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Chapter 9
9
Solids and Fluids
PROBLEM SOLUTIONS
9.1
The elastic limit is the maximum stress, F A where F is the tension in the wire, that the wire can withstand and
still return to its original length when released. Thus, if the wire is to experience a tension equal to the weight of the
performer without exceeding the elastic limit, the minimum cross-sectional area is
Amin 
2
 Dmin
4

F
mg

elastic limit
elastic limit
and the minimum acceptable diameter is
Dmin 
9.2
(a)

4  70 kg  9.8 m s2
4mg

  elastic limit 
  5.0 
108
Pa


 1.3  10 3 m  1.3 mm
In order to punch a hole in the steel plate, the
superhero must punch out a plug with crosssectional area, Acs , equal to that of his fist and
a height t equal to the thickness of the steel plate. The area Ashear of the face that is sheared as the plug is removed is the cylindrical surface with radius r and height t as shown in the sketch. Since Acs = r2, then
r 
A cs  and
Ashear   2 r  t  2 t
Acs

 2  2.00 cm 
1.00  102 cm 2

 70.9 cm 2
If the ultimate shear strength of steel (i.e., the maximum shear stress it can withstand before shearing) is,
2.50  108 Pa  2.50  108 N m2 the minimum force required to punch out this plug is

F  Ashear  stress   70.9 cm2

 1 m2   
N 
8
6
 4
   2.50  10
  1.77  10 N
2
2
10
cm
m 

  
Page 9.1
Chapter 9
(b)
By Newton’s third law, the wall would exert a force of equal magnitude in the opposite direction on the
superhero, who would be thrown backward at a very high recoil speed.
9.3
Two cross-sectional areas in the plank, with one directly above the rail and one at the outer end of the plank, separated by distance h  2.00 m and each with area, A   2.00 cm  15.0 cm   30.0 cm2 move a distance
 x  5.00  102 m parallel to each other. The force causing this shearing effect in the plank is the weight of the
man F  mg applied perpendicular to the length of the plank at its outer end. Since the shear modulus S is given
by
shear stress
F A
Fh


shear strain
x h
 x A
S 
we have
S 
9.4

5.00 
 80.0 kg   9.80
m s2
m  30.0
cm2
10 2
 
  2.00 m 
 1 m2 104 cm2  
 1.05  107 Pa
As a liquid, the water occupied some volume Vl . As ice, the water would occupy volume 1.090Vl if it were not
compressed and forced to occupy the original volume. Consider the pressure change required to squeeze ice back
into volume Vl . Then, V0  1.09Vl and V   0.090Vl , so
 V 
N   0.090 Vl 

P   B 
   2.00  109
 1.65  108 Pa  1600 atm


m 2   1.09 Vl 
 V0 
9.5
Using Y  F L0 A (L) with A   d2 4 and F  mg , we get


4   90 kg  9.80 m s2   50 m 
Y 
 3.5  108 Pa
2
 1.0  10 2 m 1.6 m 

9.6

From Y  F L0 A (L) the tension needed to stretch the wire by 0.10 mm is
F 

Y A  L 
L0


Y  d2
  L 
4 L0
18  1010 Pa   0.22  10 3 m 2  0.10  10 3 m 
4  3.1  10 2 m 
Page 9.2
 22 N
Chapter 9
The tension in the wire exerts a force of magnitude F on the tooth in each direction along the length of the wire as shown
in the above sketch. The resultant force exerted on the tooth has an x-component of
Rx  Fx  F cos 30 F cos 30 0 , and a y-component of Ry  Fy  F sin 30º  F sin 30º  F  22 N .
Thus, the resultant force is
R  22 N directed down the page in the diagram .
9.7
From Y  (F A) (L0 L)  (stress) (L0 L) , the maximum compression the femur can withstand is
L 
9.8
(a)
 stress   L0 
Y

160  106 Pa   0.50 m 
18  109 Pa
 4.4  10 3 m  4.4 mm
When at rest, the tension in the cable equals the weight of the 800-kg object, 7.84  103 N .
Thus, from Y  F L0 A (L) , the initial elongation of the cable is
L 
(b)



7.84  103 N  25.0 m 
F  L0

 2.45  103 m = 2.5 mm
AY
4.00  104 m2 20  1010 Pa


When the load is accelerating upward, Newton’s second law gives
F  mg  may

F  m g  ay

[1]
If m  800 kg and ay  3.0 m s2 , the elongation of the cable will be
L 
  800 kg  9.80  3.0  m s2   25.0 m 
F  L0
 
 3.2  103 m  3.2 mm
AY
4.00  104 m2 20  1010 Pa



Thus, the increase in the elongation has been
increase    L     L  initial  3.20 mm  2.45 mm  0.75 mm
(c)
From the definition of the tensile stress, stress  F A , the maximum tension the cable can withstand is
Page 9.3
Chapter 9



Fmax  A   stress max  4.00  104 m2 2 .2  108 Pa  8.8  104 N
Then, Equation [1] above gives the mass of the maximum load as
mmax 
9.9
Fmax
8.8  10 4 N

 6.9  103 kg
ga
 9.8  3.0  m s2
From the defining equation for the shear modulus, we find the displacement,  x , as
x 
h F A
S




5.0  10 3 m  20 N   10 4 cm 2 
hF

SA
3.0  106 Pa 14 cm 2  1 m 2 


 2.4  10 5 m  0.024 mm
9.10
The shear modulus is given by
S 
shear stress
stress

shear strain
 x h
Hence, the stress is
 x 
 5.0 m 
stress  S 
 1.5  1010 Pa 
 7.5  106 Pa

 h 
 10  103 m 

9.11

The tension and cross-sectional area are constant through the entire length of the rod, and the total elongation is the
sum of that of the aluminum section and that of the copper section.
 Lrod   LAl   LCu 
F  L0  Al
AYAl

F  L0  Cu
AYCu

 L0 Cu 
F   L0  Al



A  YAl
YCu 


where A   r 2 with r  0.20 cm  2.0  103 m . Thus,
 Lrod 
 5.8  103 N   1.3 m  2.6 m   1.9  102 m 

2 
10
10
  2.0  103 m   7.0  10 Pa 11  10 Pa 
Page 9.4
1.9 cm
Chapter 9
9.12
The acceleration of the forearm has magnitude
v
a 
t
80

km  103 m   1 h 
h  1 km   3 600 s 
 4.4  103 m s2
5.0  10 3 s
The compression force exerted on the arm is F  ma and the compressional stress on the bone material is
Stress 


 3.0 kg  4.4  103 m s2
F

A
2.4 cm2 104 m2 1 cm2


 5.5  107 Pa
Since the stress is less than the allowed maximum, the arm should survive.
9.13
The average density of either of the two original worlds was
0 
M
M
3M


V
4 R3 3
4 R3
The average density of the combined world is
 
M total

V
4
3
2M
3 
 4 R 
3

42  2 M 

32


R3
32 M
9 R3
so

128
 32 M   4 R3 
 

 4.74



3
 9 R   3 M 
0
27
9.14
(a)
  4.740
or
The mass of gold in the coin is
mAu 
#
karats  mtotal
24

