Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.

Chapter 11: Fluid Statics

47 problems · 6 figures

11-1 §11.2 source ↗

Gold is sold by the troy ounce (31.103 g). What is the volume of 1 troy ounce of pure gold?

Answer

1 . 610 cm 3

11-3 §11.2 source ↗

(a) What is the mass of a deep breath of air having a volume of 2.00 L? (b) Discuss the effect taking such a breath has on your body’s volume and density.

Answer

(a) 2.58 g

(b) The volume of your body increases by the volume of air you inhale. The average density of your body decreases when you take a deep breath, because the density of air is substantially smaller than the average density of the body before you took the deep breath.

11-4 §11.2 source ↗

A straightforward method of finding the density of an object is to measure its mass and then measure its volume by submerging it in a graduated cylinder. What is the density of a 240-g rock that displaces 89.0cm3 of water? (Note that the accuracy and practical applications of this technique are more limited than a variety of others that are based on Archimedes’ principle.)

Answer

2 . 70 g/cm 3

11-6 §11.2 source ↗

(a) A rectangular gasoline tank can hold 50.0 kg of gasoline when full. What is the depth of the tank if it is 0.500-m wide by 0.900-m long? (b) Discuss whether this gas tank has a reasonable volume for a passenger car.

Answer

(a) 0.163 m

(b) Equivalent to 19.4 gallons, which is reasonable

11-8 §11.2 source ↗

A 2.50-kg steel gasoline can holds 20.0 L of gasoline when full. What is the average density of the full gas can, taking into account the volume occupied by steel as well as by gasoline?

Answer

7 . 9 × 10 2 kg/m 3

11-9 §11.2 source ↗

What is the density of 18.0-karat gold that is a mixture of 18 parts gold, 5 parts silver, and 1 part copper? (These values are parts by mass, not volume.) Assume that this is a simple mixture having an average density equal to the weighted densities of its constituents.

Answer

15.8 g/cm 3

11-10 §11.2 source ↗

There is relatively little empty space between atoms in solids and liquids, so that the average density of an atom is about the same as matter on a macroscopic scale—approximately 103kg/m3. The nucleus of an atom has a radius about 10−5 that of the atom and contains nearly all the mass of the entire atom. (a) What is the approximate density of a nucleus? (b) One remnant of a supernova, called a neutron star, can have the density of a nucleus. What would be the radius of a neutron star with a mass 10 times that of our Sun (the radius of the Sun is 7×108m)?

Answer

(a) 1018kg/m3

(b) 2×104m

11-11 §11.3 source ↗

As a woman walks, her entire weight is momentarily placed on one heel of her high-heeled shoes. Calculate the pressure exerted on the floor by the heel if it has an area of 1.50cm2 and the woman’s mass is 55.0 kg. Express the pressure in Pa. (In the early days of commercial flight, women were not allowed to wear high-heeled shoes because aircraft floors were too thin to withstand such large pressures.)

Answer

3.59×106Pa; or 521 lb/in 2

11-13 §11.3 source ↗

Nail tips exert tremendous pressures when they are hit by hammers because they exert a large force over a small area. What force must be exerted on a nail with a circular tip of 1.00 mm diameter to create a pressure of 3.00×109N/m2?(This high pressure is possible because the hammer striking the nail is brought to rest in such a short distance.)

Answer

2.36 × 10 3 N

11-14 §11.4 source ↗

What depth of mercury creates a pressure of 1.00 atm?

Answer

0.760 m

11-16 §11.4 source ↗

Verify that the SI unit of hρg is N/m2.

Answer

hρ g units = ( m ) kg/m 3 m/s 2 = kg ⋅ m 2 / m 3 ⋅ s 2 = kg ⋅ m/s 2 1/m 2 = N/m 2

11-18 §11.4 source ↗

The aqueous humor in a person’s eye is exerting a force of 0.300 N on the 1.10-cm2 area of the cornea. (a) What pressure is this in mm Hg? (b) Is this value within the normal range for pressures in the eye?

Answer

(a) 20.5 mm Hg

(b) The range of pressures in the eye is 12–24 mm Hg, so the result in part (a) is within that range

11-20 §11.4 source ↗

What pressure is exerted on the bottom of a 0.500-m-wide by 0.900-m-long gas tank that can hold 50.0 kg of gasoline by the weight of the gasoline in it when it is full?

