lecture 2 mech 211 spring 2014

8/13/2019 Lecture 2 Mech 211 Spring 2014

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Lecture 2

Chapter 1: Example ProblemsChapter 2: Concurrent Force Systems

Sections 2.1-2.2,2.4,2.5


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TA Help SessionsTeaching Assistants:

Eleazar MarquezDavid Trevino-Garcia

Lecture 2-Kirkpatrick 2

Wednesdays from 5:00PM-6:00PM

Location:

Mechanical Laboratory Building (MEL) 251


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Important! Trigonometry

Law of Sines:

sin sin sin

a b c

= =

b

a


Law of Cosines:

c

2 2 2 2 cosc a b ab = +


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Newtons Law of Gravitation(From Lecture 1)

m2

1 22

m mF G r

=

Fr

Lecture 2-Kirkpatrick

m1

FG = Universal Gravitational Constant

Newtons Law of Gravity is a mathematical result from atheoretical model of how bodies interact.

4


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Some Important

Numerical Values8 3 2 11 3 23.439(10 ) /( ) 6.673(10 ) /( )G ft slug s m kg s

= = (Page 9 oftextbook)



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Calculation Of

Acceleration of Gravity

23 24

7 6

4.095(10 ) 5.976(10 )

2.090(10 ) 6.371(10 )

e

e

m slug kg

r ft m

= =

= =

8 3 2 11 3 23.439(10 ) /( ) 6.673(10 ) /( )G ft slug s m kg s = =

Given


e ear were rea y a sp ere( )

238 3 2

2 7 2 2

2 2

2

4.095(10 )3.439(10 ) /

(2.090(10 ))

(3.439)(4.095)(10)/ 32.24 /

(2.090)

e

e

m slugg G ft slug s

r ft

g ft s f s

= =

= =

Note: 2 232.17 / 9.807 / g ft s m s= =

Are the usual or official values used.


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Example Problem: Problem 1.5/p14Given: The planet Venus has a diameter of 7700 miles and a mass of

3.34(1023)slug.Find: Determine the gravitational acceleration gat the surface of Venus.

Other Information:2

venus

venus

venus

mg G

r=


8 3 23.439(10 ) /( )G ft slug s

= where

Therefore

2

238 3 2

2

2

3.34(10 )(3.439(10 ) / ( ))

((7700 )(.5)(5280 / ))

27.7965 / sec

venusvenus

venus

venus

venue

mg G

r

slugg ft slug s

miles ft mile

g ft

=

=

=

(See page 9 of text)


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Example Problem: Problem 1.21/p14Given: The first U.S. satellite, Explorer 1, had a mass of approximately 1

slug. Its perigee (low altitude) was 175 miles and its apogee (highaltitude) was 2200 miles.

Find: Determine the force exerted on the satellite by the earth at the lowand high points of the orbit.

Forces to calculate: expearth lorer m m

F G=


( )

( )

8 3 2 23

exp

7

3.439(10 ) / ( ); 4.095 10 ; 1

2.090 10 ; 175 5280 ; 2200 5280

e lorer

e perigee apogee

G ft slug s m slug m slug

ft ftr ft h mi h mimi mi

= = =

= = =

earth

altitude

r h+

where

Results

29.57 and 13.32perigee apogeeF lb F lb= =


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Let MATLAB Do The Numbers



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Example Problem: Problem 1.24/p14At what distance, in kilometers, from the surface of the earth on a line fromcenter to center would the gravitational force of the earth on a body beexactly balanced by the gravitational force of the moon on the body.

a b


Earth MoonBody

Given

( )

/ /2 2

/ /

8

;

;

3.844 10

e b moon bbody earth body moon

body earth body moon

m m m mF G F G

a b

F F a b R

R m

= =

= + =

=


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Example Problem: Problem 1.24/p14

Earth Moon

a b

Body

Givene b moon b

m m m m


( )

/ /2 2

/ /

8

;

3.844 10



a bF F a b R

R m

= + =

=

Find and . Then calculate and .ba eartha r moonb r


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Example Problem: Problem 1.24/p14Therefore

2 20e b moon b e

moon

m m m m mG G b a

a b m= =

And, we must solve simultaneously

e em m


1 1

1 01

1 11 1 1

moon moon

e e

moon moon

e e

moon moon

m m b Ra b R

m ma R

m mb Rm m

m m

=

= + =

= = + +


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Example Problem: Problem 1.24/p14Given

( )

( )

( )

8

88

24

1 1 3.4603 10

0.38375 103.844 10

e

moon

e

moon

ma R

mb m

m ma

b mR m

= +

=

=


( )22

.

