Topics in textbook

1

Topics in textbook

Actual Work Definition of Work Work on Particle

Potential Energy Definition of Energy Potential Energy

Kinetics Energy Definition of Energs

Work-Energy Equation

2

Three approaches for solving dynamics1) Direct Method

1 F am

2) Work and Energy

Kinematics Eq:( , , , )i iF F a v s t

2 22 2

( ) ( )

1 1 2 2

(change of kinetics energy)

t tm a dr m v dv

mv mv

2 1

( ) ( )

( )

( )

m a dt m dv

d mv mv mv

linear momentum

iF ma

3) Impulse and Momentumlinear impulse of Force i

work (and potential energy) of Force i along the path

Newton’s2nd Law

From 2nd Law(kinetics Eq)

From 2nd Law

From 2nd Law

( , , ) , a v s t v s

A

1F

2F

dr

3F

path

( )ii

F dr m a dr

( )iis s

F dr m a dr

( )iis s

F dt m a dt

( ) ii

F dt m a dt

( )ii s

F dr m a dr

3

Work and Energy

( )t tm v dvWork of Force i along the path

( )ii s

F dr m a dr

From 2nd Law

iF ma Newton’s

2nd Law

Principle of work and Energy21

2T mv

A

1F

2F

dr

3F

path1

Work of a force during small displacement

iF

i idU F dr

1 0dU 2 0dU 3 0dU

kinetic energy

def

i is

U F dr

B A B AT T T

cosi iF ds

( ) cosi iF ds A B B AU T T T

Usually convenient when F = F(s), and you want to find velocity at final state(without finding acc. first).

( ) ( )tm a dr m a ds

change of kinetic energy

kinetic energy at A

kinetic energy at B

B2 21 1

2 2B Amv mv

5

Work done on Particle

P

iF

2F

dr

3F

Work by a force iF

i idU F dr

path

1 0dU

2 0dU

3 0dU

idU ( ) iF dr

Work done over particle A

i

sum of works done by all forces over the particle A

( )iF dr

1F

( ) ( ) cos i iF dr

( ) cos ( ) i iF dr

( ) cos idr

( ) cos iF

Since , the total work done on object is ……

(inactive force)

F ma

6

is positive when

Note on work

1F

dr

3F

path

i idU F dr

(cos ) and i itF F s

2F

Active force is the force that does the work

idU

has the same direction.

Reactive force = constrain force that does not do the work

Unit of work is N-m or Joule (J).

The 10-kg block rest on a smooth incline. If the spring is originally stretched 0.5 m, determine the total work done by all forces acting on the block when a horizontal force P = 400 N pushes the block up the plane s = 2 m. The block is not tipping.Horizontal Force P: constant

( cos30 )(2) 692.8oPU P J

Spring Force Fs.: varying Force

))(( dxFUB

A sFs

Weight W: constant.

(2sin 30 ) 98.1WU mg J

Normal Force NB : constant

0BNU

2

0

)5.0( dxxk

A (spring stretched length is 0.5m)

B

Jxk 90)5.0(21 2

0

2

JUOverall

7.504 901.988.692

“Inactive Force”

“Active Force”

Pos A

Pos B

Work done on Particle

P

4F

2F

dr

3F

pathidU

( ) iF dr

Work done over particle A

sum of works done by all forces over the particle A

( )iF dr

1F

( ) ma dr

Pos A

Pos B B

A BA

U dU

B

A

m vdv

a

t

B

tA

m a ds

2 21 12 2B Amv mv 21

2

B

A

mv

BT AT

BT

Work done on particle P during path A->B,is to increase kinetic energy of particle

AT

212

def

T mv

B

A

m a dr

12

Kinetics Energy

Principle of work and Energy

212

T mv A

2F

3F

path

A B B AU T T

T is the work done on a particle to accelerate it from rest to the velocity v

( )A A B BT U T

Unit of T is N-m or Joule (J)

1F

No need to find acceleration first

it can be applied to system of particles with

frictionless and non-deformable links

Advantage

Get change in velocity directly from active forces.

