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INTRODUCTION
Learning Objectives
1). To introduce and define the subject of mechanics. 2). To introduce Newton's Laws, and to understand the
significance of these laws.
3). The review modeling, dimensional consistency, unit
conversions and numerical accuracy issues. 4). To review basic vector algebra (i.e., vector addition
and subtraction and scalar multiplication).
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Definitions
Mechanics: Study of forces acting on a rigid body a) Statics - body remains at rest
b) Dynamics - body moves Newton's Laws
First Law: Given no net force, a body at rest will remain at rest and a body moving at a constant velocity will continue to do so along a straight path 0, 0F M .
Second Law: Given a net force is applied, a body will experience an acceleration in the direction of the force which is proportional to the net applied force F ma . Third Law: For each action there is an equal and opposite reaction F FAB BA .
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Models of Physical Systems
Develop a model that is representative of a physical system Particle: a body of infinitely small dimensions (conceptually, a point). Rigid Body: a body occupying more than one point in space in which all the points remain a fixed distance apart. Deformable Body: a body occupying more that one point in space in which the points do not remain a fixed distance apart. Dimensional Consistency
If you add together two quantities, 1 2x C v C a
these quantities need to have the same dimensions (units); e.g., if x, v and a have units of (L), (L/T) and (L/T2), then C1 and C2 must have units of (T) and T2 to maintain dimensional consistency. Many times algebraic errors in analysis lead to dimensional inconsistency. Use dimensional consistency as a check on your
algebra.
Unit Conversions
Use a logical process in your unit conversions. For example, to convert 60 mph to ft/sec:
miles hour 5280ft ft60 88
hour 3600 s mile s
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Accuracy of Numerical Answers
Say that you do the following calculations: x
z ty
Say that you have used four significant digits for x and y but only two significant digits for t. The numerical value of z will not have more than two significant digits (and likely less than two). Examples of Significant Figures: 385.1 four significant figures 38.51 four significant figures 0.03851 four significant figures 3.851 x 107 four significant figures 7.04 x 10-4 three significant figures 25.5 three significant figures 0.51 two significant figures 0.00005 one significant figure 27.855 five significant figures 8.91 x 104 three significant figures
2200 May have two, three or four significant figures depending on the accuracy of the measurement that obtained the number. Where such doubt may exist, it is better to write the number as 2.2 x 103 to show two significant figures; or as 2.20 x 103 to show three significant figures.
55 two significant figures
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55.0 three significant figures. The zero is significant in this case, since it is not otherwise needed to show proper location of the decimal point.
Retention of Significant Figures
1. In recording measured data, only one doubtful digit is retained, and it is considered to be a significant figure.
2. In dropping figures that are not significant, the last figure retained should be increased by 1 if the first figure dropped is 5 or greater.
3. In addition and subtraction, do not carry the result beyond the first column that contains a doubtful figure. For example,
11.3
+ 31.43 42.73 = 42.7 (the 3 in the hundredth’s
place is insignificant since the 11.3 figure can only be measured to the closest tenth).
4. In multiplication and division, carry the result to the
same number of significant figures that there are in the quantity entering into the calculation that has the least number of significant figures. For example,
15.5 x 5.4
Note: Unless stated otherwise, we shall retain three significant figures (unless one or more digits are lost through additions; for example, 90.2 - 90.1 = 0.1). Thus, given a dimension of 2 ft, assume it is actually 2.00 ft. Vectors
Vectors are characterized by both a magnitude and direction. Examples of vectors: position, velocity, acceleration, force, moment. Vectors add by the parallelogram law. For example.
A B C Says that vector A has the same magnitude and direction as the sum of vectors B and C .
More the tail of C to the head of B . A is then found by connecting tail of B to the head of C . See the text for the mechanics of adding and subtracting vectors and for multiplying a scalar with a vector. Law of Cosines:
2 2C= A + B - 2AB cos c
Law of Sines:
A B C = =
sin a sin b sin c
History of Mechanics
Three divisions: Mechanics of Rigid Bodies Mechanics of Deformable Bodies Mechanics of Fluids
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Aristotle (384-322 BC) – first to investigate the statics of levers and the first to develop some of the first theories of dynamics Archimedes (287-212 BC) – explained the lever fulcrum and the theory of buoyancy
da Vinci (1452-1519) – continued Archimedes work on levers and developed a theory on moments for 3-D bodies
Brahe (1546-1601) – developed accurate astronomical measurements
Kepler (1571-1630) – developed laws of planetary motion Stevinus (1548-1620) – conceived the laws of equilibrium and the parallelogram law for vector addition Galileo (1564-1642) – studied projectile motion and took strong stance against natural philosophers and the Catholic Church Descartes (1596-1650) – developed Cartesian coordinate system and the theory of virtual work Hooke (1635-1703) – Developed theory of elasticity (Hooke’s Law) Newton (1642-1727) – Invented calculus (with Gottfried Leibniz) and wrote the Mathematical Principles of Natural Philosophy, the foundation for Newton’s Laws of Motion and classical mechanics
Newton laid the foundation for future contributors like Euler, Lagrange, Cauchy, Kolmogorov, and many others.
