Transcript Law 2

Introduction
Mechanics: deals with the responses of the bodies to the action
of forces.
Objectives:
To give students an introduction to engineering mechanics with an
emphasis on engineering problem solving
Question:
What are the mechanical issues involved in the design of a bridge
crossing a river?
Fundamental quantities of
Mechanics
-: the geometry region in which the physical events of
interest in mechanics occur.
: the interval between two events.
: a quantitative measure of inertia.
: the action of one body upon another body.
Principles --- Newton’s laws of
Motion
Law 1: - foundation for the study of statics
In the absence of external forces, a particle originally at rest or moving
with a constant velocity will remain at rest or continue to move with a
constant velocity along a straight line.
Law 2: - foundation for the study of dynamics
If an external force acts on a particle, the particle will be accelerated in
the direction of the force and the magnitude of the acceleration will be
directly proportional to the force and inversely proportional to the mass
of the particle.
Law 3: - foundation of understanding the concept of a force
For every action there is an equal and opposite reaction. The forces of
action and reaction between bodies are equal in magnitude, opposite in
direction, and collinear.
Vectors
Mechanical Quantities: scalars and vectors
Vectors:
Notation: F or F
Representations:


F  Fe

e is an unit vector




F  Fx i  Fy j  Fz k
Vectors
Addition: parallelogram law
Law of Sine:
Law of Cosine:
Dot product:
a
b
c


sin  sin  sin 
c 2  a 2  b 2  2ab cos 
 
F1  F2  F1 F2 cos 
Cross Product of Vectors
  

C  A  B  A  B  sin ec
B

Where, ec is the unit vector in a direction perpendicular

A
to the plane containing vectors A and B.







C  ( Ax i  Ay j  Az k )  ( Bx i  B y j  Bz k )



i
j
k




C  Ax Ay Az  ( Ay Bz  Az B y )i  ( Az Bx  Ax Bz ) j  ( Ax B y  Ay Bx )k
Bx
By
Bz
       

i  i  0; i  j  k ; i  k   j
      
 
j  i  k ; j  j  0; j  k  i
    
   
k  i  j ; k  j  i ; k  k  0
Particle Equilibrium
Forces:
magnitude; direction; point of application
Effects:
external effects:
internal effects:
Equilibrium of a particle:
R  F  0
Rx  0 R y  0 Rz  0
Example 1: In a Cartesian coordinate system illustrated as follows, point
A is located at (5, 10, 2). A force F with a magnitude of 10 N is
applied at a body located at the origin (O), along the direction of
vector OA. Determine the x, y, and z scalar components of the
force.
z
o
y
F
x
A
Equilibrium of a System of Particles

R  0
Key step: selection of your “free body”
1) Given W1, W2, r1, r2, ,q, what are the reaction forces
coming from the supporting wedge?
2) What is the force between two cylinders?
Example 2:
The 900-lb platform illustrated in the following figure is supported by
a light cable and pulley system. Find the tension inside the cable over
pulley A and the tension inside the cable over pulley B. Assume the mass
center location so that the platform remains horizontal.
Test
1) Draw the free-body diagram of cylinder 1
2) Draw the free-body diagram of two cylinders as one
body
Assume all surfaces are smooth.