46. The base of a solid metal cone (Sp. Gr. 6.95) is 25 cm in diameter. The
altitude of the cone is 30 cm. If placed in a basin containing mercury (Sp.
Gr. 13.60) with the apex of the cone down, how deep will the cone float?
∑Fy=0]
Given :
=W
d=25cm.
(wV)displaced mercury = (wV)cone
r=12.5cm=0.125
(9.81)(13.60)Vm = 9.81(6.95)Vcone
Vcone=4.9087x10−3
13.60Vm =
Vm =
6.95
6.95
2
3
0.125 2 (0.30)
0.30 3
3(13.60)
Vm = 2.50x10
−3
=(
)
3
3
x=
=
3
(0.30)3
0.30 3 (2.50 10 −3 )
4.91
10 −3
x = 0.24m
x =11,120
24cm N in the air, what
47. If a metal sphere 60 cm in diameter weighs
would be its weight when submerged in (a) water? (b) mercury?
Sol’n:
a.)
b.)
FB = 9.81 (4/3 πr^3)W_hy = 11,120 – 9.81(13.6)(4/3)(π)(ࠟ0.3ࠠ^2)
= 9.81(4/3)(ࠟ0.3ࠠ^2)
= -3976 N
FB = 1.11 kN
W = 11.12 – 1110
= 10.01 N
38
48. A rectangular solid piece of wood 30 cm square and 5 cm thick floats in
water to depth of 3.25 cm. How heavy an object must be placed on the
wood (Sp. Gr. 0.50) in such a way that it will just be submerged?
Given: dept=3.25cm
Req. F=?
30cm
W
5cm
Fb=w
w.s.
wv'=wsv
s=v’/v
Fb
s= (30) (30) (3.25)
w.s.
4500
S= 0.65// ans.
W
F=Fb-W
F=wv(1-s)
Fb
F=wv-wsv
F= (9.81)(4500)(1-0.65)
N
49. A hollow vessel in the shape of paraboloid of revolution F=15.45
floats in fresh
water with its axis vertical and vertex down. Find the depth to which it
must be filled with a liquid (Sp. Gr. 1.20) so that its vertex will be
submerged at 45 cm from the water surface.
Solution:
=W
By Similar Solids:
9.81Vd = 9.81(1.20)Vp
=(
)3
0.45
Vd = 1.20Vp
3
1.20
1
1.20
= 0.45 3
3
==
0.45 3
a = 0.42m
39
a = 42cm
50. A barge is 16 m long by 7 m wide 120 cm deep, outside dimensions. The
sides and bottom of the barge are made of timber having thickness of 30
cm. The timber weighs 7860 N/cu.m. If there is to be freeboard of 20 cm
in fresh water how many cubic meters of sand weighing 15700 N/cu.m
may be loaded uniformly into the barge?
Sol’n:
V_t=V_o-V_i F_b-W_t-W_s = 0
= (16)(7)(1.2) – (15.4)(16.4)(0.9)
9810(16)(7)(1) – 7860(45.7) – 15700
= 45.7 m^3
V_s = 0
V_s = 47.10 m^3
51. A brass sphere (Sp. Gr. 8.60) is placed in a body of mercury. If the
diameter of the sphere is 30 cm (a) what minimum force would be
required to hold it submerged in mercury? (b) what is the depth of
flotation of the sphere when it is floating freely?
hg.s.
F
F=Fb-W
W
F=wSmVs-wSsVs
Fb
F=wVs(Sm-Ss)
F=(9.81) (3/4)(3.14)(0.15)^3(13.60-8.6)
F=693.43N//ans.
y
Fb
40
V=4/3(3.14)(r)^3
V=4/3(3.14)(15)^3
V=14,137.17cm^3
V’=3.14/3 D^2(3r-D)
Fb=w
wSmV’=wSsV
8939.68 =3.14/3(y^2)((3x15)-y)
V’=8.60(14137.17)
13.60
y=17.10cm
V’=8939.68cm^3
52. A spherical balloon weighs 3115 N. How many newton of helium have to
be put in the balloon to cause it to rise, (a) at sea level? (b) at an
elevation of 4570 m?
