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Problem 2.101
The closed layered box in Fig. P2.101 has square horizontal cross-sections everywhere. All
fluids are at 20C. Estimate the gage pressure of the air if (a) the hydrostatic force on panel AB
is 48 kN; or if (b) the hydrostatic force on the bottom panel BC is 97 kN.
Solution 2.101
At 20C, take
oil = 891 kg/m3 and
water = 998 kg/m3. The wedding-cake shape of the box has
nothing to do with the problem. (a) the force on panel AB equals the pressure at the panel
Problem 2.102
A cubical tank is 3 m 3 m 3 m and is layered with 1 meter of fluid of specific gravity 1.0,
1 meter of fluid with SG = 0.9, and 1 meter of fluid with SG = 0.8. Neglect atmospheric pressure.
Find (a) the hydrostatic force on the bottom; and (b) the force on a side panel.
Solution 2.102
(a) The force on the bottom is the bottom pressure times the bottom area:
Problem 2.103
A solid block, of specific gravity 0.9, floats such that 75 percent of its volume is in water and
25 percent of its volume is in fluid X, which is layered above the water. What is the specific
gravity of fluid X?
Solution 2.103
The block is sketched below. A force balance is W = B, or
Problem 2.104
The can in Fig. P2.104 floats in the position shown. What is its weight in N?
Solution 2.104
The can weight simply equals the weight of the displaced water (neglecting the air above):
Problem 2.105
It is said that Archimedes discovered the buoyancy laws when asked by King Hiero of Syracuse
to determine whether his new crown was pure gold (SG = 19.3). Archimedes measured the
weight of the crown in air to be 11.8 N and its weight in water to be 10.9 N. Was it pure gold?
Solution 2.105
The buoyancy is the difference between air weight and underwater weight:
Problem 2.106
A spherical helium balloon has a total mass of 3 kg. It settles in a calm standard atmosphere at
an altitude of 5500 m. Estimate the diameter of the balloon.
Solution 2.106
From Table A.6, standard air density at 5500 m is 0.697 kg/m3. The balloon needs that same
Problem 2.107
Repeat Prob. P2.62 assuming that the 10,000 lbf weight is aluminum (SG = 2.71) and is hanging
submerged in the water.
Solution 2.107
Refer back to Prob. P2.62 for details. The only difference is that the force applied to gate AB by
Problem 2.108
A 7-cm-diameter solid aluminum ball (SG = 2.7) and a solid brass ball (SG = 8.5) balance nicely
when submerged in a liquid, as in Fig. P2.108. (a) If the fluid is water at 20C, what is the
diameter of the brass ball? (b) If the brass ball has a diameter of 3.8 cm, what is the density of
the fluid?
Solution 2.108
For water, take
= 9790 N/m3. If they balance, net weights are equal:
Problem 2.109
A hydrometer floats at a level that is a
measure of the specific gravity of the liquid. The stem is of constant diameter D , and a weight in
the bottom stabilizes the body to float vertically, as shown in Fig. P2.109. If the position h = 0 is
pure water (SG = 1.0), derive a formula for h as a function of total weight W , D , SG, and the
specific weight γ0 of water.
Solution 2.109
Let submerged volume be o when SG = 1.
Problem 2.110
A solid sphere, of diameter 18 cm, floats in 20C water with 1,527 cubic centimeters exposed
above the surface. (a) What are the weight and specific gravity of this sphere? (b) Will it float
in 20C gasoline? If so, how many cubic centimeters will be exposed?
Solution 2.110
The total volume of the sphere is (/6)(18 cm)3 = 3054 cm3. Subtract the exposed portion to find
Problem 2.111
A solid wooden cone (SG = 0.729) floats in water. The cone is 30 cm high, its vertex angle is
90º, and it floats with vertex down. How much of the cone protrudes above the water?
Solution 2.111
The cone must displace water equal
Problem 2.112
The uniform 5-m-long wooden rod in Fig. P2.112 is tied to the bottom by a string. Determine
(a) the tension in the string; and (b) the specific gravity of the wood. Is it possible for the given
information to determine the inclination angle
? Explain.
R
r
Solution 2.112
The rod weight acts at the middle, 2.5 m from point C, while the buoyancy is 2 m from C.
