ch01 - SOLUTION(1.20 Known The parameters 111 a F W s w T...

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Unformatted text preview: SOLUTION (1.20) _ Known: The parameters 111, a, F, W, s, w, T and W are identified. Find: Check the dimensional homogeneity of the following equations: (a) F = ma, (b) W = FS, (0) W = Tu). Given Data: 111 = mass 3 = acceleration F = force W = work 5 = distance (i) = angular velocity T = torque W = power Analysis: 1. Let the dimensions of length, mass, and time be given by length [=] L mass [=] M time [=] t Then, m [=] M L/t2 MLItZ I! ll || || || Haiku-l r! M t2 = M(L{t2) = Mth2 I (b) w = Fs M m = (MUIZXL) = name I (c) W = Tu) (MLthzlfllt) = MLll't3 I SOLUTION (1.21) Known: The mass and the acceleration of gravity are given. Find: Determine the weight of the object in: (a) English Engineering units (b) British Gravitational units (0) SI units Schematic and Given Data: Analysis: (21) English Engineering units F = ma/gCIEq. (l.la)] where _ 1%) 18mg )_ m—( (Mlékg —0.68551ug 01‘, m {0.1.5.35 Slug] (32-2 mm) = 22.1 lbm 1 l slug wig-inuremen-W gc = 32.2 ft-lbmflb s2 Thus, F _ (22.057 lbm] [32.2 fti’sz) _ [32.2 ft-lbm/lb 32) (b) British GravitatiOna] units F=rna [Eq.(l.1b)] F —— (0. 685 slug) (32.2 ft/SZ) = 221 slug-ftfs2 = 22.1 lb (0) SI units F=ma [Eq.(l.lc)] F = [10 kg) [9. 81 I’ll/'52] = 98.1lqg-n‘ii’s2 =93.1N =22. 1 lb Comment: The answers in (a) and (b) are equivalent to the answer in ((2), since [98.1 N)(4.£§N)=22.1 lb SOLUTION (1.24) Known: An object is suspended from a spring at a location where g 2 9.81 misl. The deflection of the spring is known. Find: Determine the mass of the object. Schematic and Given Data: Assumption: The spring has a linear force—deflection curve. Analysis: 1. The weight of the object is (30 mm) 6 N 2. Using Eq. (1.1a), F = ma =£=____6N = k m a giggling: 0.612 g SOLUTION (1.32} Known: A known mass attached to a rope wound around a pulley is falling with a constant velocity. Find: Determine the power transmitted to the pulley and the rotational speed of the pulley. Schematic and Given Data: Assumptions: 1. Wind resistance is negligible. 2. Bearing friction is negligible. 3. Mass of the rope is negligible. 4. The acceleration of gravity is a constant. Analysis: 1. First, draw a free body diagram of the mass and solve for the tension on the rope. F — mg = ma F-mg=0 F = mg = (5 lbm)(32.2 fusz) = 5 lb L I _ W=2nnT From Eq (13), 33,000 where T = FR = = 1.25 Ib-ft ng=(g]=20radfs 12 = 20rad 6OS|1rev = n l 5 min |2n rad lgl‘orpm I _ 2::(191.0)(1.25)_ 3. W — 33,000 = 0.0455 hp Comment: The power can also be calculated by using W = FV = (5 lb)(5 ftfs)a’(550 ft lb/s hp) = 0.0455 hp. SOLUTION (1.34) Known: An electric motor draws known values of current and voltage. The torque and the rotational speed of the output shaft are known. Find: (3) Determine the electric power requirement and the power developed by the output shaft. (1)) Determine the not power input to the motor. (c) Determine the amount of energy transferred to the motor by electrical work and transferred out of the motor by the shaft during two hours of operation. Schematic and Given Data: Output shaft n =1000 rpm T = 9.5 N'm Assumption: The conditions are steady state. Analysis: 1. The electric power requirement can be calculated from Welec = IV_ Wm: (10A)(110 V)(1 kwriooo VA): 1.10 kW I 2. The output shaft power can be calculated from Eq. (1.2). . _ 2:“: n T 231; (1000)(9. 5) Wont _ _ " ' ‘ 60.000 ‘ 60,000 “ 0995 kw ' 3. The net power input is Pner = Pin - Pout = 1.10 - 0.995 = 0.105 kW I 4. The amount of energy transferred to the motor by electrical work is (1.10 kW)(2 hr) = 2.20 kW-h I Applying unit conversion factors, 1 kW-h = 3.60 X 106 J , and 1 Btu = 1054 J, gives 2.20 kW-h 3-60 x 106 J 1 Btu l 1 l 1 kW-h 1054 J ‘ 7514 BI“ I 5. The amount of energy transferred out of the motor by the shaft is (0.995 kW)(2 hr) = 1.99 kW-h I 1.99 kW-h 3-60 K106 J 1 Btu l 1 I 1 kW-h 1054 J = 6797 Bl“ I 0.995 kW 2 0905 Comment: The efficiency of the motor can be written as n = 1 10 kw = 90.5%. ...
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