CHAPTER 11 FLUIDS
CONCEPTUAL QUESTIONS ______
5. REASONING AND SOLUTION The bottle of juice is sealed under a partial vacuum. Therefore, when the seal is intact, the button remains depressed, because the pressure inside the bottle is less than the atmospheric pressure outside of the bottle. The force per unit area pushing up on the button from the inside is significantly smaller than the force per unit area pushing down on the outside. When the seal is broken, air rushes inside the bottle. The force pushing up on the button increases and offsets the force pushing down. The natural springiness of the material from which the lid is made can then make the button "pop up." ______
8. REASONING AND SOLUTION When you drink through a straw, you draw the air out of the straw, and the external air pressure leads to the unbalanced force that pushes the liquid up into the straw. This action requires the presence of an atmosphere. The moon has no atmosphere, so you could not use a straw to sip a drink on the moon. ______
10. REASONING AND SOLUTION A scuba diver is below the surface of the water when a storm approaches, dropping the air pressure above the water. According to Equation 11.4, the absolute pressure at the location of the scuba diver is P P gh diver 0 where P0 is atmospheric pressure, is the density of the water, and h is the depth of the diver below the surface. Since the storm causes the atmospheric pressure to drop, the absolute water pressure at the location of the diver will also drop. If the diver were wearing a sensitive gauge designed to measure the absolute pressure, it would register the drop in pressure. ______
12. REASONING AND SOLUTION Archimedes' principle states that any fluid applies a buoyant force to an object that is partially or completely immersed in it; the magnitude of the buoyant force equals the weight of the fluid that the object displaces. A glass beaker, filled to the brim with water, is resting on a scale. A block is placed in the water, causing some of it to spill over. The water that spills is wiped away, and the beaker is still filled to the brim.
a. According to Table 11.1, the density of wood is 550 kg/m3, while the density of water at 4 °C is 1 . 0 0 0 1 0 3 k g / m 3 . Since the density of the wood is less than the density of water, the block will float with part of its volume above the surface of the water. According to Archimedes' principle, the buoyant force exerted on B u o y a n t f o r c e the bottom surface of the block must be equal in magnitude to the o n t h e b l o c k weight of water that is displaced by the block. However, since the block floats on the surface of the water, the buoyant force must be equal in magnitude and opposite in direction to the weight of the W e i g h t o f block, as suggested in the free body diagram at the right. Hence, t h e b l o c k
the weight of the block must be equal to the weight of the water displaced by the block. The weight of the contents of the beaker remains the same. Therefore, the initial and final scale readings are the same if the block is made of wood.
b. According to Table 11.1, the density of iron is 7860 kg/m3, while the density of water at 4 °C is 1 . 0 0 0 1 0 3 k g / m 3 . Since the density of iron is greater than the density of water, the iron block will sink to the bottom of the beaker. Let V represent the volume of water that spills over. According to the definition of mass density (Equation 11.1) the mass of water that spills over is equal to m w a t e r w a t e r V . The beaker is still filled to the brim, but the block now occupies the volume V where the water used to be. According to Equation 11.1, the mass that now occupies the volume V is given by m b l o c k i r o n V . Since iron > water, the mass of the block is greater than the mass of the water that spilled over. The weight of the contents of the beaker increases. Therefore, the final reading on the scale will be greater than the initial reading. ______
22. REASONING AND SOLUTION Two sheets of paper are held by adjacent corners, so that they hang downward. A i r d i r e c t e d b e t w e e n t h e s h e e t s They are oriented so that they are parallel and slightly separated by a gap through which the floor can be seen. When air is blown strongly down through the gap (see the drawing at the right), the sheets come closer together. In accord with Bernoulli's equation, the moving air between the sheets has a lower pressure than the stationary air on either side of the sheets. The greater pressure on either side of the sheets generates an inward force on the outer surface of the sheets, and the sheets are pushed together.
