ME 211 Homework 2 Solutions

82lbs answers f f x y 0 0 2tires n a 2 n b 800

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Unformatted text preview: 2 - G3 N A = 2191.18lbs Part(b) WhenthecraneisgoingtotipoverNB=0. 4.)UsetheequilibriumequationwithNBandtheweightoftheloadandsetNB=0and solveforW. 3500 * 3 ft + 900 *11 ft + 1500 *18 ft - W *10 ft Answer 2 *17 ft W = 4740lbs NB = 0 = Problem#44.31 ThesmoothpiperestsagainstthewallatthepointsofcontactA,B,andC. Determinethereactionsatthesepointsneededtosupporttheverticalforceof 45lbs.Neglectthepipe'sthicknessinthecalculation. 1.)DrawtheFBD 2.)Writetheequilibriumequations.(Fx,Fy,andM) M A = 0 = 36in * 45 * cos(30) - 8in * 45 * sin(30) - RC * 20in + RB * (8 * tan(30))in F F x y = 0 = RA + RB * sin(30) - RC cos(60) = 0 = -45 + RC * cos(30) - RB sin(60) 3.)Withtheequationsabove,wenowhave3equationsand3unknowns,sowecan solveforallthreeunknowns(RA,RB,andRC) M A = 0 = 36in * 45 * cos(30) - 8in * 45 * sin(30) - RC * 20in + RB * (8 * tan(30))in F F x y = 0 = RA + RB * sin(30) - RC cos(60) = 0 = -45 + RC * cos(30) - RB sin(60) RC cos(30) - 45 sin(60) 1.)RB = R cos(30) - 45 2.)0 = 36in * 45 * cos(30) - 8in * 45 * sin(30) - RC * 20in + C * (8 * tan(30))in sin(60) RC = 63.9lbs 63.9lbs * cos(30) - 45 3.)RB = sin(60) RB = 11.93lbs 4.)0 = RA + (11.93lbs) * sin(30) - (63.9lbs)cos(60) RA =...
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This note was uploaded on 05/12/2010 for the course MECHENG 211 taught by Professor ? during the Fall '07 term at University of Michigan.

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