Engineering Mechanics Ch 1 Force System.ppt

this is the PPT on first module of the engineering mechanics:
FORCE SYSTEM:
CONTENTS;
Introduction to Mechanics, Laws of mechanics, Newton's Laws, Law of Parallelogram, Law of transmissibility, Characteristics of force, System of Forces, Method of resolution and composition moment of a force,
Law of Moments, Varignon's Theorem, Problems on the moment, Resultant, Equivalent force& couple, properties of a couple
The following are the fundamental laws of mechanics:
(i) Newton’s first law
(ii) Newton’s second law
(iii) Newton’s third law
(iv) Newton’s gravitational law
(v) Law of transmissibility of forces
(vi) Parallelogram law of forces
Booster Questions???
1.What happens according to Newton if you let an untied balloon go????
2. Describe what happens if you are riding a skateboard and hit something (like a curb) with the front wheels???
3. Describe why you hold your gun next to your shoulder while deer hunting????
4. Why should we wear seatbelts – use one of Newton’s Laws in your answer?
5. How can Newton’s laws be used to explain how rockets are launched into space?
6. Explain how each of Newton’s laws affects a game of Tug of War.
Characteristics of force
Questions:

1. Define Mechanics. What are the different branches of mechanics?
2. What are the characteristics of force?
Force System
Coplanar
Concurrent
Collinear
Parallel
Like
Unlike
Non Concurrent & Non Parallel (General)
Non-coplanar
Concurrent
Collinear
Parallel
Like
Unlike
Non Concurrent & Non Parallel (General)
Forces added to obtain a single force which produces the same effect as the original system of forces.
This single force is known as Resultant force.
The process of finding the resultant force is called composition of forces.
There are two methods of finding resultant
Analytical method
Graphical method
Analytical methods are
Parallelogram law &
Method of Resolution

1. Prepared By
Prof. V. V. Nalawade
ENGINEERING MECHANICS
Prof. V.V.Nalawade 1

2. Module 1: Force System 8hrs
Module 2 : Equilibrium 7hrs
Module 3: Center of Gravity and Moment of Inertia 7 hrs.
Module 4: Friction 6hrs
Module 5 : Kinematics 6hrs
Module 6 : Kinetics 6hrs
C
u
r
r
i
c
u
l
u
m
2

3. Prof. V.V.Nalawade 3
Engineering Mechanics
•Dr. R.K. Bansal
•N.H. Dubey
•Beer & Johnston
•K. L. Kumar
•R. V. Kulkarni
Applied Mechanics
•R. S. Khurmi
•Sunil Deo
•R. K. Singer
Reference Books

4. Prof. V.V.Nalawade 4
CASIO
Fx-991MS

5. Prerequisite
Basic Concepts
Resolution &
composition of forces
Contents of the presentation
Prof. V.V.Nalawade 5

6. Prof. V.V.Nalawade 6

7. Coordinate Geometry
Prof. V.V.Nalawade 7
II
Quadrant
(-,+)
I
Quadrant
(+,+)
III
Quadrant
(-,-)
IV
Quadrant
(+,-)

8. Slope of line
Prof. V.V.Nalawade 8
θ
(X2-X1)
(Y2-Y1)
P
Q (X2,Y2)
(X1,Y1)
X Axis
Y Axis
m = tan θ =

9. Few more to Recall……
Trigonometry
Pythagoras Theorem
Cosine & Sine rule
Sign conversion
Unit conversion
Prof. V.V.Nalawade 9

10. 1.
Force
System
Prof. V.V.Nalawade 1
0

11. Prof. V.V.Nalawade 11
Sr.
No.
Topic Learning Objective
(TLO)
COBL
CA
Code
1
Torecallthebasicprinciplesof
mechanics
CO1 L1 1.2, 1.3
2
Todescribetheconceptson
mechanicsanditspractical
implementation
CO1 L2 1.1, 1.2
3
Toidentifytheforcesystemand
calculatetheresultantofit
CO1 L3
1.2, 2.1,
12.2
4
Toanalyzethenumericalof
differentcases
CO1 L32.1, 12.2
Learning Outcome:
At the end of the topic the student should be able to:

12. Branches of Mechanics
Engineering
Mechanics
Statics
(Rest)
Dynamics
(Motion)
KineticsKinematics
EM is the branch of physics which deals with the study of forces and their
effect on body when body is at rest or in motion.
With Reference to the
Cause of motion
Without Reference to the
Cause of motionProf. V.V.Nalawade 1
2

13. Prof. V.V.Nalawade 1
3

14. Prof. V.V.Nalawade 1
4

15. Prof. V.V.Nalawade 15
We learn
ENGINEERING
MECHANICS

16. Prof. V.V.Nalawade 16

17. automobiles,aircrafts,electricmotors,robots,
television,mobile,satellite,projectileofmissiles,
launchingofrockets,radarcommunication,trusses,
liftingmachineslikecrane,hoist,screwjack,
elevator,conveyorbelt,cargoship,submarine,etc.
Prof. V.V.Nalawade 17

18. Laws of Mechanics
The following are the fundamental laws of mechanics:
(i) Newton’s first law
(ii) Newton’s second law
(iii) Newton’s third law
(iv) Newton’s gravitational law
(v) Law of transmissibility of forces
(vi) Parallelogram law of forces
Prof. V.V.Nalawade 1
8

19. Prof. V.V.Nalawade 1
9Image source: https://issuu.com/daennagonzalez/docs/newton___s_laws_of_motion

20. Prof. V.V.Nalawade 20

21. Prof. V.V.Nalawade 21

22. Prof. V.V.Nalawade 22

23. Prof. V.V.Nalawade 23

24. Prof. V.V.Nalawade 24
Because the
amount of
acceleration of a
body is
proportional to
the acting force
and inversely
proportional to
the mass of the
body

25. Prof. V.V.Nalawade 25

26. Prof. V.V.Nalawade 26
Balloon
goes up
Air goes
down

27. 27Prof. V.V.Nalawade

28. Prof. V.V.Nalawade 28

29. Prof. V.V.Nalawade 29
1.WhathappensaccordingtoNewtonifyouletan
untiedballoongo????

30. Prof. V.V.Nalawade 30
1.WhathappensaccordingtoNewtonifyouletan
untiedballoongo????
Balloon
goes up
Air goes
down
3 rd Law
Airwillrushoutoftheballoon
forcingtheballoontomove
throughtheairintheopposite
direction,butequalinforce.

31. Prof. V.V.Nalawade 31
2. Describe what happens if you are riding a skateboard
and hit something (like a curb) with the front wheels???

32. Prof. V.V.Nalawade 32
2. Describe what happens if you are riding a skateboard
and hit something (like a curb) with the front wheels???
1 stLaw
Your body will keep moving forward and fly off your
skateboard since the curb only stops the board, not
yourself.

33. Prof. V.V.Nalawade 33
3.Describewhyyouholdyourgunnexttoyour
shoulderwhiledeerhunting????

34. Prof. V.V.Nalawade 34
3.Describewhyyouholdyourgunnexttoyour
shoulderwhiledeerhunting????
3 rd Law
When you pull the gun’s trigger, it
forces the bullet out of the gun, but
at the same time, the gun is forced
in the opposite direction of the
bullet (towards you). Your shoulder
is a new force that is introduced in
order to keep your gun from flying
away from you.

35. Prof. V.V.Nalawade 35
4.Whyshouldwewearseatbelts–useoneofNewton’s
Lawsinyouranswer?

36. Prof. V.V.Nalawade 36
4.Whyshouldwewearseatbelts–useoneofNewton’s
Lawsinyouranswer?
Weshouldwearseatbeltssoifweareinanaccidentourbody
doesn’tkeepmovingatthesamespeedandinthesamedirection
thatthecarwasgoing.Anewforcewouldbeintroducedtoour
bodies(theseatbelt)inordertokeepourbodiesinplace.