Soln:
W = Fb – Fh
W = ρ_agV - ρ_hgV
[W = V(ρ_ag - ρ_hg)]1/(ࠟ(ρࠠ_a g - ρ_h g))
V = w/(g(ρ_a-ρ_h))
= 3115/(9.81(1.29-0.179))
V = 286.1 m^3
53. The Sp. Gr. of rock used as concrete aggregate is often desirable to
know. If a rock weighed 6.15 N in the air and 3.80 N when submerged in
water, what would be the specific gravity of the rock?
Soln:
W = W_a - W_w
S = 6.15/(9810(2.4 x ࠟ10ࠠ^(-4)))
W = 6.15 – 3.8
S.g = 2.62
(9810)V = 2.35
V = 2.4 x ࠟ10ࠠ^(-4) m^3
S.W = W_a/V = (6.15 N)/(2.4 x ࠟ10ࠠ^(-4) )
S.W = 25625 N/m^3
41
54. A piece of wood weighs 17.80 N in air and piece of metal weighs 17.80
in water. Together the two weighs 13.35 N in water. What is the specific
gravity of the wood?
Solution:
W wo=17.80N (air)
W wo=17.80-FB
Wm=17.80N(water)
WT=W wo+Wm
13.35N=17.80-FB+17.80
; W T=13.85N
W wo=17.80(air)
FB=22.25N (Displaced Water)
Gs=W wo/FB
=17.80N/22.25N
Gs=0.80
55. A sphere 1.0 in diameter floats half submerged in tank of liquid (Sp. Gr.
0.80) (a) what is the weight of the sphere? (b) What is the minimum
weight of the anchor (Sp. Gr. 2.40) that will require to submerge the
sphere completely?
Given:
Find:
Sa = 7.40
Vs = / ^
= /
Sliquid = 0.80
.
Ws
Wa
^
= 0.52m³
A.) W=fb
=WsLVs/2
Ws=9.81Kn/m3(0.80)(0.52m3)/2
Ws=2.05KN
B.) Wa=Fba+Fbs–W
where: V a=Wa/Wsa
=WslVa+WslVs-WsVs
SS=Ws/wVs
42
Wa=w[0.80xwa/w2.40]+[0.80x0.32m3]-[2.04kn/9.81kn/m3]
Wa=0.33wa+4.08kn-2.04kn
Wa-0.33wa=2.04kn-4.08kn
0.67wa/0.67=2.04/0.67
Wa= 3.50kn
56. Fig. Z shows a hemispherical shell covering a circular hole 1.30 m in
diameter at the vertical side of a tank. If the shell weighs 12,450 N, what
vertical force is necessary to lift the shell considering a friction factor of
0.30 between the wall and the shell?
57. An iceberg has a specific gravity of 0.92 and floats in salt water (Sp. Gr.
1.03). If the volume of ice above the water surface is 700 cu.m, what is
the total volume of the iceberg?
given:
Si=0.92
find:
Vt
Ssw=1.03
43
W=W vƩfv=0]
W=WsiVt
vt=v1+v2
W=Fb
Fb=W ssw
=Wsssv2
v2=vt-v1
wsivt=wssw(vt-v1)
Sivt=wswvt-sswv1
sswv1=vt(ssw-si)
vt=sswv1/ssw-si
=1.03(700)/1.03-0.92
Vt=6554.55m3
58. A concrete cube 60 cm on each edge (Sp. Gr. 2.40) rests on the bottom
of a tank in which sea water stands to a depth of 5 m. The bottom edges
of the block are sealed off so that no water is admitted under the block.
Find the vertical pull required to lift the block.
Solution:
W1=23.54(0.6x7xd)
;d=1
W1=98.868
W2=197.736[1/2(24)(7)d]
X=(RM-OM)/RV
W2=197.736
=(583.3692-
204.375)/296.604
F=δhA
X=1.28m
=9.81(2.5)(5)
F=122.625
e=b/2-x
Rx=122.625d
S=Ry/b(1±6e/b)
Ry=296.604d
S=142.36992kPa
RM=98.868(3-0.3)+197.766[2/3(2.4)]
RM=583.3692kN.m
OM=122.625(5/3)
OM=204.375kN.m
44
59. A 15 cm by 15 cm by 7 m long timber weighing 6280 N/cu.m is hinged at
one end and held in horizontal position by an anchor at the other end as
shown in Fig. AA. If the anchor weighs 23450 N/cu.m, determine the
minimum total weight it must have
Solution:
Vt=(0.15)2(7)
Wa=W aVa
Vt=0.1575m2
Wa=23540Va
Wt=W tVt
=(0.1575)(62.80)
Wt=989.1
Fbt=9810(0.1575)
Fba=WVa
Fbt=1545.075
Fba=9810Va
Mh= 3.5Fbt+7Fba=3.5Wt+W a
3.5(1545.075)+7(9810)
Va=3.5(989.1)+7(23540)Va
Va=0.02m3
Wa=W aVa
=23540N/m3(0.02m3)
Wa=470.8N
45
Va=W a/23540
60. A cylinder weighing 445 N and having a diameter of 1.0 m floats in salt
water (Sp. Gr. 1.03) with its axis vertical as in Fig. BB. The anchor
consist 0f 0.0280 cu.m of concrete weighing 23450 N/cu.m. What rise in
the tide r will be required to lift the anchor off the bottom?
