# 2.0 Statics of Particles 01.04.2017 Dr. Engin Aktaş.

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2.0 Statics of Particles Dr. Engin Aktaş

Forces on a Particle 50 N A 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

Commutative P+Q=Q+P P Q P A P+Q Q A 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

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

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

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 A P Q R a 180-25=155o 25o B 20o Law of cosines R2=P2+Q2-2PQcosB R2=(40N)2+(60N)2-2(40N)(60N)cos1550 R=97.73 N Law of sines Dr. Engin Aktaş

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

Example (Beer and Johnston)
C 1 2 30o 45o 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

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

y Fy F q Fx x Dr. Engin Aktaş

Unit Vectors F Fy=Fyj j Fx=Fxi i y Magnitude=1 x 01.04.2017
Dr. Engin Aktaş

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

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

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

Equilibrium of a Particle
When the resultant of all the forces acting on a particle is zero, the particle is in equilibrium. x y F1=300 N F4=400 N 30o F2=173.2 N F4=400 N A F1=300 N F2=173.2 N F3=200 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= N -(200 N) cos30o+ (400 N) cos 30o= N N N = 0 Dr. Engin Aktaş

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

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

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

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. B C A 60o 20o Flow 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 60o j FD TAB= i+100 j TAC= (TAC) sin 20o i+(TAC) cos 20o j Unknowns; FD, TAC TAC= 0.342(TAC) i (TAC) j TAE=-(300 N) j TAE=300 N FD=FD i Dr. Engin Aktaş

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

Forces in Space V j k Vzk qy Vyj qz qx Vxi i V=Vxi+Vyj+Vzk Vx=V cos qx
Vy=V cos qy Vz=V cos qz Dr. Engin Aktaş

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

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

Rectangular Coordinates in Space
y y B B A Fy qy Fy F Fx O x O x f D f Fz Fh 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

Problems Dr. Engin Aktaş

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 x and y components of the force F. z Fx= N 3/5 = N Fy= N 4/5 = N F = Fx i+ Fy j = i j N Dr. Engin Aktaş

Problem 2-(Meriam and Kraige)
Air flow D L C 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

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 a R2= (800)(900) cos75 25o 45o R=1040 N A From the law of sines 800 N 10o 10o a 66.7o 45o 113.3o A Fx Fa 21.7o 1040 N R 65o 25o 900 N Dr. Engin Aktaş

Problem 4-(Meriam and Kraige)
C 50 m 40 m 20 m 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

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

Problem 6-(Beer and Johnston)
C 5.5 m 8 m 4 m 13.5 m y 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 k z AC=(0-8) i + (0-(-4)) j + (-13.5-(-5.5)) k = -8 i + 4 j - 8 k Dr. Engin Aktaş

Problem 6-(Beer and Johnston)
C 5.5 m 8 m 4 m 13.5 m y 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 =( ) i +( ) j + ( ) k = i j k R = 8250 N x y z Direction cosines l = 0.873 m = 0.436 n = 0.218 Dr. Engin Aktaş

Problem 7-(Beer and Johnston)
y x z B A O C D 700 mm 600 mm 1125 mm 650 mm 450 mm Let’s draw FBD first FAC FAD FAB W Unknowns : FAD, FAC and W Coordinates of points A, B, C and D A (0, -1125, 0) B (700, 0, 0) C (0, 0, -600) D (-650, 0,450) 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=( )0.5=1325 nAB=(700i+1125j+0k)/1325=0.5283i j FAB=4620(0.5283i j) N Dr. Engin Aktaş

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

Problem 7-(Beer and Johnston)
y x z B A O C D 700 mm 600 mm 1125 mm 650 mm 450 mm FAB+FAC+FAD+W=0 4620(0.5283i j) +FAC(0.8824j k)+FAD( i j k)-W j=0 FAD=0 FAD= 5160 N FAC *5160=0 FAC= 3590 N W= N W = 0 Dr. Engin Aktaş

Problem 8-(Beer and Johnston)
y D 600 mm C 200 mm A 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