# ENGR-36_Lab-05_Fa07_Lec-Notes.ppt 1 Bruce Mayer, PE Engineering-36: Vector Mechanics - Statics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer.

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ENGR-36_Lab-05_Fa07_Lec-Notes.ppt 1 Bruce Mayer, PE Engineering-36: Vector Mechanics - Statics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Engineering 36 Lab-05 Angle Problems

ENGR-36_Lab-05_Fa07_Lec-Notes.ppt 2 Bruce Mayer, PE Engineering-36: Vector Mechanics - Statics Problem  Cables AB & AC attached to Tree Trunk and fastened to Stakes in the Ground  Given Cable Tension T AC = 3.6 kN  Find Components of Force exerted by cable AC on the Tree The Space Angles θ x, θ y, θ z for Cable AC

ENGR-36_Lab-05_Fa07_Lec-Notes.ppt 3 Bruce Mayer, PE Engineering-36: Vector Mechanics - Statics

ENGR-36_Lab-05_Fa07_Lec-Notes.ppt 4 Bruce Mayer, PE Engineering-36: Vector Mechanics - Statics

ENGR-36_Lab-05_Fa07_Lec-Notes.ppt 5 Bruce Mayer, PE Engineering-36: Vector Mechanics - Statics

ENGR-36_Lab-05_Fa07_Lec-Notes.ppt 6 Bruce Mayer, PE Engineering-36: Vector Mechanics - Statics

ENGR-36_Lab-05_Fa07_Lec-Notes.ppt 7 Bruce Mayer, PE Engineering-36: Vector Mechanics - Statics Tree Prob by MATLAB >> TAC = 3.6; >> Fyv = -TAC*sind(45) Fyv = -2.5456 >> Fh = TAC*cosd(45) Fh = 2.5456 >> Fzv = Fh*cosd(25) Fzv = 2.3071 >> Fxv = -Fh*sind(25) Fxv = -1.0758 >> TACv = [Fxv Fyv Fzv] TACv = -1.0758 -2.5456 2.3071 >> uAC = TACv/TAC uAC = -0.2988 -0.7071 0.6409 >> Qxyz = acosd(uAC) Qxyz = 107.3877 135.0000 50.1443

ENGR-36_Lab-05_Fa07_Lec-Notes.ppt 8 Bruce Mayer, PE Engineering-36: Vector Mechanics - Statics Problem  Given the Geometry of the Steel FrameWork as Shown  Given EF & EG are Cables Pt-E is at MidPt of BC Tension in Cable EF is 330N  Find Angle Between EF and BC Projection on BC of the force exerted by Cable EF at Pt-E

ENGR-36_Lab-05_Fa07_Lec-Notes.ppt 9 Bruce Mayer, PE Engineering-36: Vector Mechanics - Statics û EF û BC û EF

ENGR-36_Lab-05_Fa07_Lec-Notes.ppt 10 Bruce Mayer, PE Engineering-36: Vector Mechanics - Statics

ENGR-36_Lab-05_Fa07_Lec-Notes.ppt 11 Bruce Mayer, PE Engineering-36: Vector Mechanics - Statics

ENGR-36_Lab-05_Fa07_Lec-Notes.ppt 12 Bruce Mayer, PE Engineering-36: Vector Mechanics - Statics I-Beam Prob by MATLAB >> B = [0 8.25 0]; F = [1 0 0]; C = [16 3.75, -12]; >> E = [16/2 (3.75+8.25)/2 -6] E = 8 6 -6 >> EFv = F-E EFv = -7 -6 6 >> BCv = C - B BCv = 16.0000 -4.5000 -12.0000 >> EFm = norm(EFv) EFm = 11 >> BCm = norm(BCv) BCm = 20.5000 >> uEF = EFv/EFm uEF = -0.6364 -0.5455 0.5455 >> uBC = BCv/BCm uBC = 0.7805 -0.2195 -0.5854 >> Qceg = acosd(dot(uEF,uBC)) Qceg = 134.1254

