15.1. Solve: The density of the liquid is ρ=
m 0.120 kg 0.120 kg = = = 1200 kg m 3 V 100 mL 100 × 10 −3 × 10 −3 m 3
Assess: The liquid’s density is more than that of water (1000 kg/m3) and is a reasonable number.
15.2. Solve: The volume of the helium gas in container A is equal to the volume of the liquid in container B. That is, VA = VB. Using the definition of mass density ρ = m V , the above relationship becomes
m 7000 mHe mA mB = ⇒ He = ⇒ ρ B = (7000)ρ He = (7000)(0.18 kg m 3 ) = 1260 kg m 3 ρA ρB ρ He ρB Referring to Table 15.1, we find that the liquid is glycerine.
15.3. Model: The density of water is 1000 kg/m3. Visualize:
Solve:
Volume of water in the swimming pool is
V = 6 m × 12 m × 3 m −
1 2
(6 m × 12 m × 2 m) = 144 m 3
The mass of water in the swimming pool is
m = ρV = (1000 kg m 3 )(144 m 3 ) = 1.44 × 10 5 kg
15.4. Model: The densities of gasoline and water are given in Table 15.1. Solve:
(a) The total mass is
mtotal = mgasoline + mwater = 0.050 kg + 0.050 kg = 0.100 kg The total volume is
Vtotal = Vgasoline + Vwater =
mgasoline
ρ gasoline
⇒ ρ avg =
+
mwater 0.050 kg 0.050 kg = + = 1.235 × 10 −4 m 3 ρ water 680 kg m 3 1000 kg m 3
mtotal 0.100 kg = = 810 kg m 3 Vtotal 1.235 × 10 −4 m 3
(b) The average density is calculated as follows: mtotal = mgasoline + mwater = ρ water Vwater + ρ gasoline Vgasoline ⇒ ρ avg =
ρ water Vwater + ρ gasoline Vgasoline Vwater + Vgasoline
(50 cm )(1000 kg / m 3
=
3
100 cm
3
+ 680 kg / m 3 )
= 840 kg / m 3
Assess: The above average densities are between those of gasoline and water, and are reasonable.
15.5. Model: The density of sea water is 1030 kg/m3. Solve:
The pressure below sea level can be found from Equation 15.6 as follows:
p = p0 + ρgd = 1.013 × 10 5 Pa + (1030 kg m 3 )(9.80 m s 2 )(1.1 × 10 4 m ) = 1.013 × 10 5 Pa + 1.1103 × 10 8 Pa = 1.1113 × 10 8 Pa = 1097 atm where we have used the conversion 1 atm = 1.013 × 10 5 Pa . Assess: The pressure deep in the ocean is very large.
15.6. Model: The density of water is 1,000 kg/m3 and the density of ethyl alcohol is 790 kg/m3. Solve:
(a) The volume of water that has the same mass as 8.0 m3 of ethyl alcohol is
Vwater =
mwater malcohol ρ alcohol Valcohol 790 kg m 3 3 3 = = = (8.0 m ) = 6.32 m ρ water ρ water ρ water 1000 kg m 3
(b) The pressure at the bottom of the cubic tank is p = p0 + ρ water gd :
p = 1.013 × 10 5 Pa + (1000 kg m 3 )(9.80 m s 2 )(6.32)
13
where we have used the relation d = (Vwater ) . 13
= 1.194 × 10 5 Pa
15.7.
Solve:
Visualize:
The pressure at the bottom of the vat is p = p0 + ρgd = 1.3 atm . Substituting into this equation gives
(
)
1.013 × 10 5 Pa + ρ(9.8 m s 2 )(2.0 m ) = (1.3) 1.013 × 10 5 Pa ⇒ ρ = 1550.5 kg m 3
The mass of the liquid in the vat is m = ρV = ρπ (0.5 m ) d = (1550.5 kg m 3 )π (0.5 m ) (2.0 m ) = 2440 kg 2
2
15.8. Model:
The density of oil ρ oil = 900 kg m 3 and the density of water ρ water = 1000 kg m 3 .
Visualize:
Solve: The pressure at the bottom of the oil layer is p1 = p0 + ρ oil gd1 , and the pressure at the bottom of the water layer is
p2 = p1 + ρ water gd2 = p0 + ρ oil gd1 + ρ water gd2
(
)
⇒ p2 = 1.013 × 10 5 Pa + (900 kg m 3 )(9.80 m s 2 )(0.50 m ) + (1000 kg m 3 )(9.80 m s 2 )(1.20 m ) = 1.18 × 10 5 Pa Assess: A pressure of 1.18 × 105 Pa = 1.16 atm is reasonable.
15.9. Model: The density of seawater ρseawater = 1030 kg m 2 . Visualize:
Solve: The pressure outside the submarine’s window is pout = p0 + ρ seawater gd , where d is the maximum safe depth for the window to withstand a force F. This force is F A = pout − pin , where A is the area of the window. With pin = p0, we simplify the pressure equation to
pout − p0 =
F F = ρseawater gd ⇒ d = A Aρseawater g
d=
1.0 × 10 6 N = 3153 m π (0.10 m ) (1030 kg m 2 )(9.8 m s 2 ) 2
Assess: A force of 1.0 × 106 N corresponds to a pressure of
ρ= A depth of 3 km is therefore reasonable.
F 1.0 × 10 6 N = 2 = 314 atm A π (0.10 m )
15.10.
Visualize:
We assume that the seal is at a radius of 5 cm. Outside the seal, atmospheric pressure presses on both sides of the cover and the forces cancel. Thus, only the 10 cm diameter opening inside the seal is relevant, not the 20 cm diameter of the cover. Solve: Within the 10 cm diameter area where the pressures differ,
Fto left = patmos A
Fto right = pgas A
where A = πr 2 = 7.85 × 10 −3 m 2 is the area of the opening. The difference between the forces is
(
)
Fto left − Fto right = patmos − pgas A = (101,300 Pa − 20,000 Pa )(7.85 × 10 −3 m 2 ) = 639 N Normally, the rubber seal exerts a 639 N force to the right to balance the air pressure force. To pull the cover off, an external force must pull to the right with a force ≥639 N.
The density of mercury ρ = 13, 600 kg m 3 . Visualize: Please refer to Figure Ex15.11. Solve: From the figure and the equation for hydrostatic pressure, we have pgas + ρgh = patmos
15.11. Model:
⇒ pgas + (13,600 kg m 3 )(9.8 m s 2 )(0.10 m ) = 1.013 × 10 5 Pa ⇒ pgas = 88, 000 Pa Assess: Using 1 atm = 1.013 × 10 5 Pa , the gas pressure is 0.87 atm. This is expected because the mercury level in the left tube is higher than that in the right tube.
15.12. Model: The density of water is ρ = 1000 kg m 3 . Visualize: Please refer to Figure 15.17. Solve: From the figure and the equation for hydrostatic pressure, we have
p0 + ρgh = patmos Using p0 = 0 atm , and patmos = 1.013 × 10 Pa , we get 5
0 Pa + (1000 kg m 3 )(9.8 m s 2 )h = 1.013 × 10 5 Pa ⇒ h = 10.3 m Assess: This large value of h is due to water having a much smaller density than mercury.
15.13. Model: Assume that the oil is incompressible and its density is 900 kg/m3. Visualize:
Solve:
The pressures p1 and p2 are equal. Thus,
p0 +
F1 F F F = p0 + 2 + ρgh ⇒ 1 = 2 + ρgh A1 A2 A1 A2
With F1 = m1g, F2 = 4m2g, A1 = πr12 , and A2 = πr22 , we have 12 m1 g 4 m2 g 4 m2 g m1 g = + ⇒ = ρ gh r 2 2 − ρgh 2 2 πr1 πr2 π πr1
−1 2
Using m1 = 55 kg, m2 = 110 kg, r1 = 0.08 m, ρ = 900 kg/m3, and h = 1.0 m, the calculation yields r2 = 0.276 m. The diameter is 55.2 cm. Assess: Both pistons are too small to hold the people as shown, but the ideas are correct.