PROBLEMS 2. REASONING AND SOLUTION a. We will treat the neutron star as spherical in shape, so that its volume is given by 4 3 the familiar formula, V 3 r . Then, according to Equation 11.1, the density of the neutron star described in the problem statement is m m 3m 3(2.7 1028 kg) 3.7 1018 kg / m 3 4 3 3 3 3 V 3 r 4r 4 (1.2 10 m)
b. If a dime of volume 2.0 10 –7 m 3 were made of this material, it would weigh
W mg Vg (3.7 1018 kg / m 3 )(2.0 10 –7 m 3 )(9.80 m / s2 ) = 7.31012 N
This weight corresponds to
12 F 1lb I 12 7.3 10 N 1.6 10 lb . HG4.448 N KJ ______
11. REASONING AND SOLUTION 2 a. We know that FT = PAT = PR , so that
5 –2 2 2 FT = (1.40 10 Pa)(3.20 10 m) = 4 . 5 0 1 0 N
5 –4 2 5 3 . 2 N b. Fp = (1.40 10 Pa)(3.80 10 m ) = ______
15. REASONING AND SOLUTION The least number of bricks are used when the surface area is the smallest. The area of the brick in contact with the ground has a surface area of
–3 2 A = (0.0570 m)(0.0890 m) = 5.07 10 m
The pressure exerted by a single brick is
–3 2 3 2 P = F/A = (17.8 N)/(5.07 10 m ) = 3.51 10 N/m
In order to equal or exceed atmospheric pressure
5 2 3 2 Number of bricks = (1.013 10 N/m )/(3.51 10 N/m ) = 2 9 b r i c k s ______21. REASONING The magnitude of the force that would be exerted on the window is given by Equation 11.3, F PA , where the pressure can be found from Equation
11.4: P2 P1 gh . Since P1 represents the pressure at the surface of the water, it
is equal to atmospheric pressure, Patm . Therefore, the magnitude of the force is given by
F ( Patm gh) A
where, if we assume that the window is circular with radius r, its area A is given by A r 2 .
SOLUTION a. Thus, the magnitude of the force is
F 1.013105 Pa + (1025 kg / m 3 )(9.80 m / s2 )(11 000 m) (0.10 m) 2 3.5106 N
b. The weight of a jetliner whose mass is 1.2 105 kg is W mg (1.2 105 kg)(9.80 m / s2 ) 1.2 106 N
Therefore, the force exerted on the window at a depth of 11 000 m is about three times greater than the weight of a jetliner! ______
22. REASONING As the depth h increases, the pressure increases according to Equation 11.4 (P2 = P1 + gh). In this equation, P1 is the pressure at the shallow end, P2 is the pressure at the deep end, and is the density of water 3 3 (1.00 10 kg/m , see Table 11.1). We seek a value for the pressure at the deep end minus the pressure at the shallow end.
SOLUTION Using Equation 11.4, we find
PDeep PShallow gh or PDeep PShallow gh
The drawing at the right shows that a value for h can 15 m be obtained from the 15-m length of the pool by using the tangent of the 11 angle: h 11 h tan11 or h 15 m tan11 15 m b g
PDeep PShallow gb15 mgtan11 c1.00 103 kg / m 3 hc9.80 m / s2 hb15 mgtan11 2.9 104 Pa ______
26. REASONING AND SOLUTION The difference in pressure between the ground and the roof is P = gh, so 133 Pa 13.0 mm Hg F I b gG1 mm Hg J h H K 137 m c1.29 kg / m 3 hc9.80 m / s2 h
Note that we have used the fact that 133 Pa = 1 mm Hg as a conversion factor.
______
36. REASONING Since the piston and the plunger are at the same height, Equation
11.5, F2 F1 ( A2 / A1 ) , applies, and we can find an expression for the force exerted on the spring. Then Equation 10.1, F = kx, can be used to determine the amount of compression of the spring.