Solution:
Wa=23540(0.280)
Fba=9810(1.05)(0.280)
Wa=659102N
Fba=2829.204N
Wo=445N
Fbc=9810(1.03)p(0.5)2(0.3+r)
Fbc=2380.769+7935.89866
∈Fv=0
Fba+Fbc=W a+Wc
2829.204+2380.769+7935.898r=6591.2+445
r=0.23m ; 23cm
46
61. A timber 15 cm square and 5 m long has a specific gravity of 0.50. One
end is hinged to the wall and the other is left to float in water (Fig. CC).
For a=60 cm, what is the length of the timber submerged in water?
Solution:
Wt=9.81(0.5)(5)(0.15)2
Wt=0.5518125kN
Fb=9.81(0.15)2(x)
Fb=0.220725x
Mh=0
2.5cosθW t=(5-0.5x)cosθFb
2.5(0.5518125)=(5-0.5x)(0.220725x)
0.11x2-1.1x+1.375=0
1.375=1.1x-0.11x2
=
− ±
2
2
−4
62. A metal block 30 cm square and 25 cm deep is X=1.46m
allowed to float on a
body of liquid which consist of 20 cm layer of water above a layer of
mercury. The block weighs 18,850 N/cu.m. What is the position of the
upper level of the block? If a downward vertical force of 1110 N is
applied to the centroid of the block, what is the new position of the upper
level of the block?
Solution:
a.) Fbm=wV
Fbw=9.81(0.20)(009)
=9.81(13.6)(0.09)(0.05-x)
Fbm=0.600372-12.00744x
W=18.85(0.09)(0.25)
W=0.424125kN
47
Fbw=0.17658kN
∈Fv=0
Fbw+Fbm=W
0.17658+0.600372-12.000744x=0.424125
X=0.0294m
X=2.94cm
b.) Fbm=9.81(13.6)(0.09)(0.25-x)
W=0.4241225
Fbm=3.00186-12.00744x
Wv=1.11kN
Fbw=9.81(0.09)(x)
Fbw=0.8829x
∈Fv=0
0.8829x+3.00186-12.00744x=0.4241225+1.11kN
X=0.132m
H=0.20m-0.132m
H=0.068m
H=68cm
63. Two spheres, each 1.2 m diameter, weigh 4 and 12 KN, respectively.
They are connected with a short rope and placed in water. What is the
tension in the rope and what portion of the lighter sphere produces from
the water? What should be the weight of the heavier sphere so that the
lighter sphere will float halfway out of the water?
Solution:
=
(0.60 3 )
9.81 4
3
2
= 8.8759 kN
T=
1 4 (0.60)3
= (9.81)( )(
3
)
= 4.4379 kN
-
T = Wss -
T =12 kN - 8.8759 kN
T = 3.12 kN
T = 4 kN - 4.4379 kN
T = 0.4379 kN
∑Fy=0]
=
-T
9.81Vss = 4 + 3.12
Vss = 0.72579
3
48
=T+
= 0.44 kN + 8.88 kN
Vs =
2
3
(3 − )
0.72579 = 0.60π
D = 0.85m
= 9.32 kN
2
- 0.33π
3
--- by trial & error
X = 1.20 – D
X = 0.35m
68. If the specific gravity of a body is 0.80, what proportional part of its total
volume will be submerged below the surface of a liquid (Sp. Gr. 1.20)
upon which it floats?
