) Divide the volume flow rate in the arteriole branch by the branch’s cross-sectional area: ( )

6 3

212

5.5 10 cm / s 0.78 cm/s0.0030 cm

VtV Av v

t A π

−ΔΔΔ ×

= ⇒ = = =Δ ⎡ ⎤⎣ ⎦

2. (b) Divide the volume flow rate in the capillaries (1/340th of the arteriole branch flow rate) by the cross-sectional area of the capillary:

( )

6 3

2412

5.5 10 cm /s340

0.13 cm/s4.0 10 cm

Vtv

A π

−

ΔΔ

−

⎛ ⎞×⎜ ⎟⎝ ⎠= = =⎡ ⎤×⎣ ⎦

Insight: The blood speed in the capillaries is much slower than in the other blood vessels.

56. Picture the Problem: In our sketch we label the speed of the water in the hose with v1 and the speed of the water coming out the nozzle as v2. We are given the mass flow rate through the hose and the diameters of the hose and nozzle. We need to calculate the speed of the water in the hose and in the nozzle.

Strategy: Divide the mass flow rate by the density and cross-sectional area to calculate the velocity v of the water in the hose and in the nozzle. Calculate the cross-sectional areas from the diameters.

Solution: 1. Solve the mass flow rate equation for v: 2

1 4 m m mAv vt t A t d

ρρ ρπ

Δ Δ Δ= ⇒ = =

Δ Δ Δ



15 – 19

2. (a) Insert the data for the flow in the hose: ( )

( ) ( )1 23

43.11 kg/s 3.82 m s1000 kg/m 0.0322 m

vπ

= =

3. (b) Insert the data for the flow in the nozzle: ( )

( ) ( )2 23

43.11 kg/s 73.9 m s1000 kg/m 0.00732 m

vπ

= =

4. (c) The mass flow rates in the nozzle and hose are equal, because for incompressible fluids, conservation of mass requires that “what goes in equals what comes out.”

Insight: Because the nozzle has a much smaller diameter, the speed of the water in the nozzle is much greater than the speed in the hose.

57. Picture the Problem: Water flows through a river at a constant volume flow rate. We are given the widths and depths

of the river at two points and the water speed at the initial point before the rapids. We wish to calculate the speed of the water in the rapids.

Strategy: Set the volume flow rate in the rapids equal to the flow rate before the rapids using equation 15-12. Solve the equation for the speed in the rapids. The cross-sectional areas are given by the depth of the water times the width of the river.

Solution: 1. Solve equation 15-12

for the speed in the rapids: 1 1 2 2

1 1 11 1 1 2 2 2 2

2 2

A v A vw d v

w d v w d v vw d

=

= ⇒ =

2. Insert the given data: ( )( )( )

( )( )2

12 m 2.7 m 1.2 m/s7.9 m s

5.8 m 0.85 mv = =

Insight: For the river to have the same volume flow rate at two different points, the region with a higher water velocity must have a smaller cross-sectional area than the region with a slower velocity.

58. Picture the Problem: The volume flow rate throughout the circulatory system must be equal. We are given the

diameter and velocity of blood in the aorta. We are also given the diameter and velocity of the blood in the capillaries. Using this information we need to calculate the number of capillaries.

Strategy: Use equation 15-11 to set the volume flow rate in the aorta equal to the volume flow rate in the capillaries, where the flow rate in the capillaries is equal to the flow rate in one capillary times the number of capillaries, n. Solve the resulting equation for the number of capillaries.

Solution: 1. Solve equation 15-11 for the number of capillaries: a a

a a c cc c

A v

A v nA v nA v

= ⇒ =

2. Write the cross-sectional areas in terms of the diameters:

22a a a a42

c cc c4

d v d vn

d vd v

π

π

⎛ ⎞= = ⎜ ⎟

⎝ ⎠


27

5

0.0050 m 1.0 m/s 2.5 100.01 m/s1.0 10 m

n −

⎛ ⎞ ⎛ ⎞= = ×⎜ ⎟ ⎜ ⎟×⎝ ⎠ ⎝ ⎠

Insight: The 25 million capillaries are crammed into the human body, which has a volume of about 0.5 m3. That comes to approximately 50 capillaries in every cubic centimeter!



15 – 20

59. Picture the Problem: The image shows two regions of

a blood vessel. The initial region has no plaque buildup and has a diameter d1 = 1.1 cm and pressure P1. The blood flows through this region at a rate of 15 cm/s. The second region has plaque buildup so its diameter is only d2 = 0.75 cm and its pressure is P2.

Strategy: We wish to find the speed of the blood at the plaque buildup and the difference in pressures between the two regions. Solve the volume continuity equation (equation 15-11) for the speed in the narrower region. Use the speeds through the two areas and the density of blood (1060 kg/m3 ) to solve equation 15-14 for the pressure difference PΔ .

Solution: 1. (a) Solve the continuity equation for v2:

1 1 2 222

11 4 12 1 1 12

2 224

A v A v

dA dv v v v

A dd

π

π

=

⎛ ⎞= = = ⎜ ⎟

⎝ ⎠

2. Insert the given values: ( )

2

21.1 cm 15 cm/s 32 cm/s0.75 cm

v ⎛ ⎞= =⎜ ⎟⎝ ⎠

3. (b) Solve equation 15-14 for PΔ :

( )

( ) ( ) ( )

2 21 1 2 2

2 21 2 2 1

2 23

1 12 2

121 1060 kg/m 0.323 m/s 0.15 m/s 43 Pa2

P v P v

P P v v

P

ρ ρ

ρ

+ = +

− = −

⎡ ⎤Δ = − =⎣ ⎦

Insight: The drop in pressure at the plaque means that the heart must increase its output pressure by 40 Pa in order to maintain the same flow rate as there was in the plaque-free vessel.

60. Picture the Problem: The diagram shows water flowing through a pipe with a speed v1 = 1.4 m/s. The pipe narrows to half of its initial diameter.

Strategy: We wish to find the speed and pressure in the narrow region. Solve equation 15-12 for the speed v2, given that

12 12d d= . Then use equation 15-14 to calculate the pressure P2.

Solution: 1. (a) Solve the continuity equation for the speed v2:

2 2

1 1 11 1 2 2 2 1 1 1

2 2 2

/ 4 / 4

A d dA v A v v v v vA d d

ππ⎛ ⎞ ⎛ ⎞

= ⇒ = = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2. Insert the given data: ( )

2

12 1 11

12

4 4 1.6 m/s 6.4 m/sd

v v vd

⎛ ⎞= = = =⎜ ⎟⎝ ⎠

3. (b) Solve equation 15-14 for P2:

( )

( ) ( ) ( )

2 21 1 2 2

2 22 1 1 2

2 23 3

2

1 12 2

12

1110 10 Pa+ 1000 kg/m 1.6 m/s 6.4 m/s2

91 kPa

P v P v

P P v v

P

ρ ρ

ρ

+ = +

= + −

⎡ ⎤= × −⎣ ⎦

=

Insight: The pressure is smaller in the narrow region of the pipe. The decrease in pressure provides the force necessary to accelerate the water from 1.6 m/s to 6.4 m/s.



15 – 21

61. Picture the Problem: As shown in the figure, a person blows air through

the straw at 1.5 liters per second. As the air passes over the vertical straw it decreases the pressure in the straw, which causes the water to rise. We wish to calculate the speed at which the air exits the straw and the height to which the water rises.

Strategy: Divide the volume flow rate of the air by the cross-sectional area of the straw to calculate the speed of the air. Use equation 15-14 to calculate the difference in pressure between point 1 at the top of the vertical straw and point 2 out in the room. Calculate the height of the water column from the pressure variation with depth (equation 15-7) using the pressure difference between the top of the straw and the room.

Solution: 1. (a) Solve the volume flow rate for the air velocity: ( )

( )( )

3

221 14 4

1.5 L/s 1000 L/m 53 m s

0.0060 mV V tAv vt dπ π

Δ Δ Δ= ⇒ = = =

Δ

2. (b) Solve equation 15-14 for the pressure difference: ( )( )

2 21 11 1 2 22 2

22 31 12 1 12 2 1.29 kg/m 53.1 m/s 1.82 kPa

P v P v

P P P v

ρ ρ

ρ

+ = +

− = Δ = = =

3. Solve equation 15-7 for the height of the water:

( )( )

2 1 w

2 13 2

w w

1.815 kPa 19 cm1000 kg/m 9.81 m/s

P P hgP P Ph

g g

ρ

ρ ρ

= +− Δ

= = = =

Insight: The height of the water column is inversely proportional to the fourth power of the straw’s diameter. If the diameter of the straw were decreased by a factor of two (d = 0.30 cm) the height of the water column would increase to 2.91 m, or 24=16 times the height in the problem.

62. Picture the Problem: The figure shows water flowing

through a pipe with initial diameter d1 = 2.8 cm which decreases to d2 = 1.6 cm. The pressure difference ΔP is 7.5 kPa.

Strategy: Solve equation 15-14 for the water velocity in the first tube. Use equation 15-12 to write the velocity in the second tube in terms of the velocity in the first tube.

Solution: 1. (a) The tube with the greater diameter has the higher pressure. Because the volume flow rate is the same both tubes the speed in the narrower tube must be greater. Equation 15-14 shows that the tube with higher velocity must have the lower pressure.

2. (b) The tube with the smaller diameter has the higher speed of flow.

3. (c) Solve equation 15-12 for the velocity in the second tube:

221 1 1

1 1 2 2 2 1 1 122 22

/ 4

/ 4A D D

A v A v v v v vA DD

ππ⎛ ⎞⎛ ⎞ ⎛ ⎞

= ⇒ = = =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

4. Eliminate v2 from equation 15-14:

( )

2 21 11 1 2 22 2

42 21 11 1 2 1 2 12 2

P v P v

P v P D D v

ρ ρ

ρ ρ

+ = +

+ = +

5. Solve for v1:

( )( )

( ) ( )( )

3 32 1

1 4 41 2

2 7.5 10 Pa 1000 kg/m21.3 m s

1 1 2.8 cm 1.6 cm

P Pv

D D

ρ − ×−= = =

− −

Insight: The speed in the narrow pipe is 4.1 m/s.