Scalar equation. (1 unknown)

Integral Equation (not instantaneous eq like 2nd Law)

15

18

19

How to calculate Work

iF

dr

3F

path

i i is s

U dU F dr displacement x

force component in the direction of displacement

In general

i i i ix y zU F dx F dy F dz

In xyz-coord

i i tU F ds

In nt-coord

1 i idU F dr

ˆˆ ˆ( ) ( ) ( )dr dx i dy j dz k

cos i iF ds

displacement in the direction of forcex force component

( ) cosi iF ds

2F

ˆˆ ˆi i ix y z

F F i F j F k

make sure that the direction of

is positive according to +s direction.i t

F

make sure that the direction of

is positive according to direction.i i ix y z

F F F

x y z

i rU F dr F rd

In r-coord ˆ ˆ( ) ( )rdr dr e rd e ˆ ˆr rF F e F e

scalar(be carefulOf +/-)

F=8N (const)

Does not do the work

M3/107) Calculate the work done on 10-kg object with the constant Force ( F= 8N ) during the curve path AB.

0.75

0

8 8(0.75) Jdx ,F A BU F dr

N

mg

x

y

ˆ8 (const)F i

x-y

, ( )(0.375) 36.7875 Jmg A BU mg

, 0N A BU

, , ,

42.7875 JA B F A B mg A B N A BU U U U

AnsIf F is not constant, how to calculate it?

ˆ ˆ dr dx i dy j

ˆmg mg j

0

0.375

mg dy ,mg A BU mg dr

(0.375)mg

2 2ˆ ˆ( ) ( )F x y i x y j

If F is not constant

,F A BU F dr

2 2( ) ( ) B

A

x y dx x y dy

20.375 0.667y x

2

2

( (0.375 0.667 )

0.375 0.667

B

A

B

A

x x dx

y y dy

21

ˆ ˆ( , ) ( , )x yF F x y i F x y j

More general case

,F A BU F dr

( , ( )) ( , ( )) B

x yA

F x f x dx F x f x dy

B

x yA

F dx F dy

( , ( )) ydyF x f x dxdx

( )y f x 1( ( ), ( ))yF f y f y dy

or

x

y

M3/107) Calculate the work done by F during the curve path AB.

ˆ( ) ( )t tF s F s e

,F A BU F dr

s

engine thrust

n-t

( )nF sˆ( )n nF s e

Fn doest not effect works!

( ) 2F t s

2 costF s

2 sinnF s

2 coss ds

( )tF s ds

23

M3/107) Calculate the work done by F during the curve path AB.

ˆ ˆ( ) rdr e rd e

(const)F

central force0d

r- reference point

CF ˆ C rF F e

ˆ ˆ ˆ ˆ0 ( ) ( ) r rFe e dr e rd e

B B

C CA A

C B A

F dr F dr

F r r

ˆC rF e

( )C A BF r r

,F A BU F dr

r-

Ar

Br

ˆ

ˆ ˆ ˆ ˆr

r r

r re

dr dr e r d e dr e r e

25

M3/121) The 0.2-kg slider moves freely along the fixed curved rod from A to B in the vertical plane under the action of the constant 5-N tension in the cord. If the slider is released from rest at A, calculate its velocity v as it reaches B.

A B B AU T T

Work-Energy Eq.

A B F mg NU U U U

F

B

FA

U Fdr

r- coordinatereference point

2 25( 0.6 0.25 0.15) 2.5

ˆre

( )B

A

r

B Ar

F dr F r r

212 Bmv

(0.25) 0.4905 JmgU mg

2.5 0.4905 2.0095 J

Does not do the work

N

mg F0general

position

0

What does it mean if 0?A BU

2

2 2.0095 4.4827450.2

A BB

Uv

m

( )A BF r r

33

Work on frictionless connected particles

internal force R and –R will have the same displacement.

So, the sum of these works are zero.

Only the external forces are needed to calculate the total work on a system of particles.

(If frictions exist, the sum of action and reaction of the friction may not be zero.)

34

The system starts from rest at Configuration 1. Find the velocity of A at configuration where d = 0.5 m . F is 20 N (constant)

System selection is not so good(you have to calculate Tension T for its work)

1 2A AU T

0.5

0A AFd m gd Tdy T

1 2B BU T

0.5

0B Bm gd Tdy T

initial state Final state

1 2sys AU T

2 21 12 2A B A A BFd m gd m gd T m v m v

N causes no work!