15.14. Model: Assume that the oil is incompressible. Its density is 900 kg/m3. Visualize: Please refer to Figure 15.19. Because the liquid is incompressible, the volume displaced in the left cylinder of the hydraulic lift is equal to the volume displaced in the right cylinder. Solve: Equating the two volumes, 2
2 r 0.04 m A1d1 = A2 d2 ⇒ (πr12 )d1 = (πr22 )d2 ⇒ d1 = 2 d2 = (0.20 m) = 3.2 m 0.01 m r1
15.15. Model: The buoyant force on the sphere is given by Archimedes’ principle. Visualize:
Solve:
The sphere is in static equilibrium because it is neutrally buoyant. That is,
∑F
y
= FB − w = 0 N ⇒ ρ l Vl g − ms g = 0 N
The sphere displaces a volume of liquid equal to its own volume, Vl = Vs, so
ρl =
ms m 0.0893 kg 3 = 4 s3 = 4 3 = 790 kg m Vs 3 πrs 3 π ( 0.03 m )
A density of 790 kg/m3 in Table 15.1 identifies the liquid as ethyl alcohol. Assess: If the density of the fluid and an object are equal, we have neutral buoyancy.
15.16. Model: The buoyant force on the sphere is given by Archimedes’ principle. Visualize: Please refer to Figure Ex15.16. Solve: The sphere is floating in static equilibrium, so the upward buoyant force exactly equals the sphere’s weight: FB = w. But according to Archimedes’ principle, FB is the weight of the displaced liquid. That is, the weight of the “missing” water in B is exactly matched by the weight of the “added” ball. Thus the total weights of both containers are equal.
15.17. Model: The buoyant force on the sphere is given by Archimedes’ principle. Visualize:
Solve: For the Styrofoam sphere and the mass not to sink, the sphere must be completely submerged and the buoyant force FB must be equal to the sum of the weight of the Styrofoam sphere and the attached mass. The volume of displaced water equals the volume of the sphere, so
FB = ρ water Vwater g = (1000 kg m 3 ) 43π (0.25 m ) (9.80 m s 2 ) = 641.4 N 3
[
wStyrofoam = ρ Styrofoam VStyrofoam g = (300 kg m 3 ) 43 π (0.25 m )
3
](9.80 m s ) = 192.4 N
Because wStyrofoam + mg = FB ,
m=
FB − wStyrofoam g
=
641.4 N − 192.4 N = 45.8 kg 9.80 m s 2
2
15.18. Model: The buoyant force on the wood block is given by Archimedes’ principle. The density of water is 1000 kg/m3 and the density of seawater is 1030 kg/m3. Visualize:
A floating object is in static equilibrium. Solve: The volume of displaced fluid is Vf = Ah. In static equilibrium,
FB = w ⇒ ρ f Vf g = ρ wood Vwood g ⇒ ρ f ( Ah) = ρ wood Vwood
(0.10 × 0.10 × 0.10 m V ρ ρ ⇒ h = wood wood = wood (0.10 × 0.10)m 2 ρf A ρf
3
) =ρ
wood (0.10 m ) ρf
For fresh water, h = (0.7)(0.10 m ) = 0.070 m and for seawater h = 0.068 m. The corresponding values for d are 3.0 cm and 3.2 cm. Assess: Objects float better in seawater, so the above result is reasonable.
15.19. Model: The buoyant force on the aluminum block is given by Archimedes’ principle. The density of aluminum and ethyl alcohol are ρ Al = 2700 kg m 3 and ρ ethyl alcohol = 790 kg m 3 .
Visualize:
The buoyant force FB and the tension due to the string act vertically up, and the weight of the aluminum block acts vertically down. The block is submerged, so the volume of displaced fluid equals VAl, the volume of the block. Solve: The aluminum block is in static equilibrium, so
∑F
y
= FB + T − w = 0 N ⇒ ρ f VAl g + T − ρ Al VAl g = 0 N ⇒ T = VAl g( ρ Al − ρ f )
T = (100 × 10 −6 m 3 )(9.80 m s 2 )(2700 kg m 3 − 790 kg m 3 ) = 1.87 N
(
)
3
where we have used the conversion 100 cm 3 = 100 × 10 −2 m = 10 −4 m 3 . Assess: The weight of the aluminum block is ρ Al VAl g = 2.65 N. A similar order of magnitude for T is reasonable.
15.20. Model: Treat the water as an ideal fluid. Visualize: Two separate rivers merge to form one river. Solve: The volume flow rate in the Bernoulli River is the sum of the volume flow rate of River Pascal and River Archimedes:
QB = 5.0 × 10 5 L s + 10.0 × 10 5 L s = 15.0 × 10 5 L s = 1500 m 3 /s The flowrate is related to the fluid speed and the cross section area by Q = vA. Thus
vB =
QB 1500 m 3 s = = 1.0 m s AB 150 m × 10 m 2
15.21. Model: Treat the water as an ideal fluid. The pipe is a flow tube, so the equation of continuity applies. Solve:
The volume flow rate is
Q=
300 L 300 × 10 −3 m 3 = = 1.0 × 10 −3 m 3 s 5.0 min 5.0 × 60 s
Using the definition Q = vA, we get
v=
Q 1.0 × 10 −3 m 3 s = = 3.18 m s 2 A π (0.01 m )
15.22. Model: Treat the water as an ideal fluid. The pipe itself is a flow tube, so the equation of continuity applies. Visualize:
Note that A1, A2, and A3 and v1, v2, and v3 are the cross-sectional areas and the speeds in the first, second, and third segments of the pipe. Solve: (a) The equation of continuity is
A1v1 = A2 v2 = A3 v3 ⇒ π r12 v1 = π r22 v2 = π r32 v3 ⇒ r12 v1 = r22 v2 = r32 v3 ⇒ (0.005 m ) ( 4.0 m s) = (0.01 m ) v2 = (0.0025 m ) v3 2
2
⇒ v2 =
2
2
0.005 m (4.0 m s) = 1.0 m s 0.01 m
2
v3 =
0.005 m (4.0 m s) = 16.0 m s 0.0025 m
(b) The volume flow rate through the pipe is
Q = A1v1 = π (0.005 m ) ( 4.0 m s) = 3.14 × 10 −4 m 3 s 2
15.23. Model: Treat the water as an ideal fluid so that the flow in the tube follows the continuity equation. Visualize:
Solve:
The equation of continuity is v0 A0 = v1 A1 , where A0 = L2 and A1 = π ( 12 L) . The above equation simplifies to 2
2
L 4 v0 L2 = v1π ⇒ v1 = v0 = 1.27v0 2 π
15.24. Model: Treat the oil as an ideal fluid obeying Bernoulli’s equation. Consider the path connecting point 1 in the lower pipe with point 2 in the upper pipe a streamline. Visualize: Please refer to Figure Ex15.24. Solve: Bernoulli’s equation is
p2 + 12 ρv22 + ρgy2 = p1 + 12 ρv12 + ρgy1 ⇒ p2 = p1 + 12 ρ(v12 − v22 ) + ρg( y1 − y2 ) Using p1 = 200 kPa = 2.00 × 10 5 Pa , ρ = 900 kg m 3 , y2 − y1 = 10.0 m , v1 = 2.0 m s , and v2 = 3.0 m s , we get p2 = 1.096 × 10 5 Pa = 110 kPa.