15 – 22

63. Picture the Problem: Water sprays out of a leak in a stationary garden hose and rises up to a height of 0.68 meters. Strategy: Solve equation 15-15 for the pressure P1, where point 1 is inside the garden hose with y1 = 0 and point 2 is the

highest point that the water sprays.

Solution: Solve equation 15-15 for P1:

( )( )( )

1 1 2 2

1 atm 2

5 3 2

51

1.01 10 Pa + 1000 kg/m 9.81 m/s 0.68 m

1.08 10 Pa

P g y P g yP P g y

P

ρ ρρ

+ = += +

= ×

= ×

Insight: When the water is flowing through the garden hose, the spray does not reach as high as when the water is stationary. This is because the pressure inside the hose is less when the speed is not zero.

54. Picture the Problem: Water flows out of a hole in a tank h = 2.5 m below the water level in the tank.

Strategy: We want to calculate the speed of the water. Use Torricelli’s Law to calculate the speed of the water:

Solution: Insert the given height into equation 15-17: ( )( )

2

2

2

2

2 9.81 m/s 2.7 m

7.3 m/s

v gh

v

=

=

=

Insight: As the water level decreases in the tank, the speed at which the water exits will also decrease.

65. Picture the Problem: The diagram shows an airplane wing. The air velocity over the wing is v1 = 115 m/s and the air velocity under the wing is v2 = 105 m/s. The difference in wind speeds produces a pressure difference and a resulting upward force.

Strategy: Apply Bernoulli’s equation (15-14) between a point on the upper surface and a point on the lower surface. We can neglect the difference in height between the two surfaces because the difference in static pressures between these points is negligible. Multiply the pressure difference by the wing area to calculate the net force on the wing.

Solution: 1. (a) Solve equation 15-14 for the pressure difference: ( )

( ) ( ) ( )

2 21 12 2 l l2 2

2 212 1 1 22

2 2312 1.29 kg/m 115 m/s 105 m/s 1.42 kPa

P v P v

P P v v

ρ ρ

ρ

+ = +

− = −

⎡ ⎤= − =⎣ ⎦

2. (b) Multiply by the area of the wings: ( ) ( )( )3 21.42 10 Pa 32 m 45 kNF P AΔ = Δ = × =

Insight: If the airplane weighs less than 45 kN, the net vertical force will be upward and the plane will rise in the air.

66. Picture the Problem: The figure shows an airplane window. Air rushes past the outside of the window at 170 m/s, while the air inside is at rest.

Strategy: We wish to calculate the pressure difference between inside and outside and the net force on the window. Solve equation 15-14, Bernoulli’s equation, for the pressure difference. Multiply the pressure difference by the area of the window to calculate the net force.

Solution: 1. (a) Solve Bernoulli’s equation for the pressure difference: ( )

2 21 1in in out out2 2

2 21in out out in2

P v P v

P P v v

ρ ρ

ρ

+ = +

− = −



15 – 23

2. Insert the given values: ( ) ( )23 21in out 2 1.29 kg/m 170 m/s 0 19 kPaP P ⎡ ⎤− = − =⎣ ⎦

3. (b) Multiply the pressure by the area of the window:

( )( )318.6 10 Pa 0.25 m 0.42 m 1.96 kNF PAΔ = Δ = × =

Insight: This pressure difference creates over 400 lbs of force on the window! 67. Picture the Problem: The pressure difference between the inside of

the house, where the wind speed is zero, and the top of the roof, where the wind speed is v = 47.7 m/s, creates an upward force on the roof, as shown in the figure.

Strategy: Solve Bernoulli’s equation (equation 15-14) for the change in pressure across the roof. Multiply the change in pressure by the area of the roof to calculate the net force on the roof.

Solution: 1. (a) Solve equation 15-14 for the change in pressure: ( )

2 21 1top top in in2 2

2 21in top top in2

P v P v

P P v v

ρ ρ

ρ

+ = +

− = −

2. Multiply the change in pressure by the area of the roof:

( )( ) ( ) ( )

2 21top in2

23 212 1.29 kg/m 47.7 m/s 0 668 m 980 kN

F PA v v Aρ⎡ ⎤Δ = Δ = −⎣ ⎦⎡ ⎤= − =⎣ ⎦

3. (b) The force is directed upward. Stationary air exerts a larger pressure than air moving rapidly past the outside of the roof.

Insight: This upward force is over 100 tons of force. It is no wonder that many roofs are blown off during a high wind storm!

68. Picture the Problem: As shown in the diagram, water flows through a garden hose with speed 1 0.78 m/sv = and pressure

1 1.2 atm.P = At the end of the hose a nozzle with a smaller diameter causes the water to flow at speed 2v and pressure

2P . We wish to calculate 2v and 2P .

Strategy: Use the continuity equation (15-12) to calculate the speed in the nozzle. Then solve for the pressure in the nozzle using equation 15-14.

Solution: 1. (a) Solve equation 15-12 for the speed in the nozzle:

2

1 11 1 2 2 2 1 1

2 2

A d

A v A v v v vA d

⎛ ⎞= ⇒ = = ⎜ ⎟

⎝ ⎠


2

20.63 in 0.78 m/s 5.0 m/s0.25 in

v ⎛ ⎞= =⎜ ⎟⎝ ⎠

3. (b) Solve equation 15-14 for P2: ( )2 2 2 21 1 11 1 2 2 2 1 1 22 2 2 P v P v P P v vρ ρ ρ+ = + ⇒ = + −

4. Solve numerically: ( ) ( ) ( ) ( )

( )

2 25 312 2

35

1.2 1.01 10 Pa 1000 kg/m 0.78 m/s 5.0 m/s

1 atm109 10 Pa 1.1 atm1.01 10 Pa

P ⎡ ⎤= × + −⎣ ⎦⎛ ⎞

= × =⎜ ⎟×⎝ ⎠

Insight: The pressure inside the nozzle is less than the pressure inside the hose.



15 – 24

69. Picture the Problem: Water flows through a pipe whose diameter

decreases by a factor of two. We wish to calculate the resulting change in pressure.

Strategy: Solve equation 15-14 for the change in pressure between the wide portion of the hose (point 1) and the narrow portion (point 2). Use the continuity equation (equation 15-12) to write the final speed in terms of the initial speed.

Solution: 1. (a) The pressure in the narrow region is less than the initial pressure because the fluid velocity is greater.

2. (b) Solve equation 15-14 for the change in pressure: ( )

2 21 11 1 2 22 2

2 212 1 1 22

P v P v

P P v v

ρ ρ

ρ

+ = +

− = −

3. Use the continuity equation to write v2 in terms of v1:

2221 1 1 1

2 1 1 1 1 12 12 2 12 2

/ 4 4/ 4

A D D Dv v v v v vA D DD

ππ

⎛ ⎞⎛ ⎞= = = = =⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠

4. Eliminate v2 from the change in pressure equation: ( ) ( )22 2 21 1

2 1 1 12 2154 1 162

P P v v v vρ ρ ρ⎡ ⎤− = − = − = −⎣ ⎦

Insight: The pressure difference for an arbitrary change in diameters is ( )4212 1 1 22 1P P v d dρ ⎡ ⎤− = −⎣ ⎦

70. Picture the Problem: Increasing the diameter of a blood vessel will increase the volume of blood that can pass through

the vessel per unit time. The goal is to double the blood flow by increasing the vessel diameter. Strategy: Set the final flow rate equal to twice the initial flow rate and use Poiseuille’s equation (equation 15-19) to

find final radius in terms of the initial radius. Calculate the fractional increase in radius, which is equal to the fractional increase in diameter.

Solution: 1. Set the final flow rate equal to twice the initial:

2 1

2V Vt t

Δ Δ⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟Δ Δ⎝ ⎠ ⎝ ⎠

2. Write the flow rates using Poiseuille’s equation and solve for r2:

( ) ( )4 42 1

4 42 1 2 1

28 8

2 1.189

P r P rL L

r r r r

π πη η

Δ Δ=

= ⇒ =

3. Calculate the percent increase in radius: ( )2 1 1 1

1 1

1.189100 100 19%

r r r rr r

⎛ ⎞− −× = =⎜ ⎟

⎝ ⎠

Insight: A relatively small increase in radius results in a large increase in flow rate. 71. Picture the Problem: Blood flows through the pulmonary

artery as shown in the diagram. Strategy: We wish to calculate the volume of blood that

passes through the artery every second and the decrease in flow rate if the radius were reduced by 16%. Use the data found in the figure and in Example 15-10 to solve Poiseuille’s equation (equation 15-19) for the volume flow in one second. To calculate the reduction in flow rate set the new radius equal to 2 10.82r r= (or 18% less than the initial radius) and solve for the ratio of flow rates.

Solution: 1. (a) Solve Poiseuille’s equation for :VΔ

( ) ( ) ( )( )( )

4341 2 3

2 2

450 Pa 2.4 10 m 1.0 s25 cm

8 8 0.0027 N s/m 8.5 10 m P P r t

VL

ππη

−

−

×− ΔΔ = = =

⋅ ×

2. (b) Use Poiseuille’s equation to solve for the ratio of flow rates:

( ) ( ) 44 41 2 2 1 2 1 2

2 1 18 8P P r P P r rV V

t t L L rπ π

η η− − ⎛ ⎞Δ Δ⎛ ⎞ ⎛ ⎞ = = ⎜ ⎟⎜ ⎟ ⎜ ⎟Δ Δ⎝ ⎠ ⎝ ⎠ ⎝ ⎠



15 – 25

3. Set 2 10.82r r= and simplify:

4

1

2 1 1

0.82 0.45rV Vt t r

⎛ ⎞Δ Δ⎛ ⎞ ⎛ ⎞ = =⎜ ⎟⎜ ⎟ ⎜ ⎟Δ Δ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

The flow rate is reduced by a factor of 2.2. Insight: Because the flow rate is proportional to the fourth power of the radius, a small change in radius results in a

significant change in flow rates. 72. Picture the Problem: Blood flows through an artery, the diameter of which is reduced by 15% at an occlusion.

