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hw1 answers - PROBLEM 2.70 A 350-“ load is wpponed by the...

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Unformatted text preview: PROBLEM 2.70 - A 350-“: load is wpponed by the repe-md-pulley arrangement show. Knowing [hat a = 35”; determine {a} the angle fit (1:) the magnitude of the force P which should be exerted on the free end of the rope to maintain equilibrium. (Him: The tension in the rope is the same on each side of a simple pulley. This can be proved by the methods of Chapter 4.} s P .- .- 3‘ J i A " a: @ i 3501b SOLUTION FM'BW ““5"“ 1’ “"93 A _+- EFI = o: ZPsinfi — Pcos25° = 0 Hence: (a) sinfl = £93525" or ,6 = 24.2“! (b) +1 25,20: 2Pcosfi+Psin35°+350 lb = 0 Hence: 2Pc0524.2° + Psin35° - 350 lb 2 O or P = 145.97 lb P = 146.0 lb 4 PROBLEM 2.111 A trmmlission tower is held by three guy wires attached to a pin an! and anchored by bolts at B. C, and D. If the tension in wire AB is 3.6 kN. determine the vertical force P exerted by the tower on the pin at A. The force in each cable can be Written as the product of the magnitude of the force and the unit vector along the cable. That is, with A .= (13m}i - (30111)] + (54:11)]; .. AC = (13 m]’ + Hem]: + (5.4m}1 = 35.4m E T _ _ TK = n.“ = FACE = fifils m): - {30 111)] + (5.4 m_}k] TAU = adumsfi _ 0,34751+ 0.1525k) and fi=—(fim)i-(3um}j+(7.sm}u AB = (—6m):+{—30 m): +{7.5m): = 3LSm AB T . . T” 2 T1.” = THE = fi[—(6m)l — [30 111)] + (75 Ink] 1“ = Tfl[—{}.1905i— 0.95245 + 023mm N Finally E = —{5m]i — (30111))"(22-2111JR .41) = J{-5 131]: + [—30 m)2 + [—22.2 m): - 313 m Tm = HAD = ring : _..._.THD [-[5m)i -—{30m]j — [22.2 m)k] AD 37.8 m Tm = Tm(-0.158?i - 0.79373' — 0.581%.) With P = P]. 31A: £1720: 'l"m +TAC +TAD+Pj =0 13qu the factors of L j, and II to zero, we obtain the linear algebraic equations: - ' i: 4.1905235 + 0.503er — 0.1537er = o (1) j: 4.952414, .7 0.34751}: - 0.19371'“ + P = 0 (2] In Equations (1). (2} and (3}. set T” = 3.6 kN. and. using conventional methods for solving Linear Algebraic Equations (MATLAB 01' Maple. for example). we obtain: TE : 1.963 m Tm = 1.959 W r=6.66kNT4 PROBLEM 3.21 Before 1112 trunk of a large tree is felled. cables AB and BC are attached as shown. Knowing d1at.thc tension in cables AB and BC are TI? N and 990 N, nespectiwiy. determine the moment about 0 of the resultant force exerted on the tree by the cables at B. SOLUTION Have Mu = ram} x F3 where r50: [3 4 111” F3 7' T43 + T3: - 0.9 '— 3.4 ' 7.2 Tm =kmna = ( “1,: ( Thq , “mm? H} [09]" + (8.4)” + {7.2)‘ m 5.1 '- .4 ' . TM : 1N7“: WW—WWEW N , J[5.1)' + [8.4j‘ +[121‘ In F:[- [63.[0Nli- saw}; pawns] [{SION}i-[S40N}j+[120N}k] = [447N)i 41423 N]j+[624 Njk i j k and Mo: 0 3‘4 0 N-m=[5241.6N-m]l-{3?54.SN»m]k 44? 4423 524 or Me = (524mm): -(3.75kN-m)k 1 PROBLEM 3.49 |-' To lifl a heavy crate. a man uses a block and tackle attached to the bottom ofan I-beam. at hka. Knowingthat the moments about theyami : axes of the force exerted at B by portion A3 of the rope axe. respectively, 100 lb-fl and 400 Ib-fl , dctm'mine the distance a. SOLUTION Bsscdon M0 = 11m x T“ where Mo = M1i+ Myj + M; = MJ1+(100 [b-fl}j—(400ib~fi]k w = Efifiii+(4fi}i TM 2 Marty (6ft]i — {12 an - (0)1: = -—d—T£‘ 821 i j k MJ+100j-400k= 6 4 a Si 6 —12 —a ”A = iH‘iay + {60)}- — {96M dim ' . . . mo From j-coeflimem. lDUdm = 607',“ or TM 2 6—d“ {1) a . 400 From k-coefficnmt: 41-00:!” = —96TRA or TM = Eds“ [2} Eqmting Equations (Hand [2) yields a = 10096} 6(400) or 0:4.00114 ...
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