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Topics in textbook. Actual Work Definition of Work Work on Particle. Potential Energy Definition of Energy Potential Energy. Work-Energy Equation. Kinetics Energy Definition of Energs. Newton’s 2 nd Law. Three approaches for solving dynamics. Kinematics Eq:. path.

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slide1

Topics in textbook

  • ActualWork
    • Definition of Work
    • Work on Particle
  • Potential Energy
    • Definition of Energy
    • Potential Energy

Work-Energy Equation

  • Kinetics Energy
    • Definition of Energs
three approaches for solving dynamics

Newton’s

2nd Law

Three approaches for solving dynamics

Kinematics Eq:

path

1) Direct Method

A

From 2nd Law

(kinetics Eq)

work (and potential energy)

of Force i along the path

2) Work and Energy

From 2nd Law

3) Impulse and Momentum

linear impulse of Force i

From 2nd Law

work and energy

Newton’s

2nd Law

Work and Energy

Usually convenient

when F = F(s), and you want to find velocity at final state

(without finding acc. first).

From 2nd Law

change of

kineticenergy

kinetic

energy

at B

path

B

Work of Force i along the path

kinetic

energy

at A

A

Work of a force

during small displacement

Principle of work and Energy

kineticenergy

slide4

Work done on Particle

Work by a force

path

P

(inactive force)

Work done over particle A

Since ,

the total work done on object is ……

sum of works

done by all forces

over the particle A

note on work
Noteon work

path

  • is positive when

has the same direction.

  • Unit of work is N-m or Joule (J).
  • Active force is the force that does the work
  • Reactive force = constrain force that does not do the work
slide6

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.

B

A (spring stretched length is 0.5m)

Pos B

Horizontal Force P: constant

Pos A

Weight W: constant.

“Active Force”

Spring Force Fs.: varying Force

Normal Force NB : constant

“Inactive Force”

slide7

Work done on Particle

Work done over particle A

t

path

sum of works

done by all forces

over the particle A

P

Pos B

Pos A

Work done on particle P during path A->B,

is to increase kinetic energy of particle

kinetics energy

path

Kinetics Energy

A

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

Principle of work and Energy

  • Unit of T is N-m or Joule (J)

Advantage

  • Scalar equation. (1 unknown)
  • Integral Equation (not instantaneous eq like 2nd Law)
  • No need to find acceleration first
  • Get change in velocity directly from active forces.
  • it can be applied to system of particleswith
  • frictionless and non-deformable links
how to calculate work
How to calculate Work

path

In general

displacement x force component in the direction of displacement

In xyz-coord

scalar

(be careful

Of +/-)

displacement in the direction of force

x force component

In nt-coord

In rq-coord

slide12

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

x-y

y

x

mg

F=8N (const)

N

Does not do the work

Ans

If F is not constant, how to calculate it?

slide13

If F is not constant

y

x

More general case

or

slide14

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

n-t

Fn doest not effect works!

s

engine thrust

a

slide15

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

r-q

r-q reference point

central force

slide16

M3/121) The 0.2-kg slider moves freely along the fixed curved rod from A to B in the vertical planeunder 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.

mg

F

Work-Energy Eq.

0

general

position

N

Does not

do the work

r-q coordinate

reference point

0

work on frictionless connected particles
Work on frictionless connected particles

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.)

  • internal force R and –R will have the same displacement.
  • So, the sum of these works are zero.
slide18

initial state

Final state

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

Ncauses

no work!

System selection is not so good

(you have to calculate Tension T for its work)

slide19

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.

T

does

no work

system

mg

r-q coordinate

reference point

system

power
Power

path

  • Power is defined as time rate of work

A

(scalar quantity)

  • 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)

  • Unit of power: Watt (W) = J/s = N-m/s
mechanical efficiency
Mechanical Efficiency
  • Mechanical Efficiency

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

  • Since machine consists of moving parts which may have frictions, so extra energy or power is needed to overcome the frictions.
slide22

A car has a mass of 2 Mg and an engine efficiency of e = 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.

a

x

Constant

acceleration:

slide23

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.

2F= 2(30)

B

50 N

v=?

F = 30 N

(const)

Energy

Approach

50N

s = 10 m

(start from rest)

work and energy1
Work and Energy

kinetic

energy

at B

path

B

kinetic

energy

at A

A

summation of all forces

We found that ….

It is much easier to solve dynamic problem, if we think the work done by spring and gravity forcein the form of PotentialEnergy

Work from

Gravity Force

Work from spring

Work from all other forces

(not spring & gravitation)

Gravitational Potential Energy

Elastic

Potential Energy

kineticenergy

slide25

Work of Gravity Force

Only depends on

position at

final state (2)

Only depends on

position at initial state (2)

> 0

any path

1

energy level (higher)

W=mg

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

2

energy level (lower)

h

Think in Term of “PotentialEnergy”

(for convenience)

Fixed reference line

point function

Potential Energy

- Energy from gravity field

Work done by

Gravity Force:

from position 1

to position 2

Energy “Emission”:

from position 1 to position 2

Work = “Energy in Transfer”

Work by couple is

- positive if M has the same sense as dq

- negative if M has the inverse sense as dq

Work done over Object

43

when change in g is significant
when change in g is significant

Define as negative of work done from the position to

  • the potential energy at r is
  • from
slide27

(direction may be opposite)

Work of Spring Force

Only depends on

position at initial state (1)

Only depends on

position at

final state (2)

1

L

natural length

(unstretched length)

Work done by Spring , depends only on the initial state and final state only, i.e. ,

it does not depends on actual path

2

any path

x : distance , stretched or compressed from natural length

point function

Think in Term of “Energy”

(for convenience)

Energy Emission:

from position 1 to position 2

Work done by Spring Force:

from position 1 to position 2

Work done by Gravity Force:

from position 1 to position 2

Energy Emission:

from position 1 to position 2

slide28

Work-Energy Equation

FBD **

(Use Energy Concept)

FBD

Not Recommended Method in this course

N

N

(Conservative Force)

Think of Energy

Work-Energy Equation

(2nd Form)

Work-Energy Equation

(1st Form)

Virtual work by non-conservative forces.

Energy Concept

final

location

initial

location

slide29

Work-Energy Equation

(1st Form)

FBD

Work-Energy Equation

(2nd Form)

N

(Conservative Force)

Think of Energy

The virtual work done by external active forces(other than gravitational and spring forces accounted for in the potential energy terms) on an ideal mechanical systemin equilibrium, equals the corresponding change in the total elastic and gravitational potential energy of the system for any and all virtual displacements consistent with the constraints.

slide30

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.

gravitational potential datum

At position B

gravitational

Potential datum

slide31

Advantage

  • Integral Equation (not instantaneous equation like 2nd Law)
  • Scalarequation. (easy to handle with1 unknown)
  • Get change in velocity directly.
  • (No need to find acceleration first)
  • Handle with only active forces.
  • it can be applied to system of particleswith
  • frictionless and non-deformable links

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

work on frictionless connected particle s
Work on frictionlessconnected particles

A

B

C

O

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.)

  • internal force R and –R will have the same displacement.
  • So, the sum of these works are zero.
slide33

initial state

Final state

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

T

T

T

T

F

F

Object A

Object B

system

T is internal force

F

(excluding from Work Calculation)

We have no interest in T, thus object separation

(separating object A and B) is not good in this problem.

slide35

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

up

Position A: at rest

(assume)

Position B: block B moves down as 1meter

datum

unsolvable

slide36

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

up

Position A: at rest

(assume)

Position B: block B moves down as 1meter

datum

slide37

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.

20 N

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

50 N

Position A: at rest

Position B: block B moves down as d meter

slide38

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

O2-1

O2-2

O1

r

L

r

datum

System: O1+O2+O3+4 rods

Position A: at rest with q=180

Position B: maximum compression

power1
Power

path

  • Power is defined as time rate of work

A

(scalar quantity)

  • 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)

  • Unit of power: Watt (W) = J/s = N-m/s
mechanical efficiency1
Mechanical Efficiency
  • Mechanical Efficiency

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

  • Since machine consists of moving parts which may have frictions, so extra energy or power is needed to overcome the frictions.
slide41

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.

v=?

No!

F = 30 N

(const)

50N

Energy

Approach

2F= 2(30)

s = 10 m

(start from rest)

F = 30 N

(const)

50 N

50 N

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

or

  • Scalar Equation (Only 1 unknown)
  • Equation itself is not hard to solve, but calculating work may be more difficult than you thought.
recommended problem
Recommended Problem

M3/155

M3/144

M3/160

H14/93 , H 14/96

M3/168

M3/166