# 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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