# Represent the force in each wire in c artesian vector

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The guy wires are used to support the telephone pole.Represent the force in each wire in Cartesian vector form.Neglect the diameter of the pole.SOLUTIONUnit Vector:Force Vector:Ans.Ans.={53.2i-79.8j-146k} N={53.22i-79.83j-146.36k} NFB=FBuBD=175{0.3041i+0.4562j-0.8363k} N={-43.5i+174j-174k} N={-43.52i+174.08j-174.08k} NFA=FAuAC=250{-0.1741i+0.6963j-0.6963k} NuBD=rBDrBD=2i-3j-5.5k6.576=0.3041i-0.4562j-0.8363krBD=222+(-3)2+(-5.5)2=6.576 mrBD={(2-0)i+(-3-0)j+(0-5.5)k} m={2i-3j-5.5k} muAC=rACrAC=-1i+4j-4k5.745= -0.1741i+0.6963j-0.6963krAC=2(-1)2+42+(-4)2=5.745 mrAC={(-1-0)i+(4-0)j+(0-4)k} m={-1i+4j-4k} myBCDAxz4 m4 m1.5 m1 m3 m2 mFA250 NFB175 N
2–101.The two mooring cables exert forces on the stern of a shipas shown. Represent each force as a Cartesian vector anddetermine the magnitude and coordinate direction anglesxyz50 ft30 ft40 ftACB10 ftFB= 150 lbFA= 200 lbSOLUTIONUnit Vector:Force Vector:Ans.Ans.Resultant Force:The magnitude of isAns.The coordinate direction angles of areAns.Ans.Ans.cos g= -160.00337.63g=118°cos b=131.45337.63b=67.1°cos a=266.67337.63a=37.8°FR=337.63 lb=338 lbFR=2266.672+131.452+1-160.0022FR=5266.67i+131.45j-160.00k6lb=51169.03+97.642i+133.81+97.642j+1-101.42-58.592k6lbFR=FA+FB=597.6i+97.6j-58.6k6lb=597.64i+97.64j-58.59k6lbFB=FBuCB=15050.6509i+0.6509j-0.3906k6lb=5169i+33.8j-101k6lb=5169.03i+33.81j-101.42k6lbFA=FAuCA=20050.8452i+0.1690j-0.5071k6lbuCB=rCArCA=50i+50j-30k76.81=0.6509i+0.6509j-0.3906krCB=2502+502+1-3022=76.81 ftrCB=5150-02i+150-02j+1-30-02k6ft=550i+50j-30k6ftuCA=rCArCA=50i+10j-30k59.16=0.8452i+0.1690j-0.5071krCA=2502+102+1-3022=59.16 ftrCA=5150-02i+110-02j+1-30-02k6ft=550i+10j-30k6ftof the resultant.
2–102.Each of the four forces acting at Ehas a magnitude of28 kN. Express each force as a Cartesian vector anddetermine the resultant force.SOLUTIONAns.Ans.Ans.Ans.Ans.={-96k} kNFR=FEA+FEB+FEC+FEDFED={-12i-8j-24k} kNFED=28a-614i-414j-1214kbFEC={-12i+8j-24k} kNFEC=28a-614i+414j-1214kbFEB={12i+8j-24k} kNFEB=28a614i+414j-1214kbFEA={12i-8j-24k} kNFEA=28a614i-414j-1214kb4 m6 m6 myxCABDEzFEBFEDFEAFEC4 m12 m
2–103.SOLUTIONForce Vectors:The unit vectors uA,uB,uCand uDof FA,FB,FCand FDmust bedetermined first. From Fig.a,Thus, the force vectors FA,FB,FCand FDare given byFD=FDuD=70a-37i-27j-67kb=[-30i-20j-60k] lbFC=FCuC=70a-37i+27j-67kb=[-30i+20j-60k] lbFB=FBuB=70a37i+27j-67kb=[30i+20j-60k] lbFA=FAuA=70a37i-27j-67kb=[30i-20j-60k] lbuD=rDrD=(-3-0)i+(-2-0)j+(0-6)k2(-3-0)2+(-2-0)2+(0-6)2= -37i-27j-67kuC=rCrC=(-3-0)i+(2-0)j+(0-6)k2(-3-0)2+(2-0)2+(0-6)2= -37i+27j-67kuB=rBrB=(3-0)i+(2-0)j+(0-6)k2(3-0)2+(2-0)2+(0-6)2=37i+27j-67kuA=rArA=(3-0)i+(-2-0)j+(0-6)k2(3-0)2+(-2-0)2+(0-6)2=37i-27j-67kIf the force in each cable tied to the bin is 70 lb, determinethe magnitude and coordinate direction angles of theresultant force.zBCEDAxy6 ft3 ft3 ft2 ft2 ftFCFDFAFBResultant Force:The magnitude of FRisAns.The coordinate direction angles of FRareAns.Ans.Ans.g=cos-1B(FR)zFRR=cos-1a-240240b=180°b=cos-1B(FR)yFRR=cos-1a0240b=90°a=cos-1B(FR)xFRR=cos-1a0240b=90°=20+0+(-240)2=240 lbFR=2(FR)x2+(FR)y2+(FR)z2={-240k} NFR=FA+FB+FC+FD=(30i-20j-60k)+(30i+20j-60k)+(-30i+20j-60k)+(-30i-20j-60k),,,
*2–104.If the resultant of the four forces is ,