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ENGINEERING MECHANICSUnit – I
Basics and Statics of Particles
by
S.Thanga Kasi RajanAssistant ProfessorDepartment of Mechanical EngineeringKamaraj College of Engineering & Technology,Virudhunagar – 626001.Tamil Nadu, India
Email : [email protected]
S.ThangaKasiRajan, [email protected] 3
INTRODUCTION
To deal with the above situations, we need to know about
engineering mechanics
Mechanics is the foundation of most engineering
sciences and is an indispensable prerequisite to their study.
Mechanics is the science which describes and predicts the
conditions of rest or motion of bodies under the action of forces.
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Mechanics: The actions and effects of forces on bodies.
Mechanics
Statics: Bodies at rest, or in equilibrium
Dynamics: Bodies in motion, or out of equilibrium
IN EQUILIBRIUM Will be static, OR move with constant velocity
OUT OF EQUILIBRIUM Will accelerate (velocity changing)
Velocity=0
Velocity=constant
Velocity changing with time
Engineering Mechanics
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Kinematics: how fast, how far, and how long it takes...
Kinetics: What forces were involved to produce this motion?
Dynamics: Kinematics: Study of motion without reference to forces producing motion: Relations between position, velocity, acceleration and time.
Kinetics: Relations between unbalanced forces and the changes in motion they produce.
E.g. Rollercoaster ride:
- Weight- Friction- Aerodynamic drag
What are the resulting accelerations?02/01/2017 S.ThangaKasiRajan, [email protected] 6
Six Fundamental Concept of Mechanics• Space - associated with the motion or the position of a point P given
in terms of three coordinates measured from a reference point or origin.
• Time - definition of an event requires specification of the time and position at which it occurred.
• Mass - used to characterize and compare bodies, e.g., response to earth’s gravitational attraction and resistance to change in translational motion.
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Force
A force is a push or pull. An object at rest needs a force to get it moving; a moving object needs a force to change its velocity.
The magnitude of a force can be measured using a spring scale.
Six Fundamental Concept of Mechanics
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Six Fundamental Concept of MechanicsParticle
A particle has a mass but size is neglected.
When a body is idealised as a particle, the principles of mechanics reduces to a simplified form, since the geometry of the body will not be concerned in the analysis of the problem.
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Six Fundamental Concept of MechanicsRigid Body
A combination of large number of Particles in which all the particles remain at a fixed distance from one another before and after application of load.
Here mass & size of the bodies are considered when analysing the forces.
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S.ThangaKasiRajan, [email protected] 11
Basic Laws of Mechanics
1. Newton’s Law
2. Newton’s Law of Gravitation
3. Triangle Law
4. Parallellogram Law
5. Lami’s Theorem
6. Principle of Transmissibility
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S.ThangaKasiRajan, [email protected] 12
Basic Laws of Mechanics
• Newton’s First Law: If the resultant force on a particle is zero, the particle will remain at rest or continue to move in a straight line.
• Newton’s Third Law: The forces of action and reaction between two particles have the same magnitude and line of action with opposite sense.
• Newton’s Second Law: A particle will have an acceleration proportional to a nonzero resultant applied force.
amF
• Newton’s Law of Gravitation: Two particles are attracted with equal and opposite forces,
22 ,R
GMgmgWr
MmGF 02/01/2017
S.ThangaKasiRajan, [email protected] 13
Parallelogram Law
Parallelogram Law
If two vectors acting at a point be represented in magnitude and
direction by the adjacent sides of a parallelogram, then their resultant is
represented in magnitude and direction by the diagonal of the
parallelogram passing through that point.
𝑅=√𝑃2+𝑄2+2𝑃𝑄𝑐𝑜𝑠Ө
Ө= tan−1( 𝑄𝑠𝑖𝑛 (𝛼 )𝑃+𝑄cos (𝛼 ) )
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Triangle LawIf two vectors acting at a point are represented by the two sides
of a triangle taken in order, then their resultant is represented by the
third side taken in opposite order
𝑃 𝑄
𝑅02/01/2017
S.ThangaKasiRajan, [email protected] 15
sine and cosine rulesWhen adding forces it is often useful to
consider solving the problem using geometric rules, rather than considering components
The sine rule states that
sinsinsinXBA
The cosine rule states that
cos2222 BXXBA
cos2222 AXXAB
cos2222 ABBAX
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Principle of Transmissibility
According to this law the state of rest or motion of the rigid body is
unaltered if a force acting on the body is replaced by another force of
the same magnitude and direction but acting anywhere on the body
along the line of action of the replaced force.• Principle of Transmissibility -
Conditions of equilibrium or motion are not affected by transmitting a force along its line of action.NOTE: F and F’ are equivalent forces.
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17
Coordinate System
1. An origin as the reference point
2. A set of coordinate axes with scales and labels
3. Choice of positive direction for each axis
4. Choice of unit vectors at each point in space
Coordinate system: used to describe the position of a point in space and consists of
Cartesian Coordinate System
18
Vector Representation of Forces
A vector is a quantity that has both direction and magnitude.
