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# HW04sols - Problem*3.91[Difﬁculty 2...

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Unformatted text preview: Problem *3.91 [Difﬁculty: 2] ’Ei‘éi‘Tiiéifﬁiit‘B-Tﬁxﬁae ﬁfti’ESffépéisaifiﬁﬁbéﬁiEi‘iﬁiFéd"' from a speciﬁc gravitym¢asi1r¢mennThe measurement is madcby immersing thcb'odyina tankotwarcr and measuring The net weight. DevefoP an expression for the speciﬁc; gravity of a person. innt’cnris of their weight in-a’ignct weightg iinwater, and SG =f(.T) for water; 2 Given: Speciﬁc gravity of a person is to be determined from measurements of weight in air and the met weight when totally immersed in water. Find: Expression for the speciﬁc gravity of a person from the measurements. Solution: We will apply the hydrostatics equations to this system. Governing Equation: (Buoyant force is equal to weight of displaced ﬂuid) AsSumptions: (1) Static ﬂuid (2) Incompressible ﬂuid Taking a free body diagram of the body: EFy = 0 / F is the weight measurement for the immersed body. net Therefore the weight measured in water, is: F F net = air‘ pw'g' Now in order to ﬁnd the speciﬁc gravity of the person, we need his/her density: (Fair ” Fnet) Simplifying this expression we get: Fair = (Fair — Fnet) Now ifwe call the density ofwater at4 deg c pw4c then: F - _ Solving this expression for the speciﬁc gravity of the person SG, we get: Problem *3.92 [Difﬁcultyz 2] 3.92 Quantify the statement, “Only the tip of an iceberg shows (in seawater)” Given: Iceberg ﬂoating in seawater Find: Quantify the statement, "Only the tip of an iceberg shows (in seawater)." Solution: We will apply the hydrostatics equations to this system. Governing Equations: (Buoyant force is equal to weight of displaced ﬂuid) Assumptions: (1) Static ﬂuid (2) Incompressible ﬂuid Taking a free body diagram of the iceberg: EFZ = 0 —M. g + Fbuoy = 0 Substituting in data from Tables A] and A2 we get: VF 1 ' VF = 0.1054 Only 10% of the iceberg is above wate1 Problem *3.101 [Difficultyz 3] 3.101 A block of vofume 0.0257 m3 is allowed. to sink in water as. shown. A circular rod, 5 m long and 21) cm2 in cross» section is attached to the weight and also to the wall. If the rod mass is 125 kg and the rod. makes an angle of 12. degrees Mth, horizontal at equilibrium, 'what' is the mass of the Hock? Given: Geometry ofblock and rod Find: Angle for equilibrium Assumptions: Water is static and incompressible Solution: Basic equations The free body diagram is as shown FEB and FER are the buoyancy of the block and rod, respectively; 0 is the (unknown) exposed length of the rod Taking moments about the hinge L (WE — FBB)-L-cos(9) — FBR-Egﬂ-cosw) + WR-E-cosw) = 0 with . wB = MB.g FBB = p.g.VB FER: p.g.(L _ c).A (MB _ p.VB).L ._ piA.(L _ c).(L_+c) + MKE _ Combining equations _ 2 2 l p.A.(L2 — 02) = 2-(MB — p-VB + é-MRj-L @ a p.A . 2 and Since c = M = —. L _ + .V _ _.M sin(e) B 2-L (sin(9)j p B 2 R We can solve for MB '(LZ — c2) + p-VB — -;—-MR 2 2 1 . ~ 1 MB = E X 1000-353 x 20-cm2x >< (S'm)2 — + lOOO-g x 0.025-m3 — — x 1.25-kg sin(12-deg)j 3 2 loo-cm) 5111 m Problem *3.102 _ [Difficultyz 3] 3.10.2 The stem of a glam. hydrometer med to: measure speciﬁc gravity 5 mm in diameter. The distance between on the stemisl mm per 0.1 increment of gravity. Calcuiate the magnitude and direction. of the error introduced by surface tension if the hydrometer ﬂoats in kerosene (Assume the contact angle barman kerosene and. glass is iii) Given: Glass hydrometer used to measure SG of liquids. Stem has diameter D=5 mm, distance between marks on stem is ~ d=2 mm per 0.1 SG. Hydrometer ﬂoats in kerosene (Assume zero contact angle between glass and kerosene). Find: Magnitude of error introduced by surface tension. Solution :_ We will apply the hydro statics equations to this sys em. 5 ' ' Governing Equations: Fbuoy = p.g.vd uoyant force is equal to weight ofdisplaced ﬂuid) Assumptions: (1) Static ﬂuid ' (2) Incompressible ﬂuid (3) Zero contact angle between ethyl alcohol and glass The surface tension will cause the hydrometer to sink Ah lower into the H '= hus for this change: The change in buoyant force is: The force due to surface tension is: F G = «Do-003(6) = Tr-D-O' /@ Thus, _ p-g-—.D .Ah = 1r.D.o- Upon simpliﬁcation: = 0' 4 4 Solving for Ah: ' From Table A.2, SG = 1.43 and from Table A4, 6 = 26.8 mN/m 3 35 m S 1 kg'm Ah=1.53x10_ m Therefore, Ah = 4 x 26.8 x 10— - x x — x m 1430'kg 9.81-m “10—3.m s2N ASG ‘= 1.53 x10_3-mx L3- 2x10_ .m So the change in speciﬁc gravity will be: ASG = 0.0765 From the diagram, surface tension acts to cause the hydrometer to ﬂoat lower in the liquid. Therefore, surface tension results in an indicated speciﬁc gravity smaller than the actual specific gravity. Problem *3.109 [Difficultyz 2] '3 .1 09 A bowl is. inverted synunetriically and held in a dame ﬂuid, 86 £1516, to a depth of 2100 measured along the oenterline of the bowl from the bowl The bowl height 80 mm, and the fluid rises 20 mm inside the bowl. The bow]! 100 mm inside diameter, and it is made from an old ole}r recipe, '36 = 6.1. The volume of the bowl itse'lf’is about 0.9 L. What is. the force required to hold. it in place? «l / UV“ 5 Given: Data on inverted bowl and dense ﬂuid Find: Force to hold inplace Assumption: Fluid is static and incompressible Solution: Basic equations FB = p-g-V and BF}, = 0 Hence H: FB — W FOrthe buOYamy fme FB = SGﬂuid'pH2O'g'Vsub . _ Vsub = Vbowl + Vair Now/‘6 " m For the weight whom/cl Hence 3 - 2 3 2 k . , . . F = 999-—g >< 9.81.E x 15.6x 0.9-Lx —m— + (0.08— 0.02)-m-M — 5.7x 0.9-Lx L x N S 1000L 4 1000L kg'm F = 159.4_N Problem *3.112 [Difficultyz 2] 3' fly aim/ad W 3.112 Three steel balls (each about half an inch in diam; eter) lie at the bottom. of a plastic shell ﬂoating on the water surface in a partially ﬁlled bucket. Someone removes. the steel balls the shell and carefully lets fall to the bottom of the bucket, leaving the plastic shell to ﬂoat empty. What happens to the Water level in the bucket? Does it rise, go down, or remain, unchanged? Explain, Given: Steel balls resting in ﬂoating plastic shell in a bucket of water Find: What happens to water level when balls are \$051? in water 3" Solution: Basic equation l FB = p.VdiSp.g = W ’for a oating body weight W When the balls are in the plastic shell, the shell and balls displace a volume of water equal to their own weight - a large volume because the balls are dense. When the balls are removed from the shell and dropped in the water, the shell now displaces only a small volume of water, and the balls sink, displacing only their own volume. Hence the difference in displaced water before and after moving the balls is the difference between the volume of water that is equal to the weight of the balls, and the volume of the balls themselves. .The amount of water displaced is signiﬁcantly reduced, so the water level in the bucket drops. Wplastic + Wballs p'g Volume displaced before moving balls: Vl = Volume displaced after moving balls: ' W _ SG -p-g-V Change in volume displaced AV = V — V = V _ balls = V _ M _ 2 1 balls p'g balls pg AV = Vbaus'(1 ‘ SGballs) Hence initially a large volume is displaced; ﬁnally a small volume is displaced (AV < 0 ecause SGbauS > 1) ...
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HW04sols - Problem*3.91[Difﬁculty 2...

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