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Fundamentals of Statics and Fundamentals of Statics and Dynamics - ENGR 3340Dynamics - ENGR 3340
Professor: Dr. Omar E. Meza CastilloProfessor: Dr. Omar E. Meza [email protected]
http://facultad.bayamon.inter.edu/omezaDepartment of Mechanical EngineeringDepartment of Mechanical Engineering
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Tentative Lectures Schedule
TopicTopic LectureLectureEquilibrium of a Particle in 2-DEquilibrium of a Particle in 2-D 55
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EquilibriumEquilibrium
Topic 5: Equilibrium of a Particle in 2-D
3
One thing you learn in science is that there is One thing you learn in science is that there is no perfect answer, no perfect measure.no perfect answer, no perfect measure.
A. O. BeckmanA. O. Beckman
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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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For a spool of given For a spool of given weight, what are weight, what are the the forcesforces in cables in cables AB and AC ?AB and AC ?
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For a given cable For a given cable strength, what is strength, what is the the maximum maximum weight weight that can be that can be lifted ?lifted ?
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For a given weight For a given weight of the motor, what of the motor, what are the are the forcesforces in in the cables? What the cables? What size of cable must size of cable must you use ?you use ?
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8
Static EquilibriumStatic Equilibrium
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Materials:
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Step 1:
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Materials:
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Step 1:
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Particle is said to be in EQUILIBRIUMEQUILIBRIUM if: It remains at rest (v=0) if originally at rest.
(Static Equilibrium)(Static Equilibrium) It has a constant velocity if originally in motion.
To maintain EQUILIBRIUMEQUILIBRIUM, it is necessary to satisfy Newton’s first law of motion. Resultant force Resultant force acting on a particle require to be zerozero.
where ∑F is the vector sum of all the forces acting on the particle.
0 FEquation of EquilibriumEquation of Equilibrium
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Consequently, the particle indeed moves with constant velocity or remains at rest.
For Equilibrium:For Equilibrium:
0maF ==∑ 0a =
0Fx =∑
0Fy =∑
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To apply the EQUATION OF EQUILIBRIUMEQUATION OF EQUILIBRIUM, we must account for ALL THE KNOWN ALL THE KNOWN and UNKNOWN FORCESUNKNOWN FORCES which act ONON the particle
In order to account for all the forces that act on the particle, it is necessary to draw its FREE-FREE-BODY DIAGRAMBODY DIAGRAM (FBD). (FBD).
The free-body diagram is simple a SKETCHSKETCH which shows the particle “FREE”“FREE” from its surrounding with ALLALL the forces that act on it.
FBDFBD
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Two types of connections often encountered in particle equilibrium SPRINGSSPRINGS and CABLES and CABLES and PULLEYS.PULLEYS.
SPRINGS:SPRINGS: The magnitude of force exerted on a linearly elastic spring is:
ksF =which has a stiffness which has a stiffness kk and is and is deformed (elongate or deformed (elongate or compressed) a distance compressed) a distance ss, , measured from its unloaded measured from its unloaded position.position. ol_ls =WhereWhere l loo the original length and the original length and ll the final the final lengthlength
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CABLES and PULLEYS: CABLES and PULLEYS: All cables are assumed to have negligible weight negligible weight and they cannot cannot stretch.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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Draw Outlined Shape: Draw Outlined Shape: Imagine the particle to be isolated or cut “free” from its surrounding with all the forces outlined shape.
Show All Forces: Show All Forces: Indicate on this sketch all the forces that act on the particle.
Identify Each Force: 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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The sphere has a mass of 6 kg 6 kg and is supported as shown. Draw a free-body diagram free-body diagram of the spheresphere, the cord CEcord CE, and the knot at Cknot at C.
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Sphere:Sphere: There are two forces acting on the sphere: The weight of the sphere
W=6kg(9.81m/sW=6kg(9.81m/s22)=58.9N)=58.9N The force FCE of the cord CE acting on the sphere
The Free-Body DiagramThe Free-Body Diagram
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Cord CE: 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 DiagramThe Free-Body Diagram
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FFCECE=F=FECEC
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Knot: 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 DiagramThe Free-Body Diagram
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If a particle is subjected to a SYSTEM OF SYSTEM OF COPLANAR FORCESCOPLANAR FORCES that lie in the x-y plane. Then each force can be resolvedresolved into ii and j j componentscomponents
For EQUILIBRIUMEQUILIBRIUM, these forces must SUMSUM to produce a ZERO RESULTANTZERO RESULTANT. Hence:0=F∑
0=++∑∑ jFiF yx
0=∑ xF
0=∑ yF
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Determine the tensiontension in cables BABA and BCBC necessary to support the 60-kg60-kg cylinder.
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Due to EQUILIBRIUMEQUILIBRIUM, the weightweight of the cylinder causes the tensiontension in cable BDBD to be:
The forces in cable BABA and BCBC can be determined by investigating the equilibrium of equilibrium of the ring the ring BB. Its free body diagram is then:
30
N6.588=)81.9(60=TBD
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The magnitudes of TA and TC are unknown, but their directions are known.
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Equations of equilibrium along the x and y axes:
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0=Fx∑
0=Fy∑
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Homework3 http://facultad. bayamon.inter.edu/omeza/
Omar E. Meza Castillo Ph.D.Omar E. Meza Castillo Ph.D.
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¿Preguntas?¿Preguntas?
Comentarios Comentarios
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GRACIASGRACIAS