37. Prof. V.V.Nalawade 37

38. Prof. V.V.Nalawade 38
Newton’s third law would tell us that when the rocket
pushes out fire with a specific amount of force, the rocket
will move in the opposite direction, but with the same
amount of force. This is what causes the rocket to shoot
up into the air.

39. Prof. V.V.Nalawade 39
6.ExplainhoweachofNewton’slawsaffectsagameof
TugofWar.

40. Prof. V.V.Nalawade 40
6.ExplainhoweachofNewton’slawsaffectsagameof
TugofWar.
•FirstLaw:Theropewillstayinthesameplaceuntilthetuggingstarts
(anewforceisintroduced)
•SecondLaw:Wecouldmeasureateam’sforcethattheycanpullthe
ropewithbasedontheirbodymassesandtheaccelerationthatthey
arecausingtheropetomoveat.
•ThirdLaw:1teampullstheropetowardsthemselveswithacertain
amountofforceandtheopposingteamisalsoputtingforceontherope.
Thesameamountofforceisappliedfromthegroundtothepeopleas
theyareputtingontheground.

41. 41
Concept of force and its measurements
Prof. V.V.Nalawade

42. 42Prof. V.V.Nalawade

43. 43
1. Magnitude: 2. Direction :
Prof. V.V.Nalawade

44. 44
3. Point of application :
4. Sense or
Nature :
Prof. V.V.Nalawade

45. Prof. V.V.Nalawade 45
Questions:
1.DefineMechanics.Whatarethedifferentbranchesof
mechanics?
2.Whatarethecharacteristicsofforce?

46. Prof. V.V.Nalawade 46
System of forces

47. SystemofForces
Severalforcesactingsimultaneouslyuponabody
Force System
Coplanar
Concurrent Collinear
Parallel
Like Unlike
Non Concurrent
& Non Parallel
(General)
Non-coplanar
Concurrent
Collinear
Parallel
Like Unlike
Non Concurrent
& Non Parallel
(General)
Prof. V.V.Nalawade 4
7

48. CoplanarSystemofForces2D
4
8
Prof. V.V.Nalawade

49. Non-CoplanarSystemofForces
4
9
Prof. V.V.Nalawade

50. Composition of forces
Forces added to obtain a single force which produces
the same effect as the original system of forces.
This single force is known as Resultant force.
The process of finding the resultant force is called
composition of forces.
Prof. V.V.Nalawade 50

51. Composition of forces
There are two methods of finding resultant
1.Analytical method
2.Graphical method
Analytical methods are
Parallelogram law &
Method of Resolution
Prof. V.V.Nalawade 51

52. Prof. V.V.Nalawade 5
2

53. Type I: Problems on Composition of Forces by
Parallelogram and Triangle LawSin Sin Sin
R P Q
  

Where,
R = Resultant of force P & Q
θ= Angle Between P & R
β= Angle Between P & Q
α = Angle Between Q & R
1.1 Law Of Parallelogram:- 1.2 Triangle Law :-
Prof. V.V.Nalawade 5
3

54. Ex.1. Find the resultant of the following forces
•Solution : Method i) By Parallelogram Law
Prof. V.V.Nalawade 54
3 N
4N
3 N
4N
R
α
R = 5 N

55. Continue……..
Prof. V.V.Nalawade 55
•Solution : Mathodii) By Triangle Law
3 N
4N
R
α
By Cosine rule
R = 5 N
By Sine rule

56. Prof. V.V.Nalawade 56
70 N
50 N
60ᵒ
70 N
50 N
α
R
α
70 N
50 N
120ᵒ 60ᵒ
R

57. Prof. V.V.Nalawade 57

58. Prof. V.V.Nalawade 58
500 N
50
300 N
500 N
300 N
50

59. Prof. V.V.Nalawade 59
Resolution of forces
Definition
Problems

60. Resolution of forces
•The way of representing a single force into number of
forces without changing the effect of the force on
the body is called as resolution of forces.
Prof. V.V.Nalawade 60
R
Fx
Fy