Summing moments about C gives
Problem 2.113
A spar buoy is a buoyant rod weighted to float and protrude vertically, as in Fig. P2.113. It can
be used for measurements or markers. Suppose that the buoy is maple wood (SG = 0.6), 2 in by
2 in by 12 ft, floating in seawater (SG = 1.025). How many pounds of steel (SG = 7.85) should
be added to the bottom end so that h = 18 in?
Solution 2.113
The relevant volumes needed are
Problem 2.114
The uniform rod in Fig. P2.114 is hinged at B on the waterline and is in static equilibrium as
shown when 2 kg of lead (SG = 11.4) are attached at its end. What is the specific gravity of the
rod material? What is peculiar about the rest angle
= 30?
Solution 2.114
First compute buoyancies: Brod = 9790(
/4)(0.04)2(8) = 98.42 N, and
Wlead = 2(9.81) = 19.62 N, Blead = 19.62/11.4 = 1.72 N. Sum moments about B:
Problem 2.115
The 2 inch by 2 inch by 12 ft spar buoy from Fig. P2.113 has 5 lbm of steel attached and has
gone aground on a rock, as in Fig. P2.115. Compute the angle θ at which the buoy will lean,
assuming that the rock exerts no moments on the spar. Figure P2.115
Solution 2.115
Let
be the submerged length of spar. The relevant forces are:
cos
=
Problem 2.116
The bathysphere of the chapter-opener photo is steel, SG ≈ 7.85, with inside diameter 54 inches
and wall thickness 1.5 inches. Will the empty sphere float in seawater?
Solution 2.116
Take the density of steel as 7.85(1000) = 7850 kg/m3. The outside diameter is 54+2(1.5) =
Problem 2.117
The solid sphere in Fig. P2.117 is iron (SG ≈7.9). The tension in the cable is 600 lbf. Estimate
the diameter of the sphere, in cm.
Solution 2.117
For water take γ = 9790 N/m3. Then the buoyant force and sphere weight are
Problem 2.118
An intrepid treasure-salvage group has discovered a steel box, containing gold doubloons and other
valuables, resting in 80 ft of seawater. They estimate the weight of the box and treasure (in air) at
7000 lbf. Their plan is to attach the box to a sturdy balloon, inflated with air to 3 atm pressure. The
empty balloon weighs 250 lbf. The box is 2 ft wide, 5 ft long, and 18 in high. What is the proper
diameter of the balloon to ensure an upward lift force on the box that is 20 percent more than
required?
Solution 2.118
The specific weight of seawater is approximately 64 lbf/ft3. The box volume is
(2ft)(5ft)(1.5ft) = 12 ft3, hence the buoyant force on the box is (64)(12) = 768 lbf. Thus the
Problem 2.119
When a 5-lbf weight is placed on the end of the uniform floating wooden beam in Fig. P2.119,
the beam tilts at an angle θ with its upper right corner at the surface, as shown. Determine (a) the
angle θ and (b) the specific gravity of the wood. Hint: Both the vertical forces and the moments
about the beam centroid must be balanced.
Solution 2.119
The total wood volume is (4/12)2(9) = 1 ft3. The exposed distance h = 9tan
. The vertical forces
are
Problem 2.120
A uniform wooden beam (SG = 0.65) is 10 cm by 10 cm by 3 m and hinged at A, as in
Fig. P2.120. At what angle θ will the beam float in 20C water?
Solution 2.120
The total beam volume is 3(.1)2 = 0.03 m3, and therefore its weight is
W = (0.65)(9790)(0.03) = 190.9 N, acting at the centroid, 1.5 m down from point A.
Problem 2.121
The uniform beam in Fig. P2.121, of size L by h by b and with specific weight γb , floats exactly
on its diagonal when a heavy uniform sphere is tied to the left corner, as shown. Show that this
can happen only (a) when γb = γ /3 and (b) when the sphere has size
1/3
( 1)
Lhb
DSG
=
−
.
Solution 2.121
The beam weight W =
bLhb and acts in the center, at L/2 from the left corner, while the
buoyancy, being a perfect triangle of displaced water, equals B =
Lhb/2 and acts at L/3 from the
Problem 2.122
A uniform block of steel (SG = 7.85) will “float” at a mercury–water interface as in Fig. P2.122.
What is the ratio of the distances a and b for this condition?
Solution 2.122
Let w be the block width into the paper and let
be the water specific weight. Then the vertical
force balance on the block is
