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5/22/2018 Dynamic
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Co ri ht 2010 Pearson Education South Asia Pte Ltd
Engineering Mechanics:
Dynamics in SI Units, 12e
Chapter 13
Kinetics of a Particle: Force and Acceleration
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Chapter Objectives
State Newtons Laws of Motion and Gravitational
attraction and to be able to define mass and
weight
Analyze accelerated motion of a particle using the
equation of motion
Investigate central-force motion and apply it to
problems in space mechanics
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Chapter Outline
1. Newtons Second Law of Motion
2. The Equation of Motion
3. Equation of Motion for a System of Particles
4. Equations of Motion: Rectangular Coordinates5. Equations of Motion: Normal and Tangential
Coordinates
6. Equations of Motion: Cylindrical Coordinates
7. *Central-Force Motion and Space Mechanics
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13.1 Newtons Second Law of Motion
Second Law:A particle acted upon by an unbalanced force F
experiences an acceleration athat has the same
direction as the force and a magnitude that is
directly proportional to the force.
Newtons Law of Gravitational Attraction
A law governing the mutual attractivegravitational force acting between them
221
r
mmGF
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Co ri ht 2010 Pearson Education South Asia Pte Ltd
13.1 Newtons Second Law of Motion
Newtons Law of Gravitational Attraction Massis a property of matter
Mass of the body is specified in kilograms Weight is calculated using the equation of
motion, F= ma
mgW
W= mg(N)
(g= 9.81 m/s2)
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Co ri ht 2010 Pearson Education South Asia Pte Ltd
13.2 The Equation of Motion
Equation of motion is written as Consider Pof mass m subjected to the action of
two forces, F1and F2
From free body diagram, the
resultantof these forces
producesthe vector ma
Represented graphically
on the kinetic diagram FR= F= 0, acceleration is zero
Such a condition is called static equilibrium,
Newtons First Law of Motion
maF
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Co ri ht 2010 Pearson Education South Asia Pte Ltd
13.2 The Equation of Motion
Inert ial Reference Frame Acceleration of the particle is measured with
respect to a reference frame that is either fixed
or translates with a constant velocity
Such a frame of reference is known as a
Newtonian or inertial reference frame,
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13.2 The Equation of Motion
Inert ial Reference Frame Consider the passenger who is strapped to the
seat of a rocket sled
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Co ri ht 2010 Pearson Education South Asia Pte Ltd
13.3 Equation of Motion for a System of Particles
The free body diagram for the ith particle areshown. Applying equation of motion yields
F= ma; Fi+ fi= miai
If equation of motion is applied to each of the
other particles, these equations can be added
together vectorially,
Fi+ fi= miai
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Co ri ht 2010 Pearson Education South Asia Pte Ltd
13.3 Equation of Motion for a System of Particles
The summation of internal forces will be equal tozero where
Fi= miai
If rGis a position vector which locates the center
of massGof the particles, we have
mrG= miri
Differentiating twice w.r.t time yields
maG= miai
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13.3 Equation of Motion for a System of Particles
Therefore, F= maG
The sum of the external forces acting on the
system of particles is equal to the total mass of
the particles times the acceleration of its center
of mass G
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13.4 Equation of Motion: Rectangular Coordinates
When a particle is moving relative to an inertialx,y, zframe of reference,
F= ma
Fxi+ Fy
j+ Fz
k= m(ax
i+ ay
j+ az
k)
The three scalar equations:
zz
yy
xx
maF
maFmaF
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13.4 Equation of Motion: Rectangular Coordinates
Procedure for AnalysisFree-Bod y Diagram
Select inertial coordinate system
Draw particles free body diagram (FBD) and
provides a graphical representation thataccounts for all forces (F)
Direction and sense of the particles accelerationais also be established
Acceleration is represented as mavector on thekinetic diagram
Identify the unknowns in the problem
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13.4 Equation of Motion: Rectangular Coordinates
Procedure for AnalysisEquat ion o f Mot ion
Apply equations of motion on FBD in their scalarcomponent form
Cartesian vector analysis can be used for thesolution
Kinemat ics
Apply kinematics equations once the particlesacceleration is determined from F= ma
If acceleration is a function of time, use a = dv/dtand v = ds/dt
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13.4 Equation of Motion: Rectangular Coordinates
Procedure for AnalysisKinemat ics
When acceleration is a function of displacement,
integrate a ds = v dv to find velocity as a function
of position
If acceleration is constant, use
tavv c 02
002
1tatvss c
020
2 2 ssavv c
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The 50-kg crate rests on a horizontal plane for whichthe coefficient of kinetic friction is k= 0.3. If the
crate is subjected to a 400-N towing force, determine
the velocity of the crate in 3 s starting from rest.
Example 13.1
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SolutionFree-Body DiagramWeight of the crate is W = mg= 50 (9.81) = 490.5 N.The frictional force is F = kNCand acts to the left,
There are 2 unknowns, NCand a.
Equations of Motion
Solving we get
Example 13.1
030sin4005.490;
503.030cos400;
Cyy
Cxx
NmaF
aNmaF
2/19.5,5.290 smaNNC
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Solution
Kinematics
Acceleration is constant.
Velocity of the crate in 3s is
Example 13.1
sm
tavv c
/6.15
)3(19.50
0
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The baggage truckAhas a weight of 3600 N andtows a 2200 N cart Band a 1300 N cart C. For a
short time the driving frictional force developed at
the wheels is FA= (160t) N where tis in seconds. If
the truck starts from rest, determine its speed in 2seconds. What is the horizontal force acting on the
coupling between the truck and cart Bat this
instant?
Example 13.3
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Solution
Free-Body Diagram
We have to consider all 3 vehicles.
Equations of MotionOnly horizontal motion is considered.
Example 13.3
ta
at
maF xx
221.0
81.9
130022003600160
;
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SolutionKinematics
The velocity of the truck is obtained using a = dv/dt
with the initial condition that v0= 0 at t= 0,
Free-Body Diagram
Equations of MotionWhen t = 2 s, then
Example 13.3
smtvdttdvv
/442.01105.0;)221.0(2
0
22
00
NTT
maF xx
8.157)2(221.081.9
3600)2(160
;
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The 100-kg blockAis released from rest. If themasses of the pulleys and the cord are neglected,
determine the speed of the 20-kg block Bin 2 s.
Example 13.5
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