Am g

Bm gAm g Bm g

Am gBm g

35

M3/131) The ball is released from position A with a velocity of 3m/s and swings in a vertical plane. At the bottom position, the cord strikes the fixed bar at B, and the ball continues to swing in the dashed arc. Calcuate the velocity v of the ball as it passes position C.

Work-Energy Eq.

A C mg T A CA CU U U

( ) (1.2cos60 0.8)

( 0.2) (0.2)

omg A BU mg h h mg

mg mg

A C C AU T T 212 Amv21

2 Cmv

T

mg

system

2 22 1 12.924 3.595 m/s2C mg Av U mv

m

does no work

37

Power Power is defined as time rate of work

For a machine, power tells how much work it can do in a period of time.

(small machine can deliver lots of energy given enough time)

A

1F

2F

dr

3F

path

dUPdt

d F rF v

dt

Unit of power: Watt (W) = J/s = N-m/s

(scalar quantity)

38

Mechanical Efficiency


Since machine consists of moving parts which may have frictions, so extra energy or power is needed to overcome the frictions.

power outputpower input

useful

energy outputenergy input

If energy applied to the machine occurs during the same time interval at which it is removed.

1

outputP

F v

39

A car has a mass of 2 Mg and an engine efficiency of = 0.65. The car uniformly accelerates at 5 m/s2, starting from rest. During that constant acceleration, the wind outside creates a drag resistance on the car of FD = 1.2v2 N, where v is the velocity in m/s. Find the engine output input when t=4 s.

[ : ]x xF ma 21.2CF v ma

x

21.2CF ma v

outputP F v

v at

10480 N

Constant acceleration:

10480 20 209.6 kW

4(5)(4) 20 m/s

tv

a

4 44output t ttP F v

output

input

PP

4

209.6 322.46 kW0.65

outputinput t

PP

(constant)5 a

50N

A 50-N load (B) is lifted up by the motor from rest until the distance is 10 m. The motor M has an efficiency of 0.76 and exerts a constant force of 30 N. Find the power supplied to the motor at that instant. Neglect the mass of the pulleys and cable.

s = 10 m(start from rest)

F = 30 N (const)

v=?

outputP F v

output

input

PP

2F= 2(30)

50 N

B

2 50F ma F ma

2 50 9.8150

9.81

Fa

2 2 2v u as 2(9.81)10v

21(2 )( )2

2 26.26

A B B AU T T

F s mgs mv

Fs mgsv

m

2M PL s C 2 2 2PM M P B

dsdv L v vdt dt

(30)(2 6.26) 375.6 W

375.6 494.2 W0.76

outputinput

PP

Ps

Energy Approach

42

Work and Energy

A

1F

2F

dr

3F

path

A B B AU T T kinetic energy at A

kinetic energy at B

B2 21 1

2 2B Amv mv

1 ,F A BU 2 ,F A BU 3 ,F A BU

summation of all forces

Work from spring

Work fromGravity ForceWork from all

other forces(not spring & gravitation)

We found that ….

It is much easier to solve dynamic problem, if we think the work done by spring and gravity force in the form of Potential Energy

Elastic Potential Energy

Gravitational Potential Energy

Energy “Emission”: from position 1 to position 2

Work of Gravity Force ( )WdU W dh

W=mg

Fixed reference line

h

,1 2 1 2( )WU mg h h

2

12 1 2 1( ) ( )mg h h mgh mgh

2h1h

Only depends on position at

final state (2)


initial state (2)

Work done by W , only depends on the initial state position and final state position only, i.e. , it does not depends on actual path

any path

Think in Term of “Potential Energy”(for convenience)

,1gV

,2gV

Potential Energy - Energy from gravity field

1 2 ,1 ,2 g gh h V V

def

gV mgh

1 2, W anyPathU

Work done by Gravity Force: from position 1 to position 2

,2 ,1 :

Change in Internal Energy: + (increasing)g gV V

U V

> 0

energy level (lower)

energy level (higher)

,2 ,1 ( )g gV V

VWork = “Energy in Transfer”

point function

44

when change in g is significantDefine as negative of work done from the position to