19
Application of Vectors(1) Vectors can exist at any point P in space.
(2) Vectors have direction and magnitude.
(3) Vector Equality: Any two vectors that have the same
direction and magnitude, are equal no matter where in
space they are located.
Resolution and Composition of Forces
Resolution of force is a reverse process in which a single force is expressed in terms of its components. These components are sometimes referred to as the resolved parts of the force.
F
x
y
O X
Y OX = F cos
OY = F sin
Splitting of forces into their components30°
20 N
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Examples
X = -15 Cos 42 = -11.1 N
Y = 15 Sin 42 = 10.0 N
X = 35 Cos 62 = 16.4 N
Y = -35 Sin 62 = - 30.9 N
x
y
42°
15 N
x
y
62°
35 N
x
y
32°10 N
X = -10 Sin 32 = - 5.30 N
Y = -10 Cos 32 = -8.48 N02/01/2017
S.ThangaKasiRajan, [email protected] 22
Resolution and Composition of Forces
F
F cosӨ
F sinӨ
F cosӨF cosӨ
F sinӨ F sinӨ
Ө
Ө Ө
F sinӨ
F cosӨӨ
F
FF
1. It is convenient to have Fx = F cos ӨFy = F sin Ө
and Always measure angle from horizontal reference(acute angle).
2. Assume force pointing Right and Top as positive otherwise negative
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Procedure to find Resultant Force
Procedure to find the magnitude and direction of the resultant force1. Find ƩFx
2. Find ƩFy
3. Magnitude of the resultant force is given by
4. Plot ƩFx and ƩFy with its appropriate sign5. Direction of the resultant force is given by
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Problem No 1
Two forces act on a bolt at A. Determine their resultant.
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Problem No 1 (contd…)• Graphical solution - A parallelogram with sides
equal to P and Q is drawn to scale. The magnitude and direction of the resultant or of the diagonal to the parallelogram are measured,
35N 98 R
• Graphical solution - A triangle is drawn with P and Q head-to-tail and to scale. The magnitude and direction of the resultant or of the third side of the triangle are measured,
35N 98 R
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Problem No 1 (contd…)• Trigonometric solution - Apply the triangle rule.
From the Law of Cosines,
155cosN60N402N60N40cos222
222 BPQQPR
AA
RQBA
RB
QA
2004.15
N73.97N60155sin
sinsin
sinsin
N73.97R
From the Law of Sines,
04.3502/01/2017
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Problem No 1 (contd…)Sl. No. ƩFx ƩFy
1. + 40 cos20 + 40 sin202. + 60 cos 45 + 60 cos 45
80.01 56.10
R = 97.72 N
Magnitude of the resultant force Direction of resultant force
Ʃ Fx = 80.01
Ʃ F y =
56.
01
Ө
tan Ө = ()
Ө = 35ᴼ
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Problem 2
a) the tension in each of the ropes for = 45o,
b) the value of for which the tension in rope 2 is minimum.
A barge is pulled by two tugboats. If the resultant of the forces exerted by the tugboats is a 25 kN directed along the axis of the barge, determine
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Problem 2
• Graphical solution - Parallelogram Rule with known resultant direction and magnitude, known directions for sides.
kN 12.8kN 18.5 21 TT
• Trigonometric solution - Triangle Rule with Law of Sines
105sinkN 25
30sin45sin21 TT
kN 12.94 kN 18.3 21 TT02/01/2017
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Problem 2• The angle for minimum tension in rope 2 is
determined by applying the Triangle Rule and observing the effect of variations in .
• The minimum tension in rope 2 occurs when T1 and T2 are perpendicular.
30sinkN 252T kN 12.5 2 T
30coskN 251T kN 21.7 1 T
3090 60
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Problem 3
Knowing that the tension in cable BC is 725-N, determine the resultant of the three forces exerted at point B of beam AB.
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Problem 3SOLUTION:• Resolve each force into rectangular components.
• Calculate the magnitude and direction.
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Problems for Practice
Fig 1
Fig 2
A disabled automobile is pulled by means of two ropes shown in fig 1. Determine the Magnitude and direction of Resultant by (a) parallelogram law, (b) Triangle law (c) analytical method
Determine the magnitude and direction of resultant force shown in fig 2.
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S.ThangaKasiRajan, [email protected]
Equilibrium of a Particle
2 - 35
• When the resultant of all forces acting on a particle is zero, the particle is in equilibrium.
• Newton’s First Law: If the resultant force on a particle is zero, the particle will remain at rest or will continue at constant speed in a straight line.
• Particle acted upon by two forces:- equal magnitude- same line of action- opposite sense
• Particle acted upon by three or more forces:- graphical solution yields a closed polygon- algebraic solution
00
0
yx FF
FR
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Examples for Equilibrium
Cables AB and AC carries the spool of weight
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Problem 4Given: Sack A weighs 20 N. and
geometry is as shown.
Find: Forces in the cables and weight of sack B.
