8 - 1 Kinetics of Particles So far, we have only studied the kinematics of particle motion, meaning we have studied the relationships between position,

8 - 1

Kinetics of Particles

So far, we have only studied the kinematics of particle motion, meaning we have studied the relationships between position, velocity, and acceleration, with only cursory reference to the forces which produced the motion.

In some sense, the kinematics is the study of dynamics from a purely experimental standpoint. That is, if we are able to map the positions, velocities, and accelerations of multiple particles, then we can ascertain the force which produced the motion. Note: By the word “particle”, it is implied that the object is contained entirely within a point.

In the next part of this course, we will be interested in the forward model. That is, if we know the forces on the particle, what should the resulting motion be? The reason I call this a forward model is because the classification of different forces is purely theoretical and can only be backed up by experimental measurements.

8 - 2

NEWTON’S SECOND LAW

Denoting by m the mass of a particle, by F the sum, or resultant, of the forces acting on the particle, and by a the acceleration of the particle relative to a newtonian frame of reference, we write

F = ma

Introducing the linear momentum of a particle, L = mv, Newton’s second law can also be written as

F = L.

which expresses that the resultant of the forces acting on a particle is equal to the rate of change of the linear momentum of the particle.

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To solve a problem involving the motion of a particle, F = ma should be replaced by equations containing scalar quantities. Using rectangular components of F and a, we have

Fx = max Fy = may Fz = maz

x

y

P

an

O

at

x

y

z

ax

ay

az

P

x

P

a

O

ar

Using tangential and normal components,

Ft = mat = mdvdtv2

Using radial and transverse components,

...

.. . .Fr = mar= m(r - r2)

r

Fn = man = m

F = ma = m(r + 2r)

8 - 4

Central Forces

ˆ2ˆˆˆ, 2

rrmrrrmFrFamrF r

2r

rrmF 20

In many forces, such as gravitational, electrical forces, etc., the magnitude of the force only depends on the distance away from the source (i.e. ), and the direction is always towards or away from the source . r̂

2

2

2

.

021

r

h

hconstr

rrrdt

d

r

Using these assumptions, we can derive an expression for the rotation frequency:

The result implies that under the action of a central force, the rotation frequency depends only on the radial distance away from the attractor.

8 - 5

2r

h

2

22

r

hrrmrrmFr

2

1

rmGmF eBr

The above represents a nonlinear differential equation, to which the solution will give us r as a function of time. We will return to this later in the course.

One example that we have discussed previously is the flat earth approximation. When is it valid?

Using this result, we can arrive at an expression for radial force purely in terms of the radial distance

baseball

earth

Re

hhRr e where

22

2

22

1

1

111

e

ee

e R

R

hR

hRr

wheneRh

gmR

GmmF B

e

eBr

2 2e

e

R

Gmg where

8 - 6

Example Problem

8 - 7

Free Body Diagram for Block B

yBBBABy

xBBABx

ByBxB

amgmNfF

amNfF

yFxFF

cossin

sincos

ˆˆ

gm

BAN

Nf

8 - 8

Free Body Diagram for Block A

0cossin

sincos

ˆˆ

yAAAsABAy

xAABAppliedAx

AyAxA

amgmNNfF

amNfFF

yFxFF

gmA

ABN

ABNf

AsNForce on Block A due

to substrate

AppliedF

8 - 9

What we would like to do is solve for the forces if Block B is on the verge of slipping, but has not yet done so. This allows us to develop a relationship between the accelerations of Blocks A and B

BAAB NNf

0cossin

sincos

0cossin

sincos

yAAAsABAy

xAABAppliedAx

yBBBABy

xBBABx

amgmNNfF

amNfFF

amgmNfF

amNfF

Newton’s 1st law provides us with an expression that the force balance equals zero. (i.e.)

NaNF xABapplied ,,,

These 4 equations have 4 unknowns listed below, and the solution can be solved using linear algebra

8 - 10

1tan

tan

cossin

sincos

cossin

sincos

sincoscossin

cossin

ga

ga

ga

ga

ga

ga

gmamgmam

gmamN

BBBB

BBBA

sincos

cossin

amgmNF

amgmNF

BBBABy

BBBABx

x

yy’

x’

BAN

BANf

gmB

Since we’re only interested in Block B, let’s try a change in coordinates

Thus, the critical slipping threshold can be expressed as:

8 - 11

tan

1tan

tan

ga

ga

1tan

tan

1

1tan

tan

ga

ga

Case 1. Assume that the acceleration applied to the block is zero

1

tan 1

Therefore, the Block will slide under the action of gravity if the angle exceeds θ.

Case 2. Neglect Gravity

Note: This is the result derived in the book.

