hw3 answers - PROBLEM 5.60 1" 1-=m_‘ The reflector...

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Unformatted text preview: PROBLEM 5.60 1 '" 1--=m _‘ The reflector of a small flashlight has the parabolic shape shown. ”—1— Determine the surface area of the inside of the reflector. I _ h: 125'"... SOLUTION First note that the required surface area A can be generated by rotating \ the parabolic cross section through 2:: radians about the x axis. Applying J the first theorem of Pappus-Guldinus. we have I A = QHFL | Now, since .1: = kyl, at x = a: a = Ic(7.5)2 ' | i 1 or a = 56.25 t (1) l l l 51: At x=(u+15)rnm:a+15=k{[2.5): 51 '51mm or a+lS =156.25k (2) Eq. (2)_ at +15 _ 156.25]: Then . Eq. {1) a 56.25}: or a = 8.4375 mm Eq.{l) :> k = 0.15; mm x x = 0.15);3 and fl = 0.3y a? Now as = 1+ [3] dy =./1+ 0.0911341» So A = 21W. and PL = IydL A = zetfilijyrh + 0.119qu1: n 1:.5 = 2,.[3(;](1. 0.09112)? 3 0.13 _ .25 =1013rt1m2 01' A=1013mmzi .._'a+ . PROBLEM 5.71 Determine the reactions at the beam supports for the given loading when We 2 1.5 kNJ'fm. “Tn D SOLUTION Have R, = %(9 m)(2 kam) = 9 kN Ru = (9 m)(1.5 mm) = 13.5kN Then _+.. 2FJ|r =0: Cx = [I J) 2M, =0: ‘ so kN-m — (1 m)(9 kN) , (2.5 m)(13.s kN) + (6 m]Cy = 0 or C? =15.4583kN c = 15.46kN I 1! +I 25:0: By —9kN—13.SkN+15.4583 =0 or By = 7.0417134 B = 7.04m T41 Have Now Then, for dm: Dr 01’ PROBLEM 5.77 The 9 x lZ-fl side A3 of a tank is hinged at its bottom A and is held in place by a thin rod BC. The maximum tensile force the rod can withstand without breaking is 40 lcips, and the design specifications require the force in the rod not exceed 20 percent of this value. Ifthe tank is slowly filled with water, determine the maximum allowable depth of water d in the tank. (9 fi][(0.2)(40 x 103 2:0] — fl;fi{%[(12 exam )](62.4 lblfi3)dm} = o 216 x 11::3 iii : 374.4 d3,“ d3,“ = $76.92 as am = 8.32 a 4 PROBLEM 6.60 T Determine the force in members AF and E! of the: truss shown when Jig-R P = Q = 2 kjps. (Him: Use section Ra.) I ma l SOLUTION FED Truss: Q 2M,l = o: (24 am — (13 a)(2 kips) — {36 n)(2 lcips) = o KI = 451nm —— FBD Section: (m, = o; (13 a)(2 kips) + (12 &}(4.5 kips) — (36 amp = o l__—_____— Q EMF = o: (12 a}(4.s kips} — (13 n)(2 Rips) - [36 fi)(2 kilns} + (35 avg, =0 F5, = 1.500 kips T 4 Eu.— =250kips T4 PROBLEM 6.68 I The diagonal members CF and DE of the truss shown are very slender and can act only in tension; such members are imown as counters. Determine the force in members CE and DF and in the counter which is acting. SOLUTION FBI) Section: DE mustbeintcnsion topmvidelefiward force. 50 CF is slack. —- 21F, =0t4lcips—EFDE =0 F95 = 5.00 kips T4 Q 2M0 = o: {3 11)ch — (6 n}(4 kips) = 0 F55 = 3.00 kips T 4 QZME = o: (s fi}Fop — (12 &)(4 kips] = 0 F0; = 6.00 kips c 4 ...
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