SOLUTION Substituting the right hand side of Equation 11.5 into Equation 10.1, we find that FA I F 2 kx 1 GA J H1 K
From the drawing in the text, we see that the force on the right piston must be equal
in magnitude to the weight of the rock, or F1 mg . Therefore,
FA I mg 2 kx GA J H1 K Solving for x, we obtain
mg FA I (40.0 kg)(9.80 m / s2 ) 15 cm2 x 2 F I 5.7 10 –2 m k GA J 1600 N / m G 2 J H1 K H65 cm K ______44. REASONING AND SOLUTION The buoyant force exerted by the water must at least equal the weight of the logs plus the weight of the people,
FB = WL + WP
wgV = LgV + WP
Now the volume of logs needed is M P 3 2 0 k g 3 V 3 3 3 1 . 1 6 m W L 1 . 0 0 1 0 k g / m 7 2 5 k g / m
The volume of one log is
–2 2 –2 3 VL = (8.00 10 m) (3.00 m) = 6.03 10 m
The number of logs needed is
–2 N = V/VL = (1.16)/(6.03 10 ) = 19.2
Therefore, a t l e a s t 2 0 l o g s a r e n e e d e d ______
47. SSM REASONING The height of the cylinder that is in the oil is given by 2 V hoil Voil / (r ) , where oil is the volume of oil displaced by the cylinder and r is the radius of the cylinder. We must, therefore, find the volume of oil displaced by the cylinder. After the oil is poured in, the buoyant force that acts on the cylinder is equal to the sum of the weight of the water displaced by the cylinder and the weight of the oil displaced by the cylinder. Therefore, the magnitude of the buoyant force
is given by F water gVwater oil gVoil . Since the cylinder floats in the fluid, the net force that acts on the cylinder must be zero. Therefore, the buoyant force that supports the cylinder must be equal to the weight of the cylinder, or
water gVwater oil gVoil mg
where m is the mass of the cylinder. Substituting values into the expression above leads to
–3 3 Vwater (0.725)Voil 7.00 10 m (1) From the figure in the text, Vcylinder Vwater Voil . Substituting values into the
expression for Vcylinder gives –3 3 Vwater Voil 8.48 10 m (2)
–3 3 Subtracting Equation (1) from Equation (2) yields Voil 5.38 10 m .
SOLUTION The height of the cylinder that is in the oil is, therefore,
V 5.38 10 –3 m 3 h oil 7.6 10 –2 m oil r 2 (0.150 m) 2 ______
53. REASONING AND SOLUTION a. The volume flow rate is given by –4 2 – 5 3 Q = Av = (2.0 10 m )(0.35 m/s) = 7 . 0 1 0 m / s
b. We know that A1v1 = A2v2 so that –4 2 2 – 4 v2 = v1(A1/A2) = (0.35 m/s)(2.0 10 m )/(0.28 m ) = 2 . 5 1 0 m / s ______
59. SSM REASONING AND SOLUTION Let the speed of the air below and
above the wing be given by v1 and v2 , respectively. According to Equation 11.12,
the form of Bernoulli's equation with y1 y2 , we have
1.29 kg / m 3 P P 1 v 2 v 2 (251 m / s) 2 (225 m / s) 2 7.98 10 3 Pa 1 2 2 c2 1 h 2
From Equation 11.3, the lifting force is, therefore,
F P P A (7.98 103 Pa)(24.0 m2 ) 1.92 105 N c1 2 h ______
61. REASONING AND SOLUTION a. Using Bernoulli's equation we have
P1 = P2 + gh 5 3 3 2 = (1.01 10 Pa) + (1.00 10 kg/m )(9.80 m/s )(15.0 m) 2 . 4 8 1 0 5 P a
b. The pressure becomes equal to atmospheric pressure when the valve is opened, so 5 P 1 1 . 0 1 1 0 P a .
c. The volume flow rate is given by Q = Av. We can find the speed v by using 2 Bernoulli’s equation, P2 + gh = P1 + (1/2)v , with P2 = P1. The result is
v 2gh 2c9.80 m / s2 hb15.0 mg 17.1 m / s
–2 2 3 The volume flow rate is Q = Av = (2.00 10 m )(17.1 m/s) = 0 . 3 4 2 m / s ______