Solution:
=
(
=(
)
(9.81)(1.20)
(1.20)
=
2
3
)
= (9.81)(0.80)
= (0.80)
2
3
of the total Volume
69. A vertical cylinder tank, open at the top, contains 45.50 cu.m of water. It
has a horizontal sectional area of 7.40 sq.m and its sides are 12.20 m
high. Into its lowered another similar tank, having a sectional area of 5.60
sq.m and a height of 12.20 m. The second tank is inverted so that its
open end is down, and it is allowed to rest on the bottom of the first. Find
the maximum hoop tension in the outer tank. Neglect the thickness of the
inner tank.
49
70. A small metal pan of length of 1.0 m, width 20 cm and depth 4 cm floats
in water. When a uniform load of 15 N/m is applied as shown in Fig. DD,
the pan assumes the figure shown. Find the weight of the pan and the
magnitude of the righting moment developed.
Solution:
= 0.04m (0.20m) (1m)
= 8x10−3
Θ=
3
2
=
1=
4x10−3
3
−1 0.04
( )
0.20
T=
Θ = 11.31
15 = 1
F = 15
=
+T=
-T
= 9810(4x10−3 ) – 15
= 39.24 - 15
= 24.24 N
71. A ship of 39,140 KN displacement floats in sea water with the axis of
symmetry vertical when a weight of 490 KIN is mid ship. Moving a weight
3 m toward one side of the deck cause a plumb bob, suspended at the
end of a string 4 m long, to move 24 cm. Find the metacentric height.
Given:
W=39140kn
Ta Φ= .
Φ= .
/
C=W x
˚
=
Si Φ=X/MG
x
X=MGSi Φ
=
MG=0.63m
50
MGSi
.
˚
72. A rectangular scow 9.15 m wide by 15.25 m long and 3.65 m high has a
draft of 2.44 m in sea water. Its center of gravity is 2.75 m above the
bottom of the scow. (a) Determine the initial metacentric height. (b) If the
scow tilts until one of the longitudinal sides is just at the point of
submergence, determine the righting couple or the overturning couple.
Soln:
a.)
b.)
GB_o = 2.75 – 1.22
tanθ = 1.21/4.575
= 1.53 m
θ = 14.81°
MG = MB_o - GB_o
MB_o = B^2/12D(1 + ࠟtanࠠ^2θ/2)
= 2.86 – 1.53 = ࠟ9.15ࠠ^2/(12(2.46))(1 + ࠟtanࠠ
MG = 1.33 m^(2(14.81))/2)
MB_o = 2.96 m
∑Fv = 0 ; FB = W
FB = wV
= 9.15(15.25)(2.44)(9.81)(1.03)
FB = W
W = 3490.23 kN
RM = W(MGsinθ) = 3490.23(1.34sin14.81°)
RM = 1257.7 kN.m
73. A cylindrical caisson has an outside diameter of 6 m and floats in fresh
water with its axis vertical. Its lower end is submerged to a depth of 6 m
below the water surface. Find: (a) the initial metacentric height; (b) the
righting couple when the caisson is tipped through an angle of 10
degrees.
Soln:
a.)
b.)
MB_o = I/V ; I=(π(6^2))/(12(4)) ; V=(π(6)(6))/4
MG = MB_o+GB_o = 0.375 + 0.5
M_o = 0.375 m
MG = 0.875 m
51
74. A rectangular scow 9.15 m wide by 15.25 m long has a draft of 2.44 m in
fresh water. Its center of gravity is 4.60 m above the bottom. Determine
the height of the scow if, with one side just at the point of submergence,
the scow is in unstable position.
75. A rectangular raft 3 m wide 6 m long has a thickness of 60 cm and is
made of solid timbers (Sp. Gr. 0.60). If a man weighing 890 N steps on
the edge of the raft at the middle of one side, how much will the original
water line on that side be depressed below the water surface?
Find: RM
V’= .
.
.
W=wV
=383.73m3=(9.81kn/m3)(383.73m3)
W=3764.39kn
Ta Φ= .
Φ=
/ .
.
V=1/2(15.25m)(1.85m)(4.573)
˚
V=
Mbo=VL/V’Si Φ
=64.54m3(6.1)/383.73m3 Si
.
3
G bo=2.30-1.375m
.
˚ Gbo=0.925m
Mbo=2.74m
MG=Mbo-Gbo
=2.74m- .
MG=1.815m
RM=w MGSi Φ
=
.
k
RM=1242.60kn.m
52
.
si Φ
.
˚