Download presentation

Presentation is loading. Please wait.

Published byOmarion Bow Modified over 3 years ago

1
2.0 Statics of Particles Dr. Engin Aktaş

2
**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

3
**Properties of Vector Addition**

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

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
**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

7
**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

8
**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

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 F Fy=Fyj j Fx=Fxi i y Magnitude=1 x 01.04.2017**

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
**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

14
**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

15
**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

16
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

17
**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

18
**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

19
**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

20
**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

21
**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

22
**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

23
**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

24
**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

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 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

27
**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

28
**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

29
**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

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
**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

32
**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

33
**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

34
**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

35
**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

36
**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

Similar presentations

Presentation is loading. Please wait....

OK

MCV4UW Vectors.

MCV4UW Vectors.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on natural resources and conservation degree Ppt on solar system in hindi Ppt on earth hour toronto How to download slideshare ppt on negotiations Ppt on international financial management Ppt on computer languages list Ppt on global warming for class 9 download Ppt on 4g wireless technology Free ppt on natural disasters Ppt on business plan presentation