Answer

1 . 09 × 10 3 N/m 2

11-22 §11.4 source ↗

The left side of the heart creates a pressure of 120 mm Hg by exerting a force directly on the blood over an effective area of 15.0 cm2. What force does it exert to accomplish this?

Answer

24.0 N

11-24 §11.5 source ↗

How much pressure is transmitted in the hydraulic system considered in Example 11.6? Express your answer in pascals and in atmospheres.
Example 11.6 Calculating Force of Wheel Cylinders: Pascal Puts on the Brakes in the book ↗

Answer

2.55×107Pa; or 251 atm

11-26 §11.5 source ↗

A crass host pours the remnants of several bottles of wine into a jug after a party. He then inserts a cork with a 2.00-cm diameter into the bottle, placing it in direct contact with the wine. He is amazed when he pounds the cork into place and the bottom of the jug (with a 14.0-cm diameter) breaks away. Calculate the extra force exerted against the bottom if he pounded the cork with a 120-N force.

Answer

5.76×103N extra force

11-28 §11.5 source ↗

(a) Verify that work input equals work output for a hydraulic system assuming no losses to friction. Do this by showing that the distance the output force moves is reduced by the same factor that the output force is increased. Assume the volume of the fluid is constant. (b) What effect would friction within the fluid and between components in the system have on the output force? How would this depend on whether or not the fluid is moving?

Answer

(a) V= d i A i = d o A o ⇒ d o = d i A i A o .

Now, using equation:

F 1 A 1 = F 2 A 2 ⇒ F o = F i A o A i .

Finally,

W o = F o d o = F i A o A i d i A i A o = F i d i = W i .

In other words, the work output equals the work input.

(b) If the system is not moving, friction would not play a role. With friction, we know there are losses, so that Wout=Win−Wf; therefore, the work output is less than the work input. In other words, with friction, you need to push harder on the input piston than was calculated for the nonfriction case.

11-29 §11.6 source ↗

Find the gauge and absolute pressures in the balloon and peanut jar shown in Figure 11.14, assuming the manometer connected to the balloon uses water whereas the manometer connected to the jar contains mercury. Express in units of centimeters of water for the balloon and millimeters of mercury for the jar, taking h=0.0500 m for each.
Figure 11.14 in the book ↗
An open-tube manometer has one side open to the atmosphere. (a) Fluid depth must be the same on both sides, or the pressure each side exerts at the bottom will be unequal and there will be flow from the deeper side. (b) A positive gauge pressure Pg=hρg transmitted to one side of the manometer can support a column of fluid of height h. (c) Similarly, atmospheric pressure is greater than a negative gauge pressure Pg by an amount hρg. The jar’s rigidity prevents atmospheric pressure from being transmitted to the peanuts.

Answer

Balloon:

P g = 5.00 cm H 2 O, P abs = 1.035 × 10 3 cm H 2 O.

Jar:

P g = − 50.0 mm Hg , P abs = 710 mm Hg.

11-31 §11.6 source ↗

How tall must a water-filled manometer be to measure blood pressures as high as 300 mm Hg?

Answer

4.08 m

11-33 §11.6 source ↗

Suppose you measure a standing person’s blood pressure by placing the cuff on his leg 0.500 m below the heart. Calculate the pressure you would observe (in units of mm Hg) if the pressure at the heart were 120 over 80 mm Hg. Assume that there is no loss of pressure due to resistance in the circulatory system (a reasonable assumption, since major arteries are large).

Answer

ΔP = 38.7 mm Hg, Leg blood pressure = 159 119 .

11-35 §11.6 source ↗

Assuming bicycle tires are perfectly flexible and support the weight of bicycle and rider by pressure alone, calculate the total area of the tires in contact with the ground. The bicycle plus rider has a mass of 80.0 kg, and the gauge pressure in the tires is 3.50×105Pa.

Answer

22 . 4 cm 2

11-36 §11.7 source ↗

What fraction of ice is submerged when it floats in freshwater, given the density of water at 0°C is very close to 1000 kg/m3?

Answer

91 . 7%

11-38 §11.7 source ↗

Find the density of a fluid in which a hydrometer having a density of 0.750 g/mL floats with 92.0% of its volume submerged.

Answer

815 kg /m 3

11-40 §11.7 source ↗

Bird bones have air pockets in them to reduce their weight—this also gives them an average density significantly less than that of the bones of other animals. Suppose an ornithologist weighs a bird bone in air and in water and finds its mass is 45.0 g and its apparent mass when submerged is 3.60 g (the bone is watertight). (a) What mass of water is displaced? (b) What is the volume of the bone? (c) What is its average density?