22
11
mtotal 
7.988  10 3 kg  7.322  10 3 kg
24
12


and the mass of copper is
mCu 
1
1
mtotal 
7.988  10 3 kg  6.657  10 4 kg
12
12


Page 9.5
Chapter 9
(b)
The volume of the gold present is
VAu 
mAu
Au

7.322  10 3 kg
 3.79  10 7 m 3
19.3  103 kg m 3
and the volume of the copper is
VCu 
(c)
(a)
Cu

6.657  10 4 kg
 7.46  10 8 m 3
8.92  103 kg m 3
The average density of the British sovereign coin is
av 
9.15
mCu
mtotal
mtotal
7.988  10 3 kg


 1.76  10 4 kg m 3
Vtotal
VAu  VCu
3.79  10 7 m 3  7.46  10 8 m 3
The total normal force exerted on the bottom acrobat’s shoes by the floor equals the total weight of the acro
bats in the tower. That is


n  mtotal g    75.0  68.0  62.0  55.0  kg  9.80 m s2  2.55  103 N
(b)
P 
n
Atotal

n
2.55  103 N

 3.00  104 Pa
2 Ashoe
2  425 cm2 1 m2 104 cm2 
sole


(c) If the acrobats are rearranged so different ones are at the bottom of the tower, the total weight supported, and
hence the total normal force n, will be unchanged. However, the total area Atotal  2 Ashoe sole , and hence the
pressure, will change unless all the acrobats wear the same size shoes.
9.16
We shall assume that each chair leg supports one-fourth of the total weight so the normal force each leg exerts on
the floor is n  mg 4 . The pressure of each leg on the floor is then
Pleg 
9.17
(a)


 95.0 kg  9.80 m s2
n
mg 4


 2.96  106 Pa
2
Aleg
r2

2
4 0.500  10 m


If the particles in the nucleus are closely packed with negligible space between them, the average nuclear
Page 9.6
Chapter 9
density should be approximately that of a proton or neutron. That is
nucleus 
(b)
mproton
Vproton

mproton
4 r 3
3

3 1.67  10 27 kg

4 1  10 15 m


4  1017 kg m3
3
The density of iron is Fe  7.86  103 kg m3 kg/m3 and the densities of other solids and liquids are on
the order of 103 kg m3 . Thus, the nuclear density is about 1014 times greater than that of common solids
and liquids, which suggests that atoms must be mostly empty space. Solids and liquids, as well as gases, are
mostly empty space.
9.18
Let the weight of the car be W. Then, each tire supports W 4 , and the gauge pressure is
P 
F
W

A
4A

Thus, W  4 A P  4 0.024 m2
9.19
  2.0  105 Pa  
1.9  104 N .
The volume of concrete in a pillar of height h and cross-sectional area A is V  Ah , and its weight is


Fg   Ah  5.0  104 N m3 . The pressure at the base of the pillar is then
P 
Fg
A

 Ah  5.0  104
A
N m3


 h 5.0  10 4 N m 3

Thus, if the maximum acceptable pressure on the base is, Pmax  1.7  107 Pa , the maximum allowable height is
hmax 
9.20
Pmax
1.7  107 Pa

 3.4  102 m
4
3
5.0  10 N m
5.0  10 4 N m 3
Assuming the spring obeys Hooke’s law, the increase in force on the piston required to compress the spring an additional amount  x is  F  F  F0   P  P0  A  k   x  . The gauge pressure at depth h beneath the surface
of a fluid is P  P0  gh , so we have, ghA  k   x  or the required depth is h  k   x  gA , If


k  1 250 N m , A  r2 with r  1.20  102 m , with   1.00  103 kg m3 , and the fluid is water (
= 1.00  103 kg/m3), the depth required to compress the spring an additional is 0.750 cm  7.50  103 m is
Page 9.7
Chapter 9
h 
9.21
1.00  103

1 250
N m  7.50  10 3 m
kg m3
  9.80

 
m s2   1.20  10 2 m


(a)
P  P0  gh  101.3  103 Pa  1.00  103 kg m3
(b)
The inward force the water will exert on the window is
2 
  9.80
 35.0  10 2 m 
F  PA  P  r 2  3.71  10 5 Pa  

2


9.22
 
 2.11 m

m s2
  27.5 m  
3.71  105 Pa
2
 3.57  10 4 N
The gauge pressure in a fluid at a level distance h below where Pgauge  0 is Pgauge  gh with h being positive
when measured in the downward direction. The difference in gauge pressures at two levels is
(Pgauge )2  (Pgauge )1  g (h) or (Pgauge)2 = (Pgauge)1 + g (h) with  h being positive if, in going from level 1
to level 2, one is moving lower in the fluid.
(a)
In moving from the heart to the head, one is moving higher in the blood column so h  0 and we find
P 
 Pgauge
P 
 8.11  103 Pa  8.11 kPa
gauge head


heart

  g   h   13.3  103 Pa  1 060 kg m3
  9.80
m s2
  0.500 m 
or
gauge head
(b)
In going from heart to feet, one moves deeper in the blood column, so h  0 and
P 
 Pgauge
P 
 26.8  103 Pa  26.8 kPa
gauge feet


heart

 g   h   13.3  103 Pa  1 060 kg m3
  9.80
m s2
  1.30 m 
or
gauge feet
9.23
The density of the solution is   1.02 water  1.02  103 kg m3 . The gauge pressure of the fluid at the level
of the needle must equal the gauge pressure in the vein, so Pgauge   gh  1.33  103 Pa , and
Page 9.8
Chapter 9
h 
9.24
Pgauge
g

1.33  103 Pa
1.02  103
kg m 3
  9.80
m s2

 0.133 m
From the definition of bulk modulus, B   P  V V0  , the change in volume of the 1.00 m3 of seawater
(a)
will be
V  
V0  P 
Bwater
 
1.00 m3  1.13  108 Pa  1.013  105 Pa 
0.210  1010 Pa
  0.053 8 m 3
The quantity of seawater that had volume V0  1.00 m3 at the surface has a mass of 1 030 kg. Thus, the
(b)
density of this water at the ocean floor is
 
(c)
m
m
1 030 kg


 1.09  103 kg m 3
V
V0  V
 1.00  0.053 8  m 3
Considering the small fractional change in volume (about 5%) and enormous change in pressure
generated, we conclude that it is a good approximation to think of water as incompressible.
9.25
We first find the absolute pressure at the interface between oil and water.
P1  P0  oil ghoil



 1.013  105 Pa  700 kg m 3 9.80 m s 2  0.300 m   1.03  10 5 Pa
This is the pressure at the top of the water. To find the absolute pressure at the bottom, we use, P2 = P1 + waterghwater,
or

