Ch
14.1 #3
A 50 kg woman
balances on one heel of a pair of high heeled shoes. If the heel is circular and has a radius of
0.500 cm, what pressure does she exert on the floor?
P = F/A 
P = m
g / πr^{2} P = 50kg(9.8m/s^{2})
/ π(0.005)^{2} P =
6.24 x 10^{6} Pascals 
Ch
14.2 Dam In the example
on our class notes, let’s say a hatch 5 meters in width was placed near the
bottom of a 30 meter dam. The hinged
hatch is located 20 to 22 meters below the surface. The hinge is along the top of the hatch at
20 meters. What is the torque exerted
by the water about the hinge? 


dF = P
dA dA = w dh dF = ρgh (5
dh) w = 5 meters F =
5ρg ∫h dh dτ
= F •
(h – 20) dτ
= 5ρg h dh • (h – 20) 
∫dτ
= 5ρg (∫h^{2} dh  20∫h dh) τ = 5 1000 9.8 [(h^{3}/3  20
h^{2}/2)] τ = 50000 [(22^{3}–20^{3})/3
– 10(22^{2}–20^{2})] τ = 2,090,000 Nm CCW 
Lever arm is h  20 dh is from 20 to 22 meters deep. 

Ch
14.3 #20 A Utube of
uniform cross sectional area, open to the atmosphere, is partially filled
with mercury. Water is then poured
into both arms. If the equilibrium
configuration of the tube is with h_{2} = 1.00 cm determine the value
of h_{1}. 


P_{left} = P_{right} (A_{left}
= A_{right}) F_{left} A_{right} = F_{right} A_{left} F_{left} = F_{right} ρ g
h = ρ g h +
ρ g h 1“g”(h+h_{1}+h_{2})
= 1“g”h + 13.6“g”h_{2} 
1“g”(h+h_{1}+h_{2})
= 1“g”h + 13.6“g”h_{2} h_{1}+h_{2}
= 13.6h_{2} h_{1} =
12.6h_{2} h_{2}
= 1.00 cm h_{1}
= 12.6 cm 

Ch
14.4 Ball
At the beach by
the ocean, what is the force required to submerge a 30.0
F_{Net} = F_{down}
+ F_{W} F_{down} + mg F_{down} + 30 F_{down} + 30 
= F_{B} = m_{saltwater} g = ρ_{ saltwater}_{ }(V) g = 1030(0.0335) g 
ρ_{ saltwater}_{ }≈ 1030 kg/m^{3} V = (4/3)πr^{3} V =
(4/3)π(0.2)^{3} V =
0.0335 m^{3} 
F_{down}
= 308 
Ch
14.5 #41
Water flows
through a fire hose of diameter 6.35 cm at a rate of 0.0120 m^{3}/sec. The fire hose ends in a nozzle of inner
diameter (id) of 2.20 cm. What is the
speed with which the water exits the nozzle?
We know mass/time is constant…so V_{1}
/ t = V_{2} / t Initial diameter (and radius) is irrelevant 
A_{1}
Δx_{1}/t = V_{2}
/ t πr^{2}
(v) = 0.0120 m^{3}/s v = 0.012/ π(0.011)^{2} v = 31.6 m/s 
Ch
14.6 #45
Through a pipe
15.0 cm in diameter, water is pumped from the
(a) P
+ ½ρv^{2 }+ ρgh
=
P_{2} + ½ρv_{2}^{2} +
ρ g h_{2} (1atm + P_{g})
+ ½ρv_{2}^{2}
+ 0 = 1atm + ½ρv_{2}^{2} + 1000(9.8)(2096564) (v_{init} ≈ v_{final}…if
10 gpm at one point in the hose…must be 10 gpm at any other point) I’m also
assuming the 564 meter elevation is the bottom so the 2096 meter elevation is
the top, so h_{2} = (2096564). (1atm + P_{g})
+ 0 + 0 = 1atm
+ 0 + 1000(9.8)(2096564) P = 1atm + 1000(9.8)(2096564) P = 1atm + 150 x 10^{5} Pa 

(b) flow rate =
0.052083 m^{3}/sec flow rate = A
v .052083 = π¼d^{2} v .052083 =
π¼ .15^{2} v v =
2.95 m/s 
(c) This time we need give
some extra velocity…so (1atm + P_{g})
= 1atm + ½ρv_{2}^{2} + 1000(9.8)1532 P = 1 atm + ½ 1000(2.95)^{2} + 150 x 10^{5} Pa P = 1 atm + 4340 Pa + 150 x 10^{5} Pa P =
1.013 x 10^{5} Pa + 4340 Pa + 150 x 10^{5} Pa P = 151.05
x 10^{5} Pa 
Ch
14.7 #53
A hypodermic
syringe contains water. The barrel of
the syringe has a crosssectional area of 2.50 x 10^{5} m^{2},
and the needle has a crosssectional area of 10^{8} m^{2}. In the absence of a force on the plunger, the
pressure everywhere is 1 atm. A force of magnitude 2.00 N acts on the
plunger, making water squirt horizontally from the needle. Determine the speed of the medicine as it
leaves the needle’s tip. ans 12.6m/s
A_{1}v_{1}
= A_{2}v_{2} 2.50 x 10^{5} v_{1} = 10^{8} v_{2} v_{2} =
2500 v_{1} So v_{1}^{2} is neglible
(below) 
P + ½ρv^{2 }+ ρgh = P_{2} + ½ρv_{2}^{2 }+ ρgh_{2} PP_{2} + 0^{ }+
n/a =
+ ½ρv_{2}^{2 }+ n/a ΔP
= ½ρv^{2} 2/(2.50
x 10^{5}) = ½ 1000 v^{2} v =
12.6 m/s 