1. Apply Equilibrium condition at Point E and solve for the unknowns (TEG & TEC).
2. Repeat this process at C.
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Problem 4
+ Fx = TEG sin 30º – TEC cos 45º = 0
+ Fy = TEG cos 30º – TEC sin 45º – 20 N = 0
Solving these two simultaneous equations for the two unknowns, we get
TEC = 38.6 N
TEG = 54.6 N
Note that the assumed directions for the forces in the two cables EG and EC are tensile in nature.
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Problem 4
Fx = 38.64 cos 45 – (4/5) TCD = 0
Fy = (3/5) TCD + 38.64 sin 45 – WB = 0
Solving the first equation and then the second we get
TCD = 34.2 N and WB = 47.8 N .
Apply Equilibrium Condition
Now move on to the point C and consider equilibrium at C
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Problem 5
It is desired to determine the drag force at a given speed on a prototype sailboat hull. A model is placed in a test channel and three cables are used to align its bow on the channel centerline. 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.
SOLUTION:• Choosing the hull as the free body,
draw a free-body diagram.
• Express the condition for equilibrium for the hull by writing that the sum of all forces must be zero.
• Resolve the vector equilibrium equation into two component equations. Solve for the two unknown cable tensions.
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Problem 5
• Express the condition for equilibrium for the hull by writing that the sum of all forces must be zero.
0 DAEACAB FTTTR
SOLUTION:• Choosing the hull as the free body, draw a
free-body diagram.
25.60
75.1m 4m 7
tan
56.20
375.0m 4m 1.5
tan
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Problem 5• Resolve the vector equilibrium equation into
two component equations. Solve for the two unknown cable tensions.r
jT
iFT
R
iFF
iT
jTiT
jTiTT
ji
jiT
AC
DAC
DD
ACAC
ACACAC
AB
rr
r
rrrr
rrrrr
rrrr
3009363.021.99
3512.066.173
0
N 300
9363.03512.0
56.20cos56.20sin
N 21.99N 66.173
26.60cosN 20026.60sinN 200
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Problem 5
3009363.021.9900
3512.066.17300
ACy
DACx
TF
FTF
N 98.35N 214.45
D
AC
FT
jT
iFT
R
AC
DAC
3009363.021.99
3512.066.173
0
This equation is satisfied only if each component of the resultant is equal to zero
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Problem 6A sailor is being rescued using a boatswain’s chair that is suspended from a pulley that can roll freely on the support cable ACB and is pulled at a constant speed by cable CD. Knowing that a = 25ᴼ and b = 15ᴼ and that the tension in cable CD is 80 N, determine (a) the combined weight of the boatswain’s chair and the sailor, (b) in tension in the support cable ACB.
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Forces in Space
• The vector is contained in the plane OBAC.
F • Resolve into
horizontal and vertical components.
yh FF sin
F
yy FF cos
• Resolve into rectangular components
hF
sinsin
sin
cossincos
y
hy
y
hx
F
FF
FFF
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Forces in Space
• With the angles between and the axes,F
kji
F
kjiF
kFjFiFF
FFFFFF
zyx
zyx
zyx
zzyyxx
coscoscos
coscoscos
coscoscos
• is a unit vector along the line of action ofand are the direction cosines for
F
F
zyx cos and,cos,cos
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Problem 6
The tension in the guy wire is 2500 N. Determine:
a) components Fx, Fy, Fz of the force acting on the bolt at A,
b) the angles x, y, z defining the direction of the force
SOLUTION:• Based on the relative locations of the
points A and B, determine the unit vector pointing from A towards B.
• Apply the unit vector to determine the components of the force acting on A.
• Noting that the components of the unit vector are the direction cosines for the vector, calculate the corresponding angles.
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Problem 6SOLUTION:• Determine the unit vector pointing from A
towards B.
m 3.94
m30m80m40
m30m80m40222
AB
kjiAB
• Determine the components of the force.
kji
kji
FF
N 795N 2120N1060
318.0848.0424.0N 2500
kji
kji
318.0848.0424.03.94
303.94
803.94
40
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Problem 6• Noting that the components of the unit vector are
the direction cosines for the vector, calculate the corresponding angles.
kji
kji zyx
318.0848.0424.0
coscoscos
5.71
0.32
1.115
z
y
x
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Problem 7
A crate is supported by three cables as shown. Determine the weight of the crate knowing that the tension in cable AB is 3750 N.
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References1. Ferdinand P Beer & E.Russell Johnston “VECTOR MECHANICS FOR ENGINEERS STATICS & Dynamics”, (Ninth Edition) Tata McGraw Hill Education Private Limited, New Delhi.2. Engineering Mechanics – Statics & Dynamics by S.Nagan, M.S.Palanichamy, Tata McGraw-Hill (2001).
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S.ThangaKasiRajan, [email protected] 57
Thank you
Any Queries contact
S.Thanga Kasi RajanAssistant ProfessorDepartment of Mechanical EngineeringKamaraj College of Engineering & Technology,Virudhunagar – 626001.Tamil Nadu, India
Email : [email protected]
02/01/2017