P2  1.03  105 Pa  103 kg m3
9.26
  9.80
m s2
  0.200 m  
1.05  105 Pa
If we assume a vacuum exists inside the tube above the wine column, the pressure at the base of the tube (that is, at
the level of the wine in the open container) is Patmo = 0 + gh = gh. Thus,
h 
Patmo
1.013  10 5 Pa

g
984 kg m 3 9.80 m s 2



 10.5 m
Page 9.9
Chapter 9
Some alcohol and water will evaporate, degrading the vacuum above the column.
9.27
Pascal’s principle, F1 A1  F2 A2 , or Fpedal AMaster  Fbrake Abrake
cylinder
cylinder
, gives
 Abrake cylinder 
 6.4 cm2 
Fbrake  
 44 N   156 N
 Fpedal  
 1.8 cm2 
 Amaster cylinder 
This is the normal force exerted on the brake shoe. The frictional force is
f  k n  0.50 156 N   78 N
and the torque is   f  rdrum   78 N   0.34 m   27 N  m .
9.28
First, use Pascal’s principle, F1/A1 = F2/A2, to find the force
piston 1 will exert on the handle when a 500-lb force
pushes downward on piston 2.
 A 
  d2 4 
 d2 
F1   1  F2   12  F2   12  F2
 A2 
  d2 4 
 d2 

 0.25 in  2 500 lb


 1.5 in  2
 14 lb
Now, consider an axis perpendicular to the page, passing through the left end of the jack handle.  = 0 yields
 14 lb  2 .0 in   F  12 in   0 ,
9.29
F  2.3 lb
When held underwater, the ball will have three forces acting on it: a downward gravitational force, mg; an upward
buoyant force, B = waterV = 4waterr3/3; and an applied force, F. If the ball is to be in equilibrium, we have (taking
upward as positive) Fy  F  B  mg  0 , or

 4 r 3 
 4 r 3  
F  mg  B  mg  water 
g

m



water
 3   g
 3 


giving
Page 9.10
Chapter 9
3

4  0.250 m  
2
F   0.540 kg  1.00  103 kg m3
  9.80 m s  74.9 N
3 
2






so the required applied force is . F  74.9 N directed downward
9.30
(a)
To float, the buoyant force acting on the person must equal the weight of that person, or the weight of the
water displaced by the person must equal the person’s own weight. Thus,
B  mg

sea gVsubmerged  body gVtotal
Vsubmerged
or
Vtotal

body
sea
After inhaling,
Vsubmerged
Vtotal

945 kg m 3
 0.768  76.8%
1 230 kg m 3
leaving .23.2% above surface.
After exhaling,
Vsubmerged
Vtotal

1 020 kg m 3
 0.829  82.9%
1 230 kg m 3
leaving .17.1% above surface.
(b)
In general, “sinkers” would be expected to be thinner with heavier bones, whereas “floaters” would have
lighter bones and more fat.
9.31
The boat sinks until the weight of the additional water displaced equals the weight of the truck. Thus,
wtruck   water  V   g
kg 
m

  4.00 m  6.00 m  4.00  10 2 m   9.80
  103


m 3  
s2 

or
wtruck  9.41  103 N  9.41 kN
Page 9.11

Chapter 9
9.32
(a)
(b)
Since the system is in equilibrium, Fy  B  w  wr  0 .
(c)
B  w gVsubmerged  w g  d  A 

 1 025 kg m 3
(d)
  9.80
m s2
  0.024 0 m   4.00 m 2  
964 N
From B  w  wr  0 ,


wr  B  w  B  ms g  964 N   62.0 kg  9.80 m s2  356 N
mr
w g
 r

Vr
tA
9.80 m s2
(e)
foam 
(f)
Bmax  w gVr  w g  t  A 


 1 025 kg m 3
(g)
(a)
m s2

 0.090 m  4.00 m 2

 101 kg m 3
  0.090 0 m   4.00 m 2  
3.62  10 3 N
The maximum weight of survivors the raft can support is wmax  mmax g  Bmax  wr so
mmax 
9.33
  9.80

356 N
Bmax  wr
3.62  103 N  356 N

 333 kg
g
9.80 m s2
While the system floats, B  wtotal  wblock  wsteel .or w g Vsubmerged  b g Vb  msteel g .
When msteel  0.310 kg , Vsubmerged  Vb  5.24  104 m3 giving
b 
wVb  msteel
Vb
 w 
msteel
0.310 kg
 1.00  103 kg m 3 
 408 kg m 3
Vb
5.24  10 4 m3
(b) If the total weight of the block + steel system is reduced, by having msteel  0.310 kg, a smaller buoyant force
is needed to allow the system to float in equilibrium. Thus, the block will displace a smaller volume of water
and will be only partially submerged in the water.
The block is fully submerged when msteel = 0.310 kg. The mass of the steel object can increase slightly above
this value without causing it and the block to sink to the bottom. As the mass of the steel object is gradually in-
Page 9.12
Chapter 9
creased above 0.310 kg, the steel object begins to submerge, displacing additional water, and providing a slight
increase in the buoyant force. With a density of about eight times that of water, the steel object will be able to
displace approximately 0.310 kg/8 = 0.039 kg of additional water before it becomes fully submerged. At this
point, the steel object will have a mass of about 0.349 kg and will be unable to displace any additional water.
Any further increase in the mass of the object causes it and the block to sink to the bottom. In conclusion, the
block + steel system will sink if mstee  0.350 kg.
9.34
(a)
(b)
Since the balloon is fully submerged in air, Vsubmerged = Vb = 325 m3, and

B  air gVb  1.29 kg m3
(c)
  9.80
m s2
  325 m3  
4.11  103 N
Fy  B  wb  wHe  B  mb g  He gVb  B   mb  HeV  g


 

 4.11  103 N   226 kg  0.179 kg m3 325 m 3  9.80 m s 2  1.33  103 N
Since Fy = may  0, ay will be positive (upward), and the balloon rises.
(d)
If the balloon and load are in equilibrium, Fy   B  wb  wHe   wload  0 and
wload   B  wb  wHe   1.33  103 N . Thus, the mass of the load is
mload 
(e)
wload
1.33  103 N

 136 kg
g
9.80 m s2
If mload  136 kg, then the net force acting on the balloon + load system is upward and
the balloon and its load will accelerate upward.
(f)
The density of the surrounding air, temperature, and pressure all decrease as the balloon rises. Because of
Page 9.13
Chapter 9
these effects, the buoyant force will decrease until at some height the balloon will come to equilibrium and go
no higher.
9.35
(a)
(b)
 4 r 3 
B  air gVballoon  air g 
 1.29 kg m3
 3 
 1.43  103 N  1.43 kN

  9.80

 4 
3
m s2 
3.00 m 


 3 
 Fy  B  wtotal  1.43  103 N   15.0 kg  9.80 m s 2


 1.28  103 N  1.28 kN upward
(c)
The balloon expands as it rises because the external pressure (atmospheric pressure) decreases with increasing altitude.
9.36
(a)
Taking upward as positive, Fy  B  mg  may , or may  w gV  mg .
(b)
Since m = V, we have  V ay  w g V   V g , or


ay   w  1 g
 

(c)
 1.00  103 m kg3

ay  
 1 9.80 m s2  0.467 m s2  0.467 m s 2 downward
 1 050 m kg3

(d)
From y  v0 y t  ay t 2 2 with 0y = 0, we find

t 
2  y 
ay

2  8.00 m 
0.467 m s2

 5.85 s
Page 9.14
Chapter 9
9.37
(a)

