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Problem 1.1 [Difficulty: 3]
Given: Common Substances
Tar Sand
“Silly Putty” Jello
Modeling clay Toothpaste
Wax Shaving cream
Some of these substances exhibit characteristics of solids and fluids under different conditions.
Find: Explain and give examples.
Solution: Tar, Wax, and Jello behave as solids at room temperature or below at ordinary pressures. At high
pressures or over long periods, they exhibit fluid characteristics. At higher temperatures, all three
liquefy and become viscous fluids.
Modeling clay and silly putty show fluid behavior when sheared slowly. However, they fracture
under suddenly applied stress, which is a characteristic of solids.
Toothpaste behaves as a solid when at rest in the tube. When the tube is squeezed hard, toothpaste
“flows” out the spout, showing fluid behavior. Shaving cream behaves similarly.
Sand acts solid when in repose (a sand “pile”). However, it “flows” from a spout or down a steep
incline.
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Problem 1.2 [Difficulty: 2]
Given: Five basic conservation laws stated in Section 1-4.
Write: A word statement of each, as they apply to a system.
Solution: Assume that laws are to be written for a system.
a. Conservation of mass — The mass of a system is constant by definition.
b. Newton's second law of motion — The net force acting on a system is directly proportional to the product of the
system mass times its acceleration.
c. First law of thermodynamics — The change in stored energy of a system equals the net energy added to the
system as heat and work.
d. Second law of thermodynamics — The entropy of any isolated system cannot decrease during any process
between equilibrium states.
e. Principle of angular momentum — The net torque acting on a system is equal to the rate of change of angular
momentum of the system.
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Problem 1.
(Difficulty: 1)
Given: Data on oxygen tank.
Find: Mass of oxygen.
Solution: Compute tank volume, and then us e oxygen density to find the mass.
The given or available data is:
D = 16 ⋅ft p = 1000 ⋅psi T = (77 + 460)⋅R T = 537 ⋅R
For oxygen the critical temperature and pressure are: (^) Tc = 279 ⋅R pc = 725.2⋅psi (data from NIST WebBook)
so the reduced temperature and pressure are:
Using a compressiblity factor chart: (^) Z = 0.948 Since this number is close to 1, we can assume ideal gas behavior.
Therefore, the governing equation is the ideal gas equation (^) p = ρ⋅RO2⋅T (^) and (^) ρ = M V
where V is the tank volume (^) V = π⋅D^3 6
V = π^ × (16⋅ ft) 3 6
V = 2144.7⋅ ft^3
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Hence:
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Problem 1.12 [Difficulty: 3]
mg
kVt
Given: Data on sphere and terminal speed.
Find: Drag constant k , and time to reach 99% of terminal speed.
Solution: Use given data; integrate equation of motion by separating variables.
The data provided are: (^) M = 1 × 10 −^13 ⋅slug Vt 0.2 ft s
Newton's 2nd law for the general motion is (ignoring buoyancy effects) (^) M dV dt
⋅ = M g⋅ −k V⋅ (1)
Newton's 2nd law for the steady state motion becomes (ignoring buoyancy effects) (^) M g⋅ = k V⋅ t so (^) k M g⋅ Vt
k 1 × 10 −^13 ⋅ slug× 32.2 ft s^2
⋅ s 0.2 ft⋅
× lbf s
⋅^2
slug ft⋅
= × k 1.61 × 10 −^11 lbf s⋅ ft
dV g k M
− ⋅V
To find the time to reach 99% of V (^) t , we need V ( t ). From 1, separating variables =^ dt
Integrating and using limits (^) t M k
− ln 1 k M g⋅
⎛⎜ − ⋅V
We must evaluate this when (^) V = 0.99 V⋅ t V 0.198 ft s
t − 1 × 10 −^13 ⋅ slug ft 1.61 × 10 −^11 ⋅ lbf⋅s
× lbf s
⋅^2
slug ft⋅
× ln 1 1.61 × 10 −^11 lbf s⋅ ft
1 × 10 −^13 ⋅slug
× s
2 32.2 ft⋅
× 0.198 ft⋅ s
× slug ft⋅ lbf s ⋅^2
− ×
t =0.0286 s
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Problem 1.
(Difficulty: 2)
1.7 A rocket payload with a weight on earth of 2000 𝑙𝑙𝑙 is landed on the moon where the acceleration
due to the moon’s gravity 𝑔𝑚 ≈ 𝑔 6 𝑒. Find the mass of the payload on the earth and the moon and the
payload’s moon weight.
