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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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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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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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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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Inert ial Reference Frame Consider the passenger who is strapped to the

seat of a rocket sled

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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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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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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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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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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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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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