61. Prof. V.V.Nalawade 61
Fx
Fy
R = 10
= 90

62. Ex. 1. Two Forces act at an angle of 120°. The bigger force is of 40N and the resultant
is perpendicular to the smaller one. Find the smaller force.
Prof. V.V.Nalawade 6
2
F2 = 20 N

63. Ex. 2. Resolve the 100 N force acting a 30°to horizontal into two component one along
horizontal and other along 120°to horizontal.
Prof. V.V.Nalawade 6
3

64. Resolution of a force into two mutually perpendicular
components (Rectangular Components)
•LetaforceFbeinclinedatanangleasshowninfig.
Wehavetoresolveitintotwomutuallyperpendicular
componentsFxalongX-AxisandFyalongY-Axis.
Prof. V.V.Nalawade 64
A
B
O
•FrompointAonthelineofactionof
aforce,drawperpendicularABon
X-Axis.
•Nowwehavetocalculatethe
lengthsOB&AB.
•Length OB represents the
magnitudeofXcomponenti.e.(Fx)
•&SimilarlyABrepresents(Fy)

65. Prof. V.V.Nalawade 65
•In ∆ AOB,
OB = OA cos θ
But OA = F
Therefore, OB = F cos θ
Lets say OB = Fx, as it is the magnitude of
x-component
Hence, Fx= F cos θ
A
B
O
Fy
Fx

66. Prof. V.V.Nalawade 66
•In ∆ OBA,
AB = OA sin θ
But OA = F
Therefore, AB = F sin θ
Lets say AB = Fy, as it is the magnitude of y-
component
Hence, Fy= F sin θ
A
B
O
Fy
Fx

67. Resolution of a force into two non perpendicular
components (Oblique Components)
•Aforcecanalsoberesolvedalongthetwo
directionswhicharenotatrightanglesto
eachother.
•In∆OAC,Applyingsinerule,weget
Prof. V.V.Nalawade 67
α
β
(α+β)
F1
F2
F1
F2
O
B
C
A
F
F2
F1
α
β

68. Resolution of Force By Perpendicular componentcos
sin
x
y
FF
FF




1
st
Quad = Fx (+ve) & Fy (-ve)
2
nd
Quad = Fx (-ve) & Fy (+ve)
3
rd
Quad = Fx (-ve) & Fy (-ve)
4
th
Quad = Fx (+ve) & Fy (-ve)
Where,
Fx = Horizontal component of
Force
Fy = Vertical Component of
force
Prof. V.V.Nalawade 6
8

69. Q. 2 Find the Component of force 100 N passing
through the points (0,2) & (-1,2)
Prof. V.V.Nalawade 69
-1
1
-2
2
(-1,2)(0,2)
F = 100 N
-1
1
-2
2
(-1,2)(0,2)
F = 100 N
Θ= 0cos
sin
x
y
FF
FF



 Fx = 100N
Fy = 0
Θ= 180
Fx = -100N
Fy = 0

70. Resolution of Force By Non-Perpendicular component
1
st
Quad = Fx (+ve) & Fy (-ve)
2
nd
Quad = Fx (-ve) & Fy (+ve)
3
rd
Quad = Fx (-ve) & Fy (-ve)
4
th
Quad = Fx (+ve) & Fy (-ve)
Prof. V.V.Nalawade 7
0
F
F2
F1
α
β