Given: Rocket payload weight on earth 𝑊𝑒 = 2000 𝑙𝑙𝑙. The acceleration due to the moon’s gravity
Find: The mass of payload on earth 𝑀𝑒 and on moon 𝑀𝑚 in SI and EE units. The payload’s moon weight
Solution:
Basic equation: Newton’s law applied to mass and weight
Gravity on the moon relative to that on Earth:
The value of gravity is:
The mass on earth is:
𝑔𝑒^ =
The mass on moon is the same as it on earth:
The weight on the moon is then
6 �^ =^ 𝑀𝑒^ �
6 �^ =^
6 = 333^ 𝑙𝑙𝑙
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Problem 1.
(Difficulty: 2)
1.9 Calculate the specific weight, specific volume and density of air at 40℉ and 50 𝑝𝑝𝑝𝑝. What are the
values if the air is then compressed isentropically to 100 psia?
Given: Air temperature: 40℉, Air pressure 50 psia.
Find: The specific weight, specific volume and density at 40℉ and 50 psia and the values at 100 psia
after isentropic compression.
Assume: Air can be treated as an ideal gas
Solution:
Basic equation: 𝑝𝑝 = 𝑅𝑅
The absolute temperature is
The gas constant is
The specific volume is:
50 𝑝𝑝𝑝𝑝 ×^144 𝑝𝑖^
2
× 500°𝑅 = 119.
The density is the reciprocal of the specific volume
Using Newton’s second law, the specific weight is the density times gravity:
For the isentropic compression of air to 100 psia, we have the relation for entropy change of an ideal gas:
𝑝 2 − 𝑝 1 = 𝑐𝑝 ln 𝑇𝑇^2
1
− 𝑅 ln 𝑝𝑝^2
1
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The definition of an isentropic process is
Solving for the temperature ratio
𝑅 1 =^ �
𝑅/𝑐𝑝
The values of R and specific heat are
𝑝𝑙𝑠𝑠 ∙ °𝑅 = 53.^
𝑙𝑙 ∙ °𝑅 = 0.^
The temperature after compression to 100 psia is
𝑅/𝑐𝑝
- 0686 / 0. 24
The specific volume is computed using the ideal gas law:
100 𝑝𝑝𝑝𝑝 ×^144 𝑝𝑖^
2
× 610.00°𝑅 = 72.
The density is the reciprocal of the specific volume
𝑝 2 = 0.^
The specific weight is:
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We must evaluate this when (^) V = 0.99 V⋅ t V 0.198 ft s
t 1 × 10 −^13 ⋅ slug ft 1.61 × 10 −^11 ⋅ lbf⋅s
× lbf s
⋅^2
slug ft⋅
⋅ ln 1 1.61 × 10 −^11 lbf s⋅ ft
1 × 10 −^13 ⋅slug
× s
2 32.2 ft⋅
× 0.198 ft⋅ s
× slug ft⋅ lbf s ⋅^2
− ×
t =0.0286 s
From 2, after rearranging (^) V dy dt
= M g⋅ k
1 e
k M
− ⋅t −
Integrating and using limits (^) y M g⋅ k
t M k
e
k M
− ⋅t − 1
y 1 × 10 −^13 ⋅ slug 32.2 ft⋅ s^2
× ft 1.61 × 10 −^11 ⋅ lbf⋅s
× lbf s
⋅^2
slug ft⋅
⋅ 0.0291 s⋅
10 −^13 ⋅ slug ft 1.61 × 10 −^11 ⋅ lbf⋅s
⋅ lbf s
⋅^2
slug ft⋅
⋅ e
1.61 10× −^11 1 10× −^13
− ⋅. − 1
y =4.49 × 10 −^3 ⋅ft
0 5 10 15 20 25
5
t (ms)
y (0.001 ft)
This plot can also be presented in Excel.
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Problem 1.14 [Difficulty: 4]
Given: Data on sky diver: M = 70 kg⋅ k 0.25 N s
⋅^2
m^2
Find: Maximum speed; speed after 100 m; plot speed as function of time and distance.
Solution: Use given data; integrate equation of motion by separating variables.
