KULIAH I
OLEH:
ALIEF WIKARTA, ST
JURUSAN TEKNIK MESIN, FTI – ITS
SURABAYA, 2007

Cabang ilmu fisika yang berbicara tentang keadaan diam atau geraknya benda-benda yang mengalami kerja atau aksi gaya
Mechanics
Rigid Bodies
(Things that do not change shape)
Deformable Bodies
(Things that do change shape)
Statics Dynamics
Fluids
Incompressible Compressible

• R. C. Hibbeler, Engineering Mechanics, 7 th - 10 th
Edition, Person Prentice-Hall
• F. P. Beer and E. R. Johnston Jr., Vector
Mechanics for Engineers: Statics, SI Metric
Edition, Mcgraw-hill, 3 rd Edition
• R. C. Hibbeler, Mechanics of Material, 3 th
Edition, Person Prentice-Hall
• dll

• Tugas-Kuis : 25 %
• UTS
• UAS
: 30 %
: 45 %
Tidak mentolerir segala bentuk kecurangan
Tapi tetap boleh cross check

• Dikerjakan pada kertas A4
• Tulis nama dan NRP di sebelah kanan atas, serta tanggal dan tugas ke berapa
• Silahkan mengerjakan soal apa saja yang berkaitan dengan materi yang disampaikan
• Silahkan mengerjakan berapa pun soal yang sanggup anda selesaikan
• Soal-soal harus dari buku yang disepakati
• Mencantumkan judul buku, pengarang, dan nomer soal yang dikerjakan, plus halaman buku

• Keseimbangan partikel
• Keseimbangan benda tegar
• Diagram gaya normal, diagram gaya geser, dan diagram momen
• Konsep tegangan
• Momen inersia dan momen polar
• Teori kegagalan statis

• Particle: A very small amount of matter which may be assumed to occupy a single point in space.
• Rigid body: A combination of a large number of particles occupying fixed position with respect to each other.

Partikel:
Mempunyai suatu massa namun ukurannya dapat diabaikan, sehingga geometri benda tidak akan terlibat dalam analisis masalah
Benda Tegar:
Kombinasi sejumlah partikel yang mana semua partikel berada pada suatu jarak tetap terhadap satu dengan yang lain

• Four fundamental physical quantities. Length, Time, Mass, Force.
• We will work with two unit systems in static’s: SI & US Customary.
Bagaimana konversi dari SI ke US atau sebaliknya ?

Apa yang harus dilakukan supaya
Mekanika Teknik menjadi mudah ?
Banyak dan sering menyelesaikan soal-soal
Prosedur mengerjakan soal:
1. Baca soal dengan cermat
2. Buat free body diagram dan tabulasikan data soal
3. Tuliskan prinsip dasar / persamaan yang relevan dengan soal
4. Selesaikan persamaan sepraktis mungkin sehingga didapat hasil yang signifikan dan jangan lupa disertai sistem satuan
5. Pelajari jawaban dengan akal sehat, masuk akal atau tidak
6. Jika ada waktu, coba pikirkan cara lain untuk menyelesaikan soal tersebut.

THE WHAT, WHY AND HOW OF A
FREE BODY DIAGRAM (FBD)
Free Body Diagrams are one of the most important things for you to know how to draw and use.
What ?
- It is a drawing that shows all external forces acting on the particle.
Why ?
- It helps you write the equations of equilibrium used to solve for the unknowns (usually forces or angles).

How ?
1. Imagine the particle to be isolated or cut free from its surroundings.
2. Show all the forces that act on the particle.
Active forces: They want to move the particle.
Reactive forces: They tend to resist the motion.
3. Identify each force and show all known magnitudes and directions. Show all unknown magnitudes and / or directions as variables .
A
Note : Engine mass = 250 Kg
FBD at A

• The parallelogram law for the addition of forces: Two forces acting on a particle can be replaced by a single force, called resultant, obtained by drawing the diagonal of the parallelogram which has sides equal to the given forces f1+f2 f2 f1
• Parallelogram Law

• The principle of transmissibility: A force acting at a point of a rigid body can be replaced by a force of the the same magnitude and same direction, but acting on at a different point on the line of action f2 f1 f1 and f2 are equivalent if their magnitudes are the same and the object is rigid.
•
Principle of Transmissibility

APPLICATION OF VECTOR
ADDITION
There are four concurrent cable forces acting on the bracket.
How do you determine the resultant force acting on the bracket ?