71. Prof. V.V.Nalawade 71
60°
105°
330°
X-axis
2000 N
F1
F2

72. Prof. V.V.Nalawade 72
X-axis
F1
F2
F
30°
60°

73. Method of resolution
STEPWISE PROCEDURE OF METHOD OF
RESOLUTION:
i.Resolveallforceshorizontallyandfindthealgebraic
sumofallthehorizontalcomponents(i.e.,ΣFx)
ii.Resolveallforcesverticallyandfindthealgebraicsum
ofalltheverticalcomponents(i.e.,ΣFy).
iii.TheresultantRofthegivenforceswillbegivenbythe
equation:
iv.Theresultantforcewillbeinclinedatanangleθ,with
thehorizontal,suchthat
v.Positionoftheresultant
Prof. V.V.Nalawade 73

74. 1.4 Design steps of resolution of concurrent force system :-
Case I :-When magnitude and direction
of all forces in the force system is given
& resultant is to be determined
Step 1:-Find ΣFx (Horizontal
Component)
Step 2:-Find ΣFy (Vertical Component)
Step 3:-Find Magnitude of Resultant
Step 4:-Find Direction of Resultant
Step 5:-Find Position of Resultant22
xy
R F F    1
tan
y
x
F
F





Case II :-When resultant is horizontal
Σ Fx = R and ΣFy = 0
Case III :-When resultant is Vertical
Σ Fx = 0 and ΣFy = R
Case IV :-When resultant is Zero
Σ Fx = 0 and ΣFy = 0
Case V :-When magnitude & Direction
of resultant is given & magnitude &
direction of any one force among the
force system is to be determined
Σ Fx = R cos θand ΣFy = R sin θ
θis measured w.r.tX-axis
Prof. V.V.Nalawade 7
4

75. Prof. V.V.Nalawade 75

76. Prof. V.V.Nalawade 76
X-Axis
Y-Axis
155.8 N
76.6ᵒ

77. Prof. V.V.Nalawade 77
132.3ᵒ
45.6 N
X-Axis
Y-Axis
47.7ᵒ

78. Prof. V.V.Nalawade 78
X-Axis
Y-Axis
29.09 N
45.89ᵒ

79. Prof. V.V.Nalawade 79
132.3ᵒ
45.6 N
X-Axis
Y-Axis
47.7ᵒ
132.3ᵒ
45.6 N
X-Axis
Y-Axis
47.7ᵒ
X-Axis
Y-Axis
29.09 N
45.89ᵒ

80. Prof. V.V.Nalawade 80
X-Axis
Y-Axis
29.09 N
45.89ᵒ

81. Prof. V.V.Nalawade 81
Moment of force
Law of moments
Varignon’sTheorem
Couples
Problems

82. Moment of forces
•Therotationaleffectproducedbyforceis
knownasmomentofforce.
•Itisequaltothemagnitudeofforce
multipliedbytheperpendiculardistanceof
thepointfromthelineofactionoftheforce.
•M=Fxd
•UnitN.m,KN.m,N.mmetc.,
Prof. V.V.Nalawade 82

83. Sign Convention
Prof. V.V.Nalawade 83
Clockwise (+VE) Anti-Clockwise (-VE)
F
dO
F
d
O
F
d O
F
d
O
M@O = F x d

84. Law of Moments
•It states that, “ In equilibrium when no of
coplanar forces act on a body, the sum of the
clockwise moments@ any point in their plane is
equal to the sum of the anticlockwise
moments @ the same point.
•Algebraic sum Clockwise moments = Algebraic
sum Anti-Clockwise moments @ same point
Prof. V.V.Nalawade 84

85. •Moment about pivot
•500*2 = 1000*1
•1000 = 1000
•Algebraic sum Clockwise moments = Algebraic
sum Anti-Clockwise moments @ same point
Prof. V.V.Nalawade 85

86. To find beam
reaction
To find forces in
frames
Prof. V.V.Nalawade 86
Use of Law of Moments

87. Varignon’s theorem of moments
•It status that, “ The algebraic sum of moments of all
forces about any point is equal to the moments of
their resultant about the same point.”
•Let ƩM
FA=Algebraic sum of moments of all forces
about any point A
•ƩM
RA=Moment of resultant force about same point A
•Then ƩM
FA= ƩM
RA
•i.e. F1.x1 + F2.x2 + F3.x3+……..+Fn.Xn = R.x
Prof. V.V.Nalawade 87

88. Use of Varignon’s theorem of moments
Prof. V.V.Nalawade 88
•Thistheoremisveryusefulinlocatingthe
positionoftheresultantofnon-concurrent
forces.