ENGR-36_Lab-05_Fa07_Lec-Notes.ppt 13 Bruce Mayer, PE Engineering-36: Vector Mechanics - Statics Problem  Given Geometry as Shown BungiCord PC with Tension of 30N Distance OP = 120 mm  Find Angle Between PC and OA Projection on OA of the force exerted by the BungiCord PC at Pt-P

ENGR-36_Lab-05_Fa07_Lec-Notes.ppt 14 Bruce Mayer, PE Engineering-36: Vector Mechanics - Statics

ENGR-36_Lab-05_Fa07_Lec-Notes.ppt 15 Bruce Mayer, PE Engineering-36: Vector Mechanics - Statics =

ENGR-36_Lab-05_Fa07_Lec-Notes.ppt 16 Bruce Mayer, PE Engineering-36: Vector Mechanics - Statics û OA |û| û OA

ENGR-36_Lab-05_Fa07_Lec-Notes.ppt 17 Bruce Mayer, PE Engineering-36: Vector Mechanics - Statics Bungi Prob by MATLAB >> OAm = norm([240 240 -120]) OAm = 360 >> OPm = 120; >> Ratio = OPm/OAm Ratio = 0.3333 >> A = [240 240 -120] A = 240 240 -120 >> P = Ratio*A P = 80 80 -40 >> C = [180 300 240] C = 180 300 240 >> PCv = C-P PCv = 100 220 280 >> OAv = A; >> PCm = norm(PCv) PCm = 369.8648 >> OAm = norm(OAv) OAm = 360 >> Qcpa = acosd(dot(PCv,OAv)/(PCm*OAm)) Qcpa = 71.0682 >> uPC = PCv/PCm uPC = 0.2704 0.5948 0.7570 >> Tpcv = 30*uPC Tpcv = 8.1111 17.8444 22.7110 >> FOAproj = 30*cosd(Qcpa) FOAproj = 9.7333

ENGR-36_Lab-05_Fa07_Lec-Notes.ppt 18 Bruce Mayer, PE Engineering-36: Vector Mechanics - Statics Problem  Given Geometry as Shown –Angle OAB is 90° Tension in Cable BC is 5.3 kN T BC = 5.3 kN

ENGR-36_Lab-05_Fa07_Lec-Notes.ppt 19 Bruce Mayer, PE Engineering-36: Vector Mechanics - Statics Problem T BC = 5.3 kN  Find The Angle, θ, Between the Cable and Pipe BA The Components of F BC (cable force acting at Pt-B) that are || and  to Pipe BA

ENGR-36_Lab-05_Fa07_Lec-Notes.ppt 20 Bruce Mayer, PE Engineering-36: Vector Mechanics - Statics

ENGR-36_Lab-05_Fa07_Lec-Notes.ppt 21 Bruce Mayer, PE Engineering-36: Vector Mechanics - Statics −

ENGR-36_Lab-05_Fa07_Lec-Notes.ppt 22 Bruce Mayer, PE Engineering-36: Vector Mechanics - Statics

ENGR-36_Lab-05_Fa07_Lec-Notes.ppt 23 Bruce Mayer, PE Engineering-36: Vector Mechanics - Statics

ENGR-36_Lab-05_Fa07_Lec-Notes.ppt 24 Bruce Mayer, PE Engineering-36: Vector Mechanics - Statics Pipe Prob by MATLAB >> TBC = 5.3; % in kN >> A = [2400 0 0]; C = [0 1050 1800]; >> B = [2400 -1200*sind(30) 1200*cosd(30)] B = 1.0e+03 * 2.4000 -0.6000 1.0392 >> BAv = A-B BAv = 1.0e+03 * 0 0.6000 -1.0392 >> BCv = C-B BCv = 1.0e+03 * -2.4000 1.6500 0.7608 >> BAm = norm(BAv) BAm = 1200 >> BCm = norm(BCv) BCm = 3.0102e+03 >> Qabc = acosd(dot(BAv,BCv)/(BAm*BCm)) Qabc = 86.8358 >> Fpar = TBC*cosd(Qabc) Fpar = 0.2925 >> Fper = TBC*sind(Qabc) Fper = 5.2919

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