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VECTOR DEFINITIONS; DIRECTION COSINES;
DIRECTION ANGLES
Learning Objectives
1). To determine and understand the differences between position vectors, unit vectors and force vectors.
2). To determine the direction cosines and direction
angles. 4). To do an engineering estimate of these quantities.
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Definitions
Position Vector AB(r ) : a vector used to identify the of a point in space relative to a reference point.
BAB A(x, y, z) - (x,r y, z)
Unit Vector AB(u ) : a dimensionless vector of unit magnitude that is often used to describe the of a vector of interest.
A
ABB
AB
ru
r
Force Vector AB(F ) : a vector used to represent the and of a force.
ABA BB AFF u
Let i = unit vector pointing in the x-direction j = unit vector pointing in the y-direction k = unit vector pointing in the z-direction
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Consequences
(i) The vector A can be written in terms of its "components" and the unit vectors i, j and k as (see figure):
A A i A j A kx y z (ii) The sum of two vectors is accomplished by adding
together the respective components: A B (A i A j A k) (B i B j B k)x y z x y z
(A B )i (A B )j (A B )kx x y y z z
(iii) The magnitude (length) of A is given by (using Pythagorean Theorem):
|A| A A Ax2
y2
z2
(iv) A unit vector u pointing in the same direction as A is given by dividing A its
magnitude: yx z
AA A Au i j k
|A| |A| |A| |A|
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Direction Angles 1
x xθ cos (A / |A|) = angle between + i and A vectors, measured in the i A plane
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y yθ cos (A /|A|) angle between + j and A vectors, measured in the j A plane
1
z zθ cos (A /|A|) = angle between + k and A vectors, measured in the k A plane
Direction Cosines
cos x x xθ (A / |A|) = u
cos y y yθ (A /|A|) = u
cos z z zθ (A / |A|) = u
Note: 2 2 2
x y z(cos θ ) (cos θ ) (cos θ ) = 1
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Consequences
(i) Rearranging we can see: A |Ax | cos A |Ay | cos A |Az | cos
(ii) Therefore we can write A A i A i A kx y z | | cos | | cos | | cosA i A j A k | |(cos cos cos )A i j k
(iii) Also, from before, A
u cos i cos j cos k|A|
(iv) If A thenx 0 0 90, If A thenx 0 90 180,
Final Remark
Practice making the three-dimensional vectors! Show direction angles in figures. Visualization is the first step in understanding.
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DOT PRODUCT (Scalar Product)
Learning Objectives
1). To use the dot product to determine: i) the projection of a vector in another direction, ii) the angle ( ) between two vectors and
2). To do an engineering estimate of these quantities.
Definitions
In words, A B the magnitude |A| times the magnitude of the projection of B on A (i.e., |B| cos θ ).
Projection of a Vector
uA =A u u A A u u
Angle ( ) Between Two Vectors
-1 A×Bcos
|A| |Bθ =
|
A B B A scalar quantity = A B cos θ
Mechanics
A B (A i A j A k) (B i B j B k)x y z x y z
A B A B A Bx x y y z z Recall,
i i j j k k 1
i j i k j i j k k i k j 0
Basic Properties
1. p(A B) (pA) B A (pB)
A (B C) A B A Csimilar to multiplicationUVW
2. If A B 0 , then the projection of B is in the direction
of A (i.e., 0 90 ) If A B 0, then the project of B is in the opposite direction of A (i.e., 90 180 )
3. If A B and A is perpendicular to B0 90 If A B |A| |B| θ 0 and A is parallel to B
4. A A |A|2
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Basic Trigonometric Identities
cos 0˚ =
cos 30˚ =
cos 36.87˚ =
cos 45˚ =
cos 53.13˚ =
cos 60˚ =
cos 90˚ =
sin 0˚ =
sin 30˚ =
sin 36.87˚ =
sin 45˚ =
sin 53.13˚ =
sin 60˚ =
sin 90˚ =
STATIC EQUILIBRIUM OF A PARTICLE (2-D)
Learning Objectives
1). To draw a free body diagram (FBD) of an object that is modeled as a particle.
