Basics and statics of particles unit i - GE6253 PPT

ENGINEERING MECHANICS
Unit – I
Basics and Statics of Particles
by
S.Thanga Kasi Rajan
Assistant Professor
Department of Mechanical Engineering
Kamaraj College of Engineering & Technology,
Virudhunagar – 626001.
Tamil Nadu, India
Email : stkrajan@gmail.com

INTRODUCTION
02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 2

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.

Branches of Engineering Mechanics

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

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, stkrajan@gmail.com 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.02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 7

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.

Particle
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.

Rigid 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.

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

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
GM
gmgW
r
Mm
GF 

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 + 𝑄 cos 𝛼

Triangle Law
If 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
𝑅 = 𝑃 + 𝑄
𝑃 𝑄
𝑅

sine and cosine rules
When adding forces it is often useful to
consider solving the problem using geometric
rules, rather than considering components






The sine rule states that
 sinsinsin
XBA

The cosine rule states that
cos2222
BXXBA 
cos2222
AXXAB 
cos2222
ABBAX 

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.

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
17

Vector Representation of Forces
A vector is a quantity that has
both direction and magnitude.
18

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.
19

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 components
30°
20 N

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 N

Resolution and Composition of Forces
F
F cosӨ
F sinӨ
F cosӨF cosӨ
F sinӨ F sinӨ
Ө
Ө Ө
F sinӨ
F cosӨ
Ө
F
F
F
1. It is convenient to have
Fx = F cos Ө
Fy = F sin Ө
and Always measure angle
from horizontal
reference(acute angle).
2. Assume force pointing
Right and Top as positive
otherwise negative

Procedure to find Resultant Force
Procedure to find the magnitude and direction of the resultant force
1. Find ƩFx
2. Find ƩFy
3. Magnitude of the resultant force is given by
𝑅 = (Ʃ𝐹𝑥)2+(Ʃ𝐹𝑦)2
4. Plot ƩFx and ƩFy with its appropriate sign
5. Direction of the resultant force is given by
tan Ө =
Ʃ𝐹 𝑦
Ʃ𝐹𝑥

Problems

Problem No 1
Two forces act on a bolt at A. Determine their resultant.

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,
 35N98 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,
 35N98 R

• Trigonometric solution - Apply the triangle rule.
From the Law of Cosines,
       

155cosN60N402N60N40
cos2
22
222
BPQQPR
A
A
R
Q
BA
R
B
Q
A





20
04.15
N73.97
N60
155sin
sinsin
sinsin

N73.97R
From the Law of Sines,
 04.35

Sl. No. ƩFx ƩFy
1. + 40 cos20 + 40 sin20
2. + 60 cos 45 + 60 cos 45
80.01 56.10
𝑅 = (Ʃ𝐹𝑥)2+(Ʃ𝐹𝑦)2
𝑅 = (80.01)2+(56.10)2
R = 97.72 N
Magnitude of the resultant force
Direction of resultant force
Ʃ Fx = 80.01
ƩFy=56.01
Ө
tan Ө =
Ʃ𝐹𝑦
Ʃ𝐹𝑥
tan Ө = (
56.1
80.01
)
Ө = 35ᴼ

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

Problem 2
• Graphical solution - Parallelogram Rule
with known resultant direction and
magnitude, known directions for sides.
kN12.8kN18.5 21  TT
• Trigonometric solution - Triangle Rule
with Law of Sines




 105sin
kN25
30sin45sin
21 TT
kN12.94kN18.3 21  TT

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.
   30sinkN252T kN12.52 T
   30coskN251T kN21.71 T
 3090  60

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.

Problem 3
SOLUTION:
• Resolve each force into rectangular components.
• Calculate the magnitude and direction.

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.

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

02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com

Examples for Equilibrium
Cables AB and AC
carries the spool of
weight

Problem 4
Given: 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.

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.

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

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.

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
75.1
m4
m7
tan




56.20
375.0
m4
m1.5
tan



Problem 5
• Resolve the vector equilibrium equation into
two component equations. Solve for the two
unknown cable tensions.

   
   
 
 
  jT
iFT
R
iFF
iT
jTiT
jTiTT
ji
jiT
AC
DAC
DD
ACAC
ACACAC
AB









3009363.021.99
3512.066.173
0
N300
9363.03512.0
56.20cos56.20sin
N21.99N66.173
26.60cosN20026.60sinN200










Problem 5
 
  3009363.021.9900
3512.066.17300




ACy
DACx
TF
FTF
N98.35
N214.45


D
AC
F
T
 
  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

Problem 6
A 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.

Problem 6

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
cossin
cos
y
hy
y
hx
F
FF
F
FF





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 of
and are the direction
cosines for
F

F



zyx  cosand,cos,cos

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.

Problem 6
SOLUTION:
• Determine the unit vector pointing from A
towards B.
     
     
m3.94
m30m80m40
m30m80m40
222



AB
kjiAB

• Determine the components of the force.
  
     kji
kji
FF



N795N2120N1060
318.0848.0424.0N2500


 
kji
kji


318.0848.0424.0
3.94
30
3.94
80
3.94
40


















 


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




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.

Problem 7
Net Force acting at point A is

Problem 7

Problems for practice

References
1. 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).

Thank you
Any Queries contact
S.Thanga Kasi Rajan
Assistant Professor
Department of Mechanical Engineering
Kamaraj College of Engineering & Technology,
Virudhunagar – 626001.
Tamil Nadu, India
Email : stkrajan@gmail.com

Basics and statics of particles unit i - GE6253 PPT

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Basics and statics of particles unit i - GE6253 PPT