Treat the sky diver as a system; apply Newton's 2nd law:
Newton's 2nd law for the sky diver (mass M) is (ignoring buoyancy effects): (^) M dV dt
⋅ = M g⋅ −k V ⋅^2 (1)
Mg
FD = kV^2
a = dV/dt
(a) For terminal speed V (^) t , acceleration is zero, so M g⋅ − k V ⋅^2 = 0 so (^) Vt M g⋅ k
Vt 70 kg⋅ × 9.81m s^2
⋅ m
2
0.25 N⋅ ⋅s^2
× N s
⋅^2
kg ×m
1 2 = Vt 52.4 m s
(b) For V at y = 100 m we need to find V ( y ). From (1) M dV dt
⋅ M dV dy
⋅ dy dt
= ⋅ M V⋅ dV dt
= ⋅ =M g⋅ −k V ⋅^2
Separating variables and integrating:
0
V V V
1 k V
⋅^2
M g⋅
d 0
y g y
=⎮⌡ d
so (^) ln 1 k V
⋅^2
M g⋅
2 k⋅ M
= − y or^ V^2 M g⋅ k
1 e
2 k⋅ ⋅y M
− −
Hence (^) V y( ) Vt 1 e
2 k⋅ ⋅y M
− −
1 2
= ⋅
For y = 100 m: (^) V 100 m( ⋅ ) 52.4 m s
⋅ 1 e
− 2 × 0.25N s^ ⋅^2 m^2
⋅ × 100 ⋅m 1 70 kg⋅
× kg m⋅ s 2 ⋅N
× −
1 2
= ⋅ V 100 m( ⋅ ) 37.4 m s
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Problem 1.16 [Difficulty: 3]
Given: Long bow at range, R = 100 m. Maximum height of arrow is h = 10 m. Neglect air resistance.
Find: Estimate of (a) speed, and (b) angle, of arrow leaving the bow.
Plot: (a) release speed, and (b) angle, as a function of h
Solution: Let V 0 = u i 0 + v j 0 = V 0 (cos θ 0 i +sin θ 0 j)
ΣF (^) y = m dvdt= −mg , so v = v 0 – gt, and tf = 2t (^) v=0 = 2v 0 /g
Also, mv dv dy
mg, v dv g dy, 0 v 2
(^0) gh 2 = − = − − = −
Thus h = v 02 2g (1)
ΣF m du dt
0, so u u const, and R u t 2u v 0 0 f (^) g x =^ =^ =^ =^ =^ =^0 0 (2)
From Eq. 1: v 02 = 2gh (3)
From Eq. 2: u gR 2v
gR 2 2gh
u gR (^000) 8h
= = ∴ 2 =^2
Then 2
2 1 0 2 2 02 20 0 8 2 and^28 ⎟⎟ ⎠
= + = + =⎛^ +
h gh V gh gR h V u v gR (4)
s
- 7 m 10 m 100 m^1 s
m 8 10 m^9.^81 s 2 9. 81 m^2
1 2 2 (^0 22) ⎟⎟⎠ =
V =⎛^ × × + × ×
From Eq. 3: v 2gh V sin sin
2gh (^0 0) V 1 0
R
V 0
θ 0
y
x
h
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⎟ ×
= −^ ⎛^ × × 21. 8
37.7m
10 m s s
sin 2 9. 81 m^2
1 2
θ^1
Plots of V 0 = V 0 (h) (Eq. 4) and θ 0 = θ (^) 0(h) (Eq. 5) are presented below:
Initial Speed vs Maximum Height
h (m)
V^
(m/s) 0
Initial Angle vs Maximum Height
h (m)
θ^
o^ (
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Problem 1.
(Difficulty: 1)
1.14 The density of a sample of sea water is 1.99 𝑠𝑠𝑠𝑠𝑠 𝑓𝑓⁄ 3. What are the values in SI and EE units?
Given: The density of sea water is 1.99 𝑠𝑠𝑠𝑠𝑠 𝑓𝑓⁄ 3
Find: The density of sea water in SI and EE units
Solution:
For SI unit:
The relations between the units are 1 𝑚 = 3.28 𝑓𝑓 , 1 𝑘𝑠 = 0.0685 𝑠𝑠𝑠𝑠
1.99 × 0.0685^1 𝑘𝑠
3.28^3 𝑚^
3
For EE units:
The relation between a lbm and a slug is 1 𝑠𝑙𝑚 = 0.0311 𝑠𝑠𝑠𝑠
1.99 × 0.0311^1 𝑠𝑙𝑚
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Problem 1.
(Difficulty: 1)
1.15 A pump is rated at 50 ℎ𝑝; What is the rating in 𝑘𝑘 and 𝐵𝐵𝐵 ℎ𝑟⁄^?
Given: The pump is rated at 50 ℎ𝑝.
Find: The rating in 𝑘𝑘 and 𝐵𝐵𝐵 ℎ𝑟⁄^.
Solution:
The relation between the units is
The power is then
𝑃 = 50 ℎ𝑝 = 50 ℎ𝑝 ×
1.341 ℎ𝑝 = 37.3^ 𝑘𝑘
𝑃 = 50 ℎ𝑝 = 50 ℎ𝑝 ×
0.000393 ℎ𝑝^ = 127,^
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