Ch14HW All

PHSX 215N M. Schneider & J. Jacobs

Fall 2010 Homework 14 Solutions

HRW 14.5 An office window has dimensions 3.4 m by 2.1 m. As a result air  pressure sult of the passage of a storm, the outside air  pressure drops to 0.96 atm, but inside the pressure is held held at 1.0 atm. What net window? net force pushes out out on the window? A = (3.4 m)(2.1 m) = 7.14 m2 . The pressure difference and the force are:  p −  p0 = (1.0 atm − 0.96 atm)

m2 1.013 × 105  N/  atm 1 atm

3

2

m = 4.05 × 10  N/ 

4

F = (p − p0 )A = 2.9 × 10  N

HRW 14.2 A  part  partia iall lly y evacua evacuate ted d airt airtig ig ht container has a tight-fitting lid of surface area 77 m2 and negligible mass. If  the force required required to remove the lid is 480 N and the atmospheric   pressure is 1.0×105 Pa, what is the internal air   pressure? Assuming the force required required is only to overcome overcome the pressure difference, the the gauge pressure is F = (p out −  pin )A

→

pin − pout = −

F 480 N m2 =− = −6.2  N/  A 77 m 2

The absolute pressure on the inside would would be only slightly smaller than the atmospheric   pressure on the outside. HRW 14.17 Crew members attempt to escape from from a damaged submarine 100 m below the surface. What force must  b e applied to a pop-out hatch, which is 1.2 m by 0.60 m, to push it out at that depth? Assume that the dens densii ty 3 of ocean water 1024 kg/m and the internal air pressure is 1.0 atm. Assuming atmospheric   pressure inside, the gauge pressure is 2

 p −  p0 = ρgh = (1024 kg/ m3 )(9.80 m/ s )(100 m) = 1.00 × 106 2

m  N/ 

The area is simply, A = (1.2 m)(0.60 m) = 0.72 m2 , and the force is F = (p − p0 )A = (1.00 × 106  N/ m2 )(0.72 m2 ) = 7.2 × 105  N

1

HRW 14.23 In analyzing certain geological features, it is often appropriate to assume that the pressure at some horizo ntal level of  compensatio n, deep inside the Earth, is the same over a large region and is equal to the pressure due to the gravitational force on the overlying material. Thus, the pressure on the level of  compensation is given   by the fluid pressure formula. This model requires, for one thin g, that mou ntains have roots of  continental rock extending into the denser mantle (see figure). Consider a mou ntain of height H = 6.0 km on a contin ent of thickness T = 32 km. The contine ntal rock  has a density of 2.9 g/cm 3 , and  beneath this rock the mantle has a density of 3.3 g/cm 3 . Calculate the depth D of the root. (Hint: Set the pressure at points a and b equal; the depth y of the level of compensation will cancel out.)

At a and b the pressure are:  pa −  p0 = ρ1 g(T + H + D) + ρ2 g(y − D)

p b −  p0 = ρ1 gT + ρ2 gy

where ρ1 = 2.9 g/ cm3 and ρ2 = 3.3 g/ cm3 are the densities of the continent and mantle, respectively. Setting these pressures equal: ρ1 gT + ρ1 gH + ρ1 gD + ρ2 gy − ρ2 gD = ρ1 gT + ρ2 gy Solve for  D: (ρ2 − ρ1 )D = ρ1 H

→

ρ1 ρ2 − ρ1

D =

H =

→

ρ1 H + (ρ1 − ρ2 )D = 0

2.9 g/ cm3 (6.0 km) = 43.5 km 3.3 g/ cm3 − 2.9 g/ cm3

HRW 14.26 To suck lemonade of density 1000 kg/m 3 up a straw to s maximum height of 4.0 cm, what minimum gauge   pressure (in atmospheres) must you produce in your  lungs? You want a lower pressure so that the liquid will be pushed up the straw.

p0 =  p + ρgh

→

p −  p0 = −ρgh

2

 p −  p0 = −(1.00 × 10 kg/ m3 )(9.80 m/  s )(0.040 m) = −392  N/ m2 = −3.9 × 10 3

3

−

atm

HRW 14.31 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. The volumes displaced are Vd,w ater  = 0.667V and Vd,oil = 0.90V . (a)

ρw ood gV = ρw ater  gVd,w ater 

→

ρwood =

Vd,w ater  V

ρw ater 

ρwood = 0.667(1.00 × 10 kg/ m3 ) = 667 kg/ m3 3

(b)

