ME 211 Homework 2 Solutions

# 9lbs cos30 45 3rb sin60 rb 1193lbs 40 ra

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 25.98lbs Answers Problem#54.37 Thewindlassissubjectedtoaloadof150lbs.DeterminethehorizontalforceP neededtoholdthehandleinthepositionshown,andthex,y,zcomponentsof reactionattheballandsocketAandthesmoothjournalbearingB.Thebearingis inproperalignmentandexertsonlyforcereactionsontheshaft. 1.)DrawtheFBD 2.)Writetheequilibriumequationsandsolve.(Fx,Fy,andM)andsolveforP,Bz, Bx,Ax,Ay,andAz. M y = 0 = 150lbs * (0.5) - P(1) P = 75lbs M x = 0 = Bz * (4) -150(2) Bz = 75lbs M z = 0 = Bx * (4) - 75(6) Bx = 112.5lbs F F F x = 0 = AX + 75 -112.5 Answers = 0 = Ay = 0 = Az + 75 -150 Ax = 37.5lbs y Ay = 0 z Az = 75lbs Problem#64.60 DeterminethemaximumweightWthemancanliftwithconstantvelocityusingthe pulleysystem,withoutandthenwiththe&quot;leadingblock&quot;orpulleyatA.Theman hasaweightof200lbsandthecoefficientofstaticfrictionbetweenhisfeetandthe groundiss=0.6. 1.)DrawtheFBD Part(a) 2.)Writetheequilibriumequations.(Fx,Fy,andM)Noticetherearethree changesindirectionduetothepulleys,soyoucandividetheweightby3because thatishowmuchyouareactuallylifting.Alsonoticethatinpart(a)heispullingat a45degreeangle,soitwilltakemorework. W * cos(45) + 0.6N 3 W Fy = 0 = 3 sin(45) + N - 200 W = 318lbs F x =0=- Part(b) 3.)Withthenewpulley,youcanseethatthe...
View Full Document

Ask a homework question - tutors are online