Answer

(a) 41.4 g

(b) 41.4cm3

(c) 1.09 g/cm3

11-42 §11.7 source ↗

Archimedes’ principle can be used to calculate the density of a fluid as well as that of a solid. Suppose a chunk of iron with a mass of 390.0 g in air is found to have an apparent mass of 350.5 g when completely submerged in an unknown liquid. (a) What mass of fluid does the iron displace? (b) What is the volume of iron, using its density as given in Table 11.1 (c) Calculate the fluid’s density and identify it.
Table 11.1 Densities of Various Substances in the book ↗
Substance ρ (× 10 3 kg/m 3 or g/mL ) Substance ρ ( 10 3 kg/m 3 or g/mL ) Substance ρ ( 10 3 kg/m 3 or g/mL )
Solids Liquids Gases
Aluminum 2.7 Water (4ºC) 1.000 Air 1 . 29 × 10 − 3
Brass 8.44 Blood 1.05 Carbon dioxide 1 . 98 × 10 − 3
Copper (average) 8.8 Sea water 1.025 Carbon monoxide 1 . 25 × 10 − 3
Gold 19.32 Mercury 13.6 Hydrogen 0 . 090 × 10 − 3
Iron or steel 7.8 Ethyl alcohol 0.79 Helium 0 . 18 × 10 − 3
Lead 11.3 Gasoline 0.68 Methane 0 . 72 × 10 − 3
Polystyrene 0.10 Glycerin 1.26 Nitrogen 1 . 25 × 10 − 3
Tungsten 19.30 Olive oil 0.92 Nitrous oxide 1 . 98 × 10 − 3
Uranium 18.70 Oxygen 1 . 43 × 10 − 3
Concrete 2.30–3.0 Steam 100º C 0 . 60 × 10 − 3
Cork 0.24
Glass, common (average) 2.6
Granite 2.7
Earth’s crust 3.3
Wood 0.3–0.9
Ice (0°C) 0.917
Bone 1.7–2.0
Silver 10.49

Answer

(a) 39.5 g

(b) 50cm3

(c) 0.79g/cm3

It is ethyl alcohol.

11-44 §11.7 source ↗

Some fish have a density slightly less than that of water and must exert a force (swim) to stay submerged. What force must an 85.0-kg grouper exert to stay submerged in salt water if its body density is 1015kg/m3?

Answer

8.21 N

11-46 §11.7 source ↗

(a) What is the density of a woman who floats in freshwater with 4.00% of her volume above the surface? This could be measured by placing her in a tank with marks on the side to measure how much water she displaces when floating and when held under water (briefly). (b) What percent of her volume is above the surface when she floats in seawater?

Answer

(a) 960kg/m3

(b) 6.34%

She indeed floats more in seawater.

11-48 §11.7 source ↗

A simple compass can be made by placing a small bar magnet on a cork floating in water. (a) What fraction of a plain cork will be submerged when floating in water? (b) If the cork has a mass of 10.0 g and a 20.0-g magnet is placed on it, what fraction of the cork will be submerged? (c) Will the bar magnet and cork float in ethyl alcohol?

Answer

(a) 0.24

(b) 0.68

(c) Yes, the cork will float because ρobj< ρethyl alcohol(0.678g/cm3<0.79g/cm3)

11-50 §11.7 source ↗

Scurrilous con artists have been known to represent gold-plated tungsten ingots as pure gold and sell them at prices much below gold value but deservedly far above the cost of tungsten. With what accuracy must you be able to measure the mass of such an ingot in and out of water to tell that it is almost pure tungsten rather than pure gold?

Answer

The difference is 0.006%.

11-52 §11.7 source ↗

Referring to Figure 11.19, prove that the buoyant force on the cylinder is equal to the weight of the fluid displaced (Archimedes’ principle). You may assume that the buoyant force is F2−F1 and that the ends of the cylinder have equal areas A. Note that the volume of the cylinder (and that of the fluid it displaces) equals (h2−h1)A.
Figure 11.19 in the book ↗
(a) An object submerged in a fluid experiences a buoyant force FB. If FB is greater than the weight of the object, the object will rise. If FB is less than the weight of the object, the object will sink. (b) If the object is removed, it is replaced by fluid having weight wfl. Since this weight is supported by surrounding fluid, the buoyant force must equal the weight of the fluid displaced. That is, FB=wfl,a statement of Archimedes’ principle.

