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Chapter 14 Worksheet: Fluids
Static & Dynamic Fluids:
- An open cylindrical container, 20 m high and 10 m diameter, is
completely filled with water (ρ water
= 998 kg/m
3
). The container is
standing at ground level and the atmospheric pressure (p atm
) is 1.
x 10
5
Pa.
a. What is the mass of a sphere of water with radius 0.5 m?
Ans.
3
2
kg
m
m = 998 0.5m = 524 kg
b. What is the total pressure at the bottom of the container?
Ans.
3 2
5 kg 5
m
m s
P = 1.01x10 Pa + 998 9.8 20m = 2.97x10 Pa
c. What is the total force on the bottom of the container?
Ans.
2
5 7
F = PA = 2.97x10 Pa π 5m = 2.33x10 N
d. If the container were filled with oil (ρ oil
= 925 kg/m
3
) instead of water, what would the
pressure at the bottom be?
Ans.
3 2
5 kg 5
m
m s
P = 1.01x10 Pa + 925 9.8 20m = 2.83x10 Pa
Pascal’s Principle:
- The largest barometer ever built was an oil-filled barometer constructed in Leicester, England
in 1991. At an atmospheric pressure of 1.01 x 10
5
Pa, height of the oil column was 12.2 m.
a. What was the density of the oil used in the barometer?
Ans.
3
2
5
kg atm
m
m
s
P 1.01x10 Pa
gh 9.8 12.2m
b. Were the barometer filled with water instead of oil, what would be the height of the water at
the same atmospheric pressure?
Ans.
5
atm
kg
m
m s
P 1.01x10 Pa
h = = = 10.3m
g 998 9.
- A U-shaped container has a diameter of 0.25 m at one end and 1.0 m at the other with
plungers at each end and an applied force of 1000 N is applied to the narrow end.
a. What is the resultant force generated at the wide end?
Ans. ( )
2
4 2
2 1
1
A 0.5m
F = F = 1000N = 1.60x10 N
A 0.125m
b. When 2000 J of work are performed by the 1000 N applied force, how high does the fluid level
rise (on the wide end of the container). Assume the fluid rise occurs at constant velocity.
Ans.
2 2 2 2 4
2000J
W = F h = 2000J h = = 0.125 m
1.60x10 N
20 m
10 m
Chapter 14 Worksheet: Fluids
Archimedes’ Principle:
- A turtle with volume 0.25 m
3
is floating in the ocean, 80% of the turtle’s volume is
submerged. Assume the density of the ocean salt water (ρ sea water
) is 1.024 x 10
3
kg/m
3
.
a. What is the buoyant force exerted on the turtle?
Ans. V disp
= (0.80)(0.25 m
3
) = 0.20 m
3
3 2
3 kg 3
m
B disp
m s
F = ρgV = 1.024x10 9.8 0.20 m = 2007 N
b. What is the mass of the turtle?
Ans. F net
= F
B
2
B
m
s
F 2007 N
m = = = 204.8 kg
g 9.
c. If the fluid were pure water (ρ water
= 998 kg/m
3
) instead of salt water, how much of the turtle’s
volume would be submerged in the water?
Ans.
B
disp
kg m
m s
F 2007 N
V = = = 0.205 kg
g 998 9.
- A spherical mass (14.7) kg is attached by a mass-less string to a force sensor then
submerged into a container of water.
a. What is the force sensor reading (i.e. the tension force in the mass-less string) before the
mass is submerged in the fluid?
Ans. F net
= F T
= mg = (14.7 kg)(9.8 m/s
2
) = 144 N
b. When the mass is submerged, the sensor reading is 92.9 N. What is the buoyant force
exerted on the mass?
Ans. F net
= F T
= 0
→ F
T
T
= F
B
= 144 N - 92.9 N = 51.2 N
c. What is the density of the hanging mass?
Ans.
B -2 3
disp
kg
m
m s
F 51.2 N
V = = = 5.22x10 m
g 998 9.
ρ
→
3
3 kg
mass -2 3 m
m 14.7 kg
= = = 2.81x
V 5.22x10 m
ρ
- A 60 kg person is attached by a mass-less string to a force sensor then submerged into a
container of water. Assume all of the air is out of her lungs when she is submerged.
a. What is the force sensor reading (i.e. the tension force in the mass-less string) before the
mass is submerged in the fluid?