8 - 12

1tan

tan

ga

ga

ga

ga

1tan 1

Case 3. Account for everything

Note: We still haven’t solved for the acceleration explicitly. Once the block starts to slip, we have to take into account the relative motion of Blocks A and B.

So what is the effect of applying acceleration to the block? You can think of it in two ways. One is that the acceleration delays the action of gravity. Depending on the friction coefficient, the block may slide downwards when the applied force is removed. Alternatively, one can say that the acceleration keeps the block from slipping at higher angles.

8 - 13

8 - 14

222 lyyxx BABA

AAxAABAx

BByBBABBy

xmamPTF

ymamTgmF

cos

sin

ymayFxFF yByBxB ˆˆˆ

xmayFxFF xAyAxA ˆˆˆ

Constraints: Motion of Slider A is confined to the X-axis

Motion of Slider B is confined to the Y-Axis

Distance between Sliders A and B is constant

Free Body Diagrams:

Given: sm

Ax 9.0 mxA 4.0 ml 5.0

8 - 15

Additional Constraints

BBBA

AA

BBBAAABA

B

AABBBAABA

yyyxx

x

yyyxxxyxdt

d

y

xxyyyxxl

dt

dyx

dt

d

22

2222

2

2

222

1

0

0220

BBBAA

AAAAB

BBBAB

yyyxx

mxmPT

ymTgm

22cos

sin

By BAT

In these two equations, the only thing we don’t know are the accelerations of sliders A and B, however they are interdependent

Thus, we really have 2 equations with 2 unknowns… , .

8 - 16

gm

Ty

B

BAB sin

22

2222

2

1

2

22

1

22

22

22

365.132.93.03.0

4.09.09.0

4.0

11

32.95

3

3

6.46sin

6.46

54

4.0

3.0

3

25

32

4.0403.010

3.0

4.09.09.0

4.0

2

cossin

cossin

sincos

s

m

s

mm

m

m

myyyx

xa

s

m

kg

Ng

m

Ta

NT

m

m

kg

kg

kg

mNm

s

m

m

m

m

kgT

xm

ym

m

xPgyyx

x

mT

m

xPgyyx

x

m

xm

ymT

gm

Tyyx

x

mxmPT

sm

sm

BBBAA

A

B

BAB

AB

sm

sm

AB

AB

BA

A

ABBA

A

AAB

A

ABBA

A

A

AB

BAAB

B

BABBA

A

AAAAB

Note: The answer shown in the book assumes gravity is zero. What are the correct tensions and accelerations if gravity is taken into account? What also must we do if friction between the sliders and wall are taken into account?

8 - 17

P

Problem 1

Block A has a mass of 30 kg and blockB has a mass of 15 kg. The coefficientsof friction between all plane surfaces ofcontact are s = 0.15 and k = 0.10.Knowing that = 30o and that themagnitude of the force P applied toblock A is 250 N, determine (a) theacceleration of block A , (b) the tensionin the cord.

AB

8 - 18

Problem 1

P


AB

1. Kinematics: Examine the acceleration of the particles.

2. Kinetics: Draw a free body diagram showing the appliedforces and an equivalent force diagram showing the vectorma or its components.

8 - 19

Problem 1

P


AB

3. When a problem involves dry friction: It is necessary first toassume a possible motion and then to check the validity of theassumption. The friction force on a moving surface is F = k N.The friction force on a surface when motion is impendingis F = s N.

8 - 20

Problem 1


P

AB

4. Apply Newton’s second law: The relationship between theforces acting on the particle, its mass and acceleration is givenby F = m a . The vectors F and a can be expressed in terms ofeither their rectangular components or their tangential and normalcomponents. Absolute acceleration (measured with respect toa newtonian frame of reference) should be used.

8 - 21

Problem 1 Solution

Kinematics.

P

AB

Assume motion with block A movingdown.If block A moves and accelerates downthe slope, block B moves up the slopewith the same acceleration.

ABaA

aB

aA = aB

8 - 22

Problem 1 Solution

P

AB

Kinetics; draw a free body diagram.Block A :

Block B :

mA a = 30 a

N

T

250 N

WA= 294.3 N

Fk = k N

=

WB= 147.15 N

TN

N’

Fk = k N

F’k = k N’

mB a = 15 a

=

WA = mA g

WA = (30 kg)(9.81 m/s2)

WA = 294.3 N

WB = mB g

WB = (15 kg)(9.81 m/s2)

WB = 147.15 N

8 - 23

Problem 1 Solution

Apply Newton’s second law.