15.25. Model: The hanging mass creates tensile stress in the wire. Solve: The force (F) pulling on the wire, which is simply the weight (mg) of the hanging mass, produces tensile stress given by F/A, where A is the cross-sectional area of the wire. From the definition of Young’s modulus, we have
(πr 2 )Y∆L = π (2.50 × 10 −4 m) (20 × 1010 N m 2 )(1.0 × 10 −3 m) = 2.00 kg mg A ⇒m= gL ∆L L (9.80 m s2 )(2.0 m) 2
Y=
15.26. Model: Turning the tuning screws on a guitar string creates tensile stress in the string. Solve: The tensile stress in the string is given by T/A, where T is the tension in the string and A is the crosssectional area of the string. From the definition of Young’s modulus,
Y=
T A T L ⇒ ∆L = AY ∆L L
Using T = 2000 N, L = 0.80 m, A = π(0.0005 m)2, and Y = 20 × 1010 N/m2 (from Table 15.3), we obtain ∆L = 0.010 m = 1.0 cm. Assess: 1.0 cm is a large stretch for a length of 80 cm, but 2000 N is a large tension.
15.27. Model: The load supported by a concrete column creates compressive stress in the concrete column. Solve: The weight of the load produces tensile stress given by F/A, where A is the cross-sectional area of the concrete column and F equals the weight of the load. From the definition of Young’s modulus,
Y=
200,000 kg × 9.8 m s 2 3.0 m F A F L = 1.0 mm ⇒ ∆L = = 2 10 2 ∆L L A Y π (0.25 m ) 3 × 10 N m
Assess: A compression of 1.0 mm of the concrete column by a load of approximately 200 tons is reasonable.
15.28. Model: The water pressure applies a volume stress to the sphere. Solve:
The volume change is ∆V = –1.0 × 10–3 V. The volume stress is
F ∆V −1.0 × 10 −3 V = p = −B = −(7 × 1010 N m 2 ) = 7.0 × 10 7 Pa A V V Using p = p0 + ρgh , we get
7 × 10 7 Pa = 1.013 × 10 5 Pa + (1030 kg m 3 )(9.8 m s 2 )d ⇒ d = 6.92 km Assess: A pressure of 7 × 10 7 Pa = 690 atm causes only a volume change of 0.1%. This is reasonable because liquids and solids are nearly incompressible.
15.29. Solve: The pressure p at depth d in a fluid is p = p0 + ρgd . Using 1.29 kg/m3 for the density of air, pbottom = ptop + ρ air gd ⇒ pbottom − ptop = 202 Pa = 1.99 × 10 −3 atm Assuming pbottom = 1 atm,
pbottom − ptop pbottom
=
1.99 × 10 −3 atm = 0.2% 1 atm
15.30. Model: We assume that there is a perfect vacuum inside the cylinders with p = 0 Pa. We also assume that the atmospheric pressure in the room is 1 atm. Visualize: Please refer to Figure P15.30. 2 Solve: (a) The flat end of each cylinder has an area A = π r 2 = π (0.30 m ) = 0.283 m 2 . The force on each end is thus
(
)
Fatm = p0 A = 1.013 × 10 5 Pa (0.283 m 2 ) = 2.86 × 10 4 N (b) The net vertical force on the lower cylinder when it is on the verge of being pulled apart is
∑F
y
= Fatm − w players = 0 N ⇒ w players = Fatm = 2.86 × 10 4 N
⇒ number of players =
2.86 × 10 4 N = 29.2 (100 kg)(9.8 m s 2 )
That is, 30 players are needed to pull the two cylinders apart.
15.31. Model: Assume that the oil is incompressible and its density is 900 kg/m3. Visualize: Please refer to Figure P15.31. Solve: (a) The pressure at depth d in a fluid is p = p0 + ρgd . Here, pressure p0 at the top of the fluid is due both to the atmosphere and to the weight of the piston. That is, p0 = patm + wp/A. At point A, w pA = patm + P + ρg(1.00 m − 0.30 m ) A = 1.013 × 10 5 Pa
(10 kg)(9.8 m s 2 ) π (0.02 m )
2
+ (900 kg m 3 )(9.8 m s 2 )(0.70 m ) = 185, 460 Pa
⇒ FA = pA A = (185,460 Pa )π (0.10 m ) = 5830 N (b) In the same way, w pB = patm + P + ρg(1.30 m ) = 190, 752 Pa ⇒ FB = 5990 N A Assess: FB is larger than FA, because pB is larger than pA. 2
15.32. Model: The tire flattens until the pressure force against the ground balances the upward normal force of the ground on the tire. Solve: The area of the tire in contact with the road is A = (0.15 m )(0.13 m ) = 0.0195 m 2 . The normal force on each tire is 2 w (1500 kg) 9.8 m s n= = = 3675 N 4 4 Thus, the pressure inside each tire is n 3675 N 14.7 psi pinside = = = 188, 500 Pa = 1.86 atm Ă— = 27.3 psi A 0.0195 m 2 1 atm
(
)
15.33.
Visualize:
Solve: (a) Because the patient’s blood pressure is 140/100, the minimum fluid pressure needs to be 100 mm of Hg above atmospheric pressure. Since 760 mm of Hg is equivalent to 1 atm and 1 atm is equivalent to 1.013 × 105 Pa, the minimum pressure is 100 mm = 1.333 × 104 Pa. The excess pressure in the fluid is due to force F pushing on the internal 6.0-mm-diameter piston that presses against the liquid. Thus, the minimum force the nurse needs to apply to the syringe is
[
F = fluid pressure × area of plunger = (1.333 × 10 4 Pa ) π (0.003 m )
2
] = 0.377 N
(b) The flow rate is Q = vA, where v is the flow speed of the medicine and A is the cross-sectional area of the needle. Thus,
v=
Q 2.0 × 10 −6 m 3 2.0 s = 2 = 20.4 m s A π (0.125 × 10 −3 m )
Assess: Note that the pressure in the fluid is due to F that is not dependent on the size of the plunger pad. Also note that the syringe is not drawn to scale.
15.34. Visualize:
Solve: By sucking on the top and reducing the pressure, the straw is essentially a barometer. Atmospheric pressure pushes the liquid up the straw. The length of the longest straw can be obtained from the formula p = p0 + ρgd . If one could reduce the mouth pressure to zero, the length of the straw would be
1.013 × 10 5 Pa = 0 Pa + (1000 kg m 3 )(9.8 m s 2 )d ⇒ d = 10.3 m
15.35. Solve: The fact that atmospheric pressure at sea level is 101.3 kPa = 101,300 N/m2 means that the weight of the atmosphere over each square meter of surface is 101,300 N. Thus the mass of air over each square meter is m = (101,300 N)/g = (101,300 N)/(9.80 m/s2) = 10,340 kg per m2. Multiplying by the earth’s surface area will give the total mass. Using Re = 6.27 × 106 m for the earth’s radius, the total mass of the atmosphere is
Mair = Aearth m = ( 4π Re2 )m = 4π (6.37 × 10 6 m) 2 (10, 340 kg / m 2 ) = 5.27 × 1018 kg
15.36. Visualize: Let d be the atmosphere’s thickness, p the atmospheric pressure on the earth’s surface, and p0 (= 0 atm) the pressure beyond the earth’s atmosphere. Solve: The pressure at a depth d in a fluid is p = p0 + ρgd . This equation becomes 1 atm = 0 atm + ρ air gd ⇒ d =
1 atm 1.013 × 10 5 Pa = = 7.95 km ρ air g (1.3 kg m 3 )(9.8 m s 2 )
15.37. Solve: (a) We can measure the atmosphere’s pressure by measuring the height of the liquid column in a barometer, because patmos = ρgh . In the case of the water barometer, the height of the column at a pressure of 1 atm is h=
patmos 1.013 × 10 5 Pa = 10.337 m = ρ water g (1000 kg m 3 )(9.8 m s 2 )
Because the pressure of the atmosphere can vary by 5 percent, the height of the barometer must be at least be 1.05 greater than this amount. That is, hmin = 10.85 m. (b) Using the conversion rate1 atm = 29.92 inches of Hg, we have
29.55 inches of Hg =
29.55 × 1 atm = 0.9876 atm 29.92
The height of the water in your barometer will be
h=
patmos 0.9876 × 1.013 × 10 5 Pa = 10.21 m = ρ water g (1000 kg m 3 )(9.8 m s 2 )
15.38. Model: Oil is incompressible and has a density 900 kg/m3. Visualize: Please refer to Figure P15.38. Solve: (a) The pressure at point A, which is 0.50 m below the open oil surface, is
pA = p0 + ρ oil g(1.00 m − 0.50 m ) = 101, 300 Pa + (900 kg m 3 )(9.8 m s 2 )(0.50 m ) = 105,700 Pa (b) The pressure difference between A and B is pB – pA = (p0 + ρgdB) – (p0 + ρgdA) = ρg(dB – dA) = (900 kg/m3)(9.8 m/s2)(0.50 m) = 4410 Pa Pressure depends only on depth, and C is the same depth as B. Thus pC – pA = 4410 Pa also, even though C isn’t directly under A.
