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2.0 Statics of Particles

2.0 Statics of Particles. Forces on a Particle. Line of Action. magnitude. Point of Application. 50 N. A. 30 o. direction. Since the vector has a well defined point of application, it is a fixed vector, therefore can not be moved without modifications. Properties of Vector Addition.

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2.0 Statics of Particles

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  1. 2.0 Statics of Particles Dr. Engin Aktaş

  2. Forces on a Particle Line of Action magnitude Point of Application 50 N A 30o direction Since the vector has a well defined point of application, it is a fixed vector, therefore can not be moved without modifications Dr. Engin Aktaş

  3. Properties of Vector Addition Commutative P+Q=Q+P Q P P A P+Q A Q Q+P P A Q Associative Q Q P P S S Q+S P+Q A A P+Q+S P+Q+S P+Q+S=(P+Q)+S=P+(Q+S) Dr. Engin Aktaş

  4. Resultant of several concurrent forces P Using polygon rule P A Q Q A S S P+Q+S Dr. Engin Aktaş

  5. Resolution of a force into components Q Q F F A A P P F Q A P Dr. Engin Aktaş

  6. Q R a P A Example (Beer & Johnston) Q=60 N The two forces P and Q act on a bolt A. Determine their resultant. 25o P=40 N A 20o 180-25=155o 25o B 20o R2=P2+Q2-2PQcosB Law of cosines R=97.73 N R2=(40N)2+(60N)2-2(40N)(60N)cos1550 Law of sines Dr. Engin Aktaş

  7. Example (cntd.) Law of sines A=15.04o a=20o+A=35.04o a=35.0o The answer is R=97.7 kN Dr. Engin Aktaş

  8. A 1 B 30o 45o 2 C Example (Beer and Johnston) A barge is pulled by two tugboats. If the resultant of the forces exerted by tugboats is a 25 kN force directed along the axis of barge, determine the tension in each of the ropes. Using law of sines R=25 kN 30o 45o 105o T2 T1 Dr. Engin Aktaş

  9. Rectangular Components y F Fy q x Fx Dr. Engin Aktaş

  10. y Fy F q Fx x Dr. Engin Aktaş

  11. Unit Vectors y F Magnitude=1 Fy=Fyj j x Fx=Fxi i Dr. Engin Aktaş

  12. Example (Beer and Johnston) F=(3.5 kN)i+(7.5kN)j Determine the magnitude of the force and angle q. q A y Fy=7.5 kN F q x A Fx=3.5 kN Dr. Engin Aktaş

  13. F1 sin 30o F2 cos 20o F1 cos 30o F4 cos 15o -(F2sin20o) -(F4sin 15o) Magnitude, N x Component, N y Component, N Forces F1 150 +129.9 +75.0 -F3 -27.4 +75.2 F2 80 -110.0 0 110 F3 -25.9 +96.6 F4 100 Addition of Forces by Summing x and y components y 20o F2=80 N Determine the resultant of the forces on the bolt F1=150 N 30o A x 15o F4=100 N F3=110 N Ry=+14.3 Rx=199.1 Dr. Engin Aktaş

  14. Example(cntd) R=Rxi+Ryj R=(199.1N)i+(14.30N)j R Ry=14.30 N a Rx=199.1 N Dr. Engin Aktaş

  15. 30o y 30o x Equilibrium of a Particle When the resultant of all the forces acting on a particle is zero, the particle is in equilibrium. F1=300 N F4=400 N F2=173.2 N F4=400 N F1=300 N A F3=200 N F2=173.2 N F3=200 N R=SF=0 (SFx)i+ (SFy)j=0 SFx=0 SFy=0 SFx=300 N -(200 N) sin30o-(400 N) sin 30o=300 N -100 N -200 N = 0 SFy=-173.2 N -(200 N) cos30o+ (400 N) cos 30o=-173.2 N -173.2 N +346.4 N = 0 Dr. Engin Aktaş

  16. B C A 50o 30o Space Diagram TAB TAC 50o 30o 736 N Free-Body Diagram A sketch showing the physical conditions of the problem is known as space diagram. Choose a significant particle and draw a separate diagram showing that particle and forces on it. This is the Free-Body diagram. Using law of sines TAB 40o 736 N 80o 60o TAB=647 N TAC Free-Body Diagram TAC=480 N Dr. Engin Aktaş

  17. Analytical Solution • Two unknowns TAB and TAC • Equilibrium Equations SFx=0 and SFy=0 The system of equations then Solvingthe above system (2 unknowns 2 equations) TAB=647 N TAC=480 N Dr. Engin Aktaş