2). To evaluate the forces required for static equilibrium of an object that is modeled as a particle.
3). To do an engineering estimate of these quantities.
Force Classifications
External Forces: applied forces which are typically known or prescribed (e.g., forces due to cables, springs, gravity, etc.). Reaction Forces: constraining forces at supports, intended to prevent motion (usually nonexistent unless system is externally loaded). Free Body Diagram (FBD)
Free Body Diagram (FBD): a graphical sketch of the system showing a coordinate system, all external/reaction forces and moments, and key geometric dimensions.
Benefits:
1). Provides a coordinate system to establish a solution methodology.
2). Provides a graphical display of all forces/moments acting on the rigid body.
3). Provides a record of geometric dimensions needed for establishing moments of the forces.
Newton’s First Law
Given no net force, a body at rest will remain at rest and a body moving at a constant velocity will continue to do so along a straight path ( )R F 0 .
Static Equilibrium
Vector Equation:
Component Equations:
N
i x y
i 1
R F F i F j
x y0 0F F
Problem Solving
1). Draw complete FBD. 2). Choose an xyz reference frame. 3). Evaluate the geometrical parameters. 4). Write equations of static equilibrium. 5). Count number of scalar equations and number of unknowns. 6). Solve equations of static equilibrium.
Note: Fdrag = ½ ρ CD SA ν
2 Where ρ = density of air CD = Drag Coefficient SA = Frontal Surface Area ν = velocity
STATIC EQUILIBRIUM OF A PARTICLE (3-D)
Learning Objectives
1). To draw a free body diagram (FBD) of an object that is modeled as a particle.
2). To evaluate the forces required for static equilibrium of an object that is modeled as a particle.
3). To do an engineering estimate of these quantities.
Force Classifications
External Forces: applied forces which are typically known or prescribed (e.g., forces due to cables, springs, gravity, etc.). Reaction Forces: constraining forces at supports, intended to prevent motion (usually nonexistent unless system is externally loaded). Free Body Diagram (FBD)
Free Body Diagram (FBD): a graphical sketch of the system showing a coordinate system, all external/reaction forces and moments, and key geometric dimensions.
Benefits:
1). Provides a coordinate system to establish a solution methodology.
2). Provides a graphical display of all forces/moments acting on the rigid body.
3). Provides a record of geometric dimensions needed for establishing moments of the forces.
Newton’s First Law
Given no net force, a body at rest will remain at rest and a body moving at a constant velocity will continue to do so along a straight path ( )R F 0 .
Static Equilibrium
Vector Equation:
Component Equations:
N
R i x y z
i 1
F F F i F j F k 0
x y z0 0 0F F F
Problem Solving
1). Draw complete FBD. 2). Choose an xyz reference frame. 3). Evaluate the geometrical parameters. 4). Write equations of static equilibrium. 5). Count number of scalar equations and number of unknowns. 6). Solve equations of static equilibrium.
Note: Fdrag = ½ ρ CD SA ν
2 Where ρ = density of air CD = Drag Coefficient SA = Frontal Surface Area ν = velocity
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CROSS PRODUCT (Vector Product)
Learning Objectives
1). To use cross products to: i) calculate the angle ( ) between two vectors and, ii) determine a vector normal to a plane.
2). To do an engineering estimate of these quantities.
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Purpose
Four primary uses of the cross product are to: 1) calculate the angle ( ) between two vectors, 2) determine a vector normal to a plane, 3) calculate the moment of a force about a point, and 4) calculate the moment of a force about a line.
Definition
In words, | |A B = the component of B perpendicular to A multiplied by the magnitude of A; or vice versa. Angle ( ) Between Two Vectors
Unit Normal Vector
|A B| |B A| A B sin θ
-1 |A×B|sin
|A| |B|
Unit NormalA×B
|A×B|
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Mechanics (assuming a right-handed coordinate system)
A B (A i A i A k) (B i B j B k)x y z x y z Recall,
i i j j k k 0 i k j k i j
i j k j i k j k i k j i Note: Given C A B, C is perpendicular to the plane containing vectors A and B . Basic Properties
1). A B B A, Use right hand rule to determine direction of resultant vector. 2). p(A B) (pA) B A (pB)
A (B C) (A B) (A C) (A B) C (A C) (B C)
3). If |A B| 0, then θ 0 If |A B| |A| |B|, then θ 90
4). A A 0
A B (A B B A ) i (A B B A ) j (A B B A ) ky z y z x z x z x y x y
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MOMENT ABOUT A POINT
Learning Objectives
1). To use cross products to calculate the moment of a force about a point.
