PHYS 131 Fall 2017
Week 11 Recitation: Chapter 11: 11, 27, 40, 44, 64, 69.
11. ssm An airtight box has a removable lid of area 1.3 × 10−2 m2 and negligible
weight. The box is taken up a mountain where the air pressure outside the box is 0.85 ×
105 Pa. The inside of the box is completely evacuated. What is the magnitude of the force
required to pull the lid off the box?
REASONING Since the inside of the box is completely evacuated; there is no air to
exert an upward force on the lid from the inside. Furthermore, since the weight of
the lid is negligible, there is only one force that acts on the lid; the downward force
caused by the air pressure on the outside of the lid. In order to pull the lid off the
box, one must supply a force that is at least equal in magnitude and opposite in
direction to the force exerted on the lid by the outside air.
SOLUTION According to Equation 11.3, pressure is defined as P = F / A ; therefore,
the magnitude of the force on the lid due to the air pressure is
(0.85 105 N/m 2 )(1.3 ×10 –2 m 2 ) =
1.1×103 N
F =×
27. ssm A water tower is a familiar sight in many towns. The purpose of such a tower is
to provide storage capacity and to provide sufficient pressure in the pipes that deliver the
water to customers. The drawing shows a spherical reservoir that contains 5.25 × 105 kg
of water when full. The reservoir is vented to the atmosphere at the top. For a full
reservoir, find the gauge pressure that the water has at the faucet in (a) house A and (b)
house B. Ignore the diameter of the delivery pipes.
REASONING The pressure P at a distance h beneath the water surface at the vented top
of the water tower is=
P Patm + ρ gh (Equation 11.4). We note that the value for h
in this expression is different for the two houses and must take into account the
diameter of the spherical reservoir in each case.
SOLUTION
a. The pressure at the level of house A is given by Equation 11.4 as
=
PA Patm + ρ ghA . Now the height hA consists of the 15.0 m plus the diameter d of
the tank. We first calculate the radius of the tank, from which we can infer d. Since
the tank is spherical, its full mass is given by =
M ρ=
V ρ[(4 / 3)π r 3 ] . Therefore,
1/ 3
 3M 
3M
r
or r =
=
=

4π ρ
 4π ρ 
3
1/3
 3(5.25 ×105 kg) 
5.00 m
=

3
3 
 4π (1.000 ×10 kg/m ) 
PHYS 131 Fall 2017
Week 11 Recitation: Chapter 11: 11, 27, 40, 44, 64, 69.
Therefore, the diameter of the tank is 10.0 m, and the height hA is given by
hA = 10.0 m + 15.0 m = 25.0 m
According to Equation 11.4, the gauge pressure in house A is, therefore,
PA − Patm = ρ ghA = (1.000 × 103 kg/m3 )(9.80 m/s2 )(25.0 m) = 2.45 × 105 Pa
b. The pressure at house B is =
PB Patm + ρ ghB , where
hB = 15.0 m + 10.0 m − 7.30 m = 17.7 m
According to Equation 11.4, the gauge pressure in house B is
PB − Patm = ρ ghB = (1.000 × 103 kg/m3 )(9.80 m/s2 )(17.7 m) = 1.73 × 105 Pa
40. The density of ice is 917 kg/m3, and the density of seawater is 1025 kg/m3. A
swimming polar bear climbs onto a piece of floating ice that has a volume of 5.2 m3.
What is the weight of the heaviest bear that the ice can support without sinking
completely beneath the water?
REASONING The ice with the bear on it is floating, so that the upward-acting buoyant
force balances the downward-acting weight Wice of the ice and weight Wbear of the
bear. The magnitude FB of the buoyant force is the weight WH
2
of the displaced
O
water, according to Archimedes’ principle. Thus, we have =
FB WH=
O Wice + Wbear ,
2
the expression with which we will obtain Wbear. We can express each of the weights
WH
2
O
and Wice as mass times the magnitude of the acceleration due to gravity
(Equation 4.5) and then relate the mass to the density and the displaced volume by
using Equation 11.1.
SOLUTION Since the ice with the bear on it is floating, the upward-acting buoyant force
FB balances the downward-acting weight Wice of the ice and the weight Wbear of the
bear. The buoyant force has a magnitude that equals the weight WH
2
O
of the
displaced water, as stated by Archimedes’ principle. Thus, we have
FB =
WH O =
Wice + Wbear or
2
Wbear =
WH O − Wice
(1)
2
In Equation (1), we can use Equation 4.5 to express the weights WH
2
O
and Wice as
mass m times the magnitude g of the acceleration due to gravity. Then, the each mass
PHYS 131 Fall 2017
Week 11 Recitation: Chapter 11: 11, 27, 40, 44, 64, 69.
can be expressed as m = ρ V (Equation 11.1). With these substitutions, Equation (1)
becomes
Wbear = mH O g − mice g = ( ρ H OVH
2
2
2
O )g
− ( ρiceVice ) g
(2)
When the heaviest possible bear is on the ice, the ice is just below the water surface
and displaces a volume of water that is VH O = Vice . Substituting this result into
2
Equation (2), we find that
Wbear =
( ρ H OVice ) g − ( ρiceVice ) g =
( ρH
2
(
)(
2
O
− ρice )Vice g
)(
)
=
1025 kg/m3 − 917 kg/m3 5.2 m3 9.80 m/s 2 =
5500 N
44. A paperweight, when weighed in air, has a weight of W = 6.9 N. When completely
immersed in water, however, it has a weight of Win water = 4.3 N. Find the volume of the
paperweight.
REASONING The paperweight weighs less in water than in air, because of the buoyant
force FB of the water. The buoyant force points upward, while the weight points
downward, leading to an effective weight in water of WIn water = W – FB. There is
also a buoyant force when the paperweight is weighed in air, but it is negligibly
small. Thus, from the given weights, we can obtain the buoyant force, which is the
weight of the displaced water, according to Archimedes’ principle. From the weight
of the displaced water and the density of water, we can obtain the volume of the
water, which is also the volume of the completely immersed paperweight.
SOLUTION We have
WIn water = W − FB
or
FB = W − WIn water
According to Archimedes’ principle, the buoyant force is the weight of the displaced
water, which is mg, where m is the mass of the displaced water. Using Equation
11.1, we can write the mass as the density times the volume or m = ρV. Thus, for the
buoyant force, we have
FB = W − WIn water = ρVg
Solving for the volume and using ρ = 1.00 × 103 kg/m3 for the density of water (see
Table 11.1), we find
PHYS 131 Fall 2017
Week 11 Recitation: Chapter 11: 11, 27, 40, 44, 64, 69.
64. Water flowing out of a horizontal pipe emerges through a nozzle. The radius of the
pipe is 1.9 cm, and the radius of the nozzle is 0.48 cm. The speed of the water in the pipe
is 0.62 m/s. Treat the water as an ideal fluid, and determine the absolute pressure of the
water in the pipe.
REASONING The absolute pressure in the pipe must be greater than atmospheric
pressure. Our solution proceeds in two steps. We will begin with Bernoulli’s
equation. Then we will incorporate the equation of continuity.
SOLUTION According to Bernoulli’s equation, as given by Equation 11.11, we have
The pipe and nozzle are horizontal, so that y1 = y2 and Bernoulli’s equation
simplifies to
P1 + 12 ρ v12 =+
P2 12 ρ v22
or
(
P1 =+
P2 12 ρ v22 − v12
)
where P1 is the absolute pressure of the water in the pipe. We have values for the
pressure P2 (atmospheric pressure) at the nozzle opening and the speed v1 in the
pipe. However, to solve this expression we also need a value for the speed v2 at the
nozzle opening. We obtain this value by using the equation of continuity, as given
by Equation 11.9:
=
A1v1
A2v=
or π r12v1
2
π=
r22v2 or v2
r12v1
r22
Here, we have used that fact that the area of a circle is A = π r2. Substituting this
result for v2 into Bernoulli’s equation, we find that
 r4

