MAE221 Fluid Mechanics
Homework #2 (7 problems)
Due: 10:00pm on October 2, 2015.
-----------------------------------------------------------------------------------------------1. A piston having a cross-sectional area of 0.07 m2 is located in a cylinder containing water as shown
in Fig. P1. An open U-tube manometer is connected to the cylinder as shown. For h1 = 60 mm and h =
100 mm, what is the value of the applied force, P, acting on the piston? The weight of the piston is
𝑘𝑔
𝑘𝑔
negligible. (𝜌water = 1 × 103 [𝑚3 ] , 𝜌𝑚𝑒𝑟𝑐𝑢𝑟𝑦 = 13.6 × 103 [𝑚3 ] , 𝑔 = 9.80[𝑚/𝑠 2 ])
Figure. P1
[Solution]
For equilibrium at location A
𝑃A = 𝑝𝑝𝑖𝑠𝑡𝑜𝑛 + 𝜌𝑤 𝑔ℎ1 = 𝜌𝑚 𝑔ℎ
𝑝𝑝𝑖𝑠𝑡𝑜𝑛 = 𝜌𝑚 𝑔ℎ − 𝜌𝑤 𝑔ℎ1 = 13.6 × 103 × 9.80 × 0.1 − 1 × 103 × 9.80 × 0.06
= 1.27 × 104 [𝑁/𝑚2 ]
Thus,
𝑃 = 𝑝piston × (𝐴𝑟𝑒𝑎)𝑝𝑖𝑠𝑡𝑜𝑛 = 1.27 × 104 × 0.07 = 889 [𝑁]
MAE221 Fluid Mechanics
Homework #2 (7 problems)
Due: 10:00pm on October 2, 2015.
-----------------------------------------------------------------------------------------------2. A structure is attached to the ocean floor as shown in Fig. P2. A 2-m-diameter hatch is located in an
inclined wall and hinged on one edge. Determine the minimum air pressure, p1, within the container
that will open the hatch. Neglect the weight of the hatch and friction in the hinge. (𝜌sea water = 1.03 ×
𝑘𝑔
𝑚
103 [𝑚3 ] , 𝑔 = 9.80[𝑠2 ] )
Figure. P2
[Solution]
1
2
𝐹R = 𝜌𝑔ℎ𝑐 𝐴 where ℎ𝑐 = 10 + × 2 × sin30° = 10.5[m].
Thus, 𝐹R = 1.03 × 103 × 9.80 × 10.5 × 𝜋 × 12 = 3.33 × 105 [𝑁]
To locate 𝐹𝑅 ,
𝐼
10
𝑦R = 𝑦𝑥𝑐𝐴 + 𝑦𝑐 where 𝑦c = sin 30° + 1 = 21[𝑚].
𝑐
𝜋 4
∙1
4
So that 𝑦R = 21∙𝜋
+ 21 = 21.012[𝑚]
For equilibrium
∑ 𝑀𝐻𝑎𝑡𝑐ℎ = 0 ; 𝐹𝑅 (𝑦𝑅 − 𝑦𝑐 ) = 𝑝1 𝐴𝑙 ;
𝑝1 =
𝐹𝑅 (𝑦𝑅 − 𝑦𝑐 ) 3.33 × 105 × (21.012 − 20)
=
= 1.07 × 105 [𝑃𝑎]
𝐴𝑙
𝜋×1
Thus,
𝒑𝟏 = 𝟏𝟎𝟕 [𝒌𝑷𝒂]
MAE221 Fluid Mechanics
Homework #2 (7 problems)
Due: 10:00pm on October 2, 2015.
-----------------------------------------------------------------------------------------------3. A 3-m-wide, 8-m-high rectangular gate is located at the end of a rectangular passage that is connected
to a large open tank filled with water as shown in Fig. P3. The gate is hinged at its bottom and held
closed by a horizontal force, FH, located at the center of the gate. The maximum value for FH is 3500kN.
(a) Determine the maximum water depth, h, above the center of the gate that can exist without the gate
opening. (b) Is the answer the same if the gate is hinged at the top? Explain your answer.