3). To do an engineering estimate of this quantity.
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Definition
Moment About a Point: a measure of the tendency of a force to turn a body to which the force is applied.
or or
opo op|M |=r |F|= r sin rθ |F|= F sin θ
opo op op|M |=|r | F =|r | F sin θ = r F sin θ
o op op|M | = |r ×F|=|r | |F| sin θ
|( )|r i r j) (F i F jx y x y
| |r F r F ) kx y y x
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Comments
1). Direction of the moment can be determined by the right
hand rule. 2). Point P can be any point along the line of action of the force
without altering the resultant moment O(M ) . 3). Use trigonometric relationships for calculating the moment
about a point for 2-D problems. 4). Use vectors and cross products when calculating the
moment about a point for 3-D problems.
MOMENT ABOUT A LINE Learning Objectives 1). To use cross products and dot products to evaluate the
moment about a line. 2). To do an engineering estimate of this quantity.
Definition Moment About a Line: The vector component of the moment of the force (about any point on the line) which is parallel to the line.
OM = moment of the force ( )F about any point along the line
AB (i.e., O opM r F= × ).
ABu = unit vectors along line AB. Procedure 1). Calculate the moment about any point O along line AB (i.e.,
O opM r F= × ).
2). Define a unit vector along line AB (i.e., ABu ).
3). Find the magnitude of the projection of OM onto line AB
by finding the dot product of OM and ABu ( i.e., o ABM u⋅ ). 4). Find the moment projection along line AB by multiplying the magnitude of the projection (part 3) by the same unit vector
as used in part 3 [i.e., ( O ABM u⋅ ) ABu ].
( ) ( )A AB AB B A AA BB BM u u M uM = u ⋅ = ⋅
Comments 1). The moment of the force ( O
M ) may be taken about any point along the line AB. Choose the point that will provide the simplest cross product.
2). The moment about line AB will be zero if the force F passes through any point along line AB.
3). The moment about line AB will be zero if the force ( )F if parallel to line AB.
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FORCE COUPLES
Learning Objectives
1). To determine a resultant torque of a system of force couples. 2). To determine an equivalent force-couple of a system of
forces and moments. 3). To do an engineering estimate of this quantity.
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Definition
Force Couple: a pair of forces which are (i) equal in magnitude, (ii) parallel and (iii) opposite in direction. where d = perpendicular distance between the lines of action of the forces forming the couple. where ABr = any position vector between the lines of action of the forces forming the couple. Comments
1). Force couples cause no net force (i.e., F = 0) . 2). The moment due to a force couple is the same regardless of
the point the moment is summed about. (This not true of a non-couple).
|M| |F | d
ABM r F
EQUIVALENT SYSTEMS
Learning Objectives 1). To determine an equivalent force-couple of a system of
forces and moments. 2). To do an engineering estimate of this quantity.
Definition Equivalent Systems: two force-couple systems which exert i) the same net force on a body and ii) the same net moment (or torque). Note: Selection of point P is arbitrary. Thus, choose a point that will simplify the moment equation. Force-Couple Equivalent
∴ Equivalent Force-Couple System: O
R RF , M (about point P).
Force Condition: ( )∑ = ∑F F1 2c h
Moment Condition: ( ) ( )∑ = ∑M MP P1 2
Force: 1 2 3F + F + F +F …R = ∑ = Moment:
( ) ( ) ( )1 1 2 2 3 3P 1 2r P r F +...+ r F + r + F +...+ C + CM M +...= × ×∑ =
= × +
= =∑ ∑r F Ci ii
N
ii
NF c
c h1 1
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FREE BODY DIAGRAMS (FBDs)
Learning Objectives
1). To inspect the supports of a rigid body in order to determine the nature of the reactions, and to use that information to draw a free body diagram (FBD).
Force/Moment Classifications
External Forces/Moments: applied forces/moments which are typically known or prescribed (e.g., forces/moments due to cables springs, gravity, etc.). Reaction Forces/Moments: constraining forces/moments at supports intended to prevent motion (usually nonexistent unless system is externally loaded). Free Body Diagram (FBD)
Free Body Diagram (FBD): a graphical sketch of the system showing a coordinate system, all external/reaction forces and moments, and key geometric dimensions.