Strategy: We wish to calculate the factor by which the pressure drop across the artery must increase. Use Poiseuille’s equation to set the volume flow rates equal in the normal and occluded sections of the artery. Solve for the ratio of the pressure change in terms of the change in radius.

Solution: 1. Set the volume flow rates equal and solve for the ratio of pressure drops:

4 4 41 1 2 2 2 1

41 2

8 8P r P r P r

L L P rπ πη η

Δ Δ Δ= ⇒ =

Δ

2. Set 2 10.85r r= and solve for 2 1 :P PΔ Δ :

( )

42 1

41 1

1.920.85

P rP r

Δ= =

Δ

Insight: Decreasing the radius by only 15% almost doubles the drop in pressure. 73. Picture the Problem: Water flows through a garden hose of known diameter at a given volume flow rate. We wish to

calculate the water speed and pressure drop across the hose.

Strategy: Divide the volume flow rate by the cross-sectional area of the hose to calculate the water velocity. Solve Poiseuille’s equation (15-19) for the pressure change across the hose. To calculate the effect of cutting the cross-sectional area in half on the velocity and flow rate, find the ratio of the initial to final radii and solve Poiseuille’s equation for the ratio of the flow rates. Calculate the ratio of the velocities by dividing the flow rates by their corresponding cross-sectional areas.

Solution: 1. (a) Divide flow rate by cross- sectional area to calculate velocity:

( )( )

4 3

22

5.0 10 m /s 1.0 m/s1.25 10 m

Vtv

A π

Δ −Δ

−

×= = =

×

2. (b) Solve Poiseuille’s equation for :PΔ

( )( )( )( )

4

3 2 4 3

42

8 ( / )

8 1.0055 10 N s/m 15 m 5.0 10 m /s

1.25 10 m

0.79 kPa

L V tPr

P

ηπ

π

− −

−

Δ ΔΔ =

× ⋅ ×=

×

Δ =

3. (c) Set the new area equal to half the initial area and solve for the ratio of radii:

12 12

2 212 1 2 12 2

A A

r r r rπ π

=

= ⇒ =

4. Solve Poiseuille’s equation for the ratio of flow rates:

( ) ( )4 41 2 2 1 2 1

2 144

2 1

1 1

8 8

2 14

P P r P P rV Vt t L L

r rr r

π πη η

− −Δ Δ⎛ ⎞ ⎛ ⎞ =⎜ ⎟ ⎜ ⎟Δ Δ⎝ ⎠ ⎝ ⎠

⎛ ⎞⎛ ⎞= = =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

5. Divide the flow rates by the area to calculate the ratio of velocities:

2 2 2 11

1 1 1 12

( / ) / 1 1( / ) / 4 2

v V t A Av V t A A

⎛ ⎞Δ Δ ⎛ ⎞= = =⎜ ⎟⎜ ⎟Δ Δ ⎝ ⎠⎝ ⎠

The water speed is multiplied by a factor of 1/2. 6. (d) Use the ratio of flow rates from step 4: The volume flow rate is multiplied by a factor of 1/4.

Insight: Because the volume flow rate is proportional to the fourth power of the radius, and the cross-sectional area is proportional to the square of the radius, the volume flow rate is proportional to the square of the cross-sectional area. Equivalently the water speed is proportional to the cross-sectional area.



15 – 26

74. Picture the Problem: A weather glass, as shown at right, is used to give an indication of a

change in the weather.

Strategy: Use the principles that govern the operation of a barometer to answer the conceptual question.

Solution: Different water levels in the tube and the main body of the weather glass indicate different pressures in the atmosphere and the interior of the weather glass. As a low-pressure system approaches, atmospheric pressure drops but the pressure inside the weather glass stays the same. We conclude that the water level in the neck of the weather glass will move up when a low-pressure system approaches.

Insight: A barometer (Figure 15-4) works in a similar fashion, only in reverse. For that instrument the pool of fluid is exposed to the atmosphere and the narrow tube is sealed off at a constant, low pressure (ideally zero pressure).

75. Picture the Problem: A passenger in a car holds the lower end of a string that is connected to a helium balloon. When

the car is at rest at a stop sign the string is vertical.

Strategy: Use the principle of pressure in a fluid (air) to answer the conceptual question.

Solution: The string going to the balloon will lean forward. The reason is that as the car accelerates forward, the air inside it shifts toward the rear of the car, just as passengers are pressed back into their seats. This makes the pressure of the air in the car increase as one moves from front to back. The helium-filled balloon moves in the direction of decreasing pressure, like any buoyant object, which in this case is toward the front of the car.

Insight: Another way to view the problem is to realize that air, unlike water, is compressible. The acceleration of the car in the forward direction will force the air toward the back of the car, compressing it and making it more dense. The balloon will move forward in the direction of lower density of the surrounding air.

76. Picture the Problem: A person and some bricks float in a boat in a small backyard swimming pool. The person drops

the bricks overboard and they sink to the bottom of the pool.

Strategy: Use Archimedes’ principle to answer the conceptual questions.

Solution: 1. (a) When a brick is in the boat it displaces a volume of water equal to its weight. When the brick is at the bottom of the pool it displaces a volume of water equal to its own volume. Thus, the brick displaces a greater volume of water when it is in the boat. We conclude that the water level will decrease when the bricks are thrown overboard.

2. (b) The best explanation is I. When the bricks sink they displace less water than when they were floating in the boat; hence, the water level decreases. Statement II is partly true but ignores the volume of water displaced by the bricks, and statement III is false.

Insight: Suppose the density of a brick were 3 g/cm3 and its volume is 300 cm3. When the brick is in the boat it must displace 900 g = 900 cm3 of water to supply an upward force on the brick to keep it afloat. Once the brick sinks to the bottom it will only displace 300 cm3, so the water level will go down by an amount equivalent to 600 cm3.

77. Picture the Problem: A person and some blocks of wood float in a boat in a small backyard swimming pool. The

person throws the blocks of wood into the pool, where they float.

Strategy: Use Archimedes’ principle to answer the conceptual questions.

Solution: 1. (a) The boat is now carrying less mass. Therefore, it will float higher relative to the water.

2. (b) The water level in the pool will stay the same because the blocks of wood displace the same amount of water whether they are in the boat or in the water. In either case, they displace a volume of water equivalent to their weight.

Insight: A floating boat must displace an amount of fluid equivalent to its mass so that the buoyant force equals the weight of the ship. That is why ship sizes are often given in terms of tons of displacement. In this problem when the wood blocks are removed the boat’s mass and therefore the volume of its displacement decreases and it floats higher.



15 – 27

78. Picture the Problem: The three identical containers shown at right are

open to the air and filled with water to the same level. A block of wood floats in container A; an identical block of wood floats in container B, supporting a small lead weight; container C holds only water.

Strategy: Use Archimedes’ principle and the concepts of weight and mass to rank the weights of the fluid and of the container plus its contents.

Solution: 1. (a) The containers are filled to the same level, but less water is needed for container A than C because the wood displaces some water. An even greater amount of fluid is displaced by the floating lead weight. We conclude that the ranking of the weights of water in each container is C < A < B.

2. (b). The floating block of wood displaces water equal to its own mass. If the wood were to be converted to an equivalent mass of water, the water level in container A would be exactly the same as in container B. The same argument could be made for any floating object, including the container C system. We conclude that the total masses must all be the same and the ranking of the weights of the containers plus contents is A = B = C.

Insight: Another approach to part (b) is to say each of the containers has the same pressure on its bottom surface (because the fluid levels are all the same), meaning that the downward force exerted by each container (which is its weight) is the same in each case.

79. Picture the Problem: A pan half-filled with water is placed near the rim of a rotating turntable.

Strategy: Use the principle of pressure in a fluid to answer the conceptual question.

Solution: A line perpendicular to the surface of the water in the pan is tilted inward toward the axis of rotation; that is, the water level increases as you move farther from the axis. The fluid pressure is therefore higher in the fluid at the edge of the pan (where it is deeper) and lower at the axis of rotation. As a result, any given element of water will experience a net force directed toward the axis of rotation, as it must to undergo circular motion.

Insight: Another way to view the problem is to say that the rotation has created an “artificial gravity” pointed outward from the axis of rotation. The surface of the water automatically arranges itself perpendicular to the effective direction of gravity, which is a combination of downward and outward.

80. Picture the Problem: A pan half-filled with water is placed near the rim of a rotating turntable. The temperature is

lowered below the freezing point of water as the turntable continues to rotate, and the water becomes a solid block of ice. A marble is placed on the surface of the ice and released from rest.

Strategy: Note the forces acting on the marble and use Newton’s Second Law to answer the conceptual question.

Solution: The marble will stay where it is released. The reason is that the surface of the water is perpendicular to the local “effective gravity,” which is a combination of the outward “centrifugal” force from the rotating turntable and ordinary gravity. Even the water itself does not flow inward or outward when it is tilted at this angle. The marble will stay put just as the liquid water did before it was frozen. In its frame of reference the marble is in equilibrium because the normal force from the ice is balanced by the force of the “effective gravity.” It therefore does not accelerate.

Insight: If the rotation rate of the turntable were now increased, the marble would accelerate “uphill” (away from the axis of rotation) because the direction of “effective gravity” points more in the outward direction than it did before.