  18. Sumary of Solution techniques • equilibrium under three forces may use force trianglerule • equilibrium under more than three forces may use force polygon rule • If analytical solution is desired may use equations of equilibrium Dr. Engin Aktaş

  19. B C 20o 60o A Flow Example(Beer and Johnston) As a part of the design of a new sailboat, it is desired to determine the drag force which may be expected at a given speed. To do so, a model of the proposed hull is placed in a test channel and three cables are used to keep its bow on the centerline of the channel. Dynamometer reading indicate that for a given speed, the tension is 200 N in cable AB and 300 N in cable AE. Determine the drag force exerted on the hull and the tension in cable AC. E Equilibrium condition Start with drawing Free-Body Diagram Resultant of the all forces should be zero R=TAB+TAC+TAE+FD=0 20o TAC Let’s write all the forces in x and y components TAB=200 N 60o TAB=-(200 N) sin 60o i+(200 N) cos 60oj FD TAB=-173.2 i+100 j TAC= (TAC) sin 20o i+(TAC) cos 20oj Unknowns; FD, TAC TAC= 0.342(TAC) i+0.9397(TAC) j TAE=-(300 N) j TAE=300 N FD=FDi Dr. Engin Aktaş

  20. Example(cntd.) R=TAB+TAC+TAE+FD=0 R=(-173.2 N)i + 100 N j + 0.342(TAC) i + 0.9397 (TAC) j + (-300 N) j + FDi =0 R={-173.2 + 0.342(TAC) + FD}i + {100 + 0.9397 (TAC) + (-300 N)}j = 0 -173.2 N + 0.342(TAC) + FD= 0 SFx=0 100 + 0.9397 (TAC) + (-300 N)= 0 SFy=0 TAC=213 N FD=100 N Dr. Engin Aktaş

  21. y z V x Forces in Space j k Vzk qy Vyj qz qx Vxi i V=Vxi+Vyj+Vzk Vx=V cos qx Vy=V cosqy Vz=Vcosqz Dr. Engin Aktaş

  22. Direction cosines n= cos qz m= cos qy l = cos qx Vz=nV Vy=mV Vx=l V V 2=Vx2+Vy2+Vz2 l 2 + m 2 + n 2=1 Dr. Engin Aktaş

  23. y z x B(xB, yB, zB) V n A(xA, yA, zA) (xB-xA) (yB-yA) i V= + j (zB-zA) + k Unit vector along AB Dr. Engin Aktaş

  24. Rectangular Coordinates in Space y y B B A Fy qy Fy F Fx x O x O f D Fh f Fz Fh C E C Fy=F cosqy Fx = Fh cosf = F sinqy cosf z z Fh=F sinqy Fz = Fh sinf = F sinqy sinf F 2 = Fy2 + Fh2 Fh2 = Fx2 + Fz2 Dr. Engin Aktaş

  25. Problems Dr. Engin Aktaş

  26. Problem 1-(Meriam and Kraige) y F=1800 N 4 The 1800 N force F is applied at the end of the I-beam. Express F as a vector using the unit vectors i and j. A 3 x First let’s find the xand ycomponents of the force F. z Fx= -1800 N 3/5 = -1080 N Fy= -1800 N 4/5 = -1440 N F = Fx i+ Fy j = -1080 i -1440 j N Dr. Engin Aktaş

  27. L C D Air flow Problem 2-(Meriam and Kraige) The ratio of the lift force L to the drag force D for the simple airfoil is L/D = 10. If the lift force on a short section of the airfoil is 50 N, compute the magnitude of the resultant force R and the angle q which it makes with the horizontal. L= 50 N tan q = 50/5 q C q = tan-110 =84.3o D=5 N F = (502+52)0.50 = 50.2 N Dr. Engin Aktaş

  28. Fx A 45o 66.7o 113.3o 45o 1040 N Fa 21.7o Problem 3-(Meriam and Kraige) y The gusset plate is subjected to the two forces shown. Replace them by two equivalent force, Fx in the x-direction and Fa in the a direction. Determine the magnitudes of Fx and Fa. A x 10o From the law of cosines 800 N 900 N R2=8002+9002-2(800)(900) cos75 a 45o 25o R=1040 N A From the law of sines 800 N 10o 10o a 65o R 25o 900 N Dr. Engin Aktaş