 4 3  
 600  air gVballoon   600  air g 
r 
 3

balloon

4

 600  1.29 kg m3 9.80 m s2
 0.50 m 3   4.0  10 3 N  4.0 kN
3


Btotal  600  Bsingle





(b)
Fy  Btotal  mtotal g  4.0  103 N  600  0.30 kg  9.8 m s2  2.2  103 N  2.2 kN
(c)
Atmospheric pressure at this high altitude is much lower than at Earth’s surface, so the balloons expanded
and eventually burst.
9.38
Note: We deliberately violate the rules of significant figures in this problem to illustrate a point.
(a)
The absolute pressure at the level of the top of the block is
Ptop  P0  water ghtop
kg  
m

 1.0130  105 Pa   103
9.80 2  5.00  10 2 m

m 3  
s 


 1.0179  105 Pa
and that at the level of the bottom of the block is
Pbottom  P0  water ghbottom
kg  
m

 1.0130  105 Pa   103
9.80 2  17.0  10 2 m

m 3  
s 


 1.0297  105 Pa
Thus, the downward force exerted on the top by the water is


Ftop  Ptop A  1.0179  105 Pa  0.100 m   1017.9 N
2
and the upward force the water exerts on the bottom of the block is


Fbot  Pbot A  1.0297  105 Pa  0.100 m   1029.7 N
2
(b) The scale reading equals the tension, T, in the cord supporting the block. Since the block is in equilibrium,
Fy  T  Fbot  Ftop  mg  0 , or
Page 9.15
Chapter 9


T  10.0 kg  9.80 m s2  1029.7  1017.9  N  86.2 N
(c)
From Archimedes’s principle, the buoyant force on the block equals the weight of the displaced water. Thus,
B   water Vblock  g




2
 103 kg m3   0.100 m   0.120 m   9.80 m s2  11.8 N


From part (a), Fbot  Ftop  1 029.7  1 017.9  N  11.8 N , which is the same as the buoyant force found
above.
9.39
Constant velocity means that the submersible is in equilibrium under the gravitational force, the upward buoyant
force, and the upward resistance force:
Fy  may  0


 1.20  104 kg  m g  sea water gV  1 100 N  0
where m is the mass of the added sea water and V is the sphere’s volume.
Thus,
kg    4 
1 100 N

3
m   1.03  103
1.50 m   
 1.20  10 4 kg





3
2

m   3 
9.80
m
s

or
m  2.67  103 kg
9.40
At equilibrium, Fy  B  Fspring  mg  0 , so the spring force is
Fspring  B  mg    water Vblock   m  g
where
Vblock 
m
wood

5.00 kg
 7.69  10 3 m 3
650 kg m3
Page 9.16
Chapter 9





Thus, Fspring   103 kg m 3 7.69  10 3 m 3  5.00 kg  9.80 m s2  26.4 N.
The elongation of the spring is then
x 
9.41
(a)
Fspring
k

26.4 N
 0.165 m  16.5 cm
160 N m
The buoyant force is the difference between the weight in air and the apparent weight when immersed in the
alcohol, or B = 300 N  200 N = 100 N. But, from Archimedes’s principle, this is also the weight of the
displaced alcohol, so B = (alcoholV)g. Since the sample is fully submerged, the volume of the displaced alco-
hol
is the same as the volume of the sample. This volume is
B
V 
(b)
alcohol g

 700
100 N
kg m3
  9.80
m s2

 1.46  10 2 m3
The mass of the sample is
weight in air
300 N

 30.6 kg
g
9.80 m s2
m 
and its density is
 
9.42
m
30.6 kg

 2.10  103 kg m3
V
1.46  10 2 m3
The difference between the weight in air and the apparent weight when immersed is the buoyant force exerted on
the object by the fluid.
(a)
The mass of the object is
m 
weight in air
300 N

 30.6 kg
g
9.80 m s2
The buoyant force when immersed in water is the weight of a volume of water equal to the volume of the
object, or Bw = (wV)g. Thus, the volume of the object is
Page 9.17
Chapter 9
Bw
300 N  265 N

3
w g
10 kg m 3 9.80 m s2
V 



 3.57  10 3 m 3
and its density is
object 
(b)
m
30.6 kg

 8.57  103 kg m 3
V
3.57  10 3 m3
The buoyant force when immersed in oil is equal to the weight of a volume V = 3.57  10.3 m3 of oil. Hence
Boil= (oilV)g, or the density of the oil is
oil 
9.43
Boil
300 N  275 N

Vg
3.57  10 3 m 3 9.80 m s2



 714 kg m 3
The volume of the iron block is
V 
miron
iron

2 .00 kg
 2 .54  10 4 m 3
7.86  103 kg m 3
and the buoyant force exerted on the iron by the oil is




B   oil V  g  916 kg m3 2 .54  104 m3 9.80 m s2  2 .28 N
Applying Fy = 0 to the iron block gives the support force exerted by the upper scale (and hence the reading on that
scale) as
Fupper  miron g  B  19.6 N  2.28 N  17.3 N
From Newton’s third law, the iron exerts force B downward on the oil (and hence the beaker). Applying Fy = 0 to
the system consisting of the beaker and the oil gives
Flower  B   moil  mbeaker  g  0
The support force exerted by the lower scale (and the lower scale reading) is then


Flower  B   moil  mbeaker  g  2.28 N    2.00  1.00  kg  9.80 m s2  31.7 N
Page 9.18
Chapter 9
9.44
(a)
The cross-sectional area of the hose is A   r 2   d 2 4    2.74 cm 
2
4 , and the volume flow rate
(volume per unit time) is A =25.0 L/1.50 min. Thus,
  1 min   103 cm3 
 25.0 L  
25.0 L 1.50 min
4


 


A
 1.50 min      2.74  2 cm 2   60 s   1 L 


 1m 
  47.1 cm s   2
 0.471 m s
 10 cm 
v 
(b)
d 
  d2   4 
A2
  2 2   2
A1
 4    d1 
 d1 
2
 1
  
 3
2

1
9
or
A2 
A1
9
Then from the equation of continuity, A2v2  A1v1 , we find
 A 
v 2   1  v1  9  0.471 m s   4.24 m s
 A2 
9.45
(a)
The volume flow rate is A, and the mass flow rate is
 Av  1.0 g cm3   2.0 cm2   40 cm s   80 g s
(b)
From the equation of continuity, the speed in the capillaries is
 Aaorta 


2 .0 cm2
vcapiliaries  
 40 cm s 
 vaorta  
3 cm2 
A
3.0

10


 capillaries 
or vcapiliaries  2.7  102 cm s  0.27 mm s .
9.46
(a)
From the equation of continuity, the flow speed in the second pipe is
 A 
 10.0 cm 2 
v 2   1  v1  
 2.75 m s   11.0 m s
 2.50 cm 2 
 A2 
(b)
Using Bernoulli’s equation and choosing y = 0 along the centerline of the pipes gives
P2  P1 
1
1
2
2
 v12  v 22  1.20  105 Pa  1.65  103 kg m3   2.75 m s   11.0 m s  


2
2



or P2  2.64  104 Pa .
Page 9.19

Chapter 9
9.47
From Bernoulli’s equation, choosing at y = 0 the level of the syringe and needle, so the flow speed in the needle,
P2 
1
2
v22  P1 
v2 
1
2
v12 is
2  P1  P2 
v12 