2earth

g

mg RV

r

gV r

the potential energy at r is

from

1 2 to r rgV

2 2

2 1g

mgR mgRVr r

2earthGMF m

r

2 9.81earth

earth

GMgR

2 2

2( )gr

mgR mgRV r drr r

0earthg

earth

GMVR

sF

Work of Spring Force

sF sdU F dL

L

2

,1 21

( )sF oU k L L dL

1

2

2

1

21 ( )2

L

oL

k L L

1L


final state (2)


initial state (1)

Work done by Spring , depends only on the initial state and final state only, i.e. , it does not depends on actual path

any path

Think in Term of “Energy” (for convenience)

1E

2E

212

def

eV kx

2L

x : distance , stretched or compressed from natural length

1 2, sF anyPathU ,1 ,2( )e eV V

Energy Emission: from position 1 to position 2

,2 ,1 ( )e eV V

Work done by Spring Force: from position 1 to position 2 U V

oL natural length(unstretched length)

1 2

,1 ,2

o o

e e

L L L LV V

2 22 1

1 1( ) ( )2 2o ok L L k L L

point function

Work-Energy EquationFBD

A BU T

Virtual work by non-conservative forces.

Energy Concept

* A B A BU U T

A B B AU V V V

* A BU V T

(Conservative Force)Think of Energy



FBD **(Use Energy Concept)

N

(1st Form)

(2nd Form)

Not Recommended Method in this course

N

A BU T

* A BU V T



(1st Form)

(2nd Form)

= E

FBD

(Conservative Force)Think of Energy

N

49

M3/173) The 0.6-kg slider is released from rest at A and slides down under the influence of its own weight and of the spring of k = 120 N/m. Determine the speed of the slider and the normal force at point B. The unstrecthed length of the spring is 200 mm.

*A BU V T

gravitational potential datum

2 2 2 21 1 1 12 2 2 2A A A B B Bmgh kx mv mgh kx mv

(0.25 0.2)BF k

BN

mg

22y x

32 2

2

2

1 dfdx

d ydx

5.9234 m/sBv

84.09 NBN

0

1 m4x

B s nN mg F m a

5.866mg

N F

0.5Ah 2 2( 0.5 0.25 0.2)Ax

A A B BV T V T

(0.25 0.2)Bx

At position B

nma

tma

2Bv

3

2 21 (4 )

4

x

*, 0A B N A BU U

57

Get change in velocity directly. (No need to find acceleration first)

it can be applied to system of particles with

frictionless and non-deformable links

Advantage

Handle with only active forces.

Scalar equation. (easy to handle with1 unknown)

Integral Equation (not instantaneous equation like 2nd Law)

We will see this later, when applying at system of particles

58

Work on frictionless connected particles

AB C

O

60

The system starts from rest at Configuration 1. Find the velocity of A at configuration where d = 0.5 m . F is 20 N (constant)

We have no interest in T, thus object separation(separating object A and B) is not good in this problem.

*,1 2A A AU V T

0.5

0A AFd Tdy V T

initial state Final state

*,1 2sys sys sysU V T

Am g

F

T

Bm g

T

Object A

*,1 2B A AU V T

0.5

0A ATdy V T

Am g

F

T

T

Bm g

A AFd V T

Object B

xN

yN

syste

m

F T is internal force (excluding from Work Calculation)

64

datum

*,sys A BU V T

M3/158) If the system is released from rest, determine the speeds of both masses after B have move 1 m. Neglect friction and the masses of pulleys.

System: block A + block B + cord+ 2 Pulleys

Position A: at rest

Position B: block B moves down as 1 meter

BSAS

13 2A BL S S C

3 2 0A Bv v

3 2 0A Ba a

(assume)

1Bs 2 23 3A Bs s 2

3A Bv v

221 2 110.9924 0

2 3 2A B B Bm v m v

1Bh 2 sin 203

oAh

unsolvable

2 21 10 0 02 2A A B B A A B Bm g h m g h m v m v

up

datum

*,sys A BU V T

M3/158) If the system is released from rest, determine the speeds of both masses after B have move 1 m. Neglect friction and the masses of pulleys.