81. Picture the Problem: A device known as a sphygmomanometer is used to measured a person’s blood pressure on their

arm, at approximately the same level as the heart. The measurement is now attempted on the patient’s leg instead.

Strategy: Use the principle of pressure in a fluid to answer the conceptual question.

Solution: Assuming the leg is below heart level, as in a standing person, the reading on the sphygmomanometer will be greater than when the measurement is made on the arm. This is simply a reflection of the fact that the pressure in a fluid increases with depth.

Insight: The blood pressure in a standing person’s brain, likewise, is lower than it is in their heart.



15 – 28

82. Picture the Problem: At the surface of the ocean the pressure is one atmosphere. The pressure increases with depth.

We wish to calculate the depth at which the pressure is equal to two atmospheres. Strategy: Solve equation 15-7 for the depth at which the pressure is twice the pressure at the surface. Use the density

of sea water from Table 15-1. Solution: 1. Set the pressure in equation 15-7 equal

to two atmospheres: 2 1

at at2P P ghP P gh

ρρ

= += +

2. Solve for the depth:

( )( )5

at3 2

1.01 10 Pa 10.0m1025 kg/m 9.81 m/s

Ph

gρ×

= = =

Insight: The depth for a pressure of two atmospheres in fresh water is slightly greater at 10.3 m. 83. Picture the Problem: As water exits a high pressure reservoir, its speed increases from rest to supersonic speeds. Strategy: Use equation 15-14 to calculate the speed at which the water exits the reservoir. Set point 1 inside the

reservoir where the speed is zero and the pressure is 59,500 psi. Set point 2 outside the reservoir where the pressure is atmospheric pressure and the velocity is the speed we wish to calculate. Use the density of water given in Table 15-1.

Solution: 1. Solve equation 15-14 for speed 2:

( )

2 21 11 1 2 22 2

2 1 22

P v P v

v P P

ρ ρ

ρ

+ = +

= −


( ) 5

2 3

2

s

2 59,500 psi 14.7 psi 1.01 10 Pa 904 m/s14.7 psi1000 kg/m

904 m/s 2.64 times the speed of sound343 m/s

v

vv

− ⎛ ⎞×= =⎜ ⎟

⎝ ⎠

= =

Insight: For the water to exit at the speed of sound, the pressure inside the chamber would need only be 17,100 psi. 84. Picture the Problem: Water sprays out of a water main up to a height of 8.0 ft. We wish to find the pressure inside the

water main. Strategy: Solve equation 15-15 for the pressure 1P , where point 1 is inside the water main and point 2 is the highest

point that the water sprays. Let 1 0y = correspond to the height of the water main.

Solution: Solve equation 15-15 for 1P , letting 1 0y = and 2 8.0 ft :y =

( )( )( )

1 1 2 2

1 atm 2

5 3 2

51

1 m1.01 10 Pa + 1000 kg/m 9.81 m/s 8.0 ft3.28 ft

1.25 10 Pa

P gy P gyP P gy

P

ρ ρρ

+ = += +

⎛ ⎞= × ⎜ ⎟⎝ ⎠

= ×

Insight: Equation 15-17 can be used to show that the water exits the main with a speed of 6.91 m/s. 85. Picture the Problem: The Doloriometer measures the force applied over the circular probe. We wish to measure the

pressure exerted on the skin when the scale reads 3.25 lbs. Strategy: Pressure is force divided by area (equation 15-2). Divide the force reading on the Doloriometer by the cross-

sectional area of the probe. Calculate the area of the probe using the area of a circle of diameter 1.39 cm.

Solution: 1. Calculate the area of the probe: 2 2

4 20.0139 m 1.517 10 m2 2dA π π −⎛ ⎞ ⎛ ⎞= = = ×⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

2. Divide the force by the area: 4 2

3.25 lb 1 N 95.3 kPa0.225 lb1.517 10 m

FPA −

⎛ ⎞= = =⎜ ⎟× ⎝ ⎠

Insight: A pressure of 95.3 kPa is equal to the pressure a diver would experience at a water depth of 9.7 m.



15 – 29

86. Picture the Problem: The heart expends energy at a rate given by the average blood pressure times the blood flow rate.

We wish to calculate the power output and the total energy expended in one day.

Strategy: Set the power output of the heart equal to the average blood pressure times the blood flow rate. Convert the result to units of watts. To calculate the energy usage in a day, multiply the power by the number of seconds in a day. Set this energy equal to the potential energy (equation 8-3) to calculate the height to which this energy could lift a person.

Solution: 1. (a) Multiply average blood pressure by flow rate: ( )( )2 3 1 m1.33 N/cm 105 cm /s 1.40 W

100 cmP ⎛ ⎞= =⎜ ⎟

⎝ ⎠

2. (b) Multiply power by seconds in a day: ( )( )( )1.40 W 3600 sec/hr 24 hr/day 121 kJE Pt= = =

3. (c) Set energy consumption equal to potential energy and solve for the height: ( )( )

5

2

1.21 10 J 171 m72 kg 9.81 m/s

EE mgh hmg

×= ⇒ = = =

Insight: Although 1.4 W is a small power output, the continuous pumping of the heart causes the net energy to add up to a large value.

87. Picture the Problem: A swimming pool of diameter D = 4.7 m is filled to

a depth of h = 1.8 m, as shown in the figure. We wish to calculate the average outward force that the water exerts on the pool.

Strategy: To calculate the outward force, multiply the average outward pressure by the vertical surface area of the pool. The surface area is the area of a cylinder. From equation 15-7, we know that the pressure increases linearly with depth. The average pressure is the pressure at half the depth.

Solution: 1. Calculate the vertical surface area: ( )( ) 24.8 m 1.8 m 27 mA Dhπ π= = =

2. Calculate the pressure at half depth: ( ) ( )( )( )3 21

2 1000 kg/m 9.81 m/s 0.90 m 8.8 kPaP g hρ= = =

3. Multiply pressure times area: ( )( )2 3 527 m 8.8 10 Pa 2.4 10 NF = × = ×

Insight: The water in the pool weighs 300 kN, so the force on the bottom of the pool is about 1.5 times the force on the vertical wall.



15 – 30

88. Picture the Problem: The figure shows the block

suspended from the spring scale in air, in water, and in oil. The forces acting on the block are also shown.

Strategy: Solve Newton’s Second Law for the block suspended in air to calculate the mass of the block. Solve the Second Law for the block suspended in water to calculate the volume of the block. Divide the mass by the volume to calculate its density. Solve the Second Law for the block suspended in the oil to calculate the density of the oil.

Solution: 1. (a) Calculate the mass of the block: 2

35.0 N 3.57 kg9.81 m/s

TT mg mg

= ⇒ = = =

2. Solve Newton’s Second Law for the volume of the block when it is suspended in water: ( )( )

( )( )

w

24 3

3 2w

0

3.57 kg 9.81 m/s 31.1 N3.98 10 m

1000 kg/m 9.81 m/s

T Vg mg

mg TVg

ρ

ρ−

+ − =

−−= = = ×

3. Divide the mass by the volume: 3 3

block –4 3

3.57 kg 8.97 10 kg m3.98 10 m

mV

ρ = = = ××

4. (b) Solve the Second Law when the block is suspended in oil for the density of the oil: ( )( )

oil

3oil 4 3 2

035 N 31.8 N 820 kg/m

3.98 10 m 9.81 m/s

T Vg mgmg T

Vg

ρ

ρ−

+ − =− −

= = =×

Insight: The density of the oil can also be calculated from the buoyant forces: 0.82.oil oil

w w

mg Tmg T

ρρ

−= =

−

89. Picture the Problem: The figure shows a block resting on a vertical spring.

When the spring is in air, the spring force is equal to the weight of the block. The figure also shows the block and spring submerged in water, which introduces a buoyant force on the block.

Strategy: We wish to calculate the amount that the spring is compressed or stretched when in air and when in water. Solve Newton’s Second Law for the spring force in each case. Use equation 6-4 and the spring force to calculate the amount the spring is compressed or stretched. The buoyant force is given by equation 15-9.

Solution: 1. (a) Set the forces on the block in air equal to zero:

s 0

s

W FF W

− + ==

2. Use equation 6-4 to write the spring force and solve for the compression: ( )( )( )

block

3 3 2block

710 kg/m 0.012 m 9.81 m/s0.15 m

540 N/m

kx Vg

Vgx

k

ρ

ρ

− =

−−= = = −

The spring is compressed 0.15 m.

3. (b) Set the forces on the block in water equal to zero:

s b

s b

0W F FF W F

− + + == −



15 – 31

4. Use equation 6-4 to write the spring force and solve for the spring stretch: ( )

( )( )( )

block w

w block

3 3 3 21000 kg/m 710 kg/m 0.012 m 9.81 m/s0.063 m

540 N/m

kx Vg V gV g

xk

ρ ρρ ρ

− = −

−=

−= =

The spring is stretched 0.063 m.

Insight: Because the density of the block is less than the density of the water the buoyant force is greater than the weight of the block and the block will accelerate upward. The added downward force of the spring prevents the block from reaching the water’s surface.

90. Picture the Problem: The image shows the wooden block with an iron ball attached to it. On the left side the iron ball is attached to the top surface and is not submerged. On the right side the iron ball is attached to the underside of the wood and is submerged.

Strategy: For each case, set the sum of the forces on the block and ball equal to zero. Use equation 15-9 for the

buoyant force. When the ball is above the wood, the submerged volume is only the submerged portion of the wood. When the ball is below the wood, the submerged volume is the volume of the ball plus the submerged portion of the wood.