In this situation,
P1  P2  P1  Patmo   P1  gauge 
F
2 .00 N

 8.00  10 4 Pa
A1
2 .50  10 5 m 2
Thus, assuming   0,
v2 
9.48
0
1.00 
103

 12.6 m s
kg m3
We apply Bernoulli’s equation, ignoring the very small change in vertical position, to obtain
P1  P2 
P 
9.49

2 8.00  104 Pa
(a)
1

2
 v 22

 v12 
1

2
  2 v 2  v 2  
1
1 


3
1.29 kg m3 15  10 2 m s
2


2
3
 v12 ,
2
or
 4.4  10 2 Pa
Assuming the airplane is in level flight, the net lift (the difference in the upward and downward forces exerted on the wings by the air flowing over them) must equal the weight of the plane, or
(Plower  Pupper ) Awings  mg
surface
surface
. This yields


8.66  104 kg 9.80 m s2


mg
P

P


upper 
 lower
Awings
90.0 m2
 surface
surface 
(b)

 9.43  103 Pa
Neglecting the small difference in altitude between the upper and lower surfaces of the wings and applying
Bernoulli’s equation yields
Plower 
1
1
2
2
air v lower
 Pupper  air vupper
2
2
Page 9.20
Chapter 9
so
v upper 
(c)
2
v lower


2 Plower  Pupper


air
 225
m s 
2

2 9.43  103 Pa
1.29 kg
m3

 255 m s
The density of air decreases with increasing height, resulting in a smaller pressure difference. Beyond the
maximum operational altitude, the pressure difference can no longer support the aircraft.
9.50
For level flight, the net lift (difference between the upward and downward forces exerted on the wing surfaces by
air flowing over them) must equal the weight of the aircraft, or (Plower  Pupper ) A  Mg . This gives the air
surface
surface
pressure at the upper surface as
Pupper
surface
9.51
(a)
 Plower 
surface
Mg
A
Since the pistol is fired horizontally, the emerging water stream has initial velocity components of (0x =
nozzle, 0y = 0) Then, y = 0yt + ayt2/2, with ay = g, gives the time of flight as
t 
(b)
2  y 
ay

2  1.50 m 
9.80 m s2
 0.553 s
With ax = 0, and 0x = nozzle, the horizontal range of the emergent stream is x = nozzlet where t is the time of
flight from above. Thus, the speed of the water emerging from the nozzle is
v nozzle 
(c)
x
8.00 m

 14.5 m s
t
0.553 s
From the equation of continuity,A11 = A22, the speed of the water in the larger cylinder is 1 = (A2/A1) 2 =
(A2/A1)nozzle, or
2
2
 r2 
r 
 1.00 mm 
v1   22  v nozzle   2  v nozzle  
14.5 m s   0.145 m s
 10.0 mm 
  r1 
 r1 
(d)
The pressure at the nozzle is atmospheric pressure, or P2 = 1.013  105 m/s .
(e)
With the two cylinders horizontal, y1 = y2 and gravity terms from Bernoulli’s equation can be neglected,
Page 9.21
Chapter 9
leaving P1  wv12 2  P2  wv22 2 so the needed pressure in the larger cylinder is
P1  P2 
w
2
 v22

 v12  1.013  105 Pa 
1.00  103 kg m3 
2
2
14.5 m s    0.145 m s  




2
or
P1  2.06  105 Pa
(f)
To create an overpressure of P = 2.06 105 Pa = 1.05  105 Pa in the larger cylinder, the force that must be
exerted on the piston is

 
 
F1    P  A1    P   r12  1.05  105 Pa  1.00  10 2 m
9.52
(a)
2
 33.0 N
From Bernoulli’s equation,
P1 
w v12
2
 w gy1  P2 
w v 22
2
 w gy2
or
Figure 9.29
 P  P2

v 22  v12  2  1
 g  y2  y1  
 w

and using the given data values, we obtain
 1.75  10 5 Pa  1.20  10 5 Pa
v 22  v12  2 
 9.80 m s2
3 kg m 3
1.00

10


and

  2.50 m  

[1]
v22  v12  61.0 m2 s2
From the equation of continuity,
2
A 
 r2 
r 
v2   1  v1   12  v1   1  v1
 A2 
  r2 
 r2 
Page 9.22
Chapter 9
2
 3.00 cm 
v2  
v
 1.50 cm  1
2 = 4 1
or
[2]
Substituting Equation [2] into [1] gives 16  1 v12  61.0 m2 s2 , or
61.0 m 2 s2
 2.02 m s
15
v1 
(b)
Equation [2] above now yields 2 = 4(2.02 m/s) = 8.08 m/s.
(c)
The volume flow rate through the pipe is: flow rate = A11 = A22.
Looking at the lower point:



flow rate   r12 v1   3.00  102 m
9.53
2  2.02
m s   5.71  10 3 m3 s
First, consider the path from the viewpoint of projectile motion to find the speed at which the
water emerges from the tank. From y  v 0 y t 
t 
2  y 
ay

2  1.00 m 
9.80 m s2
1
2
ay t 2 with 0y = 0, we find the time of flight as
 0.452 s
From the horizontal motion, the speed of the water coming out of the hole is
v2  v0 x 
x
0.600 m

 1.33 m s
t
0.452 s
We now use Bernoulli’s equation, with point 1 at the top of the tank and point 2 at the level of the hole. With P1 =
P2 = Patmo and, 1  0. This gives
 g y1 
1 2
 v 2   g y2
2
or
1.33 m s 
v 22

2g
2 9.80 m s2
2
h  y1  y2 


 9.00  102 m  9.00 cm
Page 9.23
Chapter 9
9.54
(a)
Apply Bernoulli’s equation with point 1 at the open top of the tank and point 2 at the opening of the hole.
1
v22
2
Then, P1 = P2 = Patmo and we assume 1  0. This gives
v2 
(b)
2 g  y1  y2  

  gy2   gy1 , or

2 9.80 m s2 16.0 m   17.7 m s
The area of the hole is found from
A2 
flow rate
2 .50  10 3 m 3 min  1 min 
6
2

 60 s   2 .35  10 m
v2
17.7 m s
The diameter is then
d2 
9.55
4 A2


4 2.35  10 6 m 2


 1.73  10 3 m  1.73 mm

First, determine the flow speed inside the larger portions from
v1 
flow rate
1.80  104 m3 s

 0.367 m s
2
A1
 2.50  102 m 4


The absolute pressure inside the large section on the left is P1 = P0 + gh1, where h1 is the height of the water in the
leftmost standpipe. The absolute pressure in the constriction is P2 = P1 + gh2, so
P1  P2   g h1  h2    g 5.00 cm 
The flow speed inside the constriction is found from Bernoulli’s equation with y1 = y2. This gives
v 22  v12 
v2 
2

 P1
 0.367
 P2   v12  2 g  h1  h2 


m s   2  9.80 m s  5.00  10 2 m  1.06 m s
2
The cross-sectional area of the constriction is then
A2 
flow rate
1.80  10 4 m 3 s

 1.71  10 4 m 2 ,
v2
1.06 m s
Page 9.24
Chapter 9
and the diameter is
d2 
9.56
(a)
4 A2


4 1.71  10 4 m 2



 1.47  10 2 m  1.47 cm
For minimum pressure, we assume the flow is very slow. Then, Bernoulli’s equation gives
1 2
1




