German University in Cairo
Physics Department
PHYS 101 Winter 2020
Dr. Amr Aboshousha
Dr. Nermeen Serag
Sheet 10: Elasticity and Fluid Mechanics
Part B: Fluid Mechanics
P1. Water and then oil (Which don’t mix) are poured into a U- shaped tube, open at both
ends. They come to equilibrium as shown in the figure, What is the density of the oil? [Hint:
Pressure at points a and b are equal. Why?].
𝑃𝑎 = 𝑃𝑏
(𝜌𝑔𝐻)𝑜𝑖𝑙 + 𝑃𝑜 = (𝜌𝑔(𝐻 − ℎ))𝑤𝑎𝑡𝑒𝑟 + 𝑃𝑜
𝐻 = 27.2𝑐𝑚 ; ℎ = 8.62𝑐𝑚
𝜌𝑜𝑖𝑙 =
𝜌𝑜𝑖𝑙 =
(𝜌𝑔(𝐻 − ℎ))𝑤𝑎𝑡𝑒𝑟
(𝑔𝐻)𝑜𝑖𝑙
1000(0.27 − 8.62)
= 683.82𝑘𝑔/𝑚3
0.272
P2. (I) The maximum gauge pressure in a hydraulic lift is 17.0atm. What is the largest- size
vehicle (kg) it can lift if the diameter of the output line is 22.5cm?.
𝑃𝑔 = 17𝑎𝑡𝑚
𝑟=
0.225
2
1 𝑎𝑡𝑚 = 1.013 × 105 𝑁/𝑚2
= 0.113𝑚
𝑃=
𝐹
𝑚𝑔
=
𝐴
𝜋 𝑟2
𝑃𝜋 𝑟 2
𝑚=
𝑔
17 × 1.013 × 105 𝜋 0.1132
𝑚=
= 7049.2𝑘𝑔
9.8
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PHYS101 Winter 20
(II) The gauge pressure in each of the four tires of an automobile is 240kPa. If each tire
has a “footprint” of 220𝑐𝑚2 , estimate the mass of the car.
𝑃𝑔 = 240 × 103 𝑝𝑎 𝑝𝑒𝑟 𝑡𝑖𝑟𝑒
𝐴 = 220 × 10−4 𝑚2
𝑃=
𝐹𝑝𝑒𝑟 𝑡𝑖𝑟𝑒
𝑚𝑔
=
𝐴
𝐴
𝑚
𝑃𝐴
=
4
𝑔
3
𝑚
240 × 10 × 220 × 10−4
=
4
9.8
𝑚 = 538.775 × 4 = 2155.1𝑘𝑔
P3. A hydraulic press for compacting powdered samples has a large cylinder which is 10.0
cm in diameter, and a small cylinder with a diameter of 2.0 cm as shown in the Figure. A
lever is attached to the small cylinder as shown. The sample, which is placed on the large
cylinder, has an area of 4.0 𝑐𝑚2.What is the pressure on the sample if F = 350 N is applied
to the lever?
(𝑅𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑙𝑎𝑟𝑔𝑒 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟)𝑅 = 0.05𝑚
𝐴𝑠𝑎𝑚𝑝𝑙𝑒 = 4 × 10−4 𝑚2
Psample = ?
(𝑅𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑠𝑚𝑎𝑙𝑙 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟)𝑟 = 0.01𝑚
at point C Σ 𝑡𝑜𝑟𝑞𝑢𝑒 = 0
𝑡𝑜𝑟𝑞𝑢𝑒𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 = 𝑡𝑜𝑟𝑞𝑢𝑒𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒
350 × 2𝑙 = 𝑓 × 𝑙
𝑓 = 2 × 350 = 700𝑁
For the hydraulic Piston:
𝐹 𝑓
=
𝐴 𝑎
𝐴
𝜋𝑅 2
𝑅2
F=𝑓 =𝑓 2 =𝑓 2
𝑎
𝜋𝑟
𝑟
F = 700 ×
0.052
= 17500𝑁
0.012
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PHYS101 Winter 20
Psample =
𝑃𝑠𝑎𝑚𝑝𝑙𝑒 =
𝐹
𝐴𝑠𝑎𝑚𝑝𝑙𝑒
17500
= 43.75 × 106 𝑁/𝑚2
4 × 10−4
P4. A 15-cm radius air duct is used to replenish the air of a room 8.2 × 5 × 3.5 𝑚 every 12
min. How fast does the air flow in the duct?
.
𝑟 = 0.15 𝑚
𝑉 = 8.2 × 5 × 3.5 𝑚3 𝑡 = 12 × 60 = 720 𝑠𝑒𝑐 𝑣 =?
𝑄 = 𝐴𝑣
𝑉
= 𝐴𝑣
𝑡
8.2 × 5 × 3.5
= 𝜋 × 0.152 𝑣
12 × 60
8.2 × 5 × 3.5
𝑣=
= 2.82 𝑚/𝑠
12 × 60 × 𝜋 × 0.152
P5. How fast does water flow from a hole at the bottom of a very wide 5.3m deep storage
tank filled with water? Ignore viscosity.
ℎ = 5.3𝑚
𝑣 =?
𝑣 = √2𝑔ℎ
𝑣 = √2 × 9.8 × 5.3
𝑣 = 10.2 𝑚/𝑠
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PHYS101 Winter 20
P6. Suppose the top surface of the vessel in the figure is subjected to an external gauge
pressure 𝑃2 (a) Derive a formula for the speed 𝑦1 , at which the liquid flows from the
opening at the bottom into atmospheric pressure, 𝑃0 . Assume the velocity of the liquid
surface 𝑉2, is approximately zero. b) If 𝑃2 = 0.85 atm and 𝑦2 − 𝑦1 = 2.4 m, determine 𝑣1
for water.
ℎ2 − ℎ1 = 2.4𝑚
𝑣2 = 0
𝑃2 = 0.85 × 1.013 × 105 𝑝𝑎 𝜌𝑤 = 1000𝑘𝑔/𝑚3
Using Bernoulli’s Equation
1
1
𝑃1 + 𝜌𝑔ℎ1 + 𝜌𝑣12 = 𝑃2 + 𝜌𝑔ℎ2 + 𝜌𝑣22
2
2
1
𝑃0 + 𝜌𝑣12 = 𝑃2 + 𝑃0 + 𝜌𝑔(ℎ2 − ℎ1 )
2
2(𝑃2 + 𝜌𝑔(ℎ2 − ℎ1 ))
𝑣= √
𝜌
𝑣= √
2(0.85 × 1.013 × 105 + 1000 × 9.8 × 2.4)
= 14.81𝑚/𝑠
1000
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PHYS101 Winter 20