Solution: 1. (a) Set the sum of the forces equal to zero and solve for Fb:

block ball b

b block ball

0W W FF W W

− − + == +

2. Use equation 15-9 for the buoyant force and solve for the submerged volume:

w sub block ball

block ballsub

w

V g m g m gm m

V

ρ

ρ

= ++

=

3. Calculate the mass of the ball from equation 15-1: ( )( )

34ball iron 3

33 437860 kg/m 0.0122 m 0.05978 kg

m rρ π

π

=

⎡ ⎤= =⎣ ⎦

4. Calculate the submerged volume: 3sub 3

1.25 kg 0.0598 kg 0.00131 m1000 kg/m

V += =

5. (b) The same total volume of water must be displaced in order to support the same total weight. Because the ball now is submerged, the volume of wood that is submerged will decrease.

6. (c) Subtract the volume of the ball from the submerged volume to obtain the submerged wood:

( )wood sub ball

33 3430.00131 m 0.0122 m 0.00130 m

V V V

π

= −

= − =

Insight: The volume of the iron weight was small compared to the submerged volume of the wood, so the change in submerged volume of the wood was less than 1%. However, the submerged volume of the wood did decrease as predicted.



15 – 32

91. Picture the Problem: When you blow across the top of the straw, the pressure

decreases above the straw, causing the water to rise in the straw.

Strategy: We want to calculate how fast you must blow the air so that the water rises in the straw a height of 1.6 cm. Solve equation 15-14 for the change in pressure when point 1 is at the top of the straw and point 2 is just above the water’s surface. Use equation 15-7 to set the change in pressure equal to the pressure due to the water that rises in the straw. Solve the resulting equation for the speed of the air.

Solution: 1. Solve equation 15-14 for the pressure difference:

21 air 2

22 1 air

12

12

P v P

P P v

ρ

ρ

+ =

− =

2. Set the change in pressure equal to the weight of the water and solve for the speed:

2 ww air

air

21 2

ghgh v v

ρρ ρ

ρ= ⇒ =


( )( )( )3 2

3

2 1000 kg/m 9.81 m/s 0.016 m16 m/s

1.29 kg/mv = =

Insight: Because it is difficult to blow this hard (16 m/s is the same as 36 mi/hr), it is not a safe bet that you can remove water from a glass in this fashion.

92. Picture the Problem: Atmospheric pressure is equal to the weight of the air above sea level to the top of the

atmosphere.

Strategy: We wish to calculate the height of the atmosphere and the pressure at the top of Mt. Everest. Use equation 15-7, the pressure variation with depth, to solve for the height of the atmosphere. Let P2 be the atmospheric pressure at sea level, and let P1 be equal to zero (the pressure of space). To find the pressure at the summit of Mt. Everest, convert the height of Mt. Everest to meters and compare with the height of the atmosphere.

Solution: 1. (a) Solve equation 15-7 for h:

( )( )

2 1 air

5at2 1

3 2air air

0 1.01 10 Pa 7.98km1.29 kg/m 9.81 m/s

P P ghPP P

hg g

ρ

ρ ρ

= +

−− ×= = = =

2. (b) Convert the height of Mt. Everest to meters.

1 m29,035 ft 8850 m3.28 ft

h ⎛ ⎞= =⎜ ⎟⎝ ⎠

Because this is higher than our model atmosphere, the pressure is zero.

Insight: The density of air actually diminishes with altitude, not remaining constant as assumed in the problem. As such, the air pressure at the top Mt. Everest is greater than zero.

93. Picture the Problem: The figure shows two containers. Container 1 has a cross-sectional area of 24 cm2 and a height of 18 cm. Container 2 also has a height of 18 cm and base cross-sectional area 24 cm2. However, the top half of this container has a cross-sectional area of 96 cm2. We want to calculate the downward force the water exerts on Container 2 and compare this force with the total weight of the water in the container.

Strategy: Set the total force equal to the sum of the hydrostatic force on the base and the force on the annular region. Use equation 15-2 to calculate the forces in terms of pressures and areas and equation 15-6 to determine the pressure.

Solution: 1. (a) Calculate the force on the base: ( )( )( )( )

base base base 1 base

3 2 4 21000 kg/m 9.81 m/s 0.18 m 24 10 m 4.24 N

F P A gh Aρ−

= =

= × =



15 – 33

2. Calculate the force on the ring:

( )( )( )( )ring ring ring 2 ring

3 2 4 21000 kg/m 9.81 m/s 0.09 m 72 10 m 6.36 N

F P A gh Aρ−

= =

= × =

3. Sum the forces: total base ring 4.24 N 6.36 N 11 NF F F= + = + =

4. (b) Set the mass equal to the density times volume:

( )1 base 2 ring

base ring

W mg Vg h A h A g

F F F

ρ ρ= = = +

= + =

Insight: The weight of Container 2 is not supported only by the water pressure, but also by the container walls. 94. Picture the Problem: The figure shows two containers.

Container 1 has a cross-sectional area of 24 cm2 and height 18 cm. Container 2 also has a height of 18 cm and base cross-sectional area of 24 cm2. However, the top half of this container has a smaller cross-sectional area of 6 cm2 (24 cm2 minus an annular region of 18 cm2 ). We want to calculate the downward force the water exerts on Container 2 and compare this force with the total weight of the water in the container. We also wish to calculate the speed at which water would exit a hole in the annular region.

Strategy: The water in Container 2 exerts a downward force on the base of the container and an upward force on the

annular region. Calculate the net downward force by subtracting these two forces. Use equation 15-6 to calculate the pressure at the bottom of the container and at the annular region. To calculate the weight of the water in the container, multiply the density of the water by the volume of the container. Calculate the speed through a hole in the annular region using equation 15-17. To calculate the height use equation 15-15 with point 1 at the surface of the water and point 2 at the maximum height of the spray. The pressure is atmospheric for both points.

Solution: 1. (a) Calculate the force on the base: ( )( )( )( )

base base base 1 base

3 2 4 21000 kg/m 9.81 m/s 0.18 m 24 10 m 4.24 N

F P A gh Aρ−

= =

= × =

2. Calculate the upward force on the ring:

( )( )( )( )ring ring ring 2 ring

3 2 4 21000 kg/m 9.81 m/s 0.09 m 18 10 m 1.59 N

F P A gh Aρ−

= =

= × =

3. Calculate the net downward force: base ring 4.24 N 1.59 N 2.7 NF F F= − = − =

4. (b) Calculate the weight of the water: ( )1 base 2 ring

1 base 2 ring

base ring

W mg Vg h A h A g

h A g h A g

W F F F

ρ ρ

ρ ρ

= = = −

= −

= − =

5. (c) Use Torricelli’s Law to calculate the speed: ( )( )22 2 9.81 m/s 0.090 m 1.3 m sv gh= = =

6. (d) Equation 15-15 shows that the water will rise to the top of the container. Because the pressure at the top of the water and the top of the spray are equal, the heights must also be equal.

Insight: The water pressure at the bottom of the container 2 in this problem and the one in problem 93 are equal. However, in problem 93 the added water created an additional weight that had to be supported by the container. In this problem the decreased volume at the top created an upward force on the annular region which decreased the net force on the container. In both problems the net force on the container is equal to the weight of the water.



15 – 34

95. Picture the Problem: The figure shows a pool of diameter D = 2.3 m that

is filled with water to a depth of h = 0.38 m. When a person enters the pool and floats in the water, the water level rises to create a buoyant force to support the person’s weight. We wish to calculate the water pressure at the bottom of the pool before the person enters the pool and while the person is floating.

Strategy: Use equation 15-7 to calculate the pressure at the bottom of the pool. When the person floats in the pool, her weight is supported by the bottom of the pool. The pressure will therefore increase by an amount equal to the person’s weight divided by the area of the pool. Use equation 15-2 to calculate the additional pressure.

Solution: 1. (a) Calculate the pressure at the bottom of the pool:

( )( )( )at w

5 3 2 51.013 10 Pa 1000 kg/m 9.81 m/s 0.38 m 1.050 10 Pa

P P ghρ= +

= × + = ×

2. (b) If a person now floats peacefully in the pool the pressure at the bottom will increase because the water depth will increase when water is displaced by the floating person.

3. (c) Add the weight of the person divided by the area of the pool:

( )( )

25

total 2 212

5 5

72 kg 9.81 m/s1.050 10 Pa

2.3 m

1.050 10 Pa 170 Pa 1.052 10 Pa

mg mgP P PA rπ π

= + = + = × +×

= × + = ×

Insight: The pressure at the bottom increases only slightly, as the170 Pa change is tiny compared to 1.050×105 Pa. You can verify that when the person enters the pool, the water level rises from a depth of 38 cm to a depth of 40 cm.

96. Picture the Problem: A rock is made of gold and granite. The mass and volume of the rock are known.

Strategy: We wish to calculate the mass of the gold and the percentage that is gold. Use equation 15-1 to write the mass of the rock in terms of the densities and volumes of the gold and granite. Write the volume of the granite as the total volume less the volume of the gold. Solve for the volume of gold. Use the volume and density of gold to calculate the mass of gold.

Solution: 1. (a) Write the mass of the rock as the sum of the masses of the gold and granite: ( )

rock Au Au gr gr

Au Au gr Au

m V V

V V V

ρ ρ

ρ ρ

= +

= + −

2. Solve for the volume of gold:

( )( )3 -4 3rock gr

Au 3 3Au gr

4 3

3.81 kg 2650 kg/m 3.55×10 m

19,300 kg/m 2650 kg/m

1.723 10 m

m VV

ρρ ρ

−

−−= =

− −

= ×

3. Solve for the mass of gold: ( )( )3 4 3Au Au Au 19,300 kg/m 1.723 10 m 3.33 kgm Vρ −= = × =

4. (b) Calculate the percent that is gold by volume:

4 3Au

4 3

1.723 10 m 0.485 48.5%3.55 10 m

VV

−

−

×= = =

×

5. (c) Calculate the percent that is gold by mass: Au 3.326 kg 0.873 87.3%

3.81 kgmm

= = =

Insight: Because the density of gold is much larger than the density of granite, most of the mass is gold, even though the volumes of gold and granite are about equal.