=  P  v 2   gy 
 P  2 v   gy 

 rim
2
river
 Priver min
 0  1 atm  0   g yrim  yriver 
 Priver  min
kg  
m

 1.013  10 5 Pa   10 3
9.80 2   2096 m  564 m 


3



m
s 
 Priver min
 1.013  105  1.50  107 Pa =1.51  107 Pa  15.1 MPa
or
(b)

The volume flow rate is flow rate =A=( d2 /4). Thus, the velocity in the pipe is
v 
(c)

4  flow rate 
 d2



4 4500 m3 d  1 d 
  2.95 m s
2 
  0.150 m   86 400 s 
We imagine the pressure being applied to stationary water at river level, so Bernoulli’s equation becomes
1 2
Priver  0  1 atm   g yrim  yriver     v rim
2
or
Priver   Priver  min 
1 2
1
kg  
m
 v rim   Priver  min   103    2.95 
2
2
m 
s
  Priver  min  4.35 kPa
Page 9.25
2
Chapter 9
The additional pressure required to achieve the desired flow rate is
P = 4.35 kPa
9.57
(a)
For upward flight of a water-drop projectile from geyser vent to fountain-top, v 2y  v02y  2 ay  y  , with y
= 0, when y = ymax, gives
v0 y 
0  2 ay  y  max 

2 9.8 m s2
  40.0 m 
 28.0 m s
(b) Because of the low density of air and the small change in altitude, atmospheric pressure at the fountain top
will be considered equal to that at the geyser vent. Bernoulli’s equation, with top = 0, then gives

1 2
 v vent  0   g ytop  yvent
2

or
v vent 
(c)

2 g ytop  yvent


2 9.80 m s2
  40.0 m 
 28.0 m s
Between the chamber and the geyser vent, Bernoulli’s equation with chamber  0 yields
 P  0   g y chamber
 Patm 
1 2
 v vent   g yvent
2
or
1 2

P  Patm    v vent
 g  yvent  ychamber  
2

2

m
 3 kg    28.0 m s 

  2.11 MPa
  10

9.80
175
m







m3  
2
s2 



or



Pgauge  P  Patmo  2.11  106 Pa 1 atm 1.013  105 Pa  20.8 atmospheres
9.58
(a)
Since the tube is horizontal, y1 = y2 and the gravity
Page 9.26
Chapter 9
terms in Bernoulli’s equation cancel, leaving
P1 
1 2
1
v1  P2  v 22
2
2
or
Figure 9.30(a)
v 22  v12 
2  P1  P2 

2 1.20  103 Pa



7.00  10 2 kg m 3
and
v22  v12  3.43 m2 s2
[1]
From the continuity equation, A11 = A22, we find
2
2
A 
r 
 2.40 cm 
v2   1  v1   1  v1  
v
 1.20 cm  1
 A2 
 r2 
or
v2  4v1
[2]
Substituting Equation [2] into [1] yields 15v12  3.43 m2 s2 and 1 = 0.478 m/s.
Then, Equation [2] gives 2 = 4(0.478 m/s) = 1.91 m/s.
(b)
The volume flow rate is



A1v1  A2 v2   r22 v2   1.20  102 m
9.59
2 1.91 m s  
8.64  10 4 m3 s
From Fy = T  mg  Fy = 0, the balance reading is found to be T = mg + Fy where Fy is the vertical component of
the surface tension force. Since this is a two-sided surface, the surface tension force is F = (2L) and its vertical
component is F = (2L)cos where  is the contact angle. Thus, T = mg + 2Lcos .
T  0.40 N when   0 
mg  2  L 0.40 N
[1]
T  0.39 N when   180 
mg  2  L  0.39 N
[2]
Subtracting Equation [2] from [1] gives
Page 9.27
Chapter 9
 
9.60
=

F
1.61  10 2 N
=
 7.32  10 2 N m
4 r 4 1.75  10 2 m


h  gr
2 cos 
 2.1  102 m  1 080
kg m 3
  9.80
m s2
  5.0  10 4 m 
2 cos0
 5.6  10 2 N m
2  0.058 N m  cos 0 
2  cos 

 5.6 m
 gr
1 050 kg m 2 9.80 m s2 2.0  10 6 m




From the definition of the coefficient of viscosity,  = FL/A, the required force is
F 
9.64
F
F
=
Ltotal
2 (circumference)
The height the blood can rise is given by
h 
9.63

From h = 2cos/gr, the surface tension is
 
9.62

Because there are two edges (the inside and outside of the ring), we have
 =
9.61
0.40 N  0.39 N
0.40 N  0.39 N

 8.3  10 2 N m
4L
4 3.0  10 2 m
 Av
L

1.79  10 3

N  s m 2   0.800 m  1.20 m    0.50 m s 
 8.6 N
0.10  10 3 m
From the definition of the coefficient of viscosity,  = FL/A, the required force is
F 
 Av
L

1 500  103

N  s m 2   0.010 m   0.040 m    0.30 m s 
 0.12 N
1.5  10 3 m
Page 9.28
Chapter 9
9.65
Poiseuille’s law gives flow rate 
 P1
 P2   R 4
8 L
and P2 = Patm in this case. Thus, the desired gauge pressure is
P1  Patm 
8 L  flow rate 

 R4



8 0.12 N  s m2  50 m  8.6  10 5 m3 s
  0.50  10 2 m 

4
or
P1  Patm  2.1  106 Pa  2.1 MPa
9.66
From Poiseuille’s law, the flow rate in the artery is
 P   R 4
flow rate 
8 L


 400 Pa    2 .6  103 m 
4

8 2 .7  10 3 N  s m2 8.4  10 2 m

 3.2  10 5 m3 s
Thus, the flow speed is
v 
9.67
flow rate
3.2  10 5 m3 s

 1.5 m s
2
A
 2.6  103 m


If a particle is still in suspension after 1 hour, its terminal velocity must be less than
 v t max
cm   1 h   1 m 

  5.0
 1.4  10 5 m s .

h   3 600 s   100 cm 
Thus, from 1 = 2r2 g(f)/9, we find the maximum radius of the particle:
rmax 

9  v t  max

2 g   f


9 1.00  10 3 N  s m 2

2 9.80 m
s2

 1.4  105
ms
  1 800  1 000  kg m 3 
Page 9.29

 2.8  10 6 m  2.8 m
Chapter 9
9.68
From Poiseuille’s law, the excess pressure required to produce a given volume flow rate of fluid with viscosity 
through a tube of radius R and length L is
P 
8 L  V  t 
 R4
If the mass flow rate is (m/t) = 1.0  103 kg/s, the volume flow rate of the water is
V
 m t
1.0  10 3 kg s


 1.0  10 6 m 3 s
t

1.0  103 kg m 3
and the required excess pressure is
P 
9.69



4
  0.15  10 3 m 
8 1.0  10 3 Pa  s 3.0  10 2 m 1.0  10 6 m3 s

 1.5  105 Pa
With the IV bag elevated 1.0 m above the needle, the pressure difference across the needle is

 P  gh  1.0  103 kg m3
  9.8 m s2  1.0 m   9.8  103 Pa
and the desired flow rate is

