A U-tube of constant cross sectional area, open to the atmosphere, is partially filled with
Hg (ρHG=13.6 g/cm3). Water (ρw=1.00 g/cm3) is then poured into both arms. If the
equilibrium configuration of the tube is as shown in the figure with h3= 1.00 cm,
determine the value of h1
Answers
Explanation:
P
1
=P
2
∴ P
0
+ρ
w
g(h
1
+h
2
+h)=P
0
+ρ
w
gh+ρ
Hg
gh
2
∴ h
1
=
ρ
w
(ρ
Hg
−ρ
w
)h
2
=
1
(13.6−1)(1.0)
=12.6 cm
Answer:
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 (∫h2 dh - 20∫h dh)
τ = 5 1000 9.8 [(h3/3 - 20 h2/2)]
τ = 50000 [(223–203)/3 – 10(222–202)]
τ = 2,090,000 Nm CCW
Lever arm is h - 20
dh is from 20 to 22 meters deep.
Ch 14.3 #20
A U-tube 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 h2 = 1.00 cm determine the value of h1.
Pleft = Pright (Aleft = Aright)
Fleft Aright = Fright Aleft
Fleft = Fright
ρ g h = ρ g h + ρ g h
1“g”(h+h1+h2) = 1“g”h + 13.6“g”h2
1“g”(h+h1+h2) = 1“g”h + 13.6“g”h2
h1+h2 = 13.6h2
h1 = 12.6h2 h2 = 1.00 cm
h1 = 12.6 cm