Ep 7.Two cables are tied together at C and loaded as shown.Determinethetension(a)incableAC,(b)incableBC..2m1.98kNLS17hSOLUTIONFree-BodyDiagramBA1.4tanα=TIBCT4.8-AC1.6m1.4mα=16.26020C1pXR1.6tanβ=3￥1.98kβ=28.073°4.8m3mForce TriangleTecLaw of sines:61.927°TAcTac1.98kN1.98kNsin44.3330sin61.9270sin73.740044.333°73.740TAc41.98 kNTACsin61.9270TAc = 2.50 kN(a)sin44.333°1.98 kN(b)Tgcsin73.740°Tsc =2.72 kNsin44.333o
Ep 7
Ep 8.Two cables are tied together at C and are loaded as shown.Knowing that P=500Nand α=60°,determine the tension in(a) incableAC, (b) incableBC.ACDE425-7145°FSOLUTIONFree-Body DiagramForceTriangleuseTACZ8c6075450125eIncα=60e500N21P=500N30!&25TACTac500NLaw of sines:sin350sin75°sin70°500NTAc=sin350TAC=305N(a)sin700500NTrc=514N(b)Tgc=sin750sin700
Ep 8
Ep 9.Determine (a) the x, y, and z components of the 600-N force,(b)the angles ,,and .that theforceforms with thecoordinate axes.60ON450NSOLUTIONF=(600N)sin25°cos30(a)F=219.60NF=220NF,=(600N)cos25F.=543.78NF,=544NF, =(380.36 N)sin 25°sin30°F, =126.785 NF, =126.8NE219.60N(b)cOse0,=68.50=-F600NF543.78 N0, = 25.00coso,=F600NE126.785N0, =77.80cos 0, =F600N
Ep 9
Ep 10.Fortheframeand cableofProblem2.85,determinethecomponents of the force exerted by the cable on thesupportatE.PROBLEM2.85AframeABC is supported inpartbycable DBE that passes through a frictionless ring at B.Knowing that thetension in thecable is 385N, determinethe components of the force exerted by the cable on thesupportatD.9250mm210mmKAD510mm400)600mm480mmSOLUTIONEB=(270mm)i(400mm)j+(600mm)kEB=(270 mm)*+(400mm)2+(600 mm)= 770 mmF=FMgB=FEBEB385N[(270mm)i-(400mm)j+(600mm)k)770mmF=(135N)i(200N)j+(300 N)kF=+135.0N,F=-200N,F.=+300N
Ep 10
Ep 11.Knowingthatthetensionis425Nin cableABand510Nincable AC, determine the magnitude and direction of the resultantoftheforcesexertedatAbythetwocables.9Ser60cmBcmSOLUTIONAB=(40 cm)i (45 cm)j + (60 cm)kAB=(40cm)°+(45cm)°+ (60 cm)2=85cmAC=(100 cm)i(45 cm)j + (60 cm)kAC= /(100 cm) +(45 cm)° +(60 cm)=125cmAB[(40 cm) -(45 cm)j + (60 cm)k=(425 N)Tan =Tans=TaD AB85cmTAB=(200N)i-(225N)j+(300N)kAC(100cm)i-(45cm)j+(60cm)k=(510N)Tac= Tachec Tac AC125cmTae = (408 N)i (183.6 N) + (244.8 N)kR=TAB+TAc=(608N)i-(408.6N)j+(544.8N)kThen:R=912.92NR=913N608Nand=0.665990,=48.20cose2912.92N-408.6N=0.447578, =116.60cose,912.92N544.8 N0, =53.40=0.59677cose,=912.92N
Ep 11