ρw ood gV = ρoil gVd,oil

→

V

ρoil =

ρw ood

Vd,water 

ρoil =

1 (667 kg/ m3 ) = 741 kg/ m3 0.90

HRW 14.38 A hollow sphere of inner radius 8.0 cm and outer radius 9.0 cm floats half-submerged in a liquid of density 800 kg/m 3 . (a) What is the mass of the sphere? (b) Calculate the density of the material of which the sphere is made. (a) The volume displaced is Vdis =

1 2

4 πR 3outer  3

=

2 π(0.090 m)3 = 1.53 × 10 3

3

m3

−

The  buoyant force supports the weight: mg = ρliq gVdis

m = ρliq Vdis = (800 kg/ m3 )(1.53 × 10

3

−

→

m3 ) = 1.22 kg

(b) The volume of the material is V =

4 4 3 ) = π (0.090 m)3 − (0.080 m)3 = 9.09 × 10 π (R 3outer  − R inner  3 3

4

−

m3

and the density is ρ=

m V

=

1.22 kg 3 m3 = 1.34 × 10 kg/  9.09 × 10 4 m3 −

HRW 14.44 A block of wood has a mass of 3.67 kg and a density of 600 kg/m 3 . It is to be loaded with lead (1.14 × 104 kg/m 3 ) so that it will float in water with 0.900 of its volume submerged. What mass of lead is needed if  the lead is atta ched to (a) the top of the wood and (b) the  bottom of the woo d?

(a) Only the wood displaces water, Vdis = 0.90Vw ood and Vw ood = mw ood /ρ w ood . (mwood + mlead )g = ρw ater  gVdis

m

mwood = 0.90ρ

− m wood =

0.90

w ater 

lead

ρw ood

mlead = 0.90ρw ater  Vw ood − mwood

→

m3 1.00 × 103 kg/  600 kg/ m3

mlead = 0.90

ρw ater  ρw ood

− 1 m wood

− 1 (3.67 kg) = 1.84 kg

(b) Both the wood and the lead displace water, Vdis = 0.90Vw ood + Vlead and Vlead = mlead /ρ lead . (mwood + mlead )g = ρw ater  gVdis

1−

m lead

ρw ater  ρlead

  0.90 =

mlead = ρw ater 

→

mlead = 3

1 .00 ×10 kg/m 600 kg/m3

1−

0.90 3

1.00 ×103 kg/m3 1.13 ×10 4 kg/m3

ρw ater  ρw ood

−1

0.90

m wood m lead + ρw ood ρlead

− 1 m wood

   (3.67 kg) = 2.01 kg

− mwood

HRW The figure at right shows an iron ball suspended by thread of negligible mass from an uprig ht cylinder  that floats  partially submerged in water. The cylinder has a height of 6.00 cm, a face area of  12.0 cm2 on the top and  bottom, and a density of 0.30 g/cm 3 , and 2.00 cm of its height is above the water surface. What is the radius of  the iron  bal l?

  Newton’s 2nd law for the iron ball (with ρw = 1000 kg/ m3 and ρi = 7900 kg/ m3 ): FT + FB − Fg = 0

→

FT + ρw gV ba ll − ρi gV ball = 0

→

FT

V ball =

(ρ i − ρw )g

To find the volume of the ball, we need the tension force. Newton’s 2nd law for  the cylinder: FB − Fg − FT = 0

→

ρw gVdis − ρc gV − FT = 0

→

2

Using Vdis = 3 V and V = (6.00 cm)(12.0 cm2 ) = 72 cm3 = 7.2 × 10

5

−

2 FT =

3

FT = (ρw Vdis − ρc V ) g

m3 , we have: 2

3

(1000 kg/  m3 ) − 300 kg/ m 3

5

−

7.2 × 10

m

9.80 m/ s

= 0.2378  N

The volume of the ball is V ball =

0.2378  N m3

(7900 kg/ 

6

−

m3 )(9.80

− 1000 kg/ 

2

m/ s )

= 3.517 × 10

m3

Finally, we can use this to obtain the radius of the  ball: 4

=

V 3 πR   ball

→

R  =

3V ba ll

1/3

=

4π

6

−

3(3.517 × 10 4π

1/3

m3 )

= 0.00943 m = 0.943 cm

3

HRW 14.51 A garden hose with internal diameter  of 1.9 cm is connected to a (stationary) lawn sprinkler  that consists of  a container with 24 holes, each 0.13 cm in diameter. If the water in the hose has a speed of 0.91 m/s, what spped does it leave the sprinkler  holes? The continuity equation tells us:

