HW5 - Problem 3-40 The pipe of mass M is supported at A by...

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Unformatted text preview: Problem 3-40 The pipe of mass M is supported at A by a system of five cords. Determine the force in each cord for equilibrium. Given: M=u30kg c=3 g .= 9.81 2 d = 4 2 s 6 = 60 deg Solution: Initial guesses: TAB = l N TAE = l N TBC = l N TBD = l N Given TABsin(0) — Mg = 0 TAE— TABCOS(9) = O TM} y d TBD ~7— + TABcos(6) — TBC = 0 Q A Tm “a: \/C +42 TAB TAE = Find(TAB, TAE, TBC» TED) Problem 3-45 The three cables are used to support the lamp of weight W. Determine the force developed in each cable for equilibrium. Units Used: kN =103 N Given: W=800N b=4m a=4m c=2m Solution: Initial Guesses: FA3=1N FAC=1N FAD=1N W Given- 0 1 F' —c O AD » FABI +FACO +-——-— -b + 0 =0 2 2 2 0 0 \/a +1; +c a -1 FAB FAB 800 Me = Find(FAB,FAc,FAD) FAC = 400 N FAD , FAD 1200 Problem 3-51 Cables AB and AC can sustain a maximum tension T max, and the pole can support a maximum compression Pmax. Determine the maximum weight of the lamp that can be supported in the position shown. The force in the pole acts along the axis of the pole. Given: Tmax=500N c=2m Pmax=300N d=1.5m a = 6 m e = 4 m b = 1.5 m f= 1.5 m Solution: Lengths A0 =1/a2+b2+02 AB =\/(12+(c+e)2+(b+d)2 AC =\/a2+c2+(b+d)2 The initial guesses: FAO = Pmax FAB = Tmax FAC = Tmax W = 300N Case 1 Assume the pole reaches maximum compression Given F c F -—c — e F —-c 0 A0 AB AC ———b +-—— b+f +— b+d + 0 =0 A0 AB AC a —a —a —1 W1 W1 138.46 FAB] = Find(W,FAB,FAc) FAB] = 10385 N FAC1 FACI 80.77 Casez éAs‘saifne thafqéfiie AB? Eéachééiinaximum tefiSibn Given F V C F -C-e F —c 0 A0 > AB‘ fl—b.+— b+f +-—A—C- b+d + 0 :0 A0 * AB AC .0 ‘a —a —1 W2 * W2 666.67 FA02 = Find(W,FAo,FAc) FA02 = 1444.44 N FAcz _ FAC2 388,89 1 ifhat cable‘Ac regiélies 4 Given ' c FAO FA FAC — —b +—B b+f +—C b+d + 0 =0 W 3 W3 857.14 FA03 =Find(W,FA0,FAB) FA03 = 1857.14 N FAB3 FA33 642.86 ml W = min(W1, Wza'W3) ...
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HW5 - Problem 3-40 The pipe of mass M is supported at A by...

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