Answer

F net = F 2 − F 1 = P 2 A− P 1 A= P 2 − P 1 A

= h 2 ρ fl g − h 1 ρ fl g A

= h 2 − h 1 ρ fl gA

where ρfl = density of fluid. Therefore,

F net = ( h 2 − h 1 ) Aρ fl g= V fl ρ fl g= m fl g= w fl

where is wfl the weight of the fluid displaced.

11-54 §11.8 source ↗

What is the pressure inside an alveolus having a radius of 2.50×10−4m if the surface tension of the fluid-lined wall is the same as for soapy water? You may assume the pressure is the same as that created by a spherical bubble.

Answer

592 N/m 2

11-56 §11.8 source ↗

What is the gauge pressure in millimeters of mercury inside a soap bubble 0.100 m in diameter?

Answer

2 . 23 × 10 − 2 mm Hg

11-58 §11.8 source ↗

Figure 11.32(a) shows the effect of tube radius on the height to which capillary action can raise a fluid. (a) Calculate the height h for water in a glass tube with a radius of 0.900 cm—a rather large tube like the one on the left. (b) What is the radius of the glass tube on the right if it raises water to 4.00 cm?
Figure 11.32 in the book ↗
(a) Capillary action depends on the radius of a tube. The smaller the tube, the greater the height reached. The height is negligible for large-radius tubes. (b) A denser fluid in the same tube rises to a smaller height, all other factors being the same.

Answer

(a) 1.65×10−3m

(b) 3.71×10–4m

11-60 §11.8 source ↗

What fluid is in the device shown in Figure 11.26 if the force is 3.16×10−3N and the length of the wire is 2.50 cm? Calculate the surface tension γ and find a likely match from Table 11.3.
Figure 11.26 in the book ↗
Sliding wire device used for measuring surface tension; the device exerts a force to reduce the film’s surface area. The force needed to hold the wire in place is F=γL=γ(2l), since there are two liquid surfaces attached to the wire. This force remains nearly constant as the film is stretched, until the film approaches its breaking point.
Table 11.3 Surface Tension of Some Liquids1 in the book ↗
Liquid Surface tension γ(N/m)
Water at 0ºC 0.0756
Water at 20ºC 0.0728
Water at 100ºC 0.0589
Soapy water (typical) 0.0370
Ethyl alcohol 0.0223
Glycerin 0.0631
Mercury 0.465
Olive oil 0.032
Tissue fluids (typical) 0.050
Blood, whole at 37ºC 0.058
Blood plasma at 37ºC 0.073
Gold at 1070ºC 1.000
Oxygen at −193ºC 0.0157
Helium at −269ºC 0.00012

Answer

6 . 32 × 10 − 2 N/m

Based on the values in table, the fluid is probably glycerin.

11-62 §11.8 source ↗

Calculate the gauge pressures inside 2.00-cm-radius bubbles of water, alcohol, and soapy water. Which liquid forms the most stable bubbles, neglecting any effects of evaporation?

Answer

P w = 14 . 6 N/m 2 , P a = 4.46 N/m 2 , P sw = 7.40 N/m 2 .

Alcohol forms the most stable bubble, since the absolute pressure inside is closest to atmospheric pressure.

11-64 §11.8 source ↗

Calculate the contact angle θ for olive oil if capillary action raises it to a height of 7.07 cm in a glass tube with a radius of 0.100 mm. Is this value consistent with that for most organic liquids?

Answer

5.1º

This is near the value of θ=0º for most organic liquids.

11-66 §11.8 source ↗

Calculate the ratio of the heights to which water and mercury are raised by capillary action in the same glass tube.

Answer

− 2 . 78

The ratio is negative because water is raised whereas mercury is lowered.

11-68 §11.9 source ↗

During forced exhalation, such as when blowing up a balloon, the diaphragm and chest muscles create a pressure of 60.0 mm Hg between the lungs and chest wall. What force in newtons does this pressure create on the 600cm2 surface area of the diaphragm?

Answer

479 N

11-70 §11.9 source ↗

One way to force air into an unconscious person’s lungs is to squeeze on a balloon appropriately connected to the subject. What force must you exert on the balloon with your hands to create a gauge pressure of 4.00 cm water, assuming you squeeze on an effective area of 50.0cm2?