A2 v2 = A1 v1

→

A1

v2 =

A2

v2 =

(1.9 cm) 2

2

v1 =

d1 24d22

v1

(0.91 m/ s) = 8.0 m/ s

24(0.13 cm) 2 HRW 14.54 The water flowing through a 1.9 cm (inside diamter)  pipe flows out through three 1.3 cm pipes. (a) If the flow rates in the three smaller pipes are 26, 19, ands 11 L/min, what is the flow rate in the 1.9 cm pipe? (b) What is the ratio of the sp eed in the 1.9 cm pipe to that in the pipe carrying 26 L/min? (a) The volume rate of flow in must equal the total volume rate of flow out. R V 0 = R V 1 + R V 2 + R V 3 = 56 L/ min (b) The volume rate of flow is equal to the velocity times the cross sectional area: A0 v0 = R V 0

→

v0 = 2

R V 0 A0

and

v1 = 2

R V 1 A1

HRW

v0 v1

=

R V 0 A1 R V 0 d1 = = R V 1 d22 A0 R V 1

56 L/ min 26 L/ min

1.3 cm 1.9 cm

= 1.00

HRW A water pipe having a 2.5 cm inside diameter  carries water into the  baseme nt of a house at a speed of  0.90 m/s and a pressure of 170 kPa. If the pipe tapers to 1.2 cm and rises to the second floor 7.6 m above the input   point, what are the (a) speed and (b) water pressure at the second floor? Coming into the  baseme nt:  p = 1.70 × 105  N/ m2 , cm

At the higher point:

p2 =?,

v2 =?,

y2 = 7.6 m,

v = 0.90 m/ s,

y = 0,

d = 2.5

d2 = 1.2 cm

(a) The continuity equation tells us: v2 =

A1

d1 d2

v1 =

A2

2

2.5 cm 1.2 cm

v1 =

2

(0.90 m/ s) = 3.91 m/ s

(b) Bernoulli’s equation tells us:  p2 +

1 2

ρv22 + ρgy2 =  p1 +

 p2 = 1.70×105  N/ m2 +(1.00×10

3

m3 ) kg/ 

1 2 ρv1 2

→

p2 =  p1 +

1 2 1 ρv1 − ρv22 − ρgy2 2 2

1 1 2 2 (0.90 m/ s) − (3.91 m/ s)2 − (9.80 m/ s )(7.6 m) = 8.83 × 104  N/ m2 2 2

HRW 14.57 A cylindrical tank area A = 6.5 cm2 flows out, in cubic area of the stream

At the top:

with a large diameter  is filled with water to a depth D = 0.30 m. A hole of cross-sectional in the  bottom of the tank allows water to drain out. (a) What is the rate at which water  meters per second? (b) At what distance below the  bottom of the tank is the cross sectional equal to one-half the area of the hole?

p = p0 ,

v ≈ 0,

At the  bottom:

y = 0.30 m

 p2 = p0 ,

v2 =?,

y2 =

0 (a) Bernoulli’s equation tells us: 1  p0 + ρv22 =  p0 + ρgy 2 The volume rate of flow is:

→

v22

= 2gy

→ 4

−

R V = A2 v2 = (6.5 × 10

q v2 = 2(9.80 m/ s2 )(0.30 m) = 2.42 m/ s

m3 )(2.42 m/ s) = 1.6 × 10

3

−

m3 / s

(b) Since R V is fixed, as v goes up, the area must shrink. A3 =

1 A2 2

→

v3 = 2v2

This additio nal kinetic energy is from gravitational potential energy. Assuming the fluid has dropped a distance d (so that y3 = −d): 1 2 1 2 ρv3 − ρgd = ρv2 2 2

→

d=

v32 − v22 (2v2 )2 − v22 3v 2 = = 2 2g 2g 2g

You can plug in the numbers or use (from above) v22 = 2gy to obtain:

d = 3y = 0.90 m

HRW In the figure at right, the fresh water behind a reservoir dam has depth D = 15 m. A horizo ntal pipe 4.0 cm in diameter   passes through the dam at depth d = 6.0 m. A plug secures the pipe opening. (a) Find the magnitu de of the frictional force between plug and pipe wall. (b) The plug is removed. What water volume exits the pipe in 3.0 h?

