HW11 – fluids: in which we solve P4, P6, P9, P20, P29, P31, P35, P54, P64, P71. Chapter 14, problem 4: Three liquids that will not mix are poured into a cylindrical container. The volumes 3 3 3 and densities of the liquids are 0.50 L, 2.6 g/cm ; 0.25 L, 1.0 g/cm ; and 0.40 L, 0.80 g/cm . What is the force 3 on the bottom of the container due to these liquids? One liter = 1L = 1000 cm . (Ignore the contribution due to the atmosphere.)
We note that the container is cylindrical, c ylindrical, the important aspect of this being that it has a uniform cross-section (as viewed from above); this allows us to relate the pressure at the bottom simply to the total weight of the 3 liquids. Using the fact that 1L = 1000 cm , we find the weight o f the first liquid to be, W1 m1 g 1V1 g ( 2.6 g / cm 3 )(0.50 L) L)(1000 cm 3 / L)(980 cm cm/s 2 ) 1.27 106 g cm/s2 12.7 N. N. (1.1)
In the last step, we have converted grams to kilograms and centimeters to meters. Similarly, for the second and the third liquids, we have
W2 m2 g 2V2 g (1.0 g/cm 3 )(0.25 L)(1000 cm cm3 L)( L)(980 cm cm s 2 ) 2.5 N and W3 m3 g 3V3 g (0. (0.80 g/cm /cm 3 )(0. (0.40 L)(10 (1000 cm3 / L)(980 cm cm/s2 ) 3.1 N.
The total force on the bottom of the container is therefore F therefore F = = W 1 + W 2 + W 3 = 18 N. Chapter 14, problem 6: You inflate the front tires on your car to 28 psi. Later, you measure your blood pressure, obtaining a reading of 120/80, the readings being in mm Hg. In metric countries (which is to say, most of the world), these pressures are customarily reported in kilopascals (kPa). In kilopascals, what are (a) your tire pressure and (b) your blood pressure?
Knowing the standard air pressure value in seve ral units allows us to set up a variety of conversion factors:
1.01 105 Pa 190 kPa . (a) P 28 lb/in. 2 14.7 lb/in 2
1.01 105 Pa 15.9 kPa , 7 6 0 m m H g
1.01 105 Pa 10.6 kPa. 7 6 0 m m H g
(b) (120 mmHg)
(80 mmHg )
Chapter 14, problem 9: Blood pressure in Argentinosaurus. (a) If this longnecked, gigantic sauropod had a head height of 21 m and a heart height of o f 9.0 m, what (hydrostatic) gauge pressure in its blood was required at the heart such that the blood pressure at the brain was 80 torr (just enough to perfuse the brain with blood)?
Assume the blood had a density of 1.06 103
kg m3
. (b) What was the blood pressure p ressure (in torr or mm Hg) at the feet?
The hydrostatic blood pressure is the gauge pressure in the column of blood between feet and brain. We calculate the gauge pressure using Eq. 14-7. (a) The gauge pressure at the heart of the Argentinosaurus the Argentinosaurus is is
pheart pbrain gh 80 torr (1.06 103 kg/m3 )(9.8 m/s2 )(21 m 9.0 m)
1 torr 133.33 Pa
1.0 10 torr. 3
(b) The gauge pressure at the feet of the Argentinosaurus is pfeet pbrain gh 80 torr (1.06 103 kg/m3 )(9.8 m/s2 )(21 m)
1torr 133.33 Pa
80 torr 1642 torr 1722 torr 1.7 103 torr. Chapter 14, problem 20: The L-shaped tank shown in Fig. 14-33 is filled with water and is open at the top. If d = 5.0 m, what is the force due to the water (a) on face A and (b) on face B?
(a) The force on face A of area A A due to the water pressure alone is 3
2 3 3 2 F A pA AA w ghA AA w g (2 d ) d 2 1.0 10 kg m 9.8 m s 5.0 m
2.5 106 N. Adding the contribution from the atmospheric pressure, F 0 = (1.0 10 Pa)(5.0 m) = 2.5 10 N, 5
2
6
we have F A F0 F A 2.5 10 6 N 2.5 106 N 5.0 106 N.
(b) The force on face B due to water pressure alone is
FW dF d PA P dA gy d ( d y) d g
3 d
2 d
y dy d g 12 ([3 d]2 [2 d]2 ) gd 3
5 3 5d 2 5 d w gd 3 1.0 103 kg m3 9.8 m s2 5.0 m 2 2 2
F B pavgB AB g
3.1 106 N. Adding the contribution from the atmospheric pressure, F 0 = (1.0 10 Pa)(5.0 m) = 2.5 10 N, 5
2
6
we obtain F B F0 F B 2.5 10 6 N 3.1 106 N 5.6 106 N.
94 2
•• chapter 14, problem 29 (||28): In Fig. 14-37, a
compress the spring by 5.00 cm?
spring of spring constant 3.00 10 4 N is between a m rigid beam and the output piston of a hydraulic lever. An empty container with negligible mass sits on the input piston. The input piston has area A ,i and the output piston has area 18.0*Ai. Initially the spring is at its rest length. How many kilograms of sand must be (slowly) poured into the container to Pascal’s principle says the fluid presses with the same pressure at all walls of its con tainer. Equating pressures,
p
Fi Ai
Fo Ao
Fo Ai Fi Ao Fi Fo
Ai Ao
mg kx
Ai 18.0 Ai
m
kx 18.0 g
(3.00 102
N cm
)(5.00cm)
18.0(9.81 sm2 )
8.50kg (1.2)
Chapter 14, problem 31: A block of wood floats in fresh water with two-thirds of its volume V submerged and in oil with 0.90V submerged. Find the density of (a) the wood and (b) the oil.
