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HW_5_solutions

# HW_5_solutions - C/gé ZOJ#A 3\Q/vgow Q 2.104 The can in...

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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 5-m-long 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 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% 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(7000-7681bf) = 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, pal-r : 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) Wag-J“ f Aziéwvf in” _____ 2.149 The waterwheel in Fig. P2.149 lifts water with 1-ft-diameter half-cylinder 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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HW_5_solutions - C/gé ZOJ#A 3\Q/vgow Q 2.104 The can in...

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