75 i 25981 j 12990 k n f 300 sin 30 sin 30 i 300 cos

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={-75i+259.81j+129.90k} NF=(-300 sin 30° sin 30°i+300 cos 30°j+300 sin 30° cos 30°k)
*2–140.Determine the length of the connecting rod ABby firstformulating a Cartesian position vector from AtoBandthen determining its magnitude.SOLUTIONAns.rAB=2(18.5)2+(4.330)2=19.0 in.={18.5 i-4.330j} in.rAB=[16-(-5 sin 30°)]i+(0-5 cos 30°) j16 in.O5 in.ABxy30
2–141.SOLUTIONAns.Ans.Ans.Ans.F2y=150 sin 30°=75 NF2x= -150 cos 30°= -130 NF1y=200 cos 45°=141 NF1x=200 sin 45°=141 NDetermine the xand ycomponents of and F2.F1yx30F1200 NF2150 N45
2–142.Determine the magnitude of the resultant force and itsdirection, measured counterclockwise from the positivexaxis.SOLUTIONAns.Ans.u=tan-1¢216.42111.518=87.0°FR=2(11.518)2+(216.421)2=217 NQ+FRy= ©Fy;FRy=150 sin 30°+200 cos 45°=216.421 N+RFRx= ©Fx;FRx= -150 cos 30°+200 sin 45°=11.518 Nyx30F1200 NF2150 N45
2–143.SOLUTIONAns.Ans.Ans.Ans.Ans.Ans.Ans.Ans.Thus,FR=0FRy=0-240+240+0=0FRy=F1y+F2y+F3y+F4yFRx= -200+320+180-300=0FRx=F1x+F2x+F3x+F4xF4y=0F4x= -300 lbF3y=300a45b=240 lbF3x=300a35b=180 lbF2y= -400a35b= -240 lbF2x=400a45b=320 lbF1y=0F1x=-200 lbDetermine the xand ycomponents of each force acting onthe gusset plateof the bridge truss.Show that the resultantforce is zero.yx345345F1200 lbF2400 lbF3300 lbF4300 lb
*2–144.Express and as Cartesian vectors.F2F1SOLUTIONAns.Ans.=-10.0 i+24.0 jkNF2= -5131262i+12131262j=5-15.0 i-26.0 j6kNF1= -30 sin 30° i-30 cos 30° jF1= 30 kNF2= 26 kN12513xy30°
2–145.SOLUTIONAns.Ans.u=180°+4.53°=185°f=tan-1a1.98125b=4.53°FR=21-2522+1-1.98122=25.1 kNFRy= -30 cos 30°+12131262= -1.981 kN+ cFRy= ©Fy;FRx= -30 sin 30°-5131262= -25 kN:+FRx= ©Fx;Determine the magnitude of the resultant force and itsdirection measured counterclockwise from the positive xaxis.F1= 30 kNF2= 26 kN12513xy30°
2–146.The cable attached to the tractor at Bexerts a force of 350 lbon the framework. Express this force as a Cartesian vector.SOLUTIONAns.F=Fu={98.1i+269j-201k} lbu=rr=(0.280i+0.770j-0.573k)r=2(17.10)2+(46.98)2+(-35)2=61.03 ftr={17.10i+46.98j-35k} ftr=50 sin 20°i+50 cos 20°j-35k20AF350 lbyzx35 ft50 ftB
xy303045F180 NF275 NF350 N2–147.Determine the magnitude and direction of the resultantof the three forcesby first findingtheresultant and then forming.Specify itsdirection measured counterclockwise from thepositive axis.xFR=F¿ +F2F¿ =F1+F3FR=F1+F2+F3SOLUTIONAns.Ans.u=75°+10.23°=85.2°sin b104.7=sin 162.46°177.7;b=10.23°FR=177.7=178 NFR=2(104.7)2+(75)2-2(104.7)(75) cos162.46°sin f80=sin 105°104.7;f=47.54°F¿ =2(80)2+(50)2-2(80)(50) cos105°=104.7 N
*2–148.If and determine the magnitude of theresultant force and its direction measured clockwise fromthe positive xaxis.F=20 kN,u=60°SOLUTIONAns.Ans.f=tan-1B15.6058.28R=15.0°FR=2(58.28)2+(-15.60)2=60.3 kN+ cFRy= ©Fy;FRy=50a35b-122(40)-20 sin 60°= -15.60 kN:+FRx= ©Fx;FRx=50a45b+122(40)-20 cos 60°=58.28 kN53411yx50 kN40 kNF2u
2–149.SOLUTIONUnit Vector:Ans.Force Vector:Ans.=-160i-180j+240klbF=FuAB=340e-817i-917j+1217kflbuAB=rABrAB=-8i-9j+12k17= -817i-917j+1217krAB=21-822+1-922+122=17.0 ft=5-8i-9j+12k6ftrAB=510-82i+10-92j+112-02k6ftThe hinged plate is supported by the cord AB. If the forcein the cord is express this force, directed fromAtoward B, as a C

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