Technical University of Sofia
Branch Plovdiv
Theoretical Mechanics
STATICS
KINEMATICS
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Notes and Recommendations: ruschev@tu-plovdiv.bg
Details of Lecturers
Course Lecturers:
Prof. Sonia Tabakova
Room Number: 4237, Faculty of Mechanical Engineering
Email: sonia@tu-plovdiv.bg
Office Hours: 14.00-16.00 Tuesday
Assoc. Prof. Dechko Ruschev
Room Number: 3226, Faculty of Mechanical Engineering
Email: ruschev@tu-plovdiv.bg
Office Hours: 12.00 -14.00 Wednesday
COURSE GOALS
(i) To introduce students to basic concepts of force, couples, moments
and mechanical motion in two and three dimensions.
(ii) To develop analytical skills relevant to the areas mentioned in (i)
above.
(iii) To provide the methodology of modeling and solving of practical
problems in the area of the Statics and Kinematics with engineering
application.
Course Content
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Introduction to Mechanics.
Fundamental Concepts.
Introduction to Statics. Axioms of Statics.
Two and Three - Dimensional Force systems.
Equilibrium.
Distributed Forces.
Structures.
Kinematics of a Particle.
Description of Motion in Vector and Cartesian Approach.
Rectilinear Kinematics.
General Curvilinear Motion.
Relative Motion of a Particle.
Planar Kinematics of a Rigid Body.
Three-DimensionaI Kinematics of a Rigid Body.
Bibliography
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Meriam J.L., L.G. Kraige, Engineering Mechanics - vol.1 Statics, vol. 2 Dynamics, 5th edition.,
John Wiley & Sons, 2002;
Hibbeler R.C., Engineering Mechanics - Statics, Dynamics,12th edition, McMillan, New York,
2009;
Beer F.P., E. R. Johnston Jr., D. F. Mozurek, E. R. Eisenberg, Vector Mechanics for Engineers
(Statics and Dynamics) , 9th edition, McGraw-Hill, 2009;
Ruina A, R. Pratap, Introduction to Statics and Dynamics, Pre-print for Oxford University
Press, January 2002.
Lecture 1
Idealizations. Models or idealizations are used in mechanics in order to simplify application of the theory.
Here we will consider three important idealizations.
A particle is a body of negligible dimensions. In the mathematical sense, a particle is a body whose
dimensions are considered to be near zero so that we may analyze it as a mass concentrated at a point. We
often choose a particle as a differential element of a body. We may treat a body as a particle when its
dimensions are irrelevant to the description of its position or the action of forces applied to it.
For example, the size of the earth is insignificant compared to the size of its orbit, and therefore the earth can
be modeled as a particle when studying its orbital motion. When a body is idealized as a particle the
principles of mechanics reduce to a rather simplified form since the geometry of the body will not be involved
in the analysis of the problem.
Rigid body. A rigid body can be considered as a combination of a large number of particles in which all the
particles remain at a fixed distance from one another, both before and after applying a load. This model is
important because the material properties of any body that is assumed to be rigid will not have to be
considered when studying the effects of forces acting on the body. In most cases the actual deformations
occurring in structures, machines, mechanisms are relatively small and the rigid-body assumption is suitable
for analysis.
For instance, the calculation of the tension in the cable, which supports the boom of a mobile crane under
load is essentially unaffected by the small internal deformations in the structural members of the boom. For
the purpose, then, of determining the external forces which act on the boom, we may treat it as a rigid body.
Mechanical system - a system of elements (particles and/or rigid-bodies) that interact on mechanical
principles.
Lecture 1
FORCE ON A PARTICLE. RESULTANT OF TWO FORCES
Experimental evidence shows that two forces P and Q acting on a particle A can be replaced by
a single force R which has the same effect on the particle. This force is called the resultant of
the forces P and Q and can be obtained, as shown, by constructing a parallelogram, using P
and Q as two adjacent sides of the parallelogram. The diagonal that passes through A
represents the resultant. This method for finding the resultant is known as the parallelogram law
for the addition of two forces.
Lecture 1
Sample Problem 1
SOLUTION:
• Graphical solution - construct a parallelogram
with sides in the same direction as P and Q
and lengths in proportion. Graphically evaluate
the resultant which is equivalent in direction
and proportional in magnitude to the diagonal.
• Trigonometric solution - use the triangle rule for
vector addition in conjunction with the law of
cosines and law of sines to find the resultant.
The two forces act on a bolt at A.
Determine their resultant.
• Analytical solution
Lecture 1
Sample Problem 1 Contd.
• Graphical solution - A parallelogram with sides equal
to P and Q is drawn to scale. The magnitude and
direction of the resultant or of the diagonal to the
parallelogram are measured,
R  98 N   35
• Graphical solution - A triangle is drawn with P and Q
head-to-tail and to scale. The magnitude and
direction of the resultant or of the third side of the
triangle are measured,
R  98 N   35
Lecture 1
Sample Problem 1 Contd.
• Trigonometric solution - Apply the triangle rule.
From the Law of Cosines,
R 2  P 2  Q 2  2 PQ cos B
 40N 2  60N 2  240N 60N  cos155
R  97.73N
From the Law of Sines,
sin A sin B

Q
R
sin A  sin B
Q
R
 sin 155
A  15.04
  20  A
  35.04
60N
97.73N
Lecture 1
Sample Problem 1 Contd.
• Analytical solution
Qx  Q.cos 450  42,43 N
Px  P cos 200  37,59 N
Rx  Qx  Px  80,02 N
Q y  Q.sin 450  42,43 N
Py  P sin 200  13,68 N
R y  Q y  Py  56,11 N
R  Rx2  R y2
R  97.73N
  arccos
Rx
 arccos 0,818786   35,040
R
  35.04
Lecture 1
Sample Problem 2
A barge is pulled by two tugboats. If the resultant of the forces exerted by the tugboats
is a 5000-lb force directed along the axis of the barge, determine :
( a ) the tension in each of the ropes knowing that a  = 45°,
( b ) the value of  for which the tension in rope 2 is minimum.
Lecture 1
SOLUTION
(a) Tension for a  = 45°.
The triangle rule can be used. We note that the triangle shown represents half of the
parallelogram shown above. Using the law of sines, we write:
T1
T2
5000 lb


sin 450 sin 300 sin1050
With a calculator, we first compute and store the value of the last quotient. Multiplying
this value successively by sin 45° and sin 30°, we obtain:
T1  3660 lb
T2  2590 lb
Lecture 1
SOLUTION
(b) Value of  for minimum T2 .
To determine the value of  for which
the tension in rope 2 is minimum, the
triangle rule is again used. In the sketch
shown, line 1-1’ is the known direction of
T1 . Several possible directions of T2 are
shown by the lines 2-2’. We note that the
minimum value of T2 occurs when T1 and
T2 are perpendicular.
The minimum value of T2 is
T2   5000 lb  sin 300  2500 lb
Corresponding values of T1 and  are:
T1   5000 lb  cos300  4330 lb
  900  300  600