Ans. F net
= F
T
T
= mg = 588 N
b. When the mass is submerged, the sensor reading is 37.2 N. What is the density of the
person?
Ans. F net
= F
T
B
T
B
T
= 0
F T
= F B
= 588 N – 37.2 N = 551 N
→
3 2
3
kg m
m s kg
m
B T2 T
w w
60 kg 998 9.
m m m
V 588N - 37.2N
F F -F
g g
ρ
ρ ρ
Chapter 14 Worksheet: Fluids
- Consider the completely full water container described in Problem 1 above.
a. What is the pressure at a depth of 15 m below the surface of the water?
Ans.
3 2
atm applied atm
5 kg 5 m
m s
P = P + gh + P P + gh
P 1.01x10 Pa + 998 9.8 15m 2.48x10 Pa
b. A small spout is opened in the container 5 m above the bottom. As the water spews from the
hole, what is the speed of the water as it leaves the hole?
Ans.
2
1
inside outside atm atm 2 outside
m
outside s
P = P - P = P + gh - P = v
2 P
v = = 2gh = 17.
ρ ρ
ρ
c. How far from the container will the water stream hit the ground?
Ans.
2
m
ox ox s
m
s
2 5m 2 y
x = v t = v = 17.1 = 17.3 m
g 9.
d. If the container were filled with oil (ρ oil
= 925 kg/m
3
) instead of water, how fast would the fluid
leave the hole in question (c)?
Ans. The same as answer (b).
- An airplane has a mass of 2.0 x 10
6
kg and the air flows past the lower surface at 100 m/s.
The wings have a combined surface area of 1200 m
2
.
a. How fast must the air flow past the upper surface of the wings if the plane is to maintain
constant elevation while in the air?
Ans.
net below above below wings above wings below above
wings
mg
F = F - F - mg = 0 P A - P A = mg P - P =
A
2 2 2
1 m
below above 2 air above below above below s
wings air wings
mg 2mg
P -P = v - v = v = + v = 192
A A
ρ
ρ
b. If the surface area of the wings were increased by 25%, what would be the upward
acceleration of the plane?
Ans.
2
below above wings
m
net below above lift
s
P - P A - mg
F = F - F - mg = ma a = = 2.
m
Chapter 14 Worksheet: Fluids
Archimedes’ Principle (one more thing…)
- A face down iron hemisphere (r = 1.0 m) is completely submerged in water, suspended by a
cable. The peak of the hemisphere is 0.5 m below the surface of the water.
a. Calculate the gauge pressure on the bottom face,
P bottom
, of the mass.
Ans.
3 2 2
bottom o
kg 4
m N
bottom
m s m
P = g(h + r)
P = 998 9.8 1.5 m = 1.4671 10
ρ
×
b. Calculate the upward force exerted on the lower
surface due to P bottom
.
Ans.
2
2
up bottom
2
4
N
up
m
F =P r
F = 1.4671 10 1m = 49,090N
π
× π
c. Derive the equation for the vertical gauge pressure, P upper-y
exerted on the upper (spherical)
surface.
Ans.
top-y o
P = ρgh + ρgr 1 - cos θ cos θ
d. Calculate the force exerted on the upper surface due to P upper-y
.
Ans. The downward fluid force on the a small area dA of the mass is:
2 2
down top-y
down top-y o
2 2
2 3 2
down o
0 0 0 0
dF =P dA where dA = r dθ r sinθd
F = P dA = gh + gr 1 - cos θ cos θ r dθ r sinθd
F = gr h + r cosθsinθdθ d - gr cos θsinθdθ d
π π
π π
2 2
down top-y o
3 2 2
2 2 3
down o
0 0 0 0
2 3 2
down o o
F = P dA = gh + gr 1 - cos θ cos θ r dθ r sinθd
cos θ
F = gr h + r sin θ - gr
1 r
F = gr h + r - gr 2 = gr h + = 25,605 N
π π
π π
e. Calculate the net force exerted on the mass due to the 2 forces above.
Ans.
net up down
F = F - F = 49,090N - 25605 N = 20,484 N (upward)
f. Apply Archimedes’ Principle to the mass and calculate the buoyant force exerted on the mass
due to the fluid. Compare this value to the answer calculate in (e).
Ans.
3 3
disp
V = (1.0 m) = 2.09 m
kg 3 m
B disp
m s
F = ρgV = 998 9.8 2.09 m = 20,484 N{the same value as (e)!}
h
o
r
θ
φ