97. Picture the Problem: The pressure on crustal rocks increases with depth, as with fluids.

Strategy: We wish to calculate the depth for which the pressure will be 91.2 10 Pa.× Solve equation 15-7 for the depth at which the pressure is 91.2 10 Pa.× The density of the rock is given in the problem.

Solution: 1. Solve the pressure equation for the depth: max at

max at earth max maxearth

P P

P P gh hg

ρρ

−= + ⇒ =



15 – 35


( )( )9 5

max 3 3 2

1.2 10 Pa 1.01 10 Pa 41 km3.0 10 kg/m 9.81 m/s

h × − ×= =

×

Insight: The atmospheric pressure could be ignored in this problem because it is four orders of magnitude smaller than the maximum pressure.

98. Picture the Problem: The image shows a piece of wood submerged in water and connected by a string to the bottom of the container. The density of the wood is 706 kg/m3 and the tension in the string is 0.89 N.

Strategy: We wish to calculate the volume of the wood and the change in water level if the string breaks and the wood rises to the surface. Set the sum of the forces on the block equal to zero. Write the buoyant force using equation 15-9 and the weight using equation 15-1. Solve for the volume of the wood. When the block rises to the surface the water level will drop by a volume equal to the volume of the wood above the surface. Divide this volume by the area of the container to determine the change in water level.

Solution: 1. (a) Sum the forces on the block:

b

w block

00

F T mgVg T Vgρ ρ

− − =− − =

2. Solve for the volume of the wood:

( ) ( )( )– 4 3

2 3w block

0.89 N 3.1 10 m9.81 m/s 1000 706 kg/m

TVg ρ ρ

= = = ×− −

3. (b) Because only part of the block will be immersed, the water level will drop.

4. (c) Calculate the submerged volume of the block:

( )

blockblock block w sub sub block

w

34 3 4 3

sub 3

706 kg/m 3.086 10 m 2.179 10 m1000 kg/m

V g V g V V

V

ρρ ρ

ρ

− −

⎛ ⎞= ⇒ = ⎜ ⎟

⎝ ⎠⎛ ⎞

= × = ×⎜ ⎟⎝ ⎠

5. Calculate the volume that is not submerged:

4 3 4 3 5 3above block sub 3.086 10 m 2.179 10 m 9.1 10 mV V V − − −= − = × − × = ×

6. Divide the volume by the area of the flask:

5 3w

– 4 2

9.1 10 m 1.5 cm62 10 m

Vh

A

−×Δ = = =

×

Insight: You can verify that the change in height is independent of the block’s density by recalculating the solution using another density.

99. Picture the Problem: The image shows an irrigation siphon. Water flows into the pipe at point 1 and out at point 3. We want to symbolically determine the speed of water as it exits the pipe and compare the speeds at points 2 and 3.

Strategy: Use Bernoulli’s equation (15-16) with the initial point at the surface of the canal and the second point where the water exits the siphon. At both of the points the pressure is atmospheric. The velocity of the water in the canal is zero. Take the height to be zero at the exit of the siphon. Use the continuity equation to compare water speeds at points 2 and 3.

Solution: 1. (a) Write equation 15-16 at points 1 and 3, and solve for 3 :v

2 21 11 1 1 3 3 32 2

21atm atm 3 32 2

P v gh P v gh

P gd P v v g d

ρ ρ ρ ρ

ρ ρ

+ + = + +

+ = + ⇒ =

2. (b) The speed of the water at point 2 is the same as point 3, assuming constant pipe diameter, since 2 2 3 3.v A v A=

Insight: The speed is the same as that given by Torricelli’s Law; the same as if the water were in free fall from the surface of the canal.



15 – 36

100. Picture the Problem: The image shows water escaping from a

hole y0 = 11 cm from the bottom of a can. The water exits at an angle θ = 36° above the horizontal.

Strategy: We wish to calculate the range of the water x and the stream’s maximum height y1. First find the velocity at the hole using Torricelli’s Law. Then use the equations of motion (equations 4-10) to solve for the time at which the water reaches the ground and calculate the range x. Use the equations of motion again to calculate the maximum height.

Solution: 1. (a) Solve for the initial velocity: ( )( )22 2 9.81 m/s 0.39 0.11 m 2.344 m/sv gh= = − =

2. Use equation 4-10 to write the height as a function of time: ( ) ( ) ( )

210 0 2

2 21land land2

2land land

0 0.11 m 2.344 m/s sin 36 9.81 m/s

0.11 1.378 4.905

y yy y v t a t

t t

t t

= + +

= + ° −

= + −

3. Solve the quadratic equation for the positive time to land:

( ) ( )( )( )

2 2

land 2

1.378 s 1.378 s 4 4.905 m/s 0.11 m0.346 s

2 4.905 m/st

− − − −= =

−

4. Solve equation 4-10 for x: ( )( )( )0 land 2.344 m/s cos36 0.346 s 0.66 mxx v t= = ° =

5. (b) Solve for the time for the stream to reach the top of its flight: ( ) ( )

0 top

0top 2

sin 0

2.344 m/s sin 36sin0.140 s

9.81 m/s

yv v gt

vt

g

θ

θ

= − =

°= = =

6. Solve for the maximum height: ( )( ) ( )( ) ( )( )

211 0 0 top top2

2212

1

sin

0.11 m 2.344 m/s sin 36 0.140 s 9.81 m/s 0.140 s

0.21 m

y y v t gt

y

θ= + −

= + ° −

=

Insight: The water stream does not reach the same height as the water in the container (y2 = 0.39 m) because the water still has a horizontal component of velocity when it is at its maximum height.

101. Picture the Problem: A person stands hip deep in water. We wish to use his change in apparent weight to calculate

volume and mass of his legs.

Strategy: Set the sum of forces on the person in the water equal to zero. Use equation 15-9 for the buoyant force and solve for the volume his legs. Divide the volume by two to determine the volume of one leg. Multiply the volume by the density to determine the mass of each leg.

Solution: 1. (a) Sum the forces to zero: b

w

00

F N mgVg N mgρ

+ − =+ − =

2. Solve for the submerged volume:

( )( )3

3 2w

685 N 497 N 0.01916 m1000 kg/m 9.81 m/s

mg NVgρ− −

= = =

3. Divide the volume by two:

3–3 3

leg0.01916 m 9.58 10 m

2 2VV = = = ×

4. (b) Multiply the volume by the density: ( )( )( )

leg leg leg w leg

3 3

1.05

1.05 1000 kg/m 0.00958 m 10.1 kg

m V Vρ ρ= =

= =

Insight: Combined together, this person’s legs constitute 28% of his total mass.



15 – 37

102. Picture the Problem: The image shows oil flowing through a horizontal

pipe of cross-sectional area A and length L = 55 m. We want to calculate the pressure difference across the pipe and the volume flow rate.

Strategy: Solve for the pressure difference using equation 15-18. Calcu-late the volume flow rate as the velocity times the cross-sectional area.

Solution: 1. Calculate the cross- sectional area of the pipe:

( )22

2

/ 4 0.052m / 4

0.002124 m

A Dπ π= =

=

2. (a) Solve equation 15-18 for the pressure difference:

( )( )( )2

1 2 2

8 0.00012 N s/m 1.2 m/s 55 m8 94 Pa

0.002124 mvLP PA

ππη

⋅− = = =

3. (b) Calculate the volume flow rate: ( )2 3volume flow rate 0.002124 m 1.2 m/s 0.0025 m /sAv= = =

Insight: Increasing the viscosity will increase the pressure needed to maintain the same velocity, but does not affect the volume flow rate.

103. Picture the Problem: A solution passes through a hypodermic needle of diameter 0.26 mm at a rate of 1.5 g/s.

Strategy: We wish to calculate the pressure difference needed across the needle due to the viscosity of the solution. Solve the mass flow rate for the velocity of the solution. Use the velocity in the equation 15-18 to calculate the pressure difference.

Solution: 1. Calculate the cross- sectional area of the needle: ( )22 8 24 0.00026 m 4 5.31 10 mA dπ π −= = = ×

2. Solve the mass flow rate for the velocity:

( )( )3 8 2

0.0015 kg/s 28.2 m/s1000 kg/m 5.31 10 m

m t Avm tv

A

ρ

ρ −

Δ Δ =Δ Δ

= = =×

3. Calculate the change in pressure: ( ) ( )( )2

–8 2

5

28.2 m/s 0.033 m8 8 0.00101 N s/m

5.31 10 m4.4 10 Pa

vLPA

πη πΔ = = ⋅×

= ×

Insight: The pressure difference is about 3 atmospheres. However, because of the small cross-sectional area, this pressure can be supplied by a force of only 0.02 N, less than a tenth of an ounce.

104. Picture the Problem: The pressure wave from an exploding meteor creates an increase in pressure across a window

pane.

Strategy: We want to calculate the force this pressure causes on a rectangular window. Use equation 15-2 to calculate the force in terms of the pressure and the window’s area. Convert the result to pounds.

Solution: 1. Solve equation 15-2 for the force: ( )( )( )

21 m 0.50 kPa 34 in 46 in 0.50 kN39.37 in

F PA ⎛ ⎞= = =⎜ ⎟⎝ ⎠

2. Convert to pounds: 2224.8 lbs0.505 kN 1.1 10 lbs1 kN

F ⎛ ⎞= = ×⎜ ⎟⎝ ⎠

Insight: A force of 110 lbs is more than sufficient to break a pane of glass.



15 – 38

105. Picture the Problem: On one episode of Mythbusters, a helium balloon is constructed using a thin lead foil with a mass

of 11 kg. The balloon was approximately cubical in shape, and 10 feet on a side. Strategy: Use the dimensions of the balloon and the known density of lead to determine the thickness of the lead foil.

