Transcript Lecture 6

ENGR 3340: Fundamentals of Statics and Dynamics
Fundamentals of Statics and
Dynamics - ENGR 3340
Professor: Dr. Omar E. Meza Castillo
[email protected]
http://facultad.bayamon.inter.edu/omeza
Department of Mechanical Engineering
of Statics and Dynamics
ENGR 3340: Fundamentals
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MSP21
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Tentative Lectures Schedule
Topic
Lecture
Equilibrium of a Particle in 2-D
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One thing you learn in science is that there is
no perfect answer, no perfect measure.
A. O. Beckman
Topic 5: Equilibrium of a Particle in
2-D
Equilibrium
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Objectives
 To introduce the concept of the free-body
diagram for a particle
 To show how to solve particle equilibrium
problems using the equations of equilibrium
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Equilibrium of a Particle Applications
For a spool of given
weight, what are the
forces in cables AB
and AC ?
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Equilibrium of a Particle Applications (continued)
For a given cable
strength, what is the
maximum weight that
can be lifted ?
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Equilibrium of a Particle Applications (continued)
For a given weight of
the motor, what are the
forces in the cables?
What size of cable
must you use ?
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Equilibrium of a Particle Applications (continued)
Static Equilibrium
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Balancing Forks
 Materials:
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Balancing Forks
 Step 1:
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Balancing Forks
 Step 2:
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Balancing Forks
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Coke Can Balancing
 Materials:
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Coke Can Balancing
 Step 1:
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Coke Can Balancing
 Step 2:
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Coke Can Balancing
 Step 3:
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Coke Can Balancing
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3.1 Condition for the Equilibrium of a Particle.
 Particle is said to be in EQUILIBRIUM if:
 It remains at rest (v=0) if originally at rest.
(Static Equilibrium)
 It has a constant velocity if originally in motion.
 To maintain EQUILIBRIUM, it is necessary to satisfy
Newton’s first law of motion. Resultant force
acting on a particle require to be zero.
F  0
Equation of Equilibrium
where ∑F is the vector sum of all the forces acting
on the particle.
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3.1 Condition for the Equilibrium of a Particle.
a =0
∑F = ma = 0
 Consequently, the particle indeed moves with
constant velocity or remains at rest.
 For Equilibrium:
∑Fx = 0
∑F y = 0
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3.2 The Free-Body Diagram (FBD)
 To apply the EQUATION OF EQUILIBRIUM, we must
account for ALL THE KNOWN and UNKNOWN
FORCES which act ON the particle
 In order to account for all the forces that act on
the particle, it is necessary to draw its FREE-BODY
DIAGRAM (FBD).
 The free-body diagram is simple a SKETCH which
shows the particle “FREE” from its surrounding with
ALL the forces that act on it.
FBD
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Types of Connections
 Two types of connections often encountered in
particle equilibrium SPRINGS and CABLES and
PULLEYS.
 SPRINGS: The magnitude of force exerted on a
linearly elastic spring is:
F = ks
which has a stiffness k and is
deformed (elongate or compressed)
a distance s, measured from its
unloaded position.
s = l _ lo
Where lo the original length and l the final length
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Types of Connections
 CABLES and PULLEYS: All cables are assumed to
have negligible weight and they cannot stretch.
 A cable can support only a tension or “pulling”
force, and this force always act in the direction of
the cable.
 The tension in a cable is constant throughout its
length.
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Procedure for Drawing a Free-Body Diagram:
 Draw Outlined Shape: Imagine the particle to be
isolated or cut “free” from its surrounding with all
the forces outlined shape.
 Show All Forces: Indicate on this sketch all the
forces that act on the particle.
 Identify Each Force: The forces which are known
should be labeled with their proper magnitudes
and directions. Letter are used to represent the
magnitudes and directions of forces that are
unknown.
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Example 1
 The sphere has a mass of 6 kg and is supported
as shown. Draw a free-body diagram of the
sphere, the cord CE, and the knot at C.
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Solution 1
 Sphere: There are two forces acting on the
sphere:
 The weight of the sphere W=6kg(9.81m/s2)=58.9N
 The force FCE of the cord CE acting on the sphere
 The Free-Body Diagram
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Solution 1
 Cord CE: When the cord CE is isolated from its
surroundings, its free-body diagram shows only
two forces acting on it:
 The force of the sphere and FCE.
 The force of the knot FEC
 The Free-Body Diagram
FCE=FEC
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Solution 1
 Knot: The knot at C is subjected to three forces.
They are caused by the cords CBA and CE and
the spring CD:
 The Free-Body Diagram
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3.3 Coplanar Force Systems
 If a particle is subjected to a SYSTEM OF COPLANAR
FORCES that lie in the x-y plane. Then each force
can be resolved into i and j components
 For EQUILIBRIUM, these forces must SUM to
produce a ZERO RESULTANT. Hence:
∑F = 0
∑F i +∑F
∑F = 0
∑F = 0
x
y
j +=0
x
y
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Example 2
 Determine the tension in cables BA and BC
necessary to support the 60-kg cylinder.
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Solution 2
 Due to EQUILIBRIUM, the weight of the cylinder
causes the tension in cable BD to be:
TBD = 60(9.81) = 588.6N
 The forces in cable BA and BC can be
determined by investigating the equilibrium of
the ring B. Its free body diagram is then:
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Solution 2
 The magnitudes of TA and TC are unknown, but
their directions are known.
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Solution 2
 Equations of equilibrium along the x and y axes:
x
y
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∑F
=0
∑F
=0
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Homework3  http://facultad. bayamon.inter.edu/omeza/
Omar E. Meza Castillo Ph.D.
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¿Preguntas?
Comentarios
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GRACIAS
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