Download Hukum Newton, Vektor

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Mata Kuliah
Kode
SKS
: Statika
: TSP – 106
: 3 SKS
Hukum Newton, Vektor
Pertemuan - 1
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• TIU :
• Mahasiswa dapat menjelaskan tentang prinsip keseimbangan, uraian,
dan penjumlahan gaya.
• TIK :
• Mahasiswa dapat menjelaskan konsep gaya
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• Sub Pokok Bahasan :
• Definisi
• Uraian dan Penjumlahan
• Text Book :
• Hibbeler, R.C. (2004). Statics and Mechanics of Materials SI Edition.
• Hibbeler, R.C. (2010). Structural Analysis. 8th edition. Prentice Hall.
ISBN : 978-0-13-257053-4
• Meriam, J.L., Kraige,L.G., (2006), Engineering Mechanics - Statics. 6th
edition. John Wiley & Sons, Inc. ISBN : 978-0471739326
Bobot Penilaian :
Tugas
Ujian Tengah Semester
Ujian Akhir Semester
: 30 %
: 30 %
: 40%
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Mechanics
Mekanika adalah sebuah divisi dari Ilmu Pengetahuan
yang mempelajari perilaku sebuah objek akibat beban
yang bekerja tehadapnya.
Mechanics is the branch of the physical sciences which
deals with the state of rest or motion of bodies that are
subjected to the action of forces.
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Subdivisions of Mechanics
Mechanics
Deformable
Bodies
Statics
(At Rest)
Rigid Bodies
Fluids
Dynamics
(In Motion)
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Newton’s Three Laws of Motion
• First Law
“A particle originally at rest, or moving in a
straight line with constant velocity, will
remain in this state provided the particle IS
NOT subjected to an unbalanced external
force”
Balanced Forces
Particle at Rest Particle in Motion
(v=0, a=0)
(v>0, a=0)
Stay at rest
Stay in motion
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• Second Law
“A particle acted upon by an unbalanced force
(F) experiences an acceleration ( a ) that has the
same direction as the force and a magnitude that
is directly proportional to the force”. If ( F ) is
applied to a particle of mass (m), this law may be
expressed mathematically as:
F = ma
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• Third Law
“The mutual forces of action and reaction
between two particles are equal in magnitude,
opposite in direction and collinear in
orientation”.
F
-F
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Newton’s
Law
of
Gravitational
Attraction
A law governing the gravitational attraction between any two particles is mathematically stated
as:
F = G [m1
Where
m2 / r2]
F = force of gravitation between the two particles
G = A universal constant of gravitation; (66.73x10-12 m3/kg s2)
m1 and m2 = mass of each of the two particles
r = the distance between the two particles
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Weight
Weight is the gravitational force between the earth and the particle. If we
assume that:
W = weight of the particle
m = m1 = is the mass of the particle
m2 = is the mass of the earth
r = is the distance between the earth’s center and the particle
Then:
W = G [m m2 / r2]
Letting:
g = G m2 / r2
Therefore, from the second law of motion ( F = ma ):
W = mg
Where
g = the acceleration due to gravity
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Units of Measurements
SI Units
SI is known as the International System of Units where
Length is in meters (m), time is in seconds (s), and
mass is in kilograms (kg) and force is in Newton (N)
(1 Newton is the force required to give 1 kilogram of
mass an acceleration of 1 m/s2).
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Conversion Factors
Force;
Mass;
Length;
1 lb = 4.4482 N
slug = 14.5938 kg
ft = 0.304 m
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Prefixes
giga = G = 109 = 1,000,000,000
mega = M = 106 = 1,000,000
kilo = k = 103 = 1,000
milli = m = 10-3 = 0.001
micro = μ = 10-6 = 0.000 001
nano = η = 10-9 = 0.000 000 001
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FORCE VECTORS
Scalars and Vectors
Scalar
• A quantity identified by positive or negative number.
• It is characterized by its magnitude only
• Elementary algebra is used when mathematical operations are
involved
• Examples include mass, length, and volume
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Vector
• A quantity by its magnitude and direction
• Examples include force, moment, and
displacement
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Graphical Representation of a Vector
• Vectors are represented by ARROWS
• Magnitude is represented by the length of the
arrow
• Direction is defined by
• Vector A is written as A
Head
A
Tail
Line of
Action
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Vector Operations
Multiplication and Division of a Vector by a Scalar
•
•
•
•
•
•
Vector A
Scalar a
A a = aA
Magnitude of aA
Direction of A if a is positive (+)
Direction of –A (opposite) if a is negative (-)
A
1.5 A
-A
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Vector Addition
• Vectors are added according to the parallelogram law
• The resultant R is the diagonal of the parallelogram
A
B
A
A
R =A+ B
B
R =A+ B
R =A+ B
B
A
B
• If two vectors are co-linear (both have the same line of
action), they are added algebraically
R =A+ B
A
B
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Vector Subtraction
•
The resultant is the difference between vectors A and B
A
-B
R
B
-B
A
R
A
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Resolution of a Vector
If lines of action are known, the resultant R can
be resolved into two components acting along
those lines (i.e. a and b).
b
B
R
A
a
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Analysis of Problems
•
•
•
•
Two procedures to be followed:
Parallelogram law
Trigonometry
sine and/or cosine laws may be used
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Sine Law
A
B
C


sin a sin b sin c
C
b
a
A
c
Cosine Law
C
A 2  B 2  2 AB cos c
B
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Example 1
• The screw eye in figure
is subjected to two
forces F1 and F2.
Determine the
magnitude and direction
of the resultant force
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Using Parallelogram Law
150 N
65o
FR
10o

180o-65o =
115o
100 N
90o-25o = 65o
15o
FR  100 2  150 2  2100150 cos 115o  212,6 N
sine law :
150
212 ,6

sin  sin 115o
 sin  
150
 sin 115o  39 ,8o
212 ,6
Direction, f = 15 +  = 54,8o from horizontal axis
f = 54,8o
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Example 2
• Resolve the 1000-N
force acting on the pipe,
into components in the :
(a) x and y direction; (b)
x’ and y direction
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y
y
Fy
50o
Fy
Fy
70o
40o
x
Fx
Fx’
50o
Fy
40o
30o
Fx
1000

 Fx  884 ,6 N
sin 50o sin 60o
40o
Fx
Fx  1000 cos 40o  766 N
Fy  1000 sin 40o  643 N
60o
Fx’
x'
Fy
sin 70
o

1000
 Fy  1085 N
sin 60o
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Example 3
• The ring show in figure is
subjected to two forces F1
and F2. If it is required that
the resultant force have a
magnitude of 1 kN and be
directed vertically
downward, determine (a)
the magnitudes of F1 and F2
provided  = 30o, (b) the
magnitude of F1 and F2 if F2
is to be minimum
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(a)
F2
30o
20o F
1
F2
30o
130o
130o
F1
20o
30o
20o
F1
1000

 F1  653 N
sin 30o sin 130o
F2
1000

 F2  446 N
sin 20o sin 130o
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(b)
F2
70o
• If  is not specified, then by the vector triangle,
F2 may be added to F1 in various ways to yield
the resultant 1000 N force.
• In particular, the minimum length or magnitude
of F2 will occur when its line of action is
perpendicular to F1.
• Hence, when  = 90o – 20o = 70o, F2 is minimum
F1  1000 sin 70o  940 N
F1
F2  1000 cos 70o  342 N
20o