Problem 4-99Replace the force at A by an equivalent force and couple moment at point P.Given:F 375N:=a 2m:=b 4m:=c 2m:=d 1m:=θ 30deg:=Solution :F 375 N=MPF cos θ()⋅ ac+()⋅ F sin θ()⋅ bd−()⋅−:=MP737Nm⋅= For the exclusive use of adopters of the Hibbeler series of books.© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material maybe reproduced, in any form or by any means, without permission in writing from the publisher.Problem 4-102Replace the force system by an equivalent force and couple moment at point O.Units Used :kip 103lb:=Given : F1430lb:= F2260lb:=a 2ft:= e 5ft:=b 8ft:= f 12:=c 3ft:= g 5:=da:= θ 60deg:=Solution:FRxF2gg2f2+⎛⎜⎝⎞⎠⋅ F1sin θ()⋅−:= FRx272.39− lb=FRyF2fg2f2+⎛⎜⎝⎞⎠⋅ F1cos θ()⋅−:= FRy25lb=FRFRx2FRy2+:= FR274lb=θ1atanFRyFRx−⎛⎜⎝⎞⎠:= θ15.24 deg=MoF1cos θ()⋅ a()⋅ F1sin θ()⋅ b()⋅+ F2fg2f2+⎛⎜⎝⎞⎠⋅ e⋅+:=Mo4.61 kip ft⋅=Problem 4-125Replace the force and couple-moment system by an equivalent resultant force and couplemoment at point O. Express the results in Cartesian vector form.Units Used:kN 1000N:=Given:MMxi⋅ Myj⋅+ Mzk⋅+=FFxi⋅ Fyj⋅+ Fzk⋅+=Mx20− kN m⋅:= Fx8kN:=My70− kN m⋅:= Fy6kN:=Mz20kN m⋅:= Fz8kN:=a 3m:=b 3m:= e 5m:=c 4m:= f 6m:=d 6m:= g 5m:=Solution:FFxFyFz⎛⎜⎜⎜⎝⎞⎟⎠:= MMxMyMz⎛⎜⎜⎜⎝⎞⎟⎠:= rf−eg⎛⎜⎜⎜⎝⎞⎟⎠:=FRF:= MRMrF×+:= FR868⎛⎜⎜⎝⎞⎠kN= MR10−1856−⎛⎜⎜⎝⎞⎠kN m⋅= For the exclusive use of adopters of the Hibbeler series of books.© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material maybe reproduced, in any form or by any means, without permission in writing from the publisher.Problem 4-127Replace the force and couple-moment system by an equivalent resultant force and couplemoment at point Q. Express the results in Cartesian vector form.Units Used:kN 1000N:=Given:MMxi⋅ Myj⋅+ Mzk⋅+=FFxi⋅ Fyj⋅+ Fzk⋅+=Mx20− kN m⋅:= Fx8kN:=My70− kN m⋅:= Fy6kN:=Mz20kN m⋅:= Fz8kN:=a3m:=b3m:= e5m:=c4m:= f6m:=d6m:= g5m:=Solution:FFxFyFz⎛⎜⎜⎜⎝⎞⎟⎠:= MMxMyMz⎛⎜⎜⎜⎝⎞⎟⎠:= r0eg⎛⎜⎜⎝⎞⎠:=FRF:= MRMrF×+:= FR868⎛⎜⎜⎝⎞⎠kN= MR10−30−20−⎛⎜⎜⎝⎞⎠kN m⋅= For the exclusive use of adopters of the Hibbeler series of books.© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material maybe reproduced, in any form or by any means, without permission in writing from the publisher.Problem 4-161Determine the coordinate direction angles of F, which is applied to the end A of the pipeassembly, so that the moment of F about O is zero.Units Used:Given:F 20lb:=a 8in:= c 6in:=d 10in:=b 6in:=Solution:Require Mo = 0. This happens when force F is directed either towards or away from point O.rcab+d⎛⎜⎜⎜⎝⎞⎟⎠:= urr:= u0.30.80.5⎛⎜⎜⎝⎞⎠=If the force points away from O, thenαβγ⎛⎜⎜⎜⎝⎞⎟⎠acos u():=αβγ⎛⎜⎜⎜⎝⎞⎟⎠70.839.856.7⎛⎜⎜⎝⎞⎠deg=If the force points towards O, thenαβγ⎛⎜⎜⎜⎝⎞⎟⎠acos u−():=αβγ⎛⎜⎜⎜⎝⎞⎟⎠109.2140.2123.3⎛⎜⎜⎝⎞⎠deg= For the exclusive use of adopters of the Hibbeler series of books.© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material maybe reproduced, in any form or by any means, without permission in writing from the publisher.Problem 4-162Determine the moment of the force F about point O. The force has coordinate direction anglesα, β, γ. Express the result as a Cartesian vector.Given:a 8in:=F 20lb:=α 60deg:= b 6in:=β 120deg:= c 6in:=γ 45deg:= d 10in:=Solution:rcab+d⎛⎜⎜⎜⎝⎞⎟⎠:= FvFcos α()cos β()cos γ()⎛⎜⎜⎜⎝⎞⎟⎠⋅:= MrFv×:= M298.015.1200.0−⎛⎜⎜⎝⎞⎠lb in⋅= For the exclusive use of adopters of the Hibbeler series of books.© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material maybe reproduced, in any form or by any means, without permission in writing from the publisher.Problem 4-166Determine the resultant couple moment of the two couples that act on the assembly. MemberOB lies in the x-z plane.Given :F1400N:=F2150N:=a 500mm:=b 600mm:=c 400mm:=θ 45deg:=Solution:F1vF100⎛⎜⎜⎜⎝⎞⎟⎠:= F2v0F20⎛⎜⎜⎜⎝⎞⎟⎠:=r1Bb cos θ()⋅0b− sin θ()⋅⎛⎜⎜⎝⎞⎠:= r1A0a0⎛⎜⎜⎝⎞⎠:= r2b cos θ()⋅c−b− sin θ()⋅⎛⎜⎜⎝⎞⎠:=Mr1BF1v× r1AF1v−()×+ r2F2v×+:= M63.64169.71−263.64⎛⎜⎜⎝⎞⎠Nm⋅= For the exclusive use of adopters of the Hibbeler series of books.© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material maybe reproduced, in any form or by any means, without permission in writing from the publisher.Problem 4-167Replace the force F having acting at point A by an equivalent force and couple moment at pointC.Given:F 50lb:=a 10ft:=b 20ft:=c 15ft:=d 10ft:=e 30ft:=Solution:rABdce−⎛⎜⎜⎜⎝⎞⎟⎠:=FvFrABrAB⋅:=rCA0ab+e⎛⎜⎜⎝⎞⎠:=FRFv:= FR14.321.442.9−⎛⎜⎜⎝⎞⎠lb=MRrCAFv×:= MR1928.6−428.6428.6−⎛⎜⎜⎝⎞⎠lb ft⋅= For the exclusive use of adopters of the Hibbeler series of books.© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material maybe reproduced, in any form or by any means, without permission in writing from the publisher.Problem 4-169The horizontal force F acts on the handle of the wrench. Determine the moment of this forceabout point O. Specify the coordinate direction angles α, β, γ of the moment axis.Given:F