Benefits:
1). Provides a coordinate system to establish a solution methodology.
2). Provides a graphical display of all forces/moments acting on the rigid body.
3). Provides a record of geometric dimensions needed for establishing moments of the forces.
50 STATIC EQUILIBRIUM OF RIGID BODIES (2-D)
Learning Objectives
1). To evaluate the unknown reactions holding a rigid body in equilibrium by solving the equations of static equilibrium.
2). To recognize situations of partial and improper constraint, as well as static indeterminacy, on the basis of the solvability of the equations of static equilibrium.
Newton’s First Law
Given no net force, a body at rest will remain at rest (and a body moving at a constant velocity will continue to do so along a straight path). Definitions
Zero-Force Members: structural members that support no
loading but aid in the stability of the truss. Two-Force Members: structural members that are: a) subject to no applied or reaction moments, and b) are loaded only at two pin joints along the member. Multi-Force Members: structural members that have a) applied or reaction moments, or b) are loaded at more than two points along the member.
51 Vector Equations
RF F 0
OR O
M M 0 where O is any arbitrary point Component Equations
There are three alternate forms of equilibrium equations for 2-D problems. (i) Two component force equations (x and y) are one moment equation (z). (ii) One component force equation (x or y) and two moment equations (both about different points in the z direction). (iii) Three moment equations (points A, B and C cannot be collinear).
x y A0 M0 0F F
0 0 0x A BF M M
0 0 0A B CM M M
52 Static Determinacy/Partial and Improper Constraints
Static Indeterminacy: occurs when a system has more constraints than is necessary to hold the system in equilibrium (i.e., the system is overconstrained and thus has redundant reactions).
Static Determinancy: occurs when a system has a sufficient number of constraints to prevent motion without any redundancy.
Partial Constraint: occurs when there is an insufficient number of reaction forces to prevent motion of the system (i.e., the system is partially constrained).
Improper Constraint: occurs when a system has a sufficient number of reaction forces but one or more are improperly
applied so as not to prevent motion of the system (i.e., the system is improperly constrained).
Comments:
1). Equations (i) are the equilibrium eqns most commonly used. 2). NEVER attempt to use MORE THAN THREE equilibrium
equations from a single planar FBD. Only three independent equations can exist for a single planar FBD.
3). If you have more than three unknown forces in your three equations, then consider breaking the system or structure into smaller systems and write down equilibrium equations for each sub-structure. If this is not possible, you may have an indeterminate structure; i.e., the evaluation of member forces requires consideration of deformation of the members resulting from the loading.
53 4). If all forces act through a single point, then the moment
equation for any point will not provide any more new information.
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STATIC EQUILIBRIUM OF RIGID BODIES (3-D)
Learning Objectives
1). To evaluate the unknown reactions holding a rigid body in equilibrium by solving the equations of static equilibrium.
2). To recognize situations of partial and improper constraint, as well as static indeterminacy, on the basis of the solvability of the equations of static equilibrium.
Newton’s First Law
Given no net force, a body at rest will remain at rest and a body moving at a constant velocity will continue to do so along a
straight path O
R R O(F F 0, M M 0).
Definitions
Zero-Force Members: structural members that support no
loading but aid in the stability of the truss. Two-Force Members: structural members that are: a) subject to
no applied or reaction moments, and b) are loaded only at two pin joints along the member.
Multi-Force Members: structural members that have a) applied
or reaction moments, or b) are loaded at more than two points along the member.
56 Vector Equations
RF F 0
OR O
M M 0 where O is any arbitrary point Component Equations
For 3-D problems, there are SIX “component” equations
0 0 0X y zF F F 0 0 0x y zM M M
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Static Determinacy/Partial and Improper Constraints
Static Indeterminacy: occurs when a system has more constraints than is necessary to hold the system in equilibrium (i.e., the system is overconstrained and thus has redundant reactions). Static Determinancy: occurs when a system has a sufficient number of constraints to prevent motion without any redundancy. Partial Constraint: occurs when there is an insufficient number of reaction forces to prevent motion of the system (i.e., the system is partially constrained). Improper Constraint: occurs when a system has a sufficient number of reaction forces but one or more are improperly
applied so as not to prevent motion of the system (i.e., the system is improperly constrained).
Problem Solving
1). Select body (or bodies) to be isolated in a FBD. 2). Choose an xyz coordinate system. 3). Complete FBD slowing all external reaction forces/moments.
4). OM 0 , select point O to eliminate some unknown
constraint forces and simplify the cross products. 5). F 0 6). Solve simultaneous equations.