Position A: at rest

Position B: block B moves down as 1 meter

BSAS

13 2A BL S S C

3 2 0A Bv v

3 2 0A Ba a

(assume)

1Bs 2 23 3A Bs s 2

3A Bv v

221 2 110.9924 0

2 3 2A B B Bm v m v

1Bh 2 sin 203

oAh

2 21 10 0 02 2A A B B A A B Bm g h m g h m v m v

up

2 0.85285 Bv 0.85285 Bv

0.61566 Av

0.85285 Bv

67

Bs

20 N

50 N

*,Sys A BU E

2A BL s s

2 21 12 ( 2) (1) ( )2 2k A B BN d m m m g d

0.3k

H14/16) Block A rest on a surface which has friction. Determine the distance d cylinder B must move down so that A has a speed of starting from rest.

As

2 0A Bv v

: 2 1A Bv v

2 m/sAv

(4 )12 ( 2 )

A B

B k A

m mdm g m g

0.1744 m

0 2A Bs s

: 2B As d s d


2 21 1 ( )2 2A A B B BE m v m v m g d

*, ( )(2 )Sys A B kU N d

Position A: at restPosition B: block B moves down as d meter

68

3/168) The system is released from rest with =180, where the uncompressed spring of stiffness k= 900 N/m is just touch the underside of 4-kg collar. Determine the angle corresponding to the maximum spring compression.

*,sys A BU V T

A A B BV T V T

O2-1 O2-2O1

System: O1+O2+O3+4 rods

Position A: at rest with =180

Position B: maximum compression* * ( 0 for all object)v

1,1802(0.2)sin 0.4

2Ah

1 2 sin2

h r

2 ( )sin2

h r L

rL

r

/ 2/ 2

2,180(0.2 0.3)sin 0.5

2Ah

2 21 1, 2 2, 1 1, 2 2,

1 1( ) 2 ( ) ( ) 2 ( )2 2A A A B B Bm g h m g h kx m g h m g h kx

*

1, 2(0.2)sin2Bh

*

2, (0.2 0.3)sin2Bh

*

2(0.2)sin 2(0.2)2Bx

2* * *14(9.81)(0.4) 1 sin (2)(3)(9.81)(0.5) 1 sin 900 0.4 1 sin

2 2 2 2

* * 45.131 sin 0 or 1 sin2 2 72

* 0 or 43.8o

datum/ 2

70

Power

Power is defined as time rate of work

For a machine, power tells how much work it can do in a period of time.

(small machine can deliver lots of energy given enough time)

A

1F

2F

dr

3F

path

dUPdt

d F rF v

dt

Unit of power: Watt (W) = J/s = N-m/s

(scalar quantity)

71



Since machine consists of moving parts which may have frictions, so extra energy or power is needed to overcome the frictions.

useful power outputpower input

energy outputenergy input

If energy applied to the machine occurs during the same time interval at which it is removed.

1

outputP

F v

50N

A 50-N load (B) is lifted up by the motor from rest until the distance is 10 m. The motor M has an efficiency of 0.76 and exerts a constant force of 30 N. Find the power supplied to the motor at that instant. Neglect the mass of the pulleys and cable.

outputP F v

output

input

PP

2F= 2(30)

50 N

2 50F ma F ma

2 50 9.8150

9.81

Fa

2 2 2v u as 2(9.81)10v

* ( ) ( )A B B A B AU V V T T

2M PL s C 2 2 2PM M P B

dsdv L v vdt dt

(30)(2 6.26) 375.6 W

375.6 494.2 W0.76

Ps

Energy Approach

( 0.76)

outputinput

PP

? (30)(6.26)outputP No!

21( )(2 20)2

2 26.26

F mgs mv

Fs mgsv

m

v=?

F = 30 N (const)

s = 10 m(start from rest) F = 30 N

(const)

50 N

xN

yN

74

Summary Make sure you write FBD (no FBD, no score)

Equation itself is not hard to solve, but calculating work may be more difficult than you thought.

*A BU V T

Scalar Equation (Only 1 unknown)

or A BU T

75

Recommended ProblemM3/144 M3/155

M3/160

M3/166 M3/168 H14/93 , H 14/96

77

Documents

Topics in textbook