Figure. P3 (a)
Figure. P3 (b)
[Solution]
(a) For gate hinged at bottom
∑ 𝑀𝐻𝑖𝑛𝑔𝑒 = 0 ;
𝐹𝑅 𝑙 − 𝐹𝐻 ∙ 4 = 0
And
FR = 𝜌𝑔ℎ𝑐 𝐴 = 1 × 103 × 9.80 × ℎ × 3 × 8 = 2.35 × 105 × ℎ
1
× 3 × 83
Ixc
5.33
12
𝑦𝑅 =
+ 𝑦𝑐 =
+ℎ =
+ℎ
yc 𝐴
ℎ×3×8
ℎ
5.33
5.33
𝑙 = ℎ + 4 − 𝑦R = ℎ + 4 − (
+ ℎ) = 4 −
h
ℎ
Thus
∑ 𝑀𝐻𝑖𝑛𝑔𝑒 = 0 ;
𝐹𝑅 𝑙 − 𝐹𝐻 ∙ 4 = 2.35 × 105 × ℎ × (4 −
So that 𝒉 = 𝟏𝟔. 𝟐 [𝐦]
5.33
) − 3500 × 103 × 4 = 0
ℎ
(b) For gate hinged at top
∑ 𝑀𝐻𝑖𝑛𝑔𝑒 = 0 ;
𝐹𝑅 𝑙1 − 𝐹𝐻 ∙ 4 = 0
And
5.33
5.33
𝑙1 = 𝑦R − ℎ + 4−= (
+ ℎ) − ℎ + 4 =
+4
h
ℎ
Thus
∑ 𝑀𝐻𝑖𝑛𝑔𝑒 = 0 ;
So that 𝒉 = 𝟏𝟑. 𝟔 [𝐦]
5.33
𝐹𝑅 𝑙1 − 𝐹𝐻 ∙ 4 = 2.35 × 105 × ℎ × (
+ 4) − 3500 × 103 × 4 = 0
ℎ
Maximum depth for gate hinged at top is less than maximum depth for gate hinged at bottom.
MAE221 Fluid Mechanics
Homework #2 (7 problems)
Due: 10:00pm on October 2, 2015.
-----------------------------------------------------------------------------------------------4. The quarter circle gate BC in Fig. P4 in hinged at C. Find the horizontal force P required to hold the
gate stationary. The width b into the paper is 2m. Neglect the weight of the gate.
Figure. P4
[Solution]
The horizontal component if water force is
𝐹𝐻 = 𝜌𝑤 𝑔ℎ𝐶𝐺 𝐴 = 1 × 103 × 9.8 × 1 × (2 × 2) = 3.92 × 104 [N]
This force acts 2/3 of the way down or 1.333m down from the surface (0.667m up from C). The vertical
is the weight of the quarter-circle of water above gate BC:
𝐹𝑉 = 𝜌𝑔(𝑉𝑜𝑙)𝑤𝑎𝑡𝑒𝑟 = 1 × 103 × 9.80 ×
𝜋 × 22
× 2 = 6.16 × 104 [N]
4
𝐹𝑉 acts down at (4R/3π)=0.849 m to the left of C. Sum moments clockwise about point C:
∑ 𝑀𝐶 = 0 = 2𝑃 − 𝐹𝐻 × 0.667 − 𝐹𝑉 × 0.849
𝑃=
FH × 0.667 + 𝐹𝑉 × 0.849 3.92 × 104 × 0.667 − 6.16 × 104 × 0.849
=
= 3.92 × 104 [𝑁]
2
2
So 𝑷 = 𝟑. 𝟗𝟐 × 𝟏𝟎𝟒 [𝐍] or 𝑷 = 𝟑𝟗. 𝟐 [𝐤𝐍]
MAE221 Fluid Mechanics
Homework #2 (7 problems)
Due: 10:00pm on October 2, 2015.
-----------------------------------------------------------------------------------------------5. A spar buoy is a buoyant rod weighted to float and protrude vertically, as in Fig. P5. It can be used
for measurements or markers. Suppose that the buoy is maple wood (SG = 0.6), 4 cm by 4 cm by 0.4
m, floating in seawater (SG = 1.025). How many weight of steel (SG = 7.85) should be added to the
bottom end so that h = 36cm?