89. Examples of Moment of forces
•Rotation of door
•Tightening of nut by spanner
•Compass etc.,
Prof. V.V.Nalawade 89

90. Couple
•Twonon-colinear,equal,unlike,parallelforces
formsacouple.
•Astheforcesareequal&oppositetheir
resultantisZERO.
•Hencecoupleproducesonlyrotarymotion
withoutproducinglinearmotion.
Prof. V.V.Nalawade 90

91. Prof. V.V.Nalawade 91
Examples of Couple
Rotation of Steering Wheel, key, tap etc.,

92. Lever Arm OR Arm of the Couple
•The distance between two forces of a couple
is known as lever arm.
•SI unit of couple is same as moment i.e. N.m,
N.mm, KN.m, KN.mm etc.,
Prof. V.V.Nalawade 92
a
P
P

93. Sign Convention
Prof. V.V.Nalawade 93
a
P
P
a
P
P
Clockwise (+VE) Anti-Clockwise (-VE)

94. Properties of Couple
•Theresultantoftheforceofacoupleisalways
ZERO.i.e.R=P–P=0
•Themomentofcoupleisequaltotheproduct
ofoneoftheforceandleverarm.
i.e.M=Pxa
Prof. V.V.Nalawade 94
a
P
P

95. Properties of Couple
•Themomentofcoupleaboutanypointis
constant.
•Momentofcouple=Pxa
•Momentofcouple@C=P*AC–P*BC=P*a
•Momentofcouple@D=-P*AD+P*BD=P*a
Prof. V.V.Nalawade 95
a
P
P
D CBA

96. Properties of Couple
•Acouplecanbebalancedonlybyanother
coupleofequalandoppositemoment.
•Twoormorecouplesaresaidtobeequal
whentheyhavesamesenseormoment
•Moment=100*1=100N.m=10*10=100N.m
=50*2=100N.m
Prof. V.V.Nalawade 96
2 m
50 N
50 N
10 m
10 N
10 N
1 m
100 N
100 N

97. Properties of Couple
•Couplecanonlyrotatethebodybutcannot
translatethebody.
•Acoupledoesnothavemomentcentre,like
momentofforce.
Prof. V.V.Nalawade 97

98. Prof. V.V.Nalawade 9
8

99. Prof. V.V.Nalawade 9
9

100. 1.5 Moment of force:-
M= F x d
Unit = N.m or KN.m
Clockwise = +ve
Anticlockwise = -ve
Couple:-
M= F x d
Unit = N.m or KN.m
Clockwise = +ve
Anticlockwise = -ve
Varignon’s Theorem:-
Moment of resultant about any point A = Σof moments
of all the forces about same point A.
R x d = ΣM
Prof. V.V.Nalawade 1
0
0

101. 1.6 Resultant of parallel force
system:-
Step 1:-Find R = ΣF
Step 2:-Find ΣM @ O
Step 3:-Apply Varignon’s theorem
ΣM @ O = R. d
Step 4:-Find position of resultant
w.r.t. O
i)ΣF (upward) = Σmo (+ve)
ii)ΣF (downward) =Σmo (-ve)
1.7 Resultant of general force
system:-
Step 1:-Find ΣFx (Horizontal Component)
Step 2:-Find ΣFy (Vertical Component)
Step 3:-Find Magnitude of Resultant
Step 4:-Find Direction of Resultant
Step 5:-Find ΣM @O
Step 6:-:-Apply Varignon’s theorem
ΣM @ O = R. d
Step 4:-Find position of resultant w.r.t. O22
xy
R F F    1
tan
y
x
F
F