Answer

1.96 N

11-71 §11.9 source ↗

Heroes in movies hide beneath water and breathe through a hollow reed (villains never catch on to this trick). In practice, you cannot inhale in this manner if your lungs are more than 60.0 cm below the surface. What is the maximum negative gauge pressure you can create in your lungs on dry land, assuming you can achieve −3.00 cm water pressure with your lungs 60.0 cm below the surface?

Answer

− 63.0 cm H 2 O

11-73 §11.9 source ↗

A full-term fetus typically has a mass of 3.50 kg. (a) What pressure does the weight of such a fetus create if it rests on the mother’s bladder, supported on an area of 90.0cm2? (b) Convert this pressure to millimeters of mercury and determine if it alone is great enough to trigger the micturition reflex (it will add to any pressure already existing in the bladder).

Answer

(a) 3.81×103N/m2

(b) 28.7 mm Hg, which is sufficient to trigger micturition reflex

11-75 §11.9 source ↗

Pressure in the spinal fluid is measured as shown in Figure 11.40. If the pressure in the spinal fluid is 10.0 mm Hg: (a) What is the reading of the water manometer in cm water? (b) What is the reading if the person sits up, placing the top of the fluid 60 cm above the tap? The fluid density is 1.05 g/mL.
Figure 11.40 in the book ↗
A water manometer used to measure pressure in the spinal fluid. The height of the fluid in the manometer is measured relative to the spinal column, and the manometer is open to the atmosphere. The measured pressure will be considerably greater if the person sits up.

Answer

(a) 13.6 m water

(b) 76.5 cm water

11-77 §11.9 source ↗

During heavy lifting, a disk between spinal vertebrae is subjected to a 5000-N compressional force. (a) What pressure is created, assuming that the disk has a uniform circular cross section 2.00 cm in radius? (b) What deformation is produced if the disk is 0.800 cm thick and has a Young’s modulus of 1.5×109N/m2?

Answer

(a) 3.98×106Pa

(b) 2.1×10−3cm

11-79 §11.9 source ↗

(a) How high will water rise in a glass capillary tube with a 0.500-mm radius? (b) How much gravitational potential energy does the water gain? (c) Discuss possible sources of this energy.

Answer

(a) 2.97 cm

(b) 3.39×10−6J

(c) Work is done by the surface tension force through an effective distance h/2 to raise the column of water.

11-81 §11.9 source ↗

Suppose you hit a steel nail with a 0.500-kg hammer, initially moving at 15.0 m/s and brought to rest in 2.80 mm. (a) What average force is exerted on the nail? (b) How much is the nail compressed if it is 2.50 mm in diameter and 6.00-cm long? (c) What pressure is created on the 1.00-mm-diameter tip of the nail?

Answer

(a) 2.01×104N

(b) 1.17×10−3m

(c) 2.56×1010N/m2

11-83 §11.9 source ↗

The hydraulic system of a backhoe is used to lift a load as shown in Figure 11.42. (a) Calculate the force F the secondary cylinder must exert to support the 400-kg load and the 150-kg brace and shovel. (b) What is the pressure in the hydraulic fluid if the secondary cylinder is 2.50 cm in diameter? (c) What force would you have to exert on a lever with a mechanical advantage of 5.00 acting on a primary cylinder 0.800 cm in diameter to create this pressure?
Figure 11.42 in the book ↗
Hydraulic and mechanical lever systems are used in heavy machinery such as this back hoe.

Answer

(a) 1.38×104N

(b) 2.81×107N/m2

(c) 283 N

11-85 §11.9 source ↗

You are pumping up a bicycle tire with a hand pump, the piston of which has a 2.00-cm radius.

(a) What force in newtons must you exert to create a pressure of 6.90×105Pa (b) What is unreasonable about this (a) result? (c) Which premises are unreasonable or inconsistent?

Answer

(a) 867 N

(b) This is too much force to exert with a hand pump.

(c) The assumed radius of the pump is too large; it would be nearly two inches in diameter—too large for a pump or even a master cylinder. The pressure is reasonable for bicycle tires.

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Problems and answers are adapted from College Physics 2e by Paul Peter Urone, Roger Hinrichs, Kim Dirks and Manjula Sharma, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/college-physics-2e/pages/1-introduction-to-science-and-the-realm-of-physics-physical-quantities-and-units. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.