15.39. Model: Assume that oil is incompressible and its density is 900 kg/m3. Visualize: Please refer to Figure P15.39. Solve: (a) The hydraulic lift is in equilibrium and the pistons on the left and the right are at the same level. Equation 15.11, therefore, simplifies to Fleft piston Aleft piston
=
Fright piston Aright piston
⇒
w ⇒ rstudent = student relephant = w elephant
(
)
wstudent
π (rstudent )
2
=
welephant
(
π relephant
)
2
(70 kg)g 1.0 m = 0.2415 m ( ) 1200 kg)g (
The diameter of the piston the student is standing on is therefore 2 × 0.2415 m = 0.483 m. (b) From Equation 15.13, we see that an additional force ∆F is required to increase the elephant’s elevation through a distance d2. That is,
(
)
∆F = ρg Aleft piston + Aright piston d2
[
]
⇒ (70 kg)(9.8 m s 2 ) = (900 kg m 3 )(9.8 m s 2 )π (0.2415 m ) + (1.0 m ) d2 ⇒ d2 = 0.0234 m = 2.34 cm
2
2
15.40. Model: Assume that the oil is incompressible. Visualize:
Solve:
When the force F1 balances the weight mg, the pressure p2 is related to the pressure p1 as
p1 = p2 + ρgh When F1 is increased to F1′ , the weight is raised higher through a distance d2 and the left piston is lowered through a distance d1. The pressures p2′ and p1′ are now related through
p1′ = p2′ + ρg(h + d1 + d2 ) Subtracting these two equations,
p1′ − p1 = ( p2′ − p2 ) + ρg( d1 + d2 ) Because p1 = F1 A1 , p1′ = F1′ A1 , and p2 = p2′ = mg A2 , the above equation simplifies to
( F1′− F1 )
A1 = ρg( d1 + d2 ) ⇒ ∆F = ρg( A1d1 + A1d2 )
Since the oil is incompressible, A1 d1 = A2 d2 . The equation for ∆F thus becomes
∆F = ρg( A2 d2 + A1d2 ) = ρg( A1 + A2 )d2
15.41. Model: Water and mercury are incompressible and immiscible liquids. Visualize:
The water in the left arm floats on top of the mercury and presses the mercury down from its initial level. Because points 1 and 2 are level with each other and the fluid is in static equilibrium, the pressure at these two points must be equal. If the pressures were not equal, the pressure difference would cause the fluid to flow, violating the assumption of static equilibrium. Solve: The pressure at point 1 is due to water of depth dw = 10 cm:
p1 = patmos + ρ w gdw Because mercury is incompressible, the mercury in the left arm goes down a distance h while the mercury in the right arm goes up a distance h. Thus, the pressure at point 2 is due to mercury of depth dHg = 2h:
p2 = patmos + ρ Hg gdHg = patmos + 2 ρ Hg gh Equating p1 and p2 gives
patmos + ρ w gdw = patmos + 2 ρ Hg gh ⇒ h =
1 ρw 1 1000 kg m 3 dw = 10 cm = 3.7 mm 2 ρ Hg 2 13,600 kg m 3
The mercury in the right arm rises 3.7 mm above its initial level.
15.42. Model: Glycerin and ethyl alcohol are incompressible and do not mix. Visualize:
Solve: The alcohol in the left arm floats on top of the denser glycerin and presses the glycerin down distance h from its initial level. This causes the glycerin to rise distance h in the right arm. Points 1 and 2 are level with each other and the fluids are in static equilibrium, so the pressures at these two points must be equal:
p1 = p2 ⇒ p0 + ρ eth gdeth = p0 + ρ gly gdgly ⇒ ρ eth g(20 cm) = ρ gly g(2 h) h=
1 ρ eth 1 790 kg / m 3 (20 cm) = (20 cm) = 6.27 cm 2 ρ gly 2 1260 kg / m 3
You can see from the figure that the difference between the top surfaces of the fluids is
∆y = 20 cm − 2 h = 20 cm − 2(6.27 cm) = 7.46 cm
15.43.
Visualize:
The figure shows a dam with water height d. We chose a coordinate system with the origin at the bottom of the dam. The horizontal slice has height dy, width w, and area dA = wdy. The slice is at the depth of d − y. Solve: (a) The water exerts a small force dF = pdA = pwdy on this small piece of the dam, where p = ρg(d − y) is the pressure at depth d − y. Altogether, the force on this small horizontal slice at position y is dF = ρgw(d – y)dy. Note that a force from atmospheric pressure is not included. This is because atmospheric pressure exerts a force on both sides of the dam. The total force of the water on the dam is found by adding up all the small forces dF for the small slices dy between y = 0 m and y = d. This summation is expressed as the integral d
Ftotal =
d
d
1 1 dF = ∫ dF = ρgw ∫ ( d − y) dy = ρgw yd − y 2 = ρgwd 2 ∫ 2 2 0 0 0 all slices
(b) The total force is
Ftotal =
1 2
(1000 kg
m 3 )(9.8 m s 2 )(100 m )(60 m ) = 1.76 × 10 9 N 2
15.44.
Visualize:
The figure shows an aquarium tank with water at a height d. We chose a coordinate system with the origin at the bottom of the tank. The slice has a height dy, length l, and area dA = ldy. It is at depth d − y. Solve: (a) Pressure in excess of atmospheric pressure of the water on the bottom is pbottom = ρ water gd . Atmospheric pressure is ignored because it exerts an equal force on both sides of the bottom. Therefore, the force of the water on the bottom is
Fbottom = pbottom (lw) = ( ρ water gd )(lw) = (1000 kg m 3 )(9.8 m s 2 )(0.40 m )(1.0 m )(0.35 m ) = 1370 N (b) The water exerts a small force dF = pdA = pldy on a small piece of the window, where p = ρg(d − y) is the excess pressure at a depth of d − y. The force on this small horizontal slice at position y is dF = ρgl( d − y)dy . The total force on the front window is d d 1 Ftotal = ∫ dF = ∫ ρ water gL( d − y) dy = ρ water gl yd − 12 y 2 = ρ water gld 2 0 2 0
(
=
1 2
(1000 kg
)
m 3 )(9.8 m s 2 )(1.00 m )(0.40 m ) = 784 N 2
15.45.
