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ne cs

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of the relations

unbalanced forces

an e c anges nmotion that they

produce.

Newtons SecondNewton's First and Second laws in

aw o mo on

Latin, from the original 1687 editionof the Principia Mathematica.

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Introduction

According to Newtons Second Law, a particle will

accelerate when it is subjected to unbalanced Forces.

There are three approaches to the solution of kinetic

problems:

Direct application of Newtons Second Law

--- orce- ass- cce era on e o

Use of Work and Energy principles

Solution by Impulse and Momentum method

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NEWTONS LAWS OF MOTION

The motion of a particle is governed by Newtons three

laws of motion.

First Law: A particle originally at rest, or moving in a

straight line at constant velocity, will remain in this

state if the resultant force acting on the particle is zero.

Second Law: If the resultant force on the particle is not

same direction as the resultant force. This acceleration

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Third Law: Mutual forces of action and reaction

between two particles are equal, opposite, and collinear.

Force-Mass-Acceleration Method

Pre-re uisite knowled e

Free-Body Diagram

Kinematics of article motion

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Newtons Law of Motion

The first and third laws were used in develo in

the concepts of statics. Newtons second law

forms the basis of the stud of d namics.

,

can be written

=

whereF

is the resultant unbalanced force acting one par c e, an a s e acce era on o e par c e.

The positive scalar m is called the mass of the

par c e.

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Newtons Law of Motion

Second Law (law of acceleration):A particle

an acceleration a that has the same direction as

proportional to the force.

Equation of motion: F = ma

.

specified in kilograms (kg) and the acceleration in

Newton (N)

1N = 1kg x 1m/s2

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Newtons Law of Motion

Mass and Weight.

Mass (Unit: kg) is a property of matter by which we

can compare the response of one body with that ofanother. It is an absolute quantity since the

measurement can be made at any location.

Weight(Unit: N) of a body is not absolute. Its

follows the equation of motion, F = ma

W= mg(N)

= 2 .

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The Equation of Motion

From the free body diagram, the resultantof these

,

direction can be represented graphically on the.

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Equation of Motion: Rectangular

Coordinates , ,

Forces and acceleration expressed in term of, ,

F = ma

Fx i + Fyj + Fz k

= x y z

= xx maF

= yy maF

= zz maF

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PROCEDURE FOR THE APPLICATION OF THE

EQUATION OF MOTION

1 Select a convenient inertial coordinate s stem. Rectan ular

normal/tangential, or cylindrical coordinates may be used.

2 Draw a free-bod dia ram showin all external forces

applied to the particle. Resolve forces into their

appropriate components.

3) Draw the kinetic diagram, showing the particles inertialforce, ma. Resolve this vector into its appropriate

components.

4) Apply the equations of motion in their scalar componentform and solve these equations for the unknowns.

5) It may be necessary to apply the proper kinematic relations

to generate additional equations.

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lesson should be learned in the study of Engineering.

The same key purpose in Dynamics as it does in Statics.

1. Isolate the particle under consideration from all

con ac ng an n uenc ng o es

2. Replace the bodies by forces

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Exam le 1

-

which the coefficient of kinetic friction is k= 0.3. If

- ,

determine the velocity of the crate in 3 s starting

. .

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Exam le 1

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Exam le 1

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Exam le 1

Kinematics. Acceleration is constant, since

the applied force P is constant. Initial velocity is

zero, the velocity of the crate in 3 s is

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EXAMPLE 2Given: WA = 10 kN

W = 20 kN

voA = 2 m/s

= 0.2Find:vA when A has moved 4 meters.

Plan: Since both forces and velocity are involved,

s pro em requ res o e equa on o mo on

and kinematics. First, draw free body diagrams of A

an . pp y e equa on o mo on . s ng

dependent motion equations, derive a relationship

e ween aA an aB an use w e equa on o

motion formulas.

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EXAMPLE 2

=

WB mBaB

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EXAMPLE 2 continued

Free-body and kinetic diagrams of A:y

=

WA

T

mAaAx

F = m N

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EXAMPLE 2 continuedNow consider the kinematics.

sADatums

sB

B

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EXAMPLE continued

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Exam le 3Given:The 400 kg mine car is

The force in the cable is

F = 3200t2 N. The carhas an initial velocity of

v = 2 m/s at t = 0.

Find: The velocit when t = 2 s.

Plan: Draw the free-body diagram of the car andapply the equation of motion to determine the

acceleration. Apply kinematics relations to

determine the velocity.

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Exam le 3

1) Draw the free-body and kinetic diagrams of theo u on:

mine car:

W= mg m

=y

x

N

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Exam le 32) Apply the equation of motion in the x-direction:

3) Use kinematics to determine the velocity: