lecture 2 mech 211 spring 2014
TRANSCRIPT
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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
Lecture 2-Kirkpatrick 3
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)
Lecture 2-Kirkpatrick 5
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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
Lecture 2-Kirkpatrick 6
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=
Lecture 2-Kirkpatrick 7
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=
Lecture 2-Kirkpatrick 8
( )
( )
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
Lecture 2-Kirkpatrick 9
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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
Lecture 2-Kirkpatrick 10
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
Lecture 2-Kirkpatrick 11
( )
/ /2 2
/ /
8
;
3.844 10
body earth body moon
body earth body moon
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
Lecture 2-Kirkpatrick 12
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
= +
=
=
Lecture 2-Kirkpatrick 13
( )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
Lecture 2-Kirkpatrick 14
[ ] [ ] [ ]
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
Lecture 2-Kirkpatrick 16
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
Lecture 2-Kirkpatrick 17
x
y
z
0
Line of Action
Sense:Denoted byarrowhead
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Representation of VectorsF
In Two DimensionsMagnitudeOne Angle
x
y
Lecture 2-Kirkpatrick 18
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
Lecture 2-Kirkpatrick 19
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
Lecture 2-Kirkpatrick 21
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
Lecture 2-Kirkpatrick 22
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
Lecture 2-Kirkpatrick 23
.
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Other Stuff About Forces Concurrent Forces
All forces act through a
common point
Lecture 2-Kirkpatrick 24
Coplaner Forces All forces lie in a single
plane
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More Stuff About Forces Parallel Forces
All forces are parallel
Lecture 2-Kirkpatrick 25
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
Lecture 2-Kirkpatrick 26
A
+ = +A B B A Commutative Property
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Vector Algebra
(Cont.)
Scalar Multiplication
Changes length or magnitude
A
A
Lecture 2-Kirkpatrick 27
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
Lecture 2-Kirkpatrick 28
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
Lecture 2-Kirkpatrick 29
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
Lecture 2-Kirkpatrick 30
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
Lecture 2-Kirkpatrick 31
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
Lecture 2-Kirkpatrick 32
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
Lecture 2-Kirkpatrick 33
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= = =
Lecture 2-Kirkpatrick 34
( ) ( ) ( )
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