Visualize:
The figure shows a small column of air of thickness dz, of cross-sectional area A = 1 m2, and of density ρ(z). The column is at a height z above the surface of the earth. Solve: (a) The atmospheric pressure at sea level is 1.013 × 10 5 Pa. That is, the weight of the air column with a 1 m2 cross section is 1.013 × 10 5 N. Consider the weight of a 1 m2 slice of thickness dz at a height z. This slice has volume dV = Adz = (1 m2)dz, so its weight is dw = (ρdV)g = ρg(1 m2)dz = ρ 0 e − z z0 g (1 m2)dz. The total weight of the 1 m2 column is found by adding all the dw. Integrating from z = 0 to z = ∞, ∞
w = ∫ ρ 0 g(1 m 2 )e − z z0 dz 0
[
= ( − ρ 0 g(1 m 2 )z0 ) e − z z0
]
∞ 0
= ρ 0 g(1 m )z0 2
Because w = 101,300 N = ρ 0 g(1 m )z0 , 2
z0 =
101, 300 N = 8.08 × 10 3 m (1.28 kg m 3 )(9.8 m s2 )(1.0 m 2 )
(b) Using the density at sea level from Table 15.1,
ρ = (1.28 kg m 3 )e This is 82% of ρ0.
(
− z 8.08×10 3 m
) = 1.28 kg m 3 e −1600 m (8.08×10 m ) = 1.05 kg m 3 ( ) 3
15.46.
Visualize:
Vf is the volume of the fish, Va is the volume of air, and Vw is the volume of water displaced by the fish + air system. The mass of the fish is mf, the mass of the air is ma, and the mass of the water displaced by the fish is mw. Solve: Neutral buoyancy implies static equilibrium, which will happen when ρ avg = ρ w . This means
mf + ma ρ f Vf + ρ a Va ρ w ⇒ ρ f Vf + ρ a Va = ρ w Vf + ρ w Va = Vf + Va Vf + Va ⇒ Va ( ρ a − ρ w ) = Vf ( ρ w − ρ f ) ⇒
Va ρ w − ρ f 1000 kg m 3 − 1080 kg m 3 = = = 8.01% ρ a − ρ w 1.28 kg m 3 − 1000 kg m 3 Vf
15.47. Model: The buoyant force on the cylinder is given by Archimedes’ principle. Visualize:
Vcyl is the volume of the cylinder and Vw is the volume of the water displaced by the cylinder. Note that the volume displaced is only from the part of the cylinder that is immersed in water. Solve: The cylinder is in static equilibrium, so FB = w. The buoyant force is the weight ρwVwg of the displaced water. Thus
FB = ρ w Vw g = w = mg = ρ cyl Vcyl g ⇒ ρ w Vw = ρ cyl Vcyl ⇒ ρ cyl = ρ w ⇒ ρ cyl = (1000 kg m 3 )
A(0.04 m ) = 667 kg m 3 A(0.06 m )
Assess: ρcyl < ρw for a cylinder floating in water is an expected result.
Vw Vcyl
15.48. Model: The buoyant force on the sphere is given by Archimedes’ principle. Visualize:
Solve:
The sphere is in static equilibrium. The free body diagram on the sphere shows that
∑F
y
⇒ ρ w Vsphere g =
= FB − T − w = 0 N ⇒ FB = T + w =
1 4 w+w= w 3 3
4 3 3 ρsphere Vsphere g ⇒ ρsphere = ρ w = (1000 kg m 3 ) = 750 kg m 3 3 4 4
15.49. Model: The buoyant force on the irregularly shaped rock is given by Archimedes’ principle. Visualize:
Solve:
The rock is in equilibrium, so
Fnet = n + FB − w = 0 N ⇒ n = w − FB = ρ rock Vrock g − ρ water Vrock g = ( ρ rock − ρ water )gVrock = ( ρ rock − ρ water )g( mrock ρ rock ) = (1 − ρ water ρ rock )gmrock 1000 kg m 3 = 1 − 9.8 m s 2 )(10 kg) = 74.7 N 3 ( 4200 kg m
15.50. Model: The buoyant force on the ceramic statue is given by Archimedes’ principle. Visualize:
Solve: The statue’s weight is the 28.4 N registered on the scale in air. In water, the weight of the statue is balanced by the sum of the buoyant force FB and the spring’s force on the statue. That is, 11.4 N mstatue wstatue = FB + Fspring on statue ⇒ 28.4 N = ρ w Vstatue g + 17.0 N ⇒ Vstatue = = gρ w ρstatue
⇒ ρstatue =
(mstatue g)ρ w = (28.4 N)(1000 kg (11.4 N)
(11.4 N)
m3 )
= 2490 kg m 3
15.51. Model: The buoyant force on the rock is given by Archimedes’ principle. Visualize:
Solve:
Because the rock is in static equilibrium, Newton’s first law is Fnet = T + FB − w rock = 0 N
m g ρ 1 1 1 ⇒ T = ρ rock Vrock g − ρ water Vrock g = ρ rock − ρ water Vrock g = ρ rock − ρ water rock = 1 − water mrock g 2 ρ rock 2 ρ rock 2 2 Using ρ rock = 4800 kg m 3 and mrock = 5.0 kg, we get T = 43.9 N.
15.52. Model: The buoyant force on the steel block is given by Archimedes’ principle. Visualize:
Solve:
(a) In air the force on the spring is the weight of the block:
Fspring on block = w = mg = ρsteel Vblock g = (7900 kg m 3 )(0.10 m ) (9.8 m s 2 ) = 77.42 N 3
(b) In oil:
Fspring on block + FB = 77.42 N ⇒ Fspring on block = 77.42 N − ρ oil Vblock g = 77.42 N − (900 kg m 3 )(0.10 m ) (9.8 m s 2 ) = 68.6 N 3
15.53. Model: The buoyant force on the steel cylinder is given by Archimedes’ principle. Visualize:
The length of the cylinder above the surface of mercury is d. Solve: The cylinder is in static equilibrium with FB = w. Thus
FB = ρ Hg VHg g = w = mg = ρ cyl Vcyl g ⇒ ρ Hg VHg = ρ cyl Vcyl ⇒ ρ Hg A(0.20 m − d ) = ρ cyl A(0.20 m ) ⇒ d = 0.20 m −
ρ cyl ρ Hg
(0.20 m) = (0.20 m)1 −
7900 kg m 3 = 0.0838 m = 8.38 cm 13,600 kg m 3
That is, the length of the cylinder above the surface of the mercury is 8.38 cm.
15.54. Model: The buoyant force on the cylinder is given by Archimedes’ principle. Visualize:
Solve: (a) Initially, as it floats, the cylinder is in static equilibrium, with the buoyant force balancing the cylinder’s weight. The volume of displaced liquid is Ah, so FB = ρliq(Ah)g = w Force F pushes the cylinder down distance x, so the submerged length is h + x and the volume of displaced liquid is A(h + x). The cylinder is again in equilibrium, but now the buoyant force balances both the weight and force F. Thus FB = ρliq(A(h + x))g = w + F Since ρliq(Ah)g = w, we’re left with F = ρliqAgx (b) The amount of work dW done by force F to push the cylinder from x to x + dx is dW = Fdx = (ρliqAgx)dx. To push the cylinder from xi = 0 m to xf = 10 cm = 0.10 m requires work
W=
∫
xf
xi
xf
F dx = ρ liq Ag ∫ x dx = 12 ρ liq Ag( x i2 − x f2 ) xi
= 12 (1000 kg / m 3 )π (0.020 m) 2 (9.8 m /s 2 )(0.10 m) 2 = 0.616 J
15.55. Model: The buoyant force on the cylinder is given by Archimedes’ principle. Visualize:
Let d1 be the length of the cylinder in the less-dense liquid with density ρ1, and d2 be the length of the cylinder in the more-dense liquid with density ρ2. Solve: The cylinder is in static equilibrium, so
∑F
y
= FB1 + FB2 − w = 0 N ⇒ ρ1 ( Ad1 )g + ρ 2 ( Ad2 )g = ρA( d1 + d2 )g ⇒
ρ1 ρ d1 + d2 = (d1 + d2 ) ρ2 ρ2
Since l = d1 + d2 ⇒ d1 = l − d2 , we can simplify the above equation to obtain
ρ − ρ1 d2 = l ρ 2 − ρ1 That is, the fraction of the cylinder in the more dense liquid is f = ( ρ − ρ1 ) ( ρ 2 − ρ1 ) . Assess: As expected f = 0 when ρ is equal to ρ1, and f = 1 when ρ = ρ2.
