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Unformatted text preview: C/gé ZOJ/ #A/ 3/ \Q/vgow Q) 2.104 The can in Fig. P2.104 ﬂoats in the position shown. What is its weight in
newtons? Solution: The can weight simply equals the weight of the displaced water (neglecting
the air above): —“'l F—D=9cm
Fig. P2.104 72'
w = Misplaced = (9790)Z(0‘09 m)2 (0.08 m) = 5.0 N Ans. 2.112 The uniform 5mlong wooden rod in the ﬁgure is tied to the bottom by a string.
Determine (a) the string tension; and (b) the speciﬁc gravity of the wood. Is it also
possible to determine the inclination angle (9? String
tension T Fig. P2.112 Solution: 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 ZMC = 0 = W(2.53in6)—B(2.05in6’), or W = 0.8B
But B = (9790)(7r/4)(0.08 my (4 m) = 196.8 N.
Thus W = 0.8B = 157.5 N = SG(9790)(7r/4)(0.08)2(5 m), or: so z 0.64 Ans. (b) Summation of vertical forces yields String tension T = B— W = 196.8 — 157.5 z 39 N Ans. (a) These results are independent of the angle 6, which cancels out of the moment balance. C/c’é 20f Hw #i’ 30/3744“ P2.118 An intrepid treasuresalvage 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% more than required? Solution: 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
balloon must develop a net upward force of 1.2(70007681bf) = 7478 lbf. The air weight in the balloon is negligible, but we can compute it anyway. The air density is: 2
L = W = 0.0071iu3g—
RT (1716ft /s — R)(520°R) ﬂ Hence the air speciﬁc weight is (0.0071)(32.2) = 0.23 lbf/ft3, much less than the water. At p=3atm, palr : Accounting for balloon weight, the desired net buoyant force on the balloon is Fm = (64—0.231bf/ft3)(7:/6)D§a,100n — 250be = 74781bf
Solve for D3 = 231.4be3 , Dbaﬂoon z 6.14ft Ans. 2.141 The same tank from Prob. 2.139 is
now accelerating while rolling up a 30°
I inclined plane, as shown. Assuming rigid—
body motion, compute (a) the acceleration a,
(b) whether the acceleration is up or down,
and (c) the pressure at point A if the ﬂuid is mercury at 20°C. Fig. P2.141
Solution: The free surface is tilted at the angle 6: —30° + 7.4l° = —22.59°. This angle
must satisfy Eq. (2.55): tan6 = tan(—22.59°) = —0.416 = aX/(g + az) But the 30° incline constrains the acceleration such that ax = 0.866a, az = 0.5a. Thus 0.866a solve for a z 330 %(down) Ans. (a, b) = —0.416 = ————,
tang 9.81+0.5a s The cartesian components are ax = —3.29 m/s2 and az = —1.90 m/s2.
(c) The distance AS normal from the surface down to point A is (28 c056) cm. Thus pA = p[a§ + (g+az)2]1/2 = (13550)[(—3.29)2 +(9.81—l.90)2]1/2(0.28cos7.41°)
z 32200 Pa (gage) Ans. (c) WagJ“ f Aziéwvf in” _____ 2.149 The waterwheel in Fig. P2.149 lifts
water with 1ftdiameter halfcylinder
blades. The Wheel rotates at 10 r/min. What
is the water surface angle :9 at pt. A? Solution: Convert Q = 10 r/min = 1.05 rad/s. Use an average radius R = 6.5 ft.
Then Thus tan 6 = aX/g = 7.13 /32.2, ax = 92R = (1.05)2(6.5) z 7.13 ft/s2 01": —+{ 1 ft P—
Fig. P2.149 toward the center a: 125° Ans. ...
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 Fall '08
 Schwartz,L

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