Prof. V.V.Nalawade 1
0
1

102. MethodofapproachtosolveCoplanar(2D)problems
Problem
2-D
Concurrent
NotEquilibrium
(Resultant)
1.ParallelogramLaw
2.Trianglelaw
3.Polygonlaw
4.Methodofprojections
Equilibrium
(Unknowns)
F=0x
F
y=0
Non-concurrent
NotEquilibrium
(Resultant)
1.ChooseareferencePoint
2.Shiftalltheforcestoapoint
3.Findtheresultantforceand
coupleatthatpoint
4.Reducetheforce-couple
systemtoasingleforce
Equilibrium
(Unknowns)
ΣF
x=0
ΣF
y=0
ΣM
z=0
Prof. V.V.Nalawade 1
0
2

103. Prof. V.V.Nalawade 1
0
3

104. Problem–2D-Concurrent-Resultant
Ex.1. Determinetheresultantofthefollowingfigure
Prof. V.V.Nalawade 1
0
4

105. Prof. V.V.Nalawade 1
0
5

106. Ex.2. Theresultantofthefourconcurrentforcesasshownin
FigactsalongY-axisandisequalto300N.Determinethe
forcesPandQ.
Problem–2D-Concurrent-Resultant
F
x
0
9
F
y
R300N
Prof. V.V.Nalawade 1
0
6

107. F
x
800380QSin45Psin500
F
y
QCos45PCos50R300
Solve quadratic equation and find
P = 511 N
Q =-40.3N
Problemsolution
Fx0
F
y
R300N
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108. Prof. V.V.Nalawade 1
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109. Determinetheresultantofthefollowingfigure
Problem–2D-NonConcurrent-Resultant
Problem
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110. ResultantofGeneralforcesinaplane–
Coplanarnon-concurrent
Step2:Shiftalltheforcestoapoint
Step3:Findtheresultantforce
Step4:Reduceresultantforce
Step1:Chooseareferencepoint
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111. Resultant–Non-concurrentgeneralforcesinaplane
Step:1:ChooseAasreferencePoint Step:2:ShiftallforcestopointA
Step:4:ReduceittoasingleforceStep3:Findresultantforceandcouple
x=1880/600
x=3.13m
Problemsolution:
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112. Determinetheresultantofthefollowingfigure
Problem–2D-NonConcurrent-Resultant
Problem
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113. Example:
Resultant–Non-concurrentgeneralforcesinaplane
Determinetheresultantforceofthenon-concurrentforcesasshownin
plate and distanceoftheresultantforcefrompoint O͛.
Step:2
Step:4
Step:3
Step:1
Problemsolution
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114. Continued……
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115. Prof. V.V.Nalawade 1
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116. Prof. V.V.Nalawade 1
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117. Prof. V.V.Nalawade 1
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118. Exercise 2
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119. Prof. V.V.Nalawade 11
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References
•Engg.Mechanics,Timoshenko&Young.
•2.Engg.Mechanics,R.K.Bansal,Laxmipublications
•3.EngineeringMechanics,Fedinand.L.Singer,Harper–Collins.
•4.EngineeringMechanicsstaticsanddynamics,ANelson,McGraHill
publications
•5.Engg.MechanicsUmeshRegl,Tayal.
•6.EngineeringMechanicsbyNHDubey
•7.EngineeringMechanics,statics–J.L.Meriam,6thEdn–WileyIndiaPvtLtd.
•8.EngineeringMechanics,dynamics–J.L.Meriam,6thEdn–WileyIndiaPvtLtd.
•9.MechanicsForEngineers,statics-F.P.Beer&E.R.Johnston5thEdnMcGraw
HillPubl.
•10.MechanicsForEngineers,dynamics-F.P.Beer&E.R.Johnston–5thEdnMc
GrawHillPubl.
•11.www.google.com
•12.http://nptel.iitm.ac.in/

120. Prof. V.V.Nalawade 12
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Thank YOU