14.1:
ρVgmgw 

 
 
   
N8.41sm80.9m1043.1m858.0mkg108.7
2
2
233


π
or 42 N to two places. A cart is not necessary.
14.2:
 
 
.mkg1033.3
m1074.1
kg1035.7
33
3
6
3
4
22
3
3
4



1
3
1
33





















m
VL
14.5:
ρπrρVm
3
















ρ
ρ
r
r
14.6: a)
 
327
30
3
8
3
4
30
sun
sun

m1000.2
kg1099.1







π
D

316
mkg1094.5 
14.7:
ρghpp 
0

m91.9
)sm80.9()mkg1030(
Pa1000.1
23
5
0





ρg

2
3
3
cm805
)Pa10205(
)N105.16(
b) With the extra weight, repeating the above calculation gives
2
cm1250
.
14.11: a)
Pa.1052.2)m250)(sm80.9)(mkg1003.1(
6
2
33
ρgh
b) The pressure
difference is the gauge pressure, and the net force due to the water and the air is
N.1078.1))m15.0()(Pa1052.2(
526
 π
14.12:
atm.61.9Pa1027.6)m640)(sm80.9)(mkg1000.1(
6233
 ρghp
14.13: a)


)m1000.7)(sm80.9)(mkg106.13(Pa10980
22332

is


mm0
ghppp
With the water to a depth
w
h
, the gauge pressure at the bottom of the
cylinder is
.
wwmm0
ghpghρpp 
If this is to be double the first value, then
m.mww
ghρghρ 

m680.0)1000.1106.13)(m0500.0()(
33
wmmw
 ρρhh
The volume of water is
33424
cm816m108.16)m10m)(12.0(0.680A 

hV
14.16: a) Gauge pressure is the excess pressure above atmospheric pressure. The
pressure difference between the surface of the water and the bottom is due to the weight
of the water and is still 2500 Pa after the pressure increase above the surface. But the
surface pressure increase is also transmitted to the fluid, making the total difference from


N.1079.1
5

14.19: The depth of the kerosene is the difference in pressure, divided by the product
,
V
mg
ρg 
m.14.4
)m250.0()smkg)(9.80205(
Pa1001.2)m(0.0700N)104.16(
32
523



h
14.20:
atm.64.1Pa1066.1
m)15.0(
)smkg)(9.801200(
)2(
5
2
2
2

πdπ
mg

N)30.6(
34
233
water




gρ
B
V
The density is
.mkg1078.2
30.6
50.17
)mkg1000.1(
3333
water
water








B
w
ρ

P
the entire object floats, and if
fluid


, none
of the object is above the surface. d) Using the result of part (c),
%.3232.0
mkg1030
)m103.04.0(5.0kg)042.0(
11
3
3-6
fluid





14.24: a)
   
N.6370m650.0sm80.9mkg1000.1
3233
water
 gVρB
b)
kg.558
2
sm9.80
N900-N6370

ghρ
b)
 
 
 
 
  
Pa.921sm80.9m0150.0mkg1000m100.0mkg790
233

c)
 
  
 
kg.822.0
sm80.9
m100.0Pa805
2
2
topbottom



g
App
g
w
m
The density of the block is
 

gw
ρ
m
V
33
m104.3or


to two figures.
b)
 
N.0.56
7.2
00.1
1N891
aluminum
water
water















The blocks have the same volume
obj
V
so experience the same
buoyant force.
c)
. so 0 BwTBwT 

.ρVgw 
The object have the same
V
but
ρ
is larger for brass than for aluminum so
w
is larger for the brass block.
B
is the same for both, so
T
is larger for the brass block,
block
B.
14.28: The rock displaces a volume of water whose weight is
N.10.8N28.4-N2.39 
The mass of this much water is thus
kg102.1sm9.80N8.10
2


2
2
2
1
cm)10.0(20,cm)80.0( πΑπΑ 

sm6.9
)10.0(20
(0.80)
s)m0.3(
2
2
2

π
π
v
14.30:

2
3
2
2
2
1
12
sm245.0)m0700.0s)(m50.3(
ΑΑΑ
Α
vv

velocity of the seawater at the top of the tank is zero, and Eq. (14.18) gives
))((2 ρpgyv 
=
))mkg10(1.03Pa)1013(3.00)(1.0m)0.11)(sm80.9((2
3
352


s.m4.28
Note that y = 0 and
a
pp 
were used at the bottom of the tank, so that p was the given
gauge pressure at the top of the tank.
14.34: a) From Eq. (14.18),
s.m6.16m)0.14)(sm80.9(22
2
 ghv
b)
s.m1069.4)m)10(0.30s)(m57.16(
3422 
 πvΑ
Note that an extra figure
was kept in the intermediate calculation.
14.35: The assumption may be taken to mean that
0
1
v
in Eq. (14.17). At the
maximum height,






)(
32
15
)()(
2
1
21
2
1121
2
2
2
112
yygυρpyyρgvvρpp








m)0.11)(sm80.9()sm00.3(
32
15

m100.355
kg355.0
33



and so the volume flow rate is
s.L1.30sm1030.1
33
mkg1000
skg30.1
3


This result may also be obtained
from
  
s.L30.1
s0.60
L355.0220

c)
24
33
m1000.2
sm1030.1
1





kPa119
14.39: The water is discharged at a rate of
s.m352.0
23
34
m1032.1
sm1065.4
1





v
The pipe is
given as horizonatal, so the speed at the constriction is
s,m95.82
2
12
 ρpvv
keeping an extra figure, so the cross-section are at the constriction is
,m1019.5
25
sm95.8
sm1065.4
34




ρpvvρpp 










 
 
Pa,1003.2sm50.2mkg1000.1
8
3
Pa1080.1
4
2
334

where the continutity relation
2
1
2
v
v

has been used.
14.41: Let point 1 be where

ρvρgypρvρgyp 
 
Pa1025.2
2
1
soPa,1040.2and
52
2
2
112
5
221
 vvρpppyy
14.42: a) The cross-sectional area presented by a sphere is
,
4
2
D
π
therefore
 
.
4
0
2
D
πppF 
b) The force on each hemisphere due to the atmosphere is
 
2

pressure are the same.
14.44: a) The weight of the water is
  
    
N,1088.5m0.3m0.4m00.5sm80.9mkg1000.1
5233
ρgV
or
N109.5
5

to two figures. b) Integration gives the expected result the force is what it
would be if the pressure were uniform and equal to the pressure at the midpoint;
2
d
gAF



  
    
N,1076.1m50.1m0.3m0.4sm80.9mkg1000.1
5233

or
N108.1
5

to two figures.
14.45: Let the width be w and the depth at the bottom of the gate be



b) The torque on a strip of
vertical thickness
dh
about the bottom is
   
,dhhHgwhhHdFdτ 
and
integrating from
Hhh  to0
gives
.66
23
ρgAHρgwHτ 
c) The force depends
on the width and the square of the depth, and the torque about the bottom depends on the
width and the cube of the depth; the surface area of the lake does not affect either result
(for a given width).
14.47: The acceleration due to gravity on the planet is
d
p
ρd
p
g
V
m




dxπRdV 
Then the integral
becomes
.
22
0
dxRCxM
L


Integrating gives
.
3
3
22
0
2
L
RCdxxRCM
L


Solving
for
.3,
32
LRMCC


b) The density at the

,0r
the model predicts
3
mkg700,12 A

and at
,Rr 
the model
predicts
.mkg103.15m)1037.6)(mkg1050.1(mkg700,12
336433


BRA

b), c)
 




























4
m)1037.6)(mkg1050.1(3
mkg700,12
3
m)1037.6(4
643
3
36
π

kg,1099.5
24

which is within 0.36% of the earth’s mass. d) If
m
)(r

3
4
2













































B
A
BA
B
A
πG
g
3
2
4
3
3
2