7.350 10

e

moon

m

m

=

=

Also ( )

( )

8

8

3.3965 10

0.36637 10

e

moon

ma r

b r m

=


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Example Problem: Problem 1.53/p24Given: The equation, where r and Rare lengths and n

is dimensionless. Also, k is the modulus of a coil spring (force per unit of

length.

4

34

Grk

R n=

Find: The dimensions of G


[ ] [ ] [ ]

33 2

4 4 24

4F

Lk R n FL FLGL L Lr

= = = =

Note: [G] reads dimensions of G


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Chapter 2

Concurrent Force Systems


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Three Characteristics of Forces Forces are Vectors

Magnitude

Direction With respect to a fixed frame of reference

Point of Application


Can think of force as a function of position whose value is avector.

Body

F

Point of Applicationz

x

y0


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More On Direction of a Vector With respect to a fixed frame of reference

Line of Action

Sense


x

y

z

0

Line of Action

Sense:Denoted byarrowhead


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Representation of VectorsF

In Two DimensionsMagnitudeOne Angle

x

y


In Three DimensionsMagnitudeTwo Angles

Will explain later!

x

y

z

0

yx

z F


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Interesting Fact Consider a force F on a Rigid Body


Since the body is rigid,

Point of action of F is anywhere on line ofaction

Sometimes called a sliding vector.


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Contact Forces Concentrated Forces

Area of contact is small compared to size ofbody

Can be treated as applied at a point


Distributed Forces Applied over an area of contact

Realistically

All contact forces are distributed Thus, contact forces are approximations of reality

Good approximations!


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Examples Concentrated forces

Automobile sitting onbridge


Distributed forces

Concrete bridge floorsitting on beam


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Frankly! You will learn about types of loads by the

many examples we will see. Most of the loads will be idealized so that


.


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Other Stuff About Forces Concurrent Forces

All forces act through a

common point


Coplaner Forces All forces lie in a single

plane


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More Stuff About Forces Parallel Forces

All forces are parallel


Collinear Forces All forces have the

same line of action


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Some Vector Algebra(Appendix A)

Addition

Parallelogram Law

+A B

B


A

+ = +A B B A Commutative Property


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Vector Algebra

(Cont.)

Scalar Multiplication

Changes length or magnitude

A

A


A


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Vector Algebra

(Cont.)

Notation

If A is a vector, its magnitude or length is written Sometimes we write

Note ifA=1, then A is a unit vector

AA = A


Dot Product cos = =A B B A A B

A

B


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Vector Algebra

(Cont.)

Cross Product

sin =A B A B n

B


A

vector perpendicular toplane ofA andB in

direction of right handrule

Note: =A A 0


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Cross Products Relationship to Area

B

n


sin =A B A BA

=Area of parallelogram

Note A B = B A


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Vector Algebra

(Cont.)

Component Representation of a Vector

z

x y zA A A= + +A i j kA = A i


x

y0

ij

ky

z

A

A

=

=

A j

A k

w ereA

( ) ( ) ( )= + + A A i i A j j A k kAn Identity


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Vector Algebra

(Cont.)z

k

1, 1, 1

0, 0, 0,

= = =

= = =

i i j j k k

i j i k j k

Because unit vectors are mutually orthogonal


x

y0

ij , ,

etc = = = i j k j k i k j i


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Vector Algebra

(Cont.)

, ,

etc

= = = i j k j k i k j i


i

k


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Vector Algebra

(Cont.)

( ) ( )x y z x y z

x x y y z z

A A A B B B

A B A B A B

= + + + +

= + +

A B i j k i j k

2 2 2= = =


( ) ( ) ( )

x y z

x y z

y z z y z x x z x y y x

A A A

B B B

A B A B A B A B A B A B

=

= + +

i j k

A B

i j k

y z

lecture 2 mech 211 spring 2014

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