Let V be the volume of the block. Then, the submerged volume in water is V s 2V / 3 . Since the block is floating, by Archimedes’ principle the weight of the displaced water is equal to the weight of the block, that is, w V s = b V , where w is the density of water, and b is the density of the block. (a) We substitute V s = 2V /3 to obtain the density of the block: b = 2 w/3 = 2(1000 kg/m )/3 6.7 10 kg/m . 3
2
3
(b) Now, if o is the density of the oil, then Archimedes’ principle yields oVs bV . Since the volume submerged in oil is V s 0.90V , the density of the oil is V V 2 3 7.4 102 kg/m3 . (6.7 10 kg/m ) 0.90V V
o b
Chapter 14, problem 35: Three children, each of weight 356 N, make a log raft by lashing together logs of diameter 0.30 m and length 1.80 m. How many logs will be needed to keep them afloat in fresh water? Take the 3 density of the logs to be 800 kg/m .
The problem intends for the children to be completely above water. The total downward pull of gravity on the system is 3 356 N N wood gV where N is the (minimum) number of logs needed to keep them afloat and V is the volume of each log: V = (0.15 m) (1.80 m) = 0.13 m . 2
3
The buoyant force is F b = water gV submerged, where we require V submerged NV . The density of water is 1000 kg/m . To obtain the minimum value of N , we set V submerged = NV and then round our “answer” for N up to the nearest integer: 3
3 356 N N wood gV water gNV N
3 356 N gV water wood
which yields N = 4.28 5 logs. Chapter 14, problem 54: The water flowing through a 1.9 cm (inside diameter) pipe flows out through three 1.3 cm pipes. (a) If the flow rates in the three smaller pipes are 26, 19, and 11 L/min, what is the flow rate in the 1.9 cm pipe? (b) What is the ratio of the speed in the 1.9 cm pipe to that in the pipe carrying 26 L/min?
(a) The equation of continuity provides (26 + 19 + 11) L/min = 56 L/min for the flow rate in the main (1.9 cm diameter) pipe. (b) Using v = R/ A and A = d /4, we set up ratios: 2
v56 v26
56 / (1.9)2 / 4 26 / (1.3) 2 / 4
Chapter 14, problem 64: In Fig. 14-49, water flows through a horizontal pipe and then out into the atmosphere at a speed v1 = 15 m/s. The diameters of the left and right sections of the pipe are 5.0 cm and 3.0 cm. (a) What volume of water flows into the atmosphere during a 10 min period?
1.0.
In the left section of the pipe, what are (b) the speed v2 and (c) the gauge pressure?
(a) The volume of water (during 10 minutes) is, 2 3 0.03m 6.4 m . 4
V v1t A1 15 m s 10 min 60s min
(b) The speed in the left section of pipe is 2
2
A d 3.0cm v2 v1 1 v1 1 15 m s 5.4 m s. A d 5.0cm 2 2 (c) Since p1 12 v12 gh1 p2 12 v22 gh2
and h1 h2 , p1 p0 , which is the atmospheric pressure, 1 1 2 2 p2 p0 v12 v22 1.01 105 Pa 1.0 103 kg m3 15 m s 5.4 m s 2 2
1.99 105 Pa 1.97 atm. Thus, the gauge pressure is (1.97 atm – 1.00 atm) = 0.97 atm = 9.8 10 Pa. 4
(1.3)
Chapter 14, problem 71: Figure 14-54 shows a stream of water flowing through a hole at depth h = 10 cm in a tank holding water to height H = 40 cm. (a) At what distance x does the stream strike the floor? (b) At what depth should a second hole be made to give the same value of x? (c) At what depth should a hole be made to maximize x?
(a) The stream of water emerges horizontally ( 0 = 0° in the notation of Chapter 4) with v0 2 gh . Setting y – y0 = –( H – h) in Eq. 4-22, we obtain the “time-of-flight” t
2 2( H h) ( H h). g g
Using this in Eq. 4-21, where x0 = 0 by choice of coordinate origin, we find x v0t 2 gh
2( H h) g
2 h( H h) 2 (10 cm)(40 cm 10 cm) 35 cm. 2
(b) The result of part (a) (which, when squared, reads x = 4h( H – h)) is a quadratic equation for h once x and H are specified. Two solutions for h are therefore mathematically possible, but are the y both physically possible? For instance, are both solutions positive and less than H ? We employ the quadratic formula: h Hh 2
x 2 4
0h
H H 2 x2 2
which permits us to see that both roots are physically possible, so long as x H . Labeling the larger root h1 (where the plus sign is chosen) and the smaller root as h2 (where the minus sign is chosen), then we note that their sum is simply h1 h2
H H 2 x 2 2
H H 2 x2 2
H .
Thus, one root is related to the other (generically labeled h' and h) by h' = H – h. Its numerical value is h 40cm 10 cm 30 cm. 2
(c) We wish to maximize the function f = x = 4h( H – h). We differentiate with respect to h and set equal to zero to obtain df H 4 H 8h 0 h dh 2 or h = (40 cm)/2 = 20 cm, as the depth from which an emerging stream of water will travel the maximum horizontal distance.