P1 =
P2 + 12 ρ  14 − 1 v12


 r2

Taking the density of water to be ρ = 1.00 × 103 kg/m3 (see Table 11.1), we find
that the absolute pressure of the water in the pipe is
 r4

P1 =
P2 + 12 ρ  14 − 1 v12
r

 2

1.01 × 105 Pa + 12
=
(
)
(
 1.9 × 10−2 m
3
3 
1.00 × 10 kg/m

−3
 4.8 × 10 m
(
) − 1 ( 0.62 m/s)2 =1.48 × 105 Pa
4
) 
4
PHYS 131 Fall 2017
Week 11 Recitation: Chapter 11: 11, 27, 40, 44, 64, 69.
*69. ssm A Venturi meter is a device that is used for measuring the speed of a fluid
within a pipe. The drawing shows a gas flowing at speed υ2 through a horizontal section
of pipe whose cross-sectional area is A2 = 0.0700 m2. The gas has a density of ρ = 1.30
kg/m3. The Venturi meter has a cross-sectional area of A1 = 0.0500 m2 and has been
substituted for a section of the larger pipe. The pressure difference between the two
sections is P2 − P1 = 120 Pa. Find (a) the speed υ2 of the gas in the larger, original pipe
and (b) the volume flow rate Q of the gas.
REASONING Since the pressure difference is known, Bernoulli’s equation can be used
to find the speed v 2 of the gas in the pipe. Bernoulli’s equation also contains the
unknown speed v 1 of the gas in the Venturi meter; therefore, we must first express
v 1 in terms of v 2 . This can be done by using Equation 11.9, the equation of
continuity.
SOLUTION
a. From the equation of continuity (Equation 11.9) it follows that
Therefore,
v1 =
0.0700 m 2
v = ( 1.40 ) v 2
0.0500 m 2 2
Substituting this expression into Bernoulli’s equation (Equation 11.12), we have
P1 + 21 ρ ( 1.40 v 2 ) 2 = P2 + 21 ρ v 22
Solving for v 2 , we obtain
v2 =
2 ( P2 − P1 )
ρ (1.40 ) − 1
2
=
2( 120 Pa)
(1.30 kg / m 3 ) (1.40) 2 − 1
= 14 m / s
b. According to Equation 11.10, the volume flow rate is
Q = A2 v 2 = (0.0700 m 2 )( 14 m / s) = 0.98 m 3 / s
.