500 cm 3 1 m 3 106 cm 3
V

t
30 min  60 s 1 min 

 2.8  10 7 m 3 s
Poiseuille’s law then gives the required diameter of the needle as
 8L  V  t  
D  2R  2 

    P  
14

or
D  4.1  104 m  0.41 mm
9.70





14
 8 1.0  10 3 Pa  s 2.5  10 2 m 2.8  10 7 m 3 s 

 2
 9.8  103 Pa


We write Bernoulli’s equation as
Page 9.30
Chapter 9
Pout 
1 2
1 2
v out   gyout  Pin  v in
  gyin
2
2
or
1 2

2
Pgauge  Pin  Pout    v out
 v in
 g  yout  yin  
2



Approximating the speed of the fluid inside the tank as in  0, we find
1
2
Pgauge  1.00  103 kg m 3   30.0 m s   9.80 m s 2
2




  0.500 m  
or
Pgauge  4.55  105 Pa  455 kPa
9.71
The Reynolds number is
RN 
1 050 kg m3  0.55 m s  2 .0  10 2 m   4.3 × 103
v d


2 .7  10 3 N  s m 2
In this region (RN > 3 000), the flow is turbulent.
9.72
From the definition of the Reynolds number, the maximum flow speed for streamlined (or laminar) flow in this
pipe is
vmax 
9.73
   RN max
d

1.0  103 N  s m2   2 000 
1 000 kg m3   2.5  102 m 
 0.080 m s  8.0 cm s
The observed diffusion rate is 8.0  1014 kg/15 s = 5.3  1015 kg/s. Then, from Fick’s law, the difference in concentration levels is found to be
C2  C1 

 diffusion rate  L
DA

 5.3  10 15
5.0 
10 10
m2

kg s  0.10 m 

s 6.0  10 4 m 2

Page 9.31
 1.8  10 3 kg m 3
Chapter 9
9.74
Fick’s law gives the diffusion coefficient as D = diffusion rate/A.(C/L), where C/L is the concentration gradient.
Thus,
D 
9.75
 2.0  10 4 m2    3.0  10 2
kg m 4

 9.5  10 10 m 2 s
Stokes’s law gives the viscosity of the air as
 
9.76
5.7  10 15 kg s
F
3.0  10 13 N

 1.4  10 5 N  s m 2
6 r v
6 2 .5  10 6 m 4.5  10 4 m s



Using 1 =2r2 g(f)/9, the density of the droplet is found to be  = f + 9t/2r2 g. Thus, if r = d/2 = 0.500 
103 m t = 1.10  102 m/s and when falling through 20 °C water ( = 1.00  103 N.s/m2) the density of the oil is
  1 000
9.77
(a)


 

9 1.00  10 3 N  s m2 1.10  10 2 m s
kg

 1.02  103 kg m3
2
m3

4
2
2 5.00  10 m 9.80 m s


Both iron and aluminum are denser than water, so both blocks will be fully submerged. Since the two blocks
have the same volume, they displace equal amounts of water and the buoyant forces acting on the two blocks
are equal.
(b)
Since the block is held in equilibrium, the force diagram at the right shows that
Fy  0  T  mg  B
The buoyant force B is the same for the two blocks, so the spring scale
reading T is largest for the iron block, which has a higher density, and
hence weight, than the aluminum block.
(c)
The buoyant force in each case is

B   waterV  g  1.0  103 kg m3
  0.20 m3   9.8 m s2  
2.0  103 N
For the iron block:

Tiron   ironV  g  B  7.86  103 kg m3
  0.20 m3   9.8 m s2   B
or
Page 9.32
Chapter 9
Tiron  1.5  104 N  2.0  103 N  13  103 N
For the aluminum block:

Taluminum   aluminumV  g  B  2.70  103 kg m3
  0.20 m3   9.8 m s2   B
or
Taluminum  5.2  103 N  2.0  103 N  3.3  103 N
9.78
In going from the ocean surface to a depth of 2.40 km, the increase in pressure is

P  P  P0  gh  1.025  103 kg m2
  9.80 m s2   2.40  103 m   2.41  107 Pa
The fractional change in volume of the steel ball is given by the defining equation for bulk modulus,
P =B(V/V), as
V
P
2.41  10 7 Pa
 
 
 1.51  10 4
V
Bsteel
16.0  1010 Pa
9.79
(a)
From Archimedes’s principle, the granite continent will
sink down into the peridotite layer until the weight
f the displaced peridotite equals the weight of the continent.
Thus, at equilibrium,
 g  At   g   p  Ad   g




or g t  p d .
(b)
If the continent sinks 5.0 km below the surface of the peridotite, then d = 5.0, and the result of part (a) gives
the first approximation of the thickness of the continent as
 p 
 3.3  103 kg m3 
t   d  
 5.0 km   5.9 km
 2.8  103 kg m3 
 g 
9.80
(a)
Starting with P = P0 + gh, we choose the reference level at the level of the heart, so P0 = PH. The pressure at
the feet, a depth hH below the reference level in the pool of blood in the body is PF = PH + ghH. The pressure
Page 9.33
Chapter 9
difference between feet and heart is then
PF = PH + ghH .
(b)
Using the result of part (a),

PF  PH  1.06  103 kg m3
9.81
  9.80

m s2 1.20 m   1.25  104 Pa
The cross-sectional area of the aorta is A1   d12 4 and that of a single capillary is Ac   d22 4 If the circulatory system has N such capillaries, the total cross-sectional area carrying blood from the aorta is
A2  NAc 
N  d22
4
From the equation of continuity,
 v   d2
v 
N  d22
  1 1 ,
A2   1  A1 , or
4
 v2  4
 v2 
which gives
v d 
N   1 1
 v2   d2 
9.82
(a)
2

  0.50  10 2 m 
1.0 m s
 

2
 1.0  10 m s   10  10 6 m 
2
 2.5  107
We imagine that a superhero is capable of producing a perfect vacuum above the water in the straw. Then
P = P0 + gh, with the reference level at the water surface inside the straw and P being atmospheric pressure
on the water in the cup outside the straw, gives the maximum height of the water in the straw as
hmax 
9.83
Patm  0
Patm
1.013  10 5 N m 2


water g
water g
1.00  103 kg m 3 9.80 m s 2


(b)
The moon has no atmosphere so Patm = 0, which yields .hmax = 0.
(a)
P  160 mm of H2O  H2O g160 mm 
kg  
m

  103
9.80 2   0.160 m   1.57 kPa


3

m 
s 
Page 9.34

 10.3 m
Chapter 9
1 atm


P  1.57  103 Pa 
 1.55  102 atm
5
 1.013  10 Pa 


The pressure is P  H2O ghH2O  Hg ghHg
 H O 


103 kg m3
hHg   2  hH2O  
160 mm   11.8 mm of Hg
3
3
 13.6  10 kg m 
 Hg 
9.84
(b)
The fluid level in the tap should rise.
(c)
Blockage of flow of the cerebrospinal fluid.
When the rod floats, the weight of the displaced fluid equals the weight of the rod, or  f gVdisplaced  0 gVrod .
But, assuming a cylindrical rod, Vrod   r 2 L . The volume of fluid displaced is the same as the volume of the rod
that is submerged, or Vdisplaced   r 2  L  h  . Thus,  f g  r 2  L  h    0 g  r 2 L  , which reduces to
 L 
 L  h 
 f  0 
9.85
Consider the diagram and apply Bernoulli’s equation to
points A and B, taking y = 0 at the level of point B,
and recognizing that A  0. This gives
PA  0  w g h  L sin  
1
 PB  w v B2  0
2
Recognize that PA = PB = Patm since both points
are open to the atmosphere. Thus, we obtain
vB 
2 g h  L sin   


2 9.80 m s2 10.0 m   2 .00 m  sin 30.0º   13.3 m s
Now the problem reduces to one of projectile motion with
v0 y  v B sin 30.0º  6.64 m s
Page 9.35
Chapter 9
At the top of the arc,y = 0, and y = ymax.