  29. A 40m B 20m C 50m 40m Problem 4-(Meriam and Kraige) The guy cables AB and AC are attached to the top of the transmission tower. The tension in cable AC is 8 kN. Determine the required tension T in cable AB such that the net effect of the two cable tensions is a downward force at point A. Determine the magnitude R of this downward force This time let’s use another approach. Since the resultant force is downward the sum of horizontal components of the two forces should add up to zero. 8 kN R The magnitude of R Dr. Engin Aktaş

  30. Problem 5-(Beer and Johnston) 500 mm 1375 mm Two cables are tied together at C and loaded as shown. Determine the tension in AC and BC. A B 1200 mm Let’s draw the Free Body Diagram (FBD) at C C y 1600 kg TBC TAC x C W=1600*9.81=15700N Writing Equilibrium Equations SFx=0 SFy=0 TAC=12470 N TCB=6370 N Dr. Engin Aktaş

  31. C 13.5 m 4 m D B A 5.5 m 8 m y Problem 6-(Beer and Johnston) A precast-concrete wall section is temporarily held by the cables shown. Knowing that tension is 4200 N in cable AB and 6000 N in cable AC, determine the magnitude and direction of the resultant of the forces exerted by cables AB and AC on stake A. x Coordinates of points A, B and C A (8, -4, -5.5) B (0, 0, 0) C (0, 0, -13.5) AB=(0-8) i + (0-(-4)) j + (0-(-5.5)) k = -8 i + 4 j + 5.5 k z AC=(0-8) i + (0-(-4)) j + (-13.5-(-5.5)) k = -8 i + 4 j - 8 k Dr. Engin Aktaş

  32. C 13.5 m 4 m D B A 5.5 m 8 m y x z y Problem 6-(Beer and Johnston) A precast-concrete wall section is temporarily held by the cables shown. Knowing that tension is 4200 N in cable AB and 6000 N in cable AC, determine the magnitude and direction of the resultant of the forces exerted by cables AB and AC on stake A. x z R = FAB + FAC R =(-3200-4002) i +(1600+1998) j + (2201-4002) k = -7200 i + 3600 j - 1800 k R = 8250 N Direction cosines n = 0.218 l = 0.873 m = 0.436 Dr. Engin Aktaş

  33. y C 700 mm 450 mm O D 600 mm x B 650 mm z 1125 mm A Problem 7-(Beer and Johnston) Let’s draw FBD first FAC FAD FAB W Unknowns : FAD, FAC and W Coordinates of points A, B, C and D B (700, 0, 0) C (0, 0, -600) D (-650, 0,450) A (0, -1125, 0) A crate is supported by three cables as shown. Determine the weight W of the crate knowing that the tension in cable AB is 4620 N. AB=((700-0)i+(0-(-1125))j+0k=700i+1125j+0k AB=(7002+11252)0.5=1325 nAB=(700i+1125j+0k)/1325=0.5283i+0.8491j FAB=4620(0.5283i+0.8491j) N Dr. Engin Aktaş

  34. y C 700 mm 450 mm O D 600 mm x B 650 mm z 1125 mm A Problem 7-(Beer and Johnston) Coordinates of points A, B, C and D B (700, 0, 0) C (0, 0, -600) D (-650, 0,450) A (0, -1125, 0) AC=(0i+(0-(-1125))j+(-600-0)k=0i+1125j-600k AC=(6002+11252)0.5=1275 nAC=(0i+1125j-600k)/1275=0.8824j-0.4706k FAC=FAC(0.8824j-0.4706k) N AD=((-650-0)i+(0-(-1125))j+(450-0)k=-650i+1125j+450k AD=(6502+11252 +4502)0.5=1375 nAD=(-650i+1125j+450k)/1375= -0.4727i+0.8182j+0.3273k FAD=FAD(-0.4727i+0.8182j+0.3273k) N W=-W j Dr. Engin Aktaş

  35. y C 700 mm 450 mm O D 600 mm x B 650 mm z 1125 mm A Problem 7-(Beer and Johnston) FAB+FAC+FAD+W=0 4620(0.5283i+0.8491j) +FAC(0.8824j-0.4706k)+FAD(-0.4727i+0.8182j+0.3273k)-W j=0 2441-0.4727FAD=0 FAD= 5160 N FAC= 3590 N -0.4706 FAC+0.3273*5160=0 W= 11320 N 3922+3170+4225-W = 0 Dr. Engin Aktaş

  36. y Problem 8-(Beer and Johnston) D 600 mm C A 200 mm x 200 mm B 200 mm 400 mm z A 16 kg triangular plate is supported by three wires as shown. Determine the tension in each wire . Dr. Engin Aktaş

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