Homework 9
Problems: 12.31, 12.32, 14.4, 14.21
Problem 12.31
Assume that if the shear stress exceeds about 4×108 N/m2 steel ruptures. Determine
the shearing force necessary (a) to shear a steel bolt 1.00 cm in diameter and (b) to
punch a 1 cm diameter hole in a steel plate 0.500 cm thick.
a)
F
-F
The above figure illustrates the formation of a shearing stress in
the bolt. The connected pieces of metal exert opposite forces on the bolt.
With good approximation we can assume that the shearing stress along
the marked line is uniform.
From the definition of shearing stress we can find the minimum
shearing force that ruptures the bolt.
F>
2
∫ dFmax
surface
⎛D⎞
= ∫ σ max dA = σ max ∫ dA = σ max A = σ max ⋅ π⎜ ⎟ =
⎝2⎠
surface
surface
= 4.0 ⋅ 108 Pa ⋅ π ⋅ (0.005m )2 = 31.4kN
b)
D
F
h
Shearing stress is created along the side surface of the punched
disk. Note that the forces exerted on the disk are exerted along the
circumference of the disk. The above figure illustrates the total force at
its center only.
With good approximation we can assume that the shearing stress
along the side surface of the disk is uniform. This approximation
simplifies the integration necessary to relate the shearing force with
stress
F>
∫ dFmax =
surface
⎛D⎞
⎝2⎠
∫ σ max dA = σ max ∫ dA = σ max A = σ max ⋅ 2π⎜ ⎟h =
surface
surface
= 4.0 ⋅108 Pa ⋅ 2π ⋅ 0.005m ⋅ 0.005m = 62.8kN
Comment. Note for comparison that the weight of a car is about 15kN.
Problem 12.32
When water freezes, it expands by about 9 %. What would be the pressure increase
inside your automobile engine block if the water in it froze? (The bulk modulus of
ice is 2 ⋅ 109 N/m2)
If water were free to expand it would
change its volume by r = 9% during the
freezing process
V1 = (1 + r ) ⋅ V0
where V0 is the original volume of water
and V1 is the volume of the ice free to
expand. With an assumption that the iced
formed in the process cannot escape from
the engine block the walls of the block
“compress” the ice to the original volume
water
water
V2 = V0
From the definition of bulk modulus, the
change in pressure in a substance ΔP is related to the relative change in
the volume (of the ice)
ΔP = −B
ΔV
V
The rest is algebra. Solving by substitution
ΔP = − B ⋅
V2 − V1
V − (1 + r )V0
r
= −B ⋅ 0
= −B ⋅
=
(1 + r )V0
V2
1+ r
= 2 ⋅ 109
N
0.09
8
⋅
≈
1
.
7
⋅
10
Pa
2
1
0
.
09
+
m
( Note. It does not matter if the change is compared with the initial or the final
volume of the substance
ΔP = B
V − V1
r
≈ Br = B 2
≈ 1.8 ⋅ 108 Pa )
V1
1+ r
Problem 14.4
Estimate the total mass of the Earth’s atmosphere. (The radius of the Earth is
6.37⋅107 m, and the atmospheric pressure at the Earth’s surface is 1.013⋅105 Pa.)
The atmosphere is relatively thin (when
compared with Earth’s radius). One can
make than a reasonable assumption that over
the entire thickness of the atmosphere the
(magnitude) of free fall acceleration has the
same value. We can therefore estimate the
atmosphere’s weight first and then determine its mass.
From the definition of pressure, the (magnitude of the) force
exerted by the atmosphere on a differential surface of the Earth
is related to the area of that (differential) surface
dF = PdA
The weight of the atmosphere (the force exerted on a (flat)
surface with area equal to the area of the Earth) is therefore
W = ∫ PdA = P ⋅ 4πR 2
surface
Hence the mass of the atmosphere is
(
W 4πPR 2 4π ⋅ 1.013 ⋅ 105 Pa ⋅ 6.37 ⋅ 107 m
m=
=
=
m
g
g
9.81 2
s
(Pretty heavy!)
)
2
= 5.27 ⋅ 1018 kg
Problem 14.21
Mercury is poured into a U-tube, as shown in Figure P14.21a. The left arm of the
tube has a cross-sectional area A1 of 10 cm2, and the right arm has a cross-sectional
area A2 of 5 cm2. One hundred grams of water are then poured into the right arm,
as shown in Figure P14.21. (a) Determine the length of the water column in the
right arm of the U-tube. (b) Given that the density of mercury is 13.6 g/cm3, what
distance h does the mercury rise in the left arm?
a) Water in the tube forms a cylinder of
height H and the base area A2. The
density of water is uniform. From the
definition of density ρ, the mass m of
water is related to its volume V, which
for a cylinder can be expressed in terms
of its height H and the area A2 of the
base
1)
A1
A2
H2O
h1
h2
H
Hg
m = ρV = ρA 2 H
In the above equation, only the height is
not given in the problem. Solving we find
2)
H=
m
100g
=
= 20cm
2
3
ρA 2 5cm ⋅ 1g / cm
b) At the level of the mercury-water interface in the right tube, the
pressure has a value determined by the water above the considered level
(and the atmospheric pressure P0)
3)
P = P0 + ρ H 2 O gH
The mercury above the considered level also determines this pressure
4)
P = P0 + ρ Hg g (h1 + h 2 )
The displacement of mercury by water has not significantly changed the
pressure distribution and therefore we can assume that this process did
not affect its volume. Therefore
5)
A1h1 = A 2 h 2
We obtained three equations with three unknowns (P-P0, h1, and h2).
Solving simultaneously we find the answer
ρH 2 O ⋅ H
1g / cm 2 ⋅ 20cm
=
h1 =
= 0.49cm
2⎞
⎛
⎛ A1 ⎞
⎟⎟ 13.6g / cm 2 ⋅ ⎜1 + 10cm ⎟
ρ Hg ⋅ ⎜⎜1 +
⎜
⎝ A2 ⎠
5cm 2 ⎟⎠
⎝