2
Then, v 2y  v02y  2 ay  y  gives 0   6.64 m s   2 9.80 m s2  ymax  0  , or
ymax  2.25 m above the level of point B .
9.86
When the balloon comes into equilibrium, the weight of the displaced air equals the weight of the filled balloon
plus the weight of string that is above ground level. If ms and L are the total mass and length of the string, the mass
of string that is above ground level is (h/L)ms. Thus,
 h
air gVballoon  mballoon g  helium gVballoon    ms g
 L
which reduces to
  air  helium Vballoon  mballoon 
h  
L
ms


This yields
h 
9.87
1.29

3
kg m3  0.179 kg m3  4  0.40 m  3   0.25 kg


 2.0 m   1.9 m
0.050 kg
When the balloon floats, the weight of the displaced air equals the combined weights of the filled balloon and its
load. Thus,
air gVballoon  mballoon g  helium gVballoon  mload g ,
or
Vballoon 
mballoon  mload
600 kg  4 000 kg

 4.14  103 m 3
air  helium
 1.29  0.179  kg m3
9.88
Page 9.36
Chapter 9
(a)
Consider the pressure at points A and B in part (b) of the figure by applying P = P0 + fgh. Looking at the left
tube gives PA  Patm  water g L  h  , and looking at the tube on the right, PB  Patm  oil gL .
Pascal’s principle says that PB = PA. Therefore, Patm  oil gL  Patm  water g L  h  , giving


 
750 kg m 3 
h   1  oil  L   1 
 5.00 cm   1.25 cm
water 
1 000 kg m 3 


(b)
Consider part (c) of the diagram showing the situation when the air flow over the left tube equalizes the fluid
levels in the two tubes. First, apply Bernoulli’s equation to points A and B. This gives
PA 
1
1
air v 2A  air g yA  PB  air v B2  air g yB
2
2
Since yA  yB , v A  v, and v B  0 , this reduces to
PB  PA 
[1]
1
air v 2
2
Now use P = P0 + fgh to find the pressure at points C and D, both at the level of the oil–water interface in
right tube. From the left tube, PC  PA  water gL , and from the right
the
tube, PD  PB  oil gL .
Pascal’s principle says that PD = Pc, and equating these two gives
PB  oil gL  PA  water gL or PB  PA   water  oil  gL
[2]
Combining Equations [1] and [2] yields
v 

9.89
2  water  oil  gL
air

2  1 000  750  9.80 m s2
  5.00  10 2 m 
1.29
 13.8 m s
While the ball is submerged, the buoyant force acting on it is B = (wV)g. The upward acceleration of the ball while
under water is
Page 9.37
Chapter 9
ay 
 Fy

m
   4 3 

B  mg
  w 
r   1 g

m
 m  3



 1 000 kg m 3  4 

3
 9.80 m s2  31 m s 2
 
0.10
m

1


 3 
1.0 kg




Thus, when the ball reaches the surface, the square of its speed is


v 2y  v02y  2 ay  y   0  2 31 m s2  2.0 m   125 m2 s2
When the ball leaves the water, it becomes a projectile with initial upward speed of v0 y 
125 m s and accel-
eration of ay  g  9.80 m s2 . Then, v 2y  v02y  2 ay  y  gives the maximum height above the surface as
ymax 
9.90
0  125 m 2 s2

2 9.80 m s2

 6.4 m
Since the block is floating, the total buoyant force
must equal the weight of the block. Thus,
oil  A 4.00 cm  x   g  water  A  x  g
 wood  A 4.00 cm   g
where A is the surface area of the top or bottom
of the rectangular block.
Solving for the distance x gives

 960  930 
 oil 
x   wood
 4.00 cm    1 000  930   4.00 cm   1.71 cm


 water  oil 
9.91
A water droplet emerging from one of the holes
becomes a projectile with 0y = 0 and 0x = 0. The
time for this droplet to fall distance h to the floor is
found from y  v 0 y t 
1
2
ay t 2 to be
Page 9.38
Chapter 9
2h
g
t 
The horizontal range is
R  vt  v
2h
.
g
If the two streams hit the floor at the same spot,
it is necessary that R1 = R2, or
v1
2 h1
2 h2
 v2
g
g
With h1 = 5.00 cm and h2 = 12.0 cm, this reduces to
v1  v 2
h2
12.0 cm
 v2
or v1  v2 2 .40
h1
5.00 cm
[1]
Apply Bernoulli’s equation to points 1 (the lower hole) and 3 (the surface of the water). The pressure is atmospheric pressure at both points and, if the tank is large in comparison to the size of the holes, 3  0. Thus, we obtain
Patm 
1 2
v1   gh1  Patm  0   gh3 or v12  2 g h3  h1  .
2
[2]
Similarly, applying Bernoulli’s equation to point 2 (the upper hole) and point 3 gives
Patm 
1 2
v 2   gh2  Patm  0   gh3 or v22  2 g h3  h2  .
2
Square Equation [1] and substitute from Equations [2] and [3] to obtain
2 g h3  h1   2 .40  2 g h3  h2  
Solving for h3 yields
h3 
2 .40  12 .0 cm   5.00 cm
2 .40 h2  h1

 17.0 cm ,
1.40
1.40
so the surface of the water in the tank is 17.0 cm above floor level.
Page 9.39
[3]
Chapter 9
9.92
When the section of walkway moves downward distance L, the cable is stretched distance L and the column is
compressed distance L. The tension force required to stretch the cable and the compression force required to compress the column this distance is
Fcable 
Ysteel Acable  L
Lcable
Fcolumn 
and
YAl Acolumn  L
Lcolumn
Combined, these forces support the weight of the walkway section:
Fcable  Fcolumn  Fg  8 500 N
Ysteel Acable  L YAl Acolumn  L

 8 500 N
Lcable
Lcolumn
or
giving
L 
8 500 N
Ysteel Acable
Y A
 Al column
Lcable
Lcolumn
The cross-sectional area of the cable is
Acable 
 D2
4

  1.27  10 2 m 
2
4
and the area of aluminum in the cross section of the column is
Acable 
2
 Douter
4

2
 Dinner
4

2
2
  Douter
 Dinner

4

2
2
   0.162 4 m    0.161 4 m  


4
Thus, the downward displacement of the walkway will be
L 
8 500 N
 20  1010 Pa   1.27  10 2 m 
4  5.75 m 
2

 7.0  1010 Pa     0.162 4 m 2   0.161 4 m 2 
4  3.25 m 
or
 L  8.6  104 m  0.86 mm
Page 9.40