ME 211 Homework 2 Solutions

1drawthefbd 2writetheequilibriumequationsfxfyandm m

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Unformatted text preview: uilibriumequations.(Fx,Fy,andM) M A = 0 = T * 5 ft + T * ( 2 ) *10 ft - 80lbs *13 ft 5 F F x = 0 = AX - T * ( 1 ) 5 2 ) - 80lbs 5 y = 0 = Ay + T + T * ( 3.)Withtheequationsabove,wenowhave3equationsand3unknowns,sowecan solveforallthreeunknowns(Ax,Ay,andT) M A = 0 = T * 5 ft + T * ( T = 74.6lbs 2 ) *10 ft - 80lbs *13 ft 5 F F x = 0 = AX - T * ( 1 ) 5 2 ) - 80lbs 5 Answers AX = 33.4lbs y = 0 = -Ay + T + T * ( Ay = 61.324lbs Problem#34.24 Thecraneconsistsofthreeparts,whichhaveweightsofW1=3500lbs,W2=900lbs, W3=1500lbsandcentersofgravityatG1,G2,andG3,respectively.Neglectingthe weightoftheboom,determine(a)thereactionsoneachofthefourtiresiftheload ishoistedatconstantvelocityandhasaweightof800lbs,and(b),withtheboom heldinthepositionshown,themaximumloadthecranecanliftwithouttipping over. 1.)DrawtheFBD Part(a) 2.)Writetheequilibriumequations.(Fx,Fy,andM) M A = 0 = 2 * N B *17 ft + 800 *10 ft - G1 * 3 ft - G2 *11 ft - G3 *18 ft F F x y =0 = 0 = 2 * N A + 2 * N B - 800 - G1 - G2 - G3 3.)Withtheequationsabove,wenowhave2equationsand2unknowns,sowecan solveforallthreeunknowns(Na,Nb) M A = 0 = 2tires * N B *17 ft + 800 *10 ft - G1 * 3 ft - G2 *11 ft - G3 *18 ft 3500 * 3 ft + 900 *11 ft + 1500 *18 ft - 800 *10 ft NB = 2tires *17 ft N B = 1158.82lbs Answers F F x y =0 = 0 = 2tires * N A + 2 * N B - 800 - G1 - G...
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