LEC 11 (Rectangular 051)

8/11/2019 LEC 11 (Rectangular 051)

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13.2 The Equation of Motion

Free-Body diagram(Force Diagram)

Kinetic diagram

(acceleration Diagram)

aF m


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13.3 Equation of Motion for a System of

Particles

iiii m

m

afF

aF

G

iii

m

m

aF

aF

Internal forces cancel each other


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13.4 Equations of Motion: Rectangular

Coordinates When the net force is

projected to separate

coordinate axes theNewtons second law

still holds

)kji(kji

aF

zyxzyx aaamFFF

m

zz

yy

xx

maF

maF

maF


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Free Body Diagram Method

Draw each object separately

Draw all the forces acting on that object

Get xand ycomponents of all the forces tocalculate the net force

Apply Newtons second law to get

acceleration

Use the acceleration in any motion analysis

and establish a Kinetic Diagram

ma


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Normal & Frictional Force

F

Ff

FN

Action-Reaction forces

- mg = FN

mg


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Static Friction ( s )

Static frictionparallel force on the

surface when there is no relative motion

between the 2 objects

Static friction force can vary from zero to

Maximum

The coefficient of static friction

is material dependent.

Static

Ff= msFN

Dynamic

Ff= mkFN

Applied external force

Frict

ion


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Kinetic Friction ( k )

Kinetic frictionparallel force on thesurface when there is relative motion

between the 2 objects

Kinetic friction force is always the same

The coefficient of Kinetic friction

is material dependent.

NkFF f

StaticFf= msFN

DynamicFf= mkFN

Applied external force

Frictio

n


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Spring Force

Spring force

k : spring stiffness (N/m)

s : stretched or

compressed length

ksFs

olls

ol l

s


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m = 2 kg

y = 1m

smooth

a = ?

maFmg s sin

2219sin

9053cosx5.0x3coscos

m/s.m

ksga

N.ksFN sc

ma

Example 13-4

0cos sc FN

1353750

1tan

507507501

1

22

..

m...S


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x

ymg

N Nkm

Problem

NN

N

369

020cos81.940

;0 yF

xx maF mamgNk m sin

0cos mgN

2

/66.5

4020sin81.940)369(25.0

sma

a

x

x

a = ?


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Example 13-5

m A= 3 kg

m B= 5 kg

From rest

vB= ? In 2 second

yy maF AaT 32981

Block A

yy maF BaT 52.196

Block B

lss BA 2 02 BA 02 BA aa

taBo

Same

Sameshould be


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mg Ama

)/(62.42 smsaV AA

)2(12525.1226

12581.9*125

A

AA

aT

aTmaFy

NTsmasma

aa

aaaa

A

A

A

c

cc

c

c

1004)/(777.1)/(888.0

500200

250200622200981)12525.1226(25.849

)1()2(

22


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Review

Example 13.1

Example 13.2

Example 13.3

Example 13.4


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LEC 11 (Rectangular 051)

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