Then subtract the weight of the helium plus foil from the weight of the displaced air to find the net upward force exerted by the balloon. The additional mass that could be lifted by the balloon could then be determined.

Solution: 1. (a) Find the surface area of the balloon, remembering that a cube has six sides: ( )22 26 6 10 ft 0.305 m/ft 55.8 mA L= = × =

2. Determine the thickness of the foil from its mass, area, and density:

( )( )5

3 2

11 kg 1.7 10 m 17 m11,300 kg/m 55.8 m

m mV A xmxA

ρ

μρ

−

= =

= = = × =

3. (b) Find the volume of the balloon: ( )33 3balloon 10 ft 0.305 m/ft 28.4 mV L= = × =

4. Subtract the weight of the helium from the weight of the air to determine whether the 11 kg foil could be lifted:

( ) ( )( )

lift b helium

lift air air helium helium

3 3lift air helium balloon 1.29 0.179 kg/m 28.4 m

31.5 kg > 11 kg, so yes the balloon can float

F F Wm g V g V g

m V

ρ ρ

ρ ρ

= −= −

= − = −

=

5. (c) determine the additional mass that can be lifted by subtracting the foil mass: add lift foil 31.5 11 kg 21 kgm m m= − = − =

Insight: The lead balloon could lift about 45 lb in addition to its own 35 lb weight (24 lb lead, 11 lb helium). This is because of the buoyant force from the 36.5 kg = 80 lb of air that was displaced.

106. Picture the Problem: The figure shows the surface of a container

of water of width W and depth d when it is not accelerating. The figure also shows the surface tilted as the container accelerates forward. As it accelerates, the front depth decreases to d − h and the back depth increases to d + h.

Strategy: Assume the pan has length L and width W, and that it contains water with depth d. When accelerating the depth at the

back wall will be d + h, while the depth at the front wall will be d − h. The surface is tilted at an angle ,θ where tan 2 .h Wθ = To calculate the height h, apply Newton’s Second Law to the water in the horizontal direction. Calculate the forces on the front and back walls from equation 15-2, where the pressure exerted by the wall is the average water pressure.

Solution: 1. (a) With no acceleration, the force on all sides of the water is the same. The water is level.

2. (b) The water tilts forward. That is, the water is deeper at the back side of the pan. The pan therefore exerts a larger force on the back side of the water than on its front side, producing a net force in the forward direction.

3. (c) Calculate the force on the back wall: ( ) ( ) ( )2 21 1b b b 2 2 2F P A g d h L d h gL d dh hρ ρ= = + + = + +⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

4. Calculate the force on the front wall: ( ) ( ) ( )2 21 1f f f 2 2 2F P A g d h L d h gL d dh hρ ρ= − = − − − = − − +⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

5. Apply Newton’s Second Law:

( ) ( ) ( )( )

2 2 2 21 12 22 2

2

b fma F F

LWd a gL d dh h gL d dh h

LWda gL dh

ρ ρ ρ

ρ ρ

= +

⎡ ⎤ ⎡ ⎤= + + − − +⎣ ⎦ ⎣ ⎦=

6. Solve for 2h / W: 2 / /h W a g=

7. Write in terms of the incline: tan 2 / /h W a gθ = =

Insight: Measuring the inclination of the surface is a method of determining the acceleration of a fluid. This method is also useful for calculating the centripetal acceleration of a rotating container.



15 – 39

107. Picture the Problem: A wooden block, when initially displaced vertically from equilibrium in a fluid will oscillate

about the equilibrium point in simple harmonic motion.

Strategy: When the block is submerged a distance hΔ beyond its equilibrium height, the submerged volume is increased by A hΔ resulting in an increase in the buoyant force. The buoyant force accelerates the block back through its equilibrium position, with the restoring force proportional to the distance .hΔ Write out the buoyant force and compare the terms to Hooke’s Law (equation 6-4). Use that analogy together with the equation for the period of a mass on a spring (equation 13-11) to write an equation for the period of the oscillation.

Solution: 1. Write the increased buoyant force in terms of the displacement:

( ) ( )2 2 2( )F V g A h g Ag hρ ρ ρ= Δ = −Δ = − Δ

2. Compare with equation 6-4, :F k x= − 2: :x h k AgρΔ …

3. Write equation 13-11 in terms of the analogous spring constant:

2

2 2m mTk Ag

π πρ

= =

4. Write the mass in terms of density and volume and simplify:

1 1

2 2

2 2AH HTAg g

ρ ρπ π

ρ ρ= =

Insight: An example of this oscillation could be an ice cube ( 31 0.917 g/cmρ = ) that is H = 3.0 cm tall and is floating in

water ( 32 1.00 g/cmρ = ). It will oscillate with a period of 0.33 seconds.

108. Picture the Problem: The image shows a log floating with half of its radius above

the surface. We want to calculate the log’s density.

Strategy: Calculate the submerged cross-sectional area of the log (Asub) by adding the yellow triangles to the green “pie.” To do this, first find the angle θ from the known sides of the right triangle. Set the weight of the log equal to the buoyant force and solve for the log’s density.

Solution: 1. Find the angle θ : 12

1

1cos2

1cos2 3

rr

θ

πθ −

= =

⎛ ⎞= =⎜ ⎟⎝ ⎠

2. Calculate the area of the two triangles: ( )( ) 231 12 2 3 8sintriA r r rπ= =

3. Calculate the area of the pie: 2 2 2pie

2 / 3 21 12 3

A r r rθ ππ π ππ π

⎛ ⎞ ⎛ ⎞= − = − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

4. Add the submerged areas: ( ) ( )2 2 23 32 2sub pie tri 3 8 3 42 2A A A r r rπ π= + = + = +

5. Set the buoyant force equal to the weight: w sub log log

w sub log log

V g V g

A L g A L g

ρ ρ

ρ ρ

=

=

6. Solve for the density:

( )

( ) ( )

2323 4sub

log w w 2log

3 3323 4 0.804 1000 kg/m 804 kg/mw

rAA r

π

πρ ρ ρ

π

ρ

+= =

= + = =

Insight: Note that the diameter (73 cm) was not needed to calculate the density of the log.



15 – 40

109. Picture the Problem: The figure shows a spherical glass shell floating in a liquid.

We want to calculate the fraction of the shell that is submerged in the water.

Strategy: To find the fractional of the shell that is submerged, we will set the buoyant force equal to the weight of the sphere and solve for the ratio of the submerged volume to the total volume. First, however, calculate the density of the sphere by dividing the mass of the sphere by its volume. The mass of the sphere is the density of the glass times the volume of the spherical shell.

Solution: 1. Calculate the mass of the sphere: ( ) ( )

( ) ( )

3 34 4sphere 3 3

3 3 34 43 3

1.2

1.2 1 0.728

g

g g

m R R

R R

ρ π π

π ρ π ρ

⎡ ⎤= −⎣ ⎦

= − =

2. Divide the mass by the volume:

( )( )

34sphere 3

sphere 34sphere 3

0.7280.4213

1.2g

g

m RV R

π ρρ ρ

π= = =

3. Set the buoyant force equal to the weight: fluid sub sphere sphere

spheresub

sphere fluid

0.42130.28

1.5g

g

V g V g

VV

ρ ρ

ρ ρρ ρ

=

= = =

Insight: If the sphere were solid, instead of a hollow shell, it would still float, but with 67% of its volume submerged. 110. Picture the Problem: A geode is a hollow spherical rock.

Strategy: Given that the weight of the geode submerged in water is half its weight in air, we wish to calculate the fraction of the volume that is hollow. Set the sum of the force on the geode in water equal to zero and solve for the geode’s mass. Because the air is much lighter than the rock, set the mass of the geode equal to the density of the rock times the solid volume. Solve the resulting equation for the solid fraction. Subtract the solid fraction from unity to calculate the hollow fraction.

Solution: 1. Sum the forces to zero and solve for the mass:

b1

w2

w

002

N F mgmg Vg mg

m Vρ

ρ

+ − =

+ − =

=

2. Write the mass in terms of the density and volume: solid solid 2 wm V Vρ ρ= =

3. Solve for the solid fraction: ( )3

solid w3

solid

2 1000 kg/m20.80

2500 kg/mV

Vρ

ρ= = =

4. Calculate the hollow fraction: air solid

geode geode

1 1 0.80 0.20V V

V V= − = − =

Insight: The smaller the apparent weight of the geode while it is submerged in water, the greater the hollow fraction. This geode will float in the water when the hollow fraction is 0.60.



15 – 41

111. Picture the Problem: The figure shows a container filled with water to a

depth d. Water escapes from a hole h above the bottom. We want to calculate the distance x from the tank where the water hits the tabletop.

Strategy: Use Torricelli’s Law to calculate the speed that the water leaves the tank. Use the equation 4-7 for horizontal projectiles to calculate the time for the water to fall a distance h, and from the time, the water’s horizontal displacement.

Solution: 1. Write the initial velocity using equation 15-17: ( )2v g d h= −

2. Solve the vertical free-fall equation for the time of flight:

212

2 hy h gt tg

= − ⇒ =

3. Insert the time into the horizontal displacement equation: ( ) ( )0

22 2hx v t g d h d h hg

= = − = −

Insight: The distance x is a maximum when / 2h d= . When / 2h d> , the lower pressure results in a lower horizontal velocity. When / 2h d< , the travel time is shorter.

112. Picture the Problem: The figure shows water escaping

from a tank from holes 3.6 m and 0.80 m from the base of the tank. The water streams hit at the same location.

Strategy: We wish to calculate the depth of the water in the tank. Use the solution from problem 111,

( )2x d h h= − , to set the horizontal distances equal

and solve for the depth of the water, d. Let th be the height of the top stream and bh of the bottom stream.