15.56. Model: The buoyant force on the cylinder is given by Archimedes’ principle. Visualize:
Solve:
The tube is in static equilibrium, so
∑F
y
= FB − wtube − wPb = 0 N ⇒ ρ liquid A(0.25 m )g = (0.030 kg)g + (0.250 kg)g ⇒ ρ liquid =
(0.280 kg) = A(0.25 m )
Assess: This is a reasonable value for a liquid.
(0.280 kg)
π (0.02 m ) (0.25 m ) 2
= 891 kg m 3
15.57. Model: The buoyant force on the can is given by Archimedes’ principle. Visualize:
The length of the can above the water level is d, the length of the can is L, and the cross-sectional area of the can is A. Solve: The can is in static equilibrium, so
∑F
y
= FB − wcan − wwater = 0 N ⇒ ρ water A( L − d )g = (0.020 kg)g + mwater g
The mass of the water in the can is mwater = ρ water
355 × 10 Vcan = 1000 kg m 3 ) 2 ( 2
−6
m3
= 0.1775 kg
⇒ ρ water A( L − d ) = 0.020 kg + 0.1775 kg = 0.1975 kg ⇒ d − L = −
0.1975 kg = 0.0654 m ρ water A
Because Vcan = π (0.031 m ) L = 355 × 10 −6 m 3 , L = 0.1176 m. Using this value of L, we get d = 0.0522 m = 5.22 cm. Assess: d L = 5.22 cm 11.76 cm = 0.444 , thus 44.4% of the length of the can is above the water surface. This is reasonable. 2
15.58. Visualize: Please refer to Figure P15.58. Solve:
(a) The pressure at a depth d in a fluid is
ptop = p0 + ρgd ⇒ Ftop = ptop A = Ap0 + ρgdA Similarly, the pressure at a depth d + h is
pbottom = p0 + ρg( d + h) ⇒ Fbottom = pbottom A = Ap0 + ρg( d + h) A (b) Subtracting the two equations gives an expression for the buoyant force:
FB = Fbottom − Ftop = p0 A + ρgA( d + h) − ( Ap0 + ρgdA) = ρgAh (c) Because the volume of the cylinder V = Ah and ρ = mass volume = m V , the result in part (b) can be rewritten as
m FB = ρgV = gV = mg V That is, the buoyant force is the weight of the liquid displaced by the cylinder.
15.59. Model: The buoyant force on the boat is given by Archimedes’ principle. Visualize:
The minimum height of the boat that will enable the boat to float in perfectly calm water is h. Solve: The boat barely floats if the water comes completely to the top of the sides. In this case, the volume of displaced water is the volume of the boat. Archimedes’ principle in equation form is FB = ρ w Vboat g . For the boat to float FB = wboat . Let us first calculate the weight of the boat:
wboat = wbottom + 2 wside 1 + 2 wside 2 , where wbottom = ρsteel Vbottom g = (7900 kg m 3 )(5 × 10 × 0.02 m 3 )g = 7900 g N wside 1 = ρsteel Vside 1g = (7900 kg m 3 )(5 × h × 0.005 m 3 )g = 197.5gh N wside 2 = ρ steel Vside 2 g = (7900 kg m 3 )(10 × h × 0.005 m 3 )g = 395gh
[
]
⇒ w boat = 7900 g + 2(197.5gh) + 2(395gh) N = (7900 + 1185h)g N Going back to the Archimedes’ equation and remembering that h is in meters, we obtain
ρ w Vboat g = (7900 + 1185h)g N ⇒ (1000)(10 × 5 × (h + 0.02)) = 7900 + 1185h ⇒ h = 14.1 cm
15.60. Model: Treat the water as an ideal fluid obeying Bernoulli’s equation. There is a streamline connecting point 1 in the wider pipe with point 2 in the narrower pipe. Visualize:
Solve:
Bernoulli’s equation, Equation 15.28, relates the pressure, water speed, and heights at points 1 and 2:
p1 + 12 ρv12 + ρgy1 = p2 + 12 ρv22 + ρgy2 For no height change y1 = y2. The flow rate is given as 5.0 L/s, which means
Q = v1 A1 = v2 A2 = 5.0 × 10 −3 m 3 s ⇒ v1 =
5.0 × 10 −3 m 3 s = 0.6366 m s 2 π (0.05 m ) 2
⇒ v2 = v1
A1 0.050 m = 2.546 m s = (0.6366 m s) 0.025 m A2
Bernoulli’s equation now simplifies to p1 + 12 ρv12 = p2 + 12 ρv22 ⇒ p1 = p2 + 12 ρ(v22 − v12 ) = 50 × 10 3 Pa +
1 2
(1000 kg
[
m 3 ) (2.546 m s) − (0.6366 m s) 2
Assess: Reducing the pipe size reduces the pressure because it makes v2 > v1.
2
] = 53.0 kPa
15.61. Model: Treat the water as an ideal fluid obeying Bernoulli’s equation. A streamline begins in the bigger size pipe and ends at the exit of the narrower pipe. Visualize: Please see Figure P15.61. Let point 1 be beneath the standing column and point 2 be where the water exits the pipe. Solve: (a) The pressure of the water as it exits into the air is p2 = patmos. (b) Bernoulli’s equation, Equation 15.28, relates the pressure, water speed, and heights at points 1 and 2:
p1 + 12 ρv12 + ρgy1 = p2 + 12 ρv22 + ρgy2 ⇒ p1 − p2 = 12 ρ(v22 − v12 ) + ρg( y2 − y1 )
From the continuity equation, v1 A1 = v2 A2 = ( 4 m s)(5 × 10 −4 m 2 ) = v1 (10 × 10 −4 m 2 ) = 20 × 10 −4 m 3 s ⇒ v1 = 2 m s Substituting into Bernoulli’s equation,
p1 − p2 = p1 − patmos =
1 2
(1000 kg
[
]
m 3 ) ( 4 m s) − (2 m s) + (1000 kg m 3 )(9.8 m s)( 4.0 m ) 2
2
= 6000 Pa + 39,200 Pa = 45,200 Pa But p1 − p2 = ρgh, where h is the height of the standing water column. Thus
h=
45,200 Pa = 4.61 m 1000 kg m 3 )(9.8 m s 2 ) (
15.62. Model: Treat the water as an ideal fluid obeying Bernoulli’s equation. A streamline begins at the faucet and continues down the stream. Visualize:
The pressure at point 1 is p1 and the pressure at point 2 is p2. Both p1 and p2 are atmospheric pressure. The velocity and the area at point 1 are v1 and A1 and they are v2 and A2 at point 2. Let the distance of point 2 below point 1 be d. Solve: Bernoulli’s equation at points 1 and 2 is p1 + 12 ρv12 + ρgy1 = p2 + 12 ρv22 + ρgy2 ⇒ ρgd = 12 ρ v22 − v12
(
)
From the continuity equation,
v1 A1 = v2 A2 ⇒ (1.0 m s)π (8.0 × 10 −3 m ) = v2π (5.0 × 10 −3 m ) ⇒ v2 = 2.56 m s 2
2
Going back to Bernoulli’s equation, we have
gd =
1 2
[(2.56 m s)
2
]
− (1.0 m s) ⇒ d = 0.283 m = 28.3 cm 2
15.63. Model: Treat the air as an ideal fluid obeying Bernoulli’s equation. Solve: (a) The pressure above the roof is lower due to the higher velocity of the air. (b) Bernoulli’s equation, with yinside ≈ youtside, is 2 130 × 1000 m 1 1 pinside = poutside + 12 ρ air v 2 ⇒ ∆p = ρ air v 2 = (1.28 kg m 3 ) = 835 Pa 3600 s 2 2 (c) The force on the roof is ( ∆p) A = (835 Pa )(6.0 m × 15.0 m ) = 7.51 × 10 4 N. The roof will blow up, because pressure inside the house is greater than pressure on the top of the roof.