B
B
C
C
• Trapezoid rule for vector addition
• Triangle rule for vector addition
• Law of cosines,
R

R
2



P
P

2 

Q
Q
2 
2 PQ cos B
• Law of sines, sin
Q
A
 sin B
R
 sin C
A
• Vector addition is commutative,

P


Q


Q


P
• Vector subtraction

The two forces act on a bolt at
A . Determine their resultant.
SOLUTION:
• Trigonometric solution - use the triangle rule for vector addition in conjunction with the law of cosines and law of sines to find the resultant.

• Trigonometric solution - Apply the triangle rule.
From the Law of Cosines,
R
2 

P
2 
Q
2 
2 PQ cos B

40 N
 
60 N

2 
2

40 N

60 N
 cos 155

R

97 .
73 N
From the Law of Sines, sin
Q
A
 sin B
R sin
A


A
 sin B
Q
R
 sin 155

60 N
97 .
73 N


15 .
20

04



35 .
04

A

ADDITION OF SEVERAL VECTORS
•
Step 1 is to resolve each force into its components
•
Step 2 is to add all the x components together and add all the y components together. These two totals become the resultant vector.
•
Step 3 is to find the magnitude and angle of the resultant vector.

Example of this process,

You can also represent a 2-D vector with a magnitude and angle.

EXAMPLE
Given: Three concurrent forces acting on a bracket.
Find: The magnitude and angle of the resultant force.
Plan: a) Resolve the forces in their x-y components.
b) Add the respective components to get the resultant vector.
c) Find magnitude and angle from the resultant components.

EXAMPLE (continued)
F
1
= { 15 sin 40° i + 15 cos 40° j } kN
= { 9.642 i + 11.49 j } kN
F
2
= { -(12/13)26 i + (5/13)26 j } kN
= { -24 i + 10 j } kN
F
3
= { 36 cos 30° i
– 36 sin 30° j } kN
= { 31.18 i
– 18 j } kN

EXAMPLE (continued)
Summing up all the i and j components respectively, we get,
F
R
= { (9.642 – 24 + 31.18) i + (11.49 + 10 – 18) j } kN
= { 16.82 i + 3.49 j } kN y
F
R F
R
= ((16.82) 2 + (3.49) 2 ) 1/2 = 17.2 kN

= tan -1 (3.49/16.82) = 11.7°
 x

Four forces act on bolt A as shown.
Determine the resultant of the force on the bolt.
SOLUTION:
• Resolve each force into rectangular components.
• Determine the components of the resultant by adding the corresponding force components.
• Calculate the magnitude and direction of the resultant.

SOLUTION:
• Resolve each force into rectangular components.
force
F


F
2

F
3

F
4
1 mag
150
80
110
100 x



 comp
129
27
96
.
.
.
9
4
0
6 y




 comp
75
75
110
25
.
.
.
.
0
2
0
9
R x
 
199 .
1 R y
 
14 .
3
• Determine the components of the resultant by adding the corresponding force components.
• Calculate the magnitude and direction.
tan
 
R y
R x

14 .
3
199 .
1
N
N
 
4 .
1
  
4 .
1

R 
14 .
3 sin
N

 199 .
6 N

READING QUIZ
1. The subject of mechanics deals with what happens to a body when ______ is / are applied to it.
A) magnetic field B) heat C) forces
D) neutrons E) lasers
2. ________________ still remains the basis of most of today’s engineering sciences.
A) Newtonian Mechanics B) Relativistic Mechanics
C) Euclidean Mechanics C) Greek Mechanics

READING QUIZ
3. Which one of the following is a scalar quantity?
A) Force B) Position C) Mass D) Velocity
4. For vector addition you have to use ______ law.
A) Newton’s Second
B) the arithmetic
C) Pascal’s
D) the parallelogram

CONCEPT QUIZ
5. Can you resolve a 2-D vector along two directions, which are not at 90° to each other?
A) Yes, but not uniquely.
B) No.
C) Yes, uniquely.
6. Can you resolve a 2-D vector along three directions (say at 0, 60, and 120°)?
A) Yes, but not uniquely.
B) No.
C) Yes, uniquely.

ATTENTION QUIZ
7. Resolve F along x and y axes and write it in vector form. F = { ___________ } N y
A) 80 cos (30°) i - 80 sin (30°) j
B) 80 sin (30°) i + 80 cos (30°) j
C) 80 sin (30°) i - 80 cos (30°) j
30°
D) 80 cos (30°) i + 80 sin (30°) j
8. Determine the magnitude of the resultant ( F
1 force in N when F
1
{ 20 i + 20 j } N .
= { 10 i + 20 j
+
} N and
F
F
2
2
)
=
A) 30 N B) 40 N C) 50 N
D) 60 N E) 70 N x
F = 80 N