(a) The friction force must overcome the net pressure force. At the depth of the plug, the gauge pressure in the fluid is 2 3 4  p −  p0 = ρgd = (1.00 × 10 kg/ m3 )(9.80 m/ s )(6.0 m) = 5.88 × 10 m2  N/  The pressure force that friction force must match is

(b) Vout

F = (p − p0 )A = (5.88 × 104  N/ m2 ) π(0.020 m)2 = 74  N q  p 2 2 s )(6.0 m)(1.08 × 104 s) = 147 m3 = R V ∆t = Av∆t = A( 2g d)∆ t = π(0.020 m) 2(9.80 m/ 

HRW 14.71 The figure 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  b e made to maximize x?

(a) The horizo ntal launch velocity of the stream is q  p 2 v0 = 2g h = 2(9.80 m/ s )(0.10 m) = 1.4 m/ s To fall from y 0 = H − h = 30 cm, it will tak e s

r 

1 0 = y0 −

2

2(0.30 m)

2

gt

→

and the fluid will travel a horizo ntal distance of:

t=

2y0 g

=

2

= 0.244 s

9.80 m/ s

x = v0 t = 0.35 m

(b) Putting in the formulas from ab ove: !

s

 p 2g h x = v0 t =

2(H − h) g

 p = 2 h(H − h)

Solving for h as a function of  x: x2 = 4h(H − h)

→

h2 − H h +

1 4

x2 = 0

→

h=

H 2

±

1 p 2 H − x2 2

Plugging in for this   particular case: h = 0.20 m ±

1 p (0.40 m)2 − (0.35 m)2 = 0.10 m or  0.30 m 2

(c) To find the maximum distance, take a derivative of x2 with respect to h and set it equal to zero:

dx2

HRW

dh = 4(H − h) − 4h = 4H − 8h

→

4H − 8h = 0

→

h=

1 H 2

HRW A venturi meter is used to measure the flow speed of a fluid in a pipe. The meter is connected between two sections of the  pip e (see figure); the cross-sectional area A of the entrance and exit of  the meter matches the pipe’s cross-sectional area. Between the entrance and exit, the fluid flows from the pipe with speed V and then through a narrow “throat” of cross-sectional area a with speed v. A manometer  connects the wider portion of the meter to the narrower portion. The change in the fluid’s speed is accompanied  by a change ∆p in the fluid’s pressure, which causes a height difference h of the liquid in the two arms of the manometer. (Here ∆p means   pressure in the throat minus pressure in the  pip e.)

(a) By applying Bernoulli’s equation and the equation of  continuity to points 1 and 2 in the figure, show that s

2a2 ∆ p ρ(a2 − A2 )

V =

where ρ is the density of the fluid. (b) Suppose that the fluid is fresh water, that the cross-sectional areas are 64 cm2 in the pipe and 32 cm2 in the throat, and that the pressure is 55 kPa in the pipe and 41 kPa in the throat. What is the rate of water flow in cubic meters per  second? (a) Continuity tells us:

AV = av

Bernoulli’s equation tells us:  p pipe +

1 2

ρV2 = pthroat +

1

1

2

2

→

∆p = p throat − p pipe =

ρv

2

2

2

ρ V −v

Plugging in v = (A/a )V , we have ∆p =

A2 1 ρ V2 − 2 V2 2 a

A2 1− 2 a

→

Solving for  V :

V

2

2∆ p

=

2

→

ρ

a −A

2

V

2

=

2a 2 ∆ p ρ

s

V =

2a2 ∆ p ρ (a2 − A2 )

(b) For this  particular  case, we have s

V =

The flow rate is then:

2(−14 × 103  N/ m2 ) s = 3.05 m/  (1000 kg/ m3 ) [1 − (2)2 ] 4

−

R V = AV = (64 × 10

m2 )(3.05 m/  s) = 2.0 × 10

2

−

m3 / s

Extra 14.1 A glass ball of radius 0.0200 m sits at the  bottom of a container  of milk  that has a density of 1.03 ×10 3 kg/m 3 . The normal force acting on the ball from the container’s lower surface has magnitude 9.48 × 10 2  N. Calculate the mass of the  ball. −

Applying Newton’s 2nd law to the ball, we have: FB + F N − m ball g = 0.

Where FB = ρmilk  gV ball , where V ball = 4 π r 3 3 This gives m ball =

ρmilk  gV ball + F N g

5

−

= 3.351 × 10

m3 .

 ball

(1.03 × 103 )(9.80)(3.351 × 10 9.80

5

−

=

2

−

) + 9.48× 10

= 0.0442 kg