Figure. P5
[Solution]
The relevant volumes needed are
𝑉𝑠𝑝𝑎𝑟 = 0.04 × 0.04 × 0.4 = 6.4 × 10−4 [𝑚3 ]
𝑉𝑠𝑡𝑒𝑒𝑙 =
𝑊𝑠𝑡𝑒𝑒𝑙
𝑊𝑠𝑡𝑒𝑒𝑙
=
[𝑚3 ]
𝑆𝐺𝑠𝑡𝑒𝑒𝑙 𝜌𝑤 𝑔 7.85 × 1000 × 9.80
𝑉𝑖𝑚𝑚𝑒𝑟𝑠𝑒𝑑 𝑠𝑝𝑎𝑟 = 0.04 × 0.04 × 0.36 = 5.76 × 10−4 [𝑚3 ]
The vertical force balance is: ∑ 𝐹𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 = 0
𝑏𝑢𝑜𝑦𝑎𝑛𝑐𝑦 𝐵 = 𝑊𝑤𝑜𝑜𝑑 + 𝑊𝑠𝑡𝑒𝑒𝑙
SGseawater 𝜌𝑤 𝑔(𝑉𝑖𝑚𝑚𝑒𝑟𝑠𝑒𝑑 𝑠𝑝𝑎𝑟 + 𝑉𝑠𝑡𝑒𝑒𝑙 ) = 𝑆𝐺𝑤𝑜𝑜𝑑 𝜌𝑤 𝑔𝑉𝑠𝑝𝑎𝑟 + 𝑊𝑠𝑡𝑒𝑒𝑙
1.025 × 1000 × 9.80 × (5.76 × 10−4 +
𝑤𝑠𝑡𝑒𝑒𝑙
)
7.85 × 1000 × 9.80
= 0.6 × 1000 × 9.80 × 6.4 × 10−4 + 𝑊𝑠𝑡𝑒𝑒𝑙
Thus,
Wsteel = 2.33[𝑘𝑔 ∙ 𝑚/𝑠 2 ] = 2.33 [𝑁]
(If you put ‘h’ as the height above the surface, then the Wsteel = −3.59 × 10−3 [𝑘𝑔 ∙ 𝑚/𝑠 2 ].)
MAE221 Fluid Mechanics
Homework #2 (7 problems)
Due: 10:00pm on October 2, 2015.
-----------------------------------------------------------------------------------------------6. A 1-m-diameter cylindrical mass, M, is connected to a 2-m-wide rectangular gate as shown in Fig.
P6. The gate is to open when the water level, h, drops below 2.5 m. Determine the required value for
M. Neglect friction at the gate hinge and the pulley.
Figure. P6
[Solution]
ℎ
FR = 𝜌𝑔ℎ𝑐 𝐴 = 𝜌𝑔 ( ) × (2 × ℎ) = 𝜌𝑔ℎ2
2
For equilibrium,
ℎ
∑ 𝑀𝑜 = 0 ; 4𝑇 = ( ) 𝐹𝑅
3
So that
𝑇=
ℎ
𝜌𝑔ℎ3
𝐹𝑅 =
12
12
For the cylindrical mass ∑ 𝐹𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 = 0 and
𝑇 = 𝑀𝑔 − 𝐹𝐵 = 𝑀𝑔 − 𝜌𝑔 𝑉𝑚𝑎𝑠𝑠
Thus,
𝑀=
𝑇 + 𝜌𝑔𝑉𝑚𝑎𝑠𝑠 𝜌ℎ3
=
+ 𝜌𝑉𝑚𝑎𝑠𝑠
𝑔
12
And for ℎ = 2.5 [𝑚]
𝑀=
So, 𝑴 = 𝟐𝟒𝟖𝟎 [𝒌𝒈]
1 × 103 × 2.53
𝜋
+ 1 × 103 × 12 (2.5 − 1) = 2.48 × 103 [𝑘𝑔]
12
4
MAE221 Fluid Mechanics
Homework #2 (7 problems)
Due: 10:00pm on October 2, 2015.
-----------------------------------------------------------------------------------------------7. The tank in Fig. P7 is filled with water and has a vent hole at point A. The tank is 1 m wide into the
paper. Inside the tank, a 10-cm balloon, filled with helium at 130 kPa, is tethered centrally by a string.
If the tank accelerates to the right at 5 m/s2 in rigid-body motion, at what angle will the balloon lean?
Will it lean to the right or to the left?
Figure. P7 (a)
Figure. P7 (b)
[Solution]
The acceleration sets up pressure isobars which slant down and to the right, in both the water and in the
helium. This means there will be a buoyancy force on the balloon up and to the right, as in Fig P7(b). It
must be balanced by a string tension down and to weight, the balloon leans up and to the right at angle
𝑎𝑥
5
θ = tan−1 ( ) = tan−1 (
) = 𝟐𝟕°
𝑔
9.80
measured from the vertical.