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Fundamentals of Statics and Dynamics - ENGR 3340
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ENGR 3340: Fundamentals of Statics and ENGR 3340: Fundamentals of Statics and Dynamics Dynamics Fundamentals of Statics and Fundamentals of Statics and Dynamics - ENGR 3340 Dynamics - ENGR 3340 Professor: Dr. Omar E. Meza Castillo Professor: Dr. Omar E. Meza Castillo [email protected] [email protected] http://facultad.bayamon.inter.edu/omeza http://facultad.bayamon.inter.edu/omeza Department of Mechanical Engineering Department of Mechanical Engineering
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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. Tentative Lectures Schedule. - PowerPoint PPT Presentation
Transcript of Fundamentals of Statics and Dynamics - ENGR 3340
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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]@bayamon.inter.edu
http://facultad.bayamon.inter.edu/omezahttp://facultad.bayamon.inter.edu/omezaDepartment of Mechanical EngineeringDepartment of Mechanical Engineering
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Tentative Lectures Schedule
TopicTopic LectureLecture
Moment of a Force – Scalar FormulationMoment of a Force – Scalar Formulation 88
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Scalar FormulationScalar Formulation
Topic 8: Moment of a Force
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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 discuss the concept of the moment of a force and show how to calculate it in two and three dimensions
To provide a method for finding the moment of a force about a specific axis
To define the moment of a couple To present methods for determining the
resultants of nonconcurrent force systems To indicate how to reduce a simple distributed
loading to a resultant force having a specified location
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What is the net effect of the two forces on the wheel?
What is the effect of the 30 N force on the lug nut?
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The moment of a force about a point provides a measure of the tendency for rotation (sometimes called a torque).
Moment about z-axis
Moment about x-axis
No moment
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Moment of a Force. Moment in 2-D – Scalar
Formulation
As shown, d is the perpendicular distance from point O to the line of action of the force.
The moment is a scalar or vector?
In the 2-D case, the magnitude of the moment isdFMO
In 2-D, the direction of MO depends on the tendency for rotation,
clockwise ,or counter-clockwise+ +
The resultant moment of a system of coplanar forces is defined as:
dFMOR +
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Formulation
+
For example,
magnitude, MO = F d
direction, counter-clockwise
bFaFM xyO +
aF
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Often it is easier to determine MO by using the components of F
a
Fy
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Note the different signs on the terms! The typical sign convention for a moment in 2-D is that counter-clockwise is considered positive.We can determine the direction of rotation by imagining the body pinned at O and deciding which way the body would rotate because of the force.
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Quiz
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1. What is the moment of the 10 N force about point A (MA)?
A) 3 N·m B) 36 N·m C) 12 N·m
D) (12/3) N·m E) 7 N·m
• Ad = 3 m
F = 12 N
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Example 1
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Example 1
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Example 2
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Example 3
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A 100-lb vertical force is applied to the end of a lever which is attached to a shaft at O.
Determine:a) Moment about O,b) Horizontal force at A which
creates the same moment,c) Smallest force at A which
produces the same moment,d) Location for a 240-lb vertical
force to produce the same moment,
e) Whether any of the forces from b, c, and d is equivalent to the original force.
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in. 12lb 100
in. 1260cosin.24
O
O
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d
FdM
a) Moment about O is equal to the product of the force and the perpendicular distance between the line of action of the force and O. Since the force tends to rotate the lever clockwise, the moment vector is into the plane of the paper.
in lb 1200 OM
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in. 8.20
in. lb 1200
in. 8.20in. lb 1200
in. 8.2060sinin. 24
F
F
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d
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b) Horizontal force at A that produces the same moment,
lb 7.57F
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in. 42
in. lb 1200
in. 42in. lb 1200
F
F
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c) The smallest force at A to produce the same moment occurs when the perpendicular distance is a maximum or when F is perpendicular to OA.
lb 50F
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in. 5cos60
in. 5lb 402
in. lb 1200
lb 240in. lb 1200
OB
d
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d) To determine the point of application of a 240 lb force to produce the same moment,
in. 10OB
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e) Although each of the forces in parts b), c), and d) produces the same moment as the 100 lb force, none are of the same magnitude and sense, or on the same line of action. None of the forces is equivalent to the 100 lb force.
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Example 4
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Plan:
Since this is a 2-D problem:
1) Resolve the 20 lb force along the handle’s x and y axes.
2) Determine MA using a scalar analysis.
Given: A 20 lb force is applied to the hammer.
Find: The moment of the force at A.
xy
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Solution:
+ Fy = 20 sin 30° lb
+ Fx = 20 cos 30° lb
xy
+ MA = {–(20 cos 30°)lb (18 in) – (20 sin 20°)lb (5 in)}
= – 351.77 lb·in = 352 lb·in (clockwise)
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Homework5 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