Solution: 1. Set the horizontal distances equal:

( ) ( )t b

t t b b2 2

x x

d h h d h h

=

− = −

2. Square both sides and solve for d: ( ) ( )( ) ( )( )

( )

2 2t b t b t b t b

t b

t t b bd h h d h h

d h h h h h h h h

d h h

− = −

− = − = + −

= +

3. Insert the heights: ( )3.6 m 0.80 m 4.4 md = + =

Insight: The water from two holes will hit at the same point if the depth of the water is equal to the sum of the heights of the holes.

113. Picture the Problem: A cubical box of sides 0.29 m in length is placed in water so that 35% of its volume is

submerged. Strategy: We want to calculate the mass of water that can be added to the box before it sinks. Since 35% of the box is

below the surface, 65% remains above the surface. Filling that 65% with water will cause the box to drop until it is completely submerged. Calculate the mass of the added water by multiplying the density of water by 65% of the volume of the box.

Solution: Multiply the density of water by 65% of the volume of the box:

( ) ( )( ) ( )

3w w box w

33

0.65 0.65

1000 kg/m 0.65 0.29 m 16 kg

m V Lρ ρ= =

⎡ ⎤= =⎣ ⎦

Insight: The mass of the box is equal to the weight of the water displaced by the 35% of its volume that is submerged, or ( )( ) ( )3 3

box 0.35 0.29 m 1000 kg/m 8.5 kg.m = =



15 – 42

114. Picture the Problem: As a doughnut is cooked, less than half of it is

submerged. This results in a white stripe around the center of the doughnut. Strategy: We want to calculate the density of the doughnut from the density

of the oil and the relative height of the stripe. Treating the doughnut as a cylinder, calculate the depth to which the doughnut is submerged from the height of the stripe. As the doughnut floats on the surface of the oil, the weight of the doughnut is supported by the buoyant force from the oil. Set the buoyant force equal to the weight and solve for the density of the doughnut.

Solution: 1. Set the submerged height equal to half of cooked height: ( ) ( )1 1

sub stripe2 2 0.22 0.39H H H H H H= − = − =

2. Set the weight equal to the buoyant force and write the volumes in terms of area and height: ( ) ( )

D D oil sub

D oil sub

V g V gAH g AH g

ρ ρρ ρ

=

=

3. Solve for the doughnut density:

( )( )

subD oil oil

3 3

0.39

919 kg/m 0.39 360 kg/m

H HH H

ρ ρ ρ⎛ ⎞ ⎛ ⎞= = ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

= =

Insight: Because the density of the doughnut is less than half the density of the oil, more than half of the doughnut remains above the surface as it cooks.

115. Picture the Problem: As the doughnut is cooked, less than half of it is

submerged. This results in a white stripe around the center of the doughnut. Strategy: We want to predict the height of the white stripe given the densities

of the doughnut and of the oil. As the doughnut floats on the surface of the oil, the weight of the doughnut is supported by the buoyant force from the oil. Set the buoyant force equal to the weight and solve for the depth to which the doughnut is submerged in the oil. Use that depth to determine the thickness of the white stripe of uncooked dough.

Solution: 1. Set the doughnut’s weight equal to the buoyant force and substitute m = ρ V: ( ) ( )

D D oil sub

D oil sub

V g V gAH g AH g

ρ ρρ ρ

=

=

2. Solve for the depth Hsub to which the doughnut will be submerged:

3D

sub 3oil

330 kg/m 0.36919 kg/m

H H H Hρρ

= = =

3. Determine the thickness of the white stripe: ( )stripe sub2 2 0.36 0.28H H H H H H= − = − =

Insight: This doughnut is less dense than the one in the previous problem. It therefore floats higher, a smaller fraction of the doughnut is cooked on each side, and the height of the white stripe increases.

116. Picture the Problem: The connection between the density of the

doughnut and the height of the white stripe is illustrated in the plot at right. On the x axis we plot the density of the doughnut as a fraction of the density of the vegetable cooking oil; and the y axis shows the height of the white stripe as a fraction of the total height of the doughnut. Notice that the height of the white stripe is plotted for both positive and negative values.

Strategy: Consider the effect of increasing the density of the doughnut to more than half the density of the oil, and use Archimede’s principle to answer the conceptual question.

Solution: The ratio of the doughnut density to oil density (x axis) gives the fraction of the doughnut’s height that is submerged. At

x = 0.75, three-quarters of the doughnut is submerged and is cooked. When the doughnut is flipped, three-quarters of the doughnut is again submerged, but that means a band of doughnut around the middle will have been submerged twice



15 – 43

and therefore cooked twice as long as the top and bottom. The height of this band is a quarter of the doughnut’s thickness on either side of the middle line, for a total height of half the thickness of the doughnut. We conclude that half the doughnut is light brown, half is dark brown. This is the appropriate comment to make for the point on the plot corresponding to (x, y) = (0.75, − 0.5).

Insight: At the point (1, −1) 100% of the doughnut is submerged and its density matches the density of the oil. If the doughnut is cooked and flipped as usual, the entire doughnut will be dark brown because it was cooked twice.

117. Picture the Problem: As the doughnut floats on the surface of the oil, the weight of

the doughnut is supported by the buoyant force from the oil.

Strategy: Set the buoyant force equal to the weight and solve for the depth to which the doughnut is submerged in the oil. Use that depth to determine the portion of the doughnut that remains above the oil.

Solution: 1. Set the doughnut’s weight equal to the buoyant force and substitute m = ρ V: ( ) ( )

D D oil sub

D oil sub

V g V gAH g AH g

ρ ρρ ρ

=

=

2. Solve for the depth Hsub to which the doughnut will be submerged:

3D

sub 3oil

550 kg/m 0.60919 kg/m

H H H Hρρ

= = =

3. Determine the portion of the doughnut above the oil: ( )above sub 0.60 0.40H H H H H H= − = − =

Insight: This doughnut is less dense than the one in the previous problem. It therefore floats higher, a smaller fraction of the doughnut is cooked on each side, and the height of the white stripe increases.

118. Picture the Problem: The U-shaped tube is shown in the figure

with all relevant distances. 5 cm of oil are inserted over the water on one side of the tube. We wish to see how h depends on the density of the oil.

Strategy: As in Example 15-4, set the pressure at points A and B equal. Solve the resulting equation for h1 in terms of h2. Then calculate h by taking the difference between h2 and h1.

Solution: 1. (a) Set the pressures at A and B equal and solve for h1:

oilatm water 1 atm oil 2 1 2

water

P g h P g h h hρ

ρ ρρ

⎛ ⎞+ = + ⇒ = ⎜ ⎟

⎝ ⎠

2. Calculate the difference in depths: oil oil

2 1 2 2 2water water

1h h h h h hρ ρρ ρ

⎛ ⎞ ⎛ ⎞= − = − = −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

3. If a higher density of oil is used, the difference in heights will decrease.

4. (b) Solve numerically: ( )

3

3

960 kg/m1 5.0 cm 0.20 cm1000 kg/m

h⎛ ⎞

= − =⎜ ⎟⎝ ⎠

Insight: The height difference in Example 15-4 was 0.40 cm. Therefore increasing the density did decrease the height difference.



15 – 44

119. Picture the Problem: The U-shaped tube is shown in the

figure with all relevant distances. Oil is inserted on one side of the tube up to a depth of h2. We wish to see how h, the difference between the heights of the two columns, depends on the height of the oil column.

Strategy: As in Example 15-4, set the pressure at points A and B equal. Solve the resulting equation for h1 in terms of h2. Then calculate h by taking the difference between h2 and h1.

Solution: 1. Set the pressures at A and B equal and solve for h1:

atm water 1 atm oil 2

oil1 2

water

P g h P g h

h h

ρ ρ

ρρ

+ = +

⎛ ⎞= ⎜ ⎟⎝ ⎠

2. Calculate the difference in depths: oil oil

2 1 2 2 2water water

1h h h h h hρ ρρ ρ

⎛ ⎞ ⎛ ⎞= − = − = −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠


3

3

920 kg/m1 7.5 cm 0.60 cm1000 kg/m

h⎛ ⎞

= − =⎜ ⎟⎝ ⎠

Insight: The height difference is linearly proportional to the height of the oil column. 120. Picture the Problem: The figure shows water flowing out of a container

filled to a depth of h. The water stream drops a distance H while traveling a horizontal distance D.

Strategy: This problem combines Torricelli’s Law and kinematics. Use Torricelli’s Law to write the velocity of the water in terms of the depth h. Then use the kinematic equations (equation 4-7) to solve for D in terms of H and h. Solve the resulting equation for H and h using the values of D given in the problem statement.

Solution: 1. Write the velocity using equation 15-17:

2v gh=

2. Solve equation 4-7 for the time to drop a distance H:

212 0

2 /

y H gt

t H g

= − =

=

3. Solve equation 4-7 for D: 2 2 / 2D vt gh H g Hh= = =

4. (a) Solve for H:

( )( )

22 0.655 m0.715 m

4 4 0.150 mDH

h= = =

5. (b) Solve for h:

( )( )

22 0.455 m0.104 m

4 4 0.500 mDhH

= = =

Insight: Increasing h or H by the same factor has the same result on the displacement. That is, doubling either h or H will increase D by the same factor, 2.



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121. Picture the Problem: Our sketch is similar to the sketch in Example 15-9,

except that the depth of the water is twice as deep and the upper tub is twice as high as in Example 15-9.

Strategy: We wish to calculate the distance D2 that the water falls as compared with D in Example 15-9. The insight to Example 15-9 shows that the distance D can be written as 2D hH= . Double the height H and depth h to determine the increase in distance.

Solution: Write the new distance in terms of the distance D:

( )( )

( )2 2 2 2

2 2 2

D h H

hH D

=

= =

The distance is increased by a factor of 2.00.

Insight: Doubling both the water depth and the height of fall will double the horizontal range of the water.



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Walker4 Ism Ch15