Chapter 18
Questions: 5, 6, 9, 11, 12, 16, 17, 18, 22, 23
Problems: 9, 13, 15, 25, 29, 35, 38, 47, 48, 51, 57
Questions
5. The pressure at the bottom only depends on the height of the contained
liquid - the pressures are all equal.
6. The elevation of the hills, towers, etc. provide hydrostatic pressure.
9. The ice is displacing its weight in water. After it melts it becomes
water, hence ignoring the change in density due to temperature (water will
become slighter cooler), the level of the water in the cup will not change.
11. The frothy bubbles are less dense and hence provide less buoyancy.
12. The pressure is greater at the bottom of dams requiring greater
structural integrity at the bottom.
16. When the barge was loaded with the steel beams, the beams displaced
their weight in water. When the steel was in the water, it only displaced its
volume. Since the steel is more dense than the water, it displaced less water
after the barge overturned and the height of the lake falls.
17. The radial acceleration of the water in a rotating cylinder is given by
ar = ! 2 r; where ! is the angular velocity and r is the radial distance from
the center of rotation. This means that the pressure forces on the surface of
an object that is further from the center is larger than the pressure forces on
the surface that is closer to the center regardless of the density of the object.
An object that ‡oats will then drift toward the center of the surface. An
object that sinks will drift toward the center on its way to the bottom.
18. From the continuity equation for the ‡ow speed to decrease its density
must increase.
22. The pump provides a change (increase) in the gauge pressure to lift
the water. Since the guage pressure is gh when the well is full it requires
almost no work for the pump to lift the water to the top of the well. When
the well is empty the guage pressure is the same as that at the top of the
well, 1 atm. Hence the work to lift the water is basically mgh:
23. So that the relative velocity of the plane and the air is increased.
From Bernoulli’s equation this greater velocity relative to the air results in
a greater lift at smaller velocities relative to the runway.
Problems
9. The area of the end of the cylinder is A = 9 = 28:3m2 : The net
internal pressure is P = :5atm ' 5 104 P a. Thus the force at each end of
the cylinder is F = P A = 1:42 106 N .
1
13. Balancing forces requires
120 9:8 = 4700 A;
A = 4700= (120 9:8) ' 4m2
15. The force required to pull in the window is
F = :5
:9
104 N = 22; 500N:
5
25. To balance the forces in the U shaped tube requires
oil g
2 = water g (2 h) ;
2 h = 2 oil = water ;
h = 2 (1
oil = water ) = :36cm
29. The maximum force is
F =
4
(:45)2 5
105 N = 79; 500N .
35. Assuming the beer is the same density as water then the volume
consumed is found by recognizing that the beer bottle displaces its weight
(including contents) in water and as a consequence,
V =
(2:6)2 2:8 ' 60cm3 = 60mL
38. Assume the minimum volume of the balloon is V and that the volume
of the mass M is negligible. Then from a free body diagram and Archimedes,
the total weight of the balloon plus gas, (M + m) g, must equal the buoyant
force, i.e. the weight of the displaced air, or:
(M + m) g =
M =
a gV
!M+
gV
V =
a
a
g
=
aV
m
g
g
g
m =
a
M:
g
47. (a) Under static conditions we know from Bernoulli’s equation that
(the tube is open)
gh1 + Pa = gh2 + Pa ;
h2 = h1 :
2
(b) When the ‡uid is ‡owing then, we know from Bernoulli’s equation
that
1 2
1
v + gh1 + Pa =
(2v)2 + gh2 + Pa ;
2
2
where the velocity in the lower section of the tube is doubled to conserve
mass ‡ow. Now solving for h2 we …nd
3v2
:
2g
h2 = h1
48. From kinematics the time required for the ‡uid to reach the ground
is
t=
p
2y=g:
Hence the velocity upon exiting the tank must be given by
p
vt = v 2y=g = y
p
gy=2:
v =
From Bernoulli’s equation
1 2
1 gy
v = g (h y) =
2
2 2
y = 4h 4y
y = 4h=5:
51. Applying Bernoulli’s equation to this problem leads to
P1 +
1
1 2
v1 = P2 + v22 !
2
2
P =
1 2
v
2 2
1 2
v :
2 1
The change in pressure is found from the change in densities of the water
and the oil over the height h = 1:4cm:
2 P = 2 gh = 2 :18 gh = v22
:36g (1:4) = :504 = v22 v12
From the continuity equation
v 2 A2 = v 1 A1 ! v 2 =
v2 =
:952
v1 :
:322
3
A1
r2
v1 = 12 v1
A2
r2
v12
Substituting this into Bernoulli’s equation we …nd
:504g = 494cm2 =s2 = v22
v12 =
:954
:324
1 v12
v1 = 2:538cm=s
The volume ‡ow rate is
V = A1 v1 = :952
2:538 = 7:2cm3 =s
57. Applying Bernoulli’s equation to this problem leads to
P+
1 2
1
v1 = :95P + v22 :
2
2
To conserve mass ‡ow we …nd
A1 v 1 = A2 v 2 :
Substituting for v2 in Bernoulli’s equation leads to
P+
1 2
1
v1 = :95P +
2
2
A1
A2
A1
A2
2
v12 ;
2
:1P= + v12 =v12 ;
q
1
=
:1P= + v12 ;
v1
1 p
=
23 + 2:25 = 3:35:
1:5m=s
=
A1
A2
A1
A2
Thus A2 ' :3A1 and the obstruction is approximately 70%::
4