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King Fahd University of Petroleum & Minerals Mechanical Engineering Dynamics ME 201 BY Dr. Meyassar N. Al-Haddad Lecture # 11

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King Fahd University of Petroleum & Minerals. Mechanical Engineering Dynamics ME 201 BY Dr. Meyassar N. Al-Haddad Lecture # 11. Sir Isaac Newton 1642-1727. Proposed fundamental laws that are the basis of modern mechanics 3 laws of motion law of gravitation. Newton ’ s laws. - PowerPoint PPT Presentation

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Page 1: King Fahd University of Petroleum & Minerals

King Fahd University of Petroleum &

MineralsMechanical Engineering

Dynamics ME 201BY

Dr. Meyassar N. Al-HaddadLecture # 11

Page 2: King Fahd University of Petroleum & Minerals

Sir Isaac Newton1642-1727

Proposed fundamental laws that are the basis of modern mechanics

3 laws of motionlaw of gravitation

Page 3: King Fahd University of Petroleum & Minerals

Newton’s lawsLaw of inertia a body in motion will stay in motion and a body at rest will

stay at rest unless acted upon by a net external force.

Law of force-acceleration A particle acted upon by an unbalanced force F experiences an acceleration a that has the same direction as the force and a magnitude that is directly proportional to the force

Law of action-reaction for every action, there is an equal and opposite reaction mg = FN

Law of gravitation - all bodies are attracted to one another with a force proportional to the product of their masses and inversely proportional to the square of the distance between them.

221

rmmGF

amF

Page 4: King Fahd University of Petroleum & Minerals

13.2 The Equation of Motion

• Free-Body diagram (Force Diagram)

• Kinetic diagram (acceleration Diagram)

aF m

Page 5: King Fahd University of Petroleum & Minerals

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

Page 6: King Fahd University of Petroleum & Minerals

13.4 Equations of Motion: Rectangular Coordinates

• When the net force is projected to separate coordinate axes the Newton’s second law still holds

)kji(kji

aF

zyxzyx aaamFFF

m

zz

yy

xx

maF

maF

maF

Page 7: King Fahd University of Petroleum & Minerals

Free Body Diagram Method

•Draw each object separately

•Draw all the forces acting on that object

•Get x and y components of all the forces to calculate the net force

•Apply Newton’s second law to get acceleration

•Use the acceleration in any motion analysis and establish a Kinetic Diagram

ma

Page 8: King Fahd University of Petroleum & Minerals

Normal & Frictional Force

F

Ff

FN

Action-Reaction forces -mg = FN

mg

Page 9: King Fahd University of Petroleum & Minerals

Static Friction ( s )

• Static friction – parallel 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.

StaticFf = sFN

DynamicFf = kFN

Applied external force

Frict

ion

Page 10: King Fahd University of Petroleum & Minerals

Kinetic Friction ( k )

• Kinetic friction – parallel force on the surface 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 μf

StaticFf = sFN

DynamicFf = kFN

Applied external force

Frict

ion

Page 11: King Fahd University of Petroleum & Minerals

Spring Force

• Spring force

• k : spring stiffness (N/m)• s : stretched or

compressed length

ksFs

olls

ol l

s

Page 12: King Fahd University of Petroleum & Minerals

• m = 2 kg• y = 1m• smooth• a = ?

maFmg s sin

2219sin

9053cosx5.0x3coscos

m/s.θmksga

N.θksθFN sc

ma

Example 13-4

0cos sc FN

13537501tan

507507501

1

22

..

θ

m...S

Page 13: King Fahd University of Petroleum & Minerals

x

ymg

N Nk

Problem

NNN

369020cos81.940

;0 yF

xx maF mamgNk sin

0cos mgN

2/66.5

4020sin81.940)369(25.0

sma

a

x

x

a? =

Page 14: King Fahd University of Petroleum & Minerals

Example 13-5

m A= 3 kgm B= 5 kgFrom restvB= ? In 2 second

yy maFAaT 32981

Block A

yy maFBaT 52.196

Block B

lss BA 2 02 BA 02 BA aa

taBo

Same

Same ٍshould be

Page 15: King Fahd University of Petroleum & Minerals

VA? =

2Sc + SA = L 0 = 2ac + aA aA = -2ac

ca

ccx

y

amgTNmaF

NNNF

20030sin25.0:

1699030cos81.9*200:0

Problem

Page 16: King Fahd University of Petroleum & Minerals

mg Ama

)/(62.42 smsaV AA

)2(12525.1226

12581.9*125

A

AA

aT

aTmaFy

NTsmasma

aaaa

aa

A

A

A

c

cc

c

c

1004)/(777.1)/(888.0

500200250200622

200981)12525.1226(25.849)1()2(

22

Page 17: King Fahd University of Petroleum & Minerals

Review

• Example 13.1• Example 13.2• Example 13.3• Example 13.4

Page 18: King Fahd University of Petroleum & Minerals