mA a = 30 a

=

+ Fy = 0: N - (294.3) cos 30o = 0 N = 254.87 N

Fk = k N = 0.10 (254.9) = 25.49 N

+ Fx = ma: 250 + (294.3) sin 30o - 25.49 - T = 30 a

371.66 - T = 30 a (1)

Block A :

N

T

250 N

WA= 294.3 N

Fk = k N

30o

8 - 24

Problem 1 SolutionBlock B :

mB a = 15 a

+ Fy = 0: N’ - N - (147.15) cos 30o = 0 N’ = 382.31 N

F’k = k N’ = 0.10 (382.31) = 38.23 N

+ Fx = ma: T - Fk - F’k - (147.15) sin 30o = 15 a

T - 137.29 = 15 a (2) Solving equations (1) and (2) gives:

T = 215 N a = 5.21 m/s2

WB= 147.15 N

TN

N’

Fk = k N

F’k = k N’

=

30o

8 - 25

Problem 1 Solution

P

AB

Check: We should verify that blocksactually move by determining the value ofthe force P for which motion is impending

Verify assumption of motion.

Find P for impending motion.For impending motion both blocks are in equilibrium:

N

T

P

WA= 294.3 N

Fm = s N

30o

WB= 147.15 N

TN

N’

Fm = s N

F’m = s N’

30o

Block BBlock A

8 - 26

Problem 1 Solution

N

T

P

WA= 294.3 N

Fm = s N

30o

WB= 147.15 N

TN

N’

Fm = s N

F’m = s N’

30o

A B

From + Fy = 0 find again

N = 254.87 N and N’ = 382.31 N,and thus

Fm = s N = 0.15 (254.87) = 38.23 N

F’m = s N’ = 0.15 (382.31) = 57.35 N

8 - 27

Problem 1 Solution

N

T

P

WA= 294.3 N

Fm = s N

30o

WB= 147.15 N

TN

N’

Fm = s N

F’m = s N’

30o

A B

For block A:

+ Fx = 0: P + (294.3) sin 30o - 38.23 - T = 0 (3)

For block B:

+ Fx = 0: T - 38.23 - 57.35 - (147.15) sin 30o = 0 (4)

Solving equations (3) and (4) gives P = 60.2 N.Since the actual value of P (250 N) is larger than the value for

impending motion (60.2 N), motion takes place as assumed.

8 - 28

Problem 2

A

B12 lb

30 lb

A 12-lb block B rests as shown onthe upper surface of a 30-lb wedgeA. Neglecting friction, determineimmediately after the system isreleased from rest (a) the accelerationof B relative to A, (b) the accelerationof B relative to A.

30o

8 - 29

Problem 2

A

B12 lb

30 lb

1. Kinematics: Examine the acceleration of the particles.

2. Kinetics: Draw a free body diagram showing the appliedforces and an equivalent force diagram showing the vectorma or its components.

30o


8 - 30

Problem 2

A

B12 lb

30 lb

3. Apply Newton’s second law: The relationship between theforces acting on the particle, its mass and acceleration is givenby F = m a . The vectors F and a can be expressed in terms ofeither their rectangular components or their tangential and normalcomponents. Absolute acceleration (measured with respect toa newtonian frame of reference) should be used.

30o


8 - 31

Problem 2 Solution

Kinematics.

A

B12 lb

30 lb

30o

A30o

aA

B

aA

aB/A

Since the wedge is constrainedto move on the inclined surface,its acceleration aA is in thedirection of the slope.The acceleration aB of the blockcan be expressed as the sum ofthe acceleration of A and theacceleration of B relative to A. aB = aA + aB/A

8 - 32

Problem 2 Solution

A

B12 lb

30 lb

30o

Kinetics; draw a free body diagram.

Block B :

A

30 lb

A

N1

N2mA aA = aA

3032.2

Block A :

B

12 lb

B

N1

mB aB/A = aB /A12

32.2

mA aA = aA12

32.2

=

=

8 - 33

Apply Newton’s second law.

Problem 2 SolutionBlock A :

A

30 lb

A

N1

N2mA aA = aA

3032.2

=

30o

x x

+ Fx = ma: (N1 + 30) sin 30o = aA (1)30

32.2

8 - 34

Problem 2 Solution

B

12 lb

N1

B

mB aB/A = aB /A12

32.2

mA aA = aA12

32.2

= 30o

+ Fy = may : 12 - N1 = aA sin 30o (2)

+ Fx = max: 0 = aB /A - aA cos 30o (3)

1232.2

1232.2

1232.2

Solving equations (1), (2), and (3) gives:

aA = 20.5 ft/s2 30o aB /A = 17.75 ft/s2

Block B :

Documents

8 - 1 Kinetics of Particles So far, we have only studied the kinematics of particle motion, meaning we have studied the relationships between position,