15.64. Model: The ideal fluid obeys Bernoulli’s equation. Visualize: Please refer to Figure P15.64. There is a streamline connecting point 1 in the wider pipe on the left with point 2 in the narrower pipe on the right. The air speeds at points 1 and 2 are v1 and v2 and the cross-sectional area of the pipes at these points are A1 and A2. Points 1 and 2 are at the same height, so y1 = y2. Solve: The volume flow rate is Q = A1v1 = A2 v2 = 1200 × 10 −6 m 3 s . Thus
v2 =
1200 × 10 −6 m 3 s = 95.49 m s π (0.0020 m ) 2
v1 =
1200 × 10 −6 m 3 s = 3.82 m s 2 π (0.010 m )
Now we can use Bernoulli’s equation to connect points 1 and 2:
p1 + 12 ρv12 + ρgy1 = p2 + 12 ρv22 + ρgy2 ⇒ p1 − p2 = 12 ρ(v22 − v12 ) + ρg( y2 − y1 ) =
1 2
(1.28 kg
[
]
m 3 ) (95.49 m s) − (3.82 m s) + 0 Pa = 5826 Pa 2
2
Because the pressure above the mercury surface in the right tube is p2 and in the left tube is p1, the difference in the pressures p1 and p2 is ρ Hg gh . That is,
p1 − p2 = 5826 Pa = ρ Hg gh ⇒ h =
5826 Pa = 4.37 cm (13600 kg m 3 )(9.8 m s2 )
15.65. Model: The ideal fluid (that is, air) obeys Bernoulli’s equation. Visualize: Please refer to Figure P15.65. There is a streamline connecting points 1 and 2. The air speeds at points 1 and 2 are v1 and v2, and the cross-sectional areas of the pipes at these points are A1 and A2. Points 1 and 2 are at the same height, so y1 = y2. Solve: (a) The height of the mercury is 10 cm. So, the pressure at point 2 is larger than at point 1 by
ρ Hg g(0.10 m ) = (13, 600 kg m 3 )(9.8 m s 2 )(0.10 m ) = 13,328 Pa ⇒ p2 = p1 + 13,328 Pa Using Bernoulli’s equation,
p1 + 12 ρ air v12 + ρ air gy1 = p2 + 12 ρ air v22 + ρ air gy2 ⇒ p2 − p1 = 12 ρ air (v12 − v22 ) ⇒ v12 − v22 =
2( p2 − p1 )
ρ air
=
2(13,328 Pa ) = 20,825 m 2 s 2 (1.28 kg m 3 )
From the continuity equation, we can obtain another equation connecting v1 and v2:
π (0.005 m ) A2 v2 = 2 v2 = 25 v2 A1 π (0.001 m ) 2
A1v1 = A2 v2 ⇒ v1 =
Substituting v1 = 25v2 in the Bernoulli equation, we get
(25 v2 )2 − v22 = 20,825 m 2
s 2 ⇒ v2 = 5.78 m s
Thus v1 = 25v2 = 144 m/s. 2 (b) The volume flow rate A2 v2 = π (0.005 m ) (5.78 m s) = 4.54 × 10 −3 m 3 s .
15.66. Model: Treat the water as an ideal fluid that obeys Bernoulli’s equation. There is a streamline from the top of the water to the hole. Visualize: Please refer to Figure P15.66. The top of the water (at y1 = h) and the hole (at y2 = y) are at atmospheric pressure. The speed of the water at the top is zero because the tank is kept filled. Solve: (a) Bernoulli’s equation connecting the two points is
0 + ρgh = 12 ρv 2 + ρgy ⇒ v = 2 g(h − y) (b) For a particle shot horizontally from a height y with speed v, the range can be found using kinematic equations. For the y-motion, using t0 = 0 s, we have
y1 = y0 + v0 y (t1 − t 0 ) + 12 a y (t1 − t 0 ) ⇒ 0 m = y + (0 m s)t1 + 2
1 2
(− g)t12 ⇒ t1 =
2y g
For the x-motion,
x1 = x 0 + v0 x (t1 − t 0 ) + 12 a x (t1 − t 0 ) ⇒ x = 0 m + vt1 + 0 m ⇒ x = v 2 y g 2
(c) Combining the results of (a) and (b), we obtain
x = 2 g(h − y) 2 y g = 4 y(h − y) To find the maximum range relative to the vertical height, 1 dx h 4(h − y) − 4 y = 0 ⇒ y = =0⇒ 2 dy 4 y(h − y)
[
]
With y = 12 h, the maximum range is
h h x max = 4 h − = h 2 2
15.67. Model: Treat water as an ideal fluid that obeys Bernoulli’s equation. There is a streamline connecting the top of the tank with the hole. Visualize: Please refer to Figure P15.67. We placed the origin of the coordinate system at the bottom of the tank so that the top of the tank (point 1) is at a height of h + 1.0 m and the hole (point 2) is at a height h. Both points 1 and 2 are at atmospheric pressure. Solve: (a) Bernoulli’s equation connecting points 1 and 2 is
p1 + 12 ρv12 + ρgy1 = p2 + 12 ρv22 + ρgy2 ⇒ patmos + 12 ρv12 + ρg(h + 1.0 m ) = patmos + 12 ρv22 + ρgh ⇒ v22 − v12 = 2 g(1.0 m ) = 19.6 m 2 s 2 Using the continuity equation A1v1 = A2 v2 ,
π (2.0 × 10 −3 m ) A v2 v1 = 2 v2 = v2 = 2 250, 000 π (1.0 m ) A1 2
Because v1 << v2, we can simply put v1 ≈ 0 m/s. Bernoulli’s equation thus simplifies to
v22 = 19.6 m 2 s 2 ⇒ v2 = 4.43 m s Therefore, the volume flow rate through the hole is
Q = A2 v2 = π (2.0 × 10 −3 m ) ( 4.43 m s) = 5.56 × 10 −5 m 3 s = 3.34 L min 2
(b) The rate at which the water level will drop is
v1 =
v2 4.43 m s = = 1.77 × 10 −2 mm s = 1.06 mm min 250, 000 250,000
Assess: Because the hole through which water flows out of the tank has a diameter of only 4.0 mm, a drop in the water level at the rate of 1.06 mm/min is reasonable.
15.68. Model: The dangling mountain climber creates tensile stress in the rope. Solve:
Young’s modulus for the rope is
Y= The tensile stress is
F A stress = ∆L L strain
(70 kg)(9.8 m s 2 ) π (0.0050 m )
2
= 8.734 × 10 6 Pa
and the strain is 0.080 m 50 m = 0.0016 . Dividing the two quantities yields Y = 5.46 × 10 9 N m 2 .
15.69. Model: The aquarium creates tensile stress. Solve:
Weight of the aquarium is
w = mg = ρ water Vg = (1000 kg m 3 )(10 m 3 )(9.8 m s 2 ) = 9.8 × 10 4 N where we have used the conversion 1 L = 10 −3 m 3 . The weight supported by each wood post is 14 (9.8 × 10 4 N) = 2 2.45 × 10 4 N . The cross-sectional area of each post is A = (0.040 m ) = 1.6 × 10 −3 m 2 . Young’s modulus for the wood is F A FL = Y = 1 × 1010 N m 2 = ∆L L A∆L
⇒ ∆L =
(2.45 × 10 4 N)(0.80 m) = 1.23 × 10 −3 m = 1.23 mm FL = AY (1.6 × 10 −3 m 2 )(1 × 1010 N m 2 )
Assess: A compression of 1.23 mm due to a weight of 2.45 × 104 N is reasonable.
15.70. Model: Water is almost incompressible and it applies a volume stress. Solve:
(a) The pressure at a depth of 5000 m in the ocean is
p = p0 + ρsea water g(5000 m ) = 1.013 × 10 5 Pa + (1030 kg m 3 )(9.8 m s 2 )(5000 m ) = 5.057 × 10 7 Pa (b) Using the bulk modulus of water,
∆V p 5.057 × 10 7 Pa =− =− = −0.025 V B 0.2 × 1010 Pa (c) The volume of a mass of water decreases from V to 0.975V. Thus the water’s density increases from ρ to ρ/0.975. The new density is
ρ 5000 m =
1030 kg / m 3 = 1056 kg / m 3 0.975
(d) More, because the mass of sea water is slightly more than if the water was not compressible. So, the actual pressure at a depth of 5000 m is slightly more than what was computed in part (a).
15.71. Model: Pressure applies a volume stress to water in the cylinder. Solve:
The volume strain of water due to the pressure applied is
∆V p 2 × 10 6 Pa =− =− = −1.0 × 10 −3 V B 0.20 × 1010 Pa
⇒ ∆V = V ′ − V = −(1.0 × 10 −3 )(1.30 m 3 ) = −1.30 × 10 −3 m 3 = −1.3 L As the safety plug on the top of the cylinder bursts, the water comes back to atmospheric pressure. The volume of water that comes out is 1.30 L.
15.72. Model: Air is an ideal gas and obeys Boyle’s law. Visualize: Please refer to Figure CP15.72. h is the length of the air column when the mercury fills the cylinder to the top. A is the cross-sectional area of the cylinder. Solve: For the column of air, Boyle’s law is p0 V0 = p1V1 , where p0 and V0 are the pressure and volume before any mercury is poured, and p1 and V1 are the pressure and volume when mercury fills the cylinder above the air. Using p1 = p0 + ρ Hg g(1.0 m − h), Boyle’s law becomes
[
]
[
]
p0 V0 = p0 + ρ Hg g(1.0 m − h) V1 ⇒ p0 A(1.0 m ) = p0 + ρ Hg g(1.0 m − h) Ah p0 (1.0 m − h) = ρ Hg g(1.0 m − h)h ⇒ h =
p0 1.013 × 10 Pa = = 0.76 m = 76 cm ρ Hg g (13,600 kg m 3 )(9.8 m s 2 ) 5
15.73. Model: The balloon displaces air, so the air exerts an upward buoyant force on the balloon. The buoyant force on the balloon is given by Archimedes’ principle. Visualize:
FB is the buoyant force on the balloon, wb is the weight of the balloon, wHe is the weight of the helium gas, and w is the weight due to the mass tied to the balloon. 3 Solve: The volume of the balloon Vballoon = 43 π (0.10 m ) = 4.1888 × 10 −3 m 3 . We have,
FB = ρ air gVballoon = (1.28 kg m 3 )(9.8 m s 2 )( 4.188 × 10 −3 m 3 ) = 0.05254 N
wHe = ρ He gVballoon = (0.18 kg m 3 )(9.8 m s 2 )( 4.188 × 10 −3 m 3 ) = 0.007387 N
wb = (0.001 kg)g = 0.0098 N For the balloon to be in static equilibrium, FB = wHe + wb + w = 0.05254 N = 0.01719 N + mg
⇒ m = 0.0036 kg = 3.6 g
15.74.
Model: Visualize:
The buoyant force on the cone is given by Archimedes’ principle.
Solve: It may seem like we need the formula for the volume of a cone. You can use that formula if you know it, but it isn’t essential. The volume is clearly the area of the base multiplied by the height multiplied by some constant. That is, the cone shown above has V = aAd, where a is some constant. But the radius of the base is r = dtanα, where α is the angle of the apex of the cone, and A = π r2, making A proportional to d2. Thus the volume of a cone of height d is V = cd3, where c is a constant. Because the cone is floating in static equilibrium, we must have FB = w. The cone’s density is ρ0, so its weight is w = ρ0Vg = ρ0cl3g. The buoyant force is the weight of the displaced fluid. The volume of displaced fluid is the full volume of the cone minus the volume of the cone of height h above the water, or Vdisp = cl3 – ch3. Thus FB = ρfVdispg = ρfc(l3 – h3)g, and the equilibrium condition is
ρ h3 h FB = w ⇒ ρ f c(l 3 − h 3 )g = ρ 0 cl 3 g ⇒ ρ f 1 − 3 = ρ 0 ⇒ = 1 − 0 l l ρf
1/ 3
15.75. Model: The buoyant force on the cylinder is given by Archimedes’ principle. Visualize:
A is cross-sectional area of the cylinder. Solve: (a) The cylinder’s weight is w = ρ 0 ( Al )g and the weight of the displaced fluid is FB = ρ f ( Ah)g . In static equilibrium, FB = w and we can write
ρ f Ahg = ρ 0 Alg ⇒ h = ( ρ 0 ρ f )l (b) Now the volume of the displaced liquid is A(h – y). Applying Newton’s second law in the y-direction,
∑F
y
= − w + FB = − ρ 0 Alg + ρ f A(h − y)g
Using ρf Ahg = ρ0 Alg from part (a), we find
( Fnet ) y = − ρ f Agy (c) The result in part (b) is F = − ky , where k = ρ f Ag. This is Hooke’s law. (d) Since the cylinder’s equation of motion is determined by Hooke’s law, the angular frequency for the resulting simple harmonic motion is ω = k m , and the period is
T=
ρ 0 Al 2π m m h = T = 2π = 2π = 2π = 2π k g ω ρ f Ag ρ f Ag
where we have used the expression for h from part (a). (e) The oscillation period for the 100-m-tall iceberg (ρice = 917 kg/m3) in sea water is
T = 2π
ρ0l h = 2π = 2π g ρf g
(917 kg
(1030 kg
m 3 )(100 m )
m 3 )(9.8 m s 2 )
= 18.9 s
15.76.
Visualize:
The column of air has a cross-sectional area A. We chose the origin of the coordinate system at z = 0 m and denote the vertical axis as the z-axis. Solve: (a) The pressure at height z is pz. The pressure at a height z + dz is
pz + dz = pz −
weight of air column of height dz ρAdzg = pz − = pz − ρgdz A A
(b) The change in pressure is dp = pz + dz − pz = − ρgdz . The sign is minus because pressure decreases as z increases. (c) From the ideal gas law,
pρ 0 p p0 = ⇒ρ= p0 ρ ρ0 The result for part (b) thus becomes dp = −( pρ 0 p0 )gdz . (d) To find the pressure p at a height z, we rewrite the result of part (c) as
ρ dp = − 0 p p0
g dz
Integrating both sides
p ρ0 g z ρ g dp = − dz ⇒ ln = − 0 z ⇒ p = p0 e − ρ0 gz ∫p p ∫ p0 0 p0 p0 0 p
where z0 = p0 ρ 0 g . (e) The scale height is z0 =
(
)
1.013 × 10 5 Pa p0 = 8076 m . = ρ 0 g (1.28 kg m 3 )(9.8 m s 2 )
p0
= p0 e − z z 0
(f) A table showing values of p = p0 e − z z0 = e − z (8076 m ) atm at selected values of z follows. z (m) p (atm) 0 1 1000 0.884 2000 0.781 4000 0.609 6000 0.476 8000 0.371 10,